chemistry...mole concept one mole is an amount of substance containing avogadro's number of...

CHEMISTRY(MODULE-I)

JEE CHEMISTRY

The IITian’s Hub I IIT/MEDICAL/FOUNDATION/OLYMPIAD

IITIAN'S HUBTH

ET

HE

CONCEPT

MOLE

JEE CHEMISTRY

The IITian’s Hub I IIT/MEDICAL/FOUNDATION/OLYMPIAD 1

IITIAN'S HUBTH

ET

HE

IIT – JEE SYLLABUS

Law of chemical combination, Atomic & molecular mass, Mole concept, Determination of

molecular formulae, Stoichiometry, Concentration terms.

CONTENTS

1. THEORY 2-30

2. SOLVED OBJECTIVE PROBLEMS 30-35

3. SOLVED SUBJECTIVE PROBLEMS 35-42

4. FOUNDATION BUILDER (OBJECTIVE) 42-48

5. FOUNDATION BUILDER (SUBJECTIVE) 48-53

6. GET EQUIPPED TO IIT-JEE 53-61

7. EXPERTISE ATTAINER 61-62

8. WINDOWS TO IIT-JEE 63-65

9. ANSWER KEY 66-67

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

1. INTRODUCTION There are a large number of objects around us which we can see and feel. Anything that occupies space and has mass is called matter.

Ancient indian and greek philospher’s believed that the wide variety of object around us are made from combination of five basic elements: Earth, Fire , Water , Air and Sky.

The Indian philosopher Kanad (600 BC) was of the view that matter was composed of very small, indivisible particle called “parmanus”.

Ancient Greek philosopher also believed that all matter was composed of tiny building blocks which were hard and indivisible.

The Greek philosopher Domocritus named these building blocks as atoms, meaning indivisible.All these people had their philosphical views about matter, these views were never put to experimental test.

It was John Dalton who firstly developed a theory on the structure of matter, latter on which was known as Dalton’s atomic theory. DALTON’S ATOMIC THEORY:

Matter is made up of very small indivisible particle called atoms. All the atoms of a element are identical in all respect i.e. mass, shape, size , etc. and atoms of

different elements are different in nature. Atoms cannot be created or destroyed by any chemical process.

Classification of matter

On the basis of physical behaviour

Liquids Gases

On the basis of chemical behaviour

Pure substances Mixtures

Element CompoundSolids

2. LAWS OF CHEMICAL COMBINATIONS

The combination of elements to form compounds is governed by the following five basic laws. 2.1 LAW OF CONSERVATION OF MASS It states that matter can neither be created nor destroyed.

This law was put forth by Antoine Lavoisier in 1789. He performed careful experimental studies for combustion reactions for reaching to the above conclusion. This law formed the basis for several later developments in chemistry. In fact, this was the result of exact measurement of masses of reactants and products, and carefully planned experiments performed by Lavoisier. Antoine Lavoisier

(1743−1794) Lavoisier stated that “during any physical or chemical change the total mass of the products produced is equal to the total mass of the reactants reacted ”. He showed that when mercuric oxide was heated the total mass of mercury and oxygen produced was equal to the total mass of mercuric oxide.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Illustration 1: When 20 g of NaHCO3 is heated, 12.62 g of Na2CO3 and 5.24g of CO2 is produced. How many grams of H2O is produced?

Solution: Total mass of NaHCO3 heated = 20 gms ; Total mass Na2CO3 produced = 12.62 gms Total mass of CO2 produced = 5.24 gms Mass of H2O produced = 20–12.62 –5.24 = 2.14 gms

2.2 LAW OF DEFINITE PROPORTIONS This law was given by, a French chemist, Joseph Proust. He stated that a given compound always contains exactly the same proportion of elements by weight.

Irrespective of the source, a given compound always contains same elements in the

same proportion. The validity of this law has been confirmed by various experiments. It is sometimes also referred to as Law of definite composition.

This law implies that irrespective of how a compound is prepared or from where the compound originates, it is always made up of the same elements combined in the same proportion by the weight.

For example, if water is taken from difference sources, such as rivers, oceans, wells etc. they all contain hydrogen and oxygen are combined in the same proportion by weight in it.

Illustration 2:When 50 g of ammonia is heated it gives 41.18 g of Nitrogen. When 10 g. of Nitrogen is combined with required amount of hydrogen it produces 12.14g ammonia. Show that the given data follows the law of constant compositions.

Solution: If 50g of Ammonia gives 41.18g of Nitrogen, then the percentage of Nitrogen in ammonia

is 10050

18.41 = 82.36%.

If 10g of Nitrogen gives 12.14 g of Ammonia then percentage of Nitrogen in ammonia is 10 100 82.37%

12.14 .

2.3 LAW OF MULTIPLE PROPORTIONS This law was proposed by Dalton in 1803. According to this law, if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers. For example, carbon and oxygen combine to form CO and CO2. In CO, 12 parts by mass of carbon combines with 16 parts by mass of oxygen while in CO2 12 parts by mass of carbon combines with 32 parts by mass of oxygen. Therefore the ratio of the masses of oxygen that combines with a fixed mass of carbon is 16:32 that is 1:2. Dalton

Illustration 3: Sodium and oxygen combine to form two compounds of which one is Na2O. The percentage of sodium in the other compound is 59%. Find the formula of this compound.

Joseph Proust ( 1754 – 1826)

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Solution: Percentage of sodium in Na2O is 10062

232

= 74.2% and percentage of oxygen is 25.8%.

Percentage of sodium in other compound is 59% while that of oxygen is 41%. This means that in the first compound (Na2O) if we take 100 gm then 25.5 gm of oxygen will be present

therefore the mass of sodium combining with 1g of oxygen would be 8.252.74 = 2.87 g.

Similarly in the second compound the mass of sodium combining with one gm of oxygen is

4159 = 1.44g. The ratio of masses of sodium combining with the fixed mass of oxygen is

2.87 : 1.44 = 2:1. Therefore formula of the other compound is Na2O2. 2.4 LAW OF RECIPROCAL PROPORTIONS

This law which was proposed by Ritcher (1792) states that “when two elements combine separately with fixed mass of third element then the ratio of their masses in which they do so is either the same or some whole number multiple of the ratio in which they combine with each other”. For example: Carbon, Sulphur and Oxygen form CO2, SO2 and CS2. In CO2 12 parts by mass of carbon combine with 32 parts by mass of oxygen while in SO2 32 parts by mass of Sulphur combine with 32 parts by mass of oxygen.Ratio of masses of carbon and sulphur which combine with fixed mass of oxygen is 12:32 or 3:8.In CS2 12 parts by mass of carbon combines with 64 parts by mass of sulphur therefore the ratio of mass of carbon to sulphur in carbon disulphide is 12:64 i.e. 3:16.

Therefore, the ratio is 83 :

163 or 2:1

2.5 GAY - LUSSAC LAW OF COMBINING VOLUMES This law which was proposed by Gay – Lussac states that, the volumes of gaseous reactants reacted and the volumes of gaseous products formed, all measured at the same temperature and pressure bear a simple ratio. For eg in the reaction involved in Haber’s Process (Nitrogen and hydrogen gases react to form ammonia)

2 2 3N g 3H g 2NH g1vol 3vol 2 vol

It is observed that the ratio of the volumes of N2 and H2 reacted and volume of NH3 produced is equal to 1:3:2 which is a simple ratio.

This law is applicable only for gaseous reactions and should not be used for non–gaseous reactants and products.

3. ATOMIC MASS & MOLECULAR MASS Analysis of water shows that it contains 88.89% oxygen and 11.11% of hydrogen by mass. Thus the ratio of masses of hydrogen and oxygen in water is 11.11: 88.89 or 1:8. Moreover the ratio of number of hydrogen and oxygen atoms in water molecule can be shown to be 2 : 1.

Therefore oxygen is 16 times heavier than hydrogen. Therefore relative atomic mass of oxygen is 16 units if we take mass of hydrogen atom as 1 unit. In 1961 International Union of Pure and Applied Chemists (IUPAC) selected the most stable isotope of carbon, C–12 as the standard for comparison of atomic masses of elements. The mass of C–12 is taken as 12 atomic mass unit.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

The scale in which the relative atomic masses of different elements are expressed is called atomic mass unit or amu.

Atomic mass Unit (amu) = 112

the mass of a carbon - 12 atom = 1.660539 x10-24 gm

126

112

Mass of one atom of theelementAtomic mass of anelementMass of one atomof C

One amu is also called one Dalton ( Da) Nowadays amu has been replaced by ‘u’ which is known as unified mass.

Molecular mass is the sum of atomic masses of the elements present in a molecule. It is obtained by

multiplying the atomic mass of each element by the number of its atoms and adding them together. For example, molecular mass of methane which contains one carbon atom and four hydrogen atoms can be obtained as follows:

Molecular mass of methane, (CH4) = (12.011) + 4 (1.008) = 16.043

Similarly, molecular mass of H2O= 2 × atomic mass of hydrogen + 1 × atomic mass of oxygen = 2 (1.008) + 16.00 = 18.016

4. MOLE CONCEPT

One mole is an amount of substance containing Avogadro's number of particles. Avogadro's number is equal to 602,214,199,000,000,000,000,000 or more simply, 6.02214199 × 1023.

A mole (symbol mol) is defined as the amount of substance that contains as many atoms, molecules, ions, electrons or any other elementary entities as there are carbon atoms in exactly 12 gm of 12C. The number of atoms in 12 gm of 12C is called Avogadro’s number (NA).

NA = 6.022 1023 Gay Lussac

One atomic mass unit (amu) = 23A

1gm 1gmN 6.022 10

= 1.66 10–24 gm = 1.66 10–27 kg

From mass spectrometer we found that there are 6.022 x 1023 atoms present in 12 gm of C – 12 isotope. The number of entities in 1 mol is so important that it is given a separate name and symbol known as Avogadro constant denoted by NA. i.e. on the whole we can say that 1 mole is the collection of 6.022 x 1023 entities. Here entities may represent atoms , ions, molecules or even pens, chairs , paper etc also include in this but as this number (NA) is very large therefore it is used only for very small things.

How big is a Mole? One mole of marbles would cover the entire Earth (oceans included) for a depth of three miles.

One mole of $100 bills stacked one on top of another would reach from the Sun to Pluto and back 7.5 million times.It would take light 9500 years to travel from the bottom to the top of a stack of 1 mole of $1 bills.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Gram Atomic Mass :

The atomic mass of an element expressed in grams is called gram atomic mass of that element. or

It is also defined as mass of 6.022 x 1023 atoms. or

It is also defined as the mass of one mole atoms. or It is also defined as the mass of 1 gram atom of the element.

For example for oxygen atom : Atomic mass of ‘O’ atom = mass of one ‘O’ atom = 16 amu Gram atomic mass = mass of 6.022 1023 ‘O’ atoms = 16 amu x 6.022 1023 = 16 x 1.66 10 -24 x 6.022 x 1023 16 gm/mole 24 231.66 10 6.022 10 1 Illustration 4: How many atoms of oxygen are their in 16 g oxygen.

Sol. 241.66 10 16 16x g 24

11.66 10 Ax N

Gram Molecular Mass : The molecular mass of a substance expressed in gram is called the gram-molecular mass of the

substance. or It is also defined as mass of 6.022 x 1023 molecules . or It is also defined as the mass of 1 mole molecules. or It is also defined as the mass of 1 gram molecule. For examples for ‘O2’ molecules : Molecular mass of ‘O2’ molecule = mass of one ‘O2’ molecule = 2 mass of one ‘O’ atom = 2 16 amu = 32 amu Gram molecular mass = mass of 6.022 1023 ‘O2’ molecules = 32 amu 6.022 1023 = 32 1.66 10-24 gm 6.022 1023 = 32 gm/mole Illustration 5: The molecular mass of H2SO4 is 98 amu. Calculate the number of moles of each element in

294 g of H2SO4. Sol. Gram molecular mass of H2SO4 = 98 gm

Moles of H2SO4 = 294 398

moles

H2SO4 H S O

one molecule 2 atoms one atom 4 atoms 1 x NA 2 x NA atoms 1 x NA atoms 4 x NA atoms one mole 2 mole one mole 4 mole 3 mole 6 mole 3 mole 12 mole

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Avogadro’s hypothesis : Equal volume of the gases have equal number of molecules ( not atoms) at same

temperature and pressure condition. S.T.P. ( Standard Temperature and Pressure) At S.T. P. condition Tempereture = 0oC or 273 K pressure = 1 atm = 760 mm of Hg and volume of one mole of gas at STP is found to be experimently equal to 22.4 litres which

is known as molar volume. Illustration 6: Calculate the volume in litre of 20 g hydrogen gas at STP.

Sol. No. of moles of hydrogen gas = Given massMolecular mass

= 20 102

gm mol

gm

Volume of hydrogen gas at STP = 10 x 22.4 lt = 224 lt.

T-map: Interconversion of mole-volume,mass and number of particles

Molex NA

NA

Number Volume at STPx22.4 lt

22.4 lt

Mass

x mol. wtx At. wt

mol. wt

At. wt

of entities

Methods of Calculations of Mole

a) Number of moles of molecules = massMolecular

gmin.Wt

b) Number of moles of atoms = massAtomicgmin.Wt

c) Number of moles of gases = Volumeat STPStandard molar Volumeat STP

(Standard molar volume at STP = 22.4 lit)

d) Number of moles of particles e.g. atoms, molecules, ions etc = numberAvogardroparticlesofNumber

e) For a compound AxBy, 1 mole of compound will have x moles of A and y moles of B.

If volume of gas is given along with its temperature (T) & pressure (P) then PVnRT

where R = 0.0821 lit-atm/mol K & P is in atmosphere & T in Kelvin. Do not use the expression for solids/liquids for eg. H2O at 10oC.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

STP or NTP conditions Standard conditions means that temperature is 0C or 273K and pressure is one atmosphere or 760mm of Hg.

1 gm - atom is same as 1 mole of an atom & hence will have wt equal to atomic wt expressed in gms. 1 gm – molecule is same as 1 mole of the molecule & hence will have wt equal to molecular wt

expressed in gms. 1 gm – Ion is same as 1 mole of an ion & hence will have wt equal to ionic wt

Now, can you differentiate b/w 1 gm – atom oxygen & 1gm – molecule oxygen. Remember 1 gm of atom & 1gm– atom are two different phrases. Former is mentioning wt (equal to

1gm) & latter is mentioning moles. e.g. (1) “ x g atom of nitrogen “ = x moles of N atom = ANx number of N atoms

(2) “ x g molecule of nitrogen” = x moles of N2 molecules = ANx molecules of N2

= AN2x number of N atom

Illustration 7: How many g atom and no. of atoms are there in (a) 60 g carbon (b) 224.4 g Cu? Given : At. Weight of C and Cu are 12 and 63.6 respectively. Avogadro’s no. = 6.02 x 1023 .

Solution : g atom .

wtat wt

and No. of atoms . . ..

wt Av Noat wt

(a) For 60 g C : g atom 60 512

No. of atoms 23

2360 6.02 10 30.1 1012

(b) For 224.4 g Cu : g atom 224.4 3.5363.6

No. of atoms 23

23224.4 6.02 10 21.24 1063.6

Illustration 8: Find the number of g atoms and weight of an element having 2 x 10 23 atoms. At.

Weight of element is 32. Solution NA atoms have 1 g atom

232 10 atoms have 23

23

2 10 0.336.023 10

g atom

NA atoms of elements weigh 32 g

232 10 atoms of element weigh 23

23

2 10 32 10.6276.023 10

g

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Illustration 9: How many mole and molecules of O2 are there in 64g O2? What is the mass of one

molecule of O2? Solution moles of O2 in 32 g O2 = 1

In 64 g O2 moles 64 1 232

mole

32 g O2 contains 6.022 x 1023 molecules

64 g O2 contains 23

236.022 10 64 12.04 1032

molecules

NA molecules of O2 weigh 32 gm

1 molecule of O2 weighs 2323

32 5.313 106.022 10

gm .

Illustration 10: From 200 mg of CO2 , 1021 molecules are removed. How many g and mole of

CO2 are left? Solution : 6.022 x 1023 molecules of CO2 = 44 g

1021 molecules of CO2 21

23

44 106.022 10

g

= 7.31 x 10 – 2 = 73. 1 mg CO2 left = 200 – 73.1 = 126.9 mg

Also Mole of CO2 left 3

3. 126.9 10 2.88 10. 44

wt

M wt

5. PERCENTAGE COMPOSITION

So far, we were dealing with the number of entities present in a given sample. But many a time, the information regarding the percentage of a particular element present in a compound is required. Suppose an unknown or new compound is given to you, the first question you would ask is: what is its formula or what are its constituents and in what ratio are they present in the given compound? For known compounds also, such information provides a check whether the given sample contains the same percentage of elements as is present in a pure sample. In other words, one can check the purity of a given sample by analysing this data.

Let us understand it by taking the example of water (H2O). Since water contains hydrogen and

oxygen, the percentage composition of both these elements can be calculated as follows:

Mass % of an element = Mass of that element × 100Molar mass

Illustration 11: A mixture of FeO and 3 4Fe O when heated in air to constant weight, gains 5% in its

weight. Find out composition of mixture.

Solution: 2 2 312 FeO + O Fe O2

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

3 4 2 2 312 Fe O + O 3Fe O2

Let weight of FeO = a g and weight of 3 4Fe O bg a + b = 100 … (1) Now, 2 72g FeO gives 2 3Fe O 160g

a g FeO gives 2 3160 aFe O g

144

Also, 2 232 g of 3 4Fe O gives 2 3Fe O 160 3g

b g 3 4Fe O gives 2 3160 3 bFe O g

464

160a 160 × 3 × b + = 105144 464

… (2)

Solving eqs. (1) and (2), a = 20.25 g FeO = 20.25 % b = 79.75 g and 3 4Fe O 79.75% Illustration12 : A sample of chalk contains impurity in form of clay which loses 14.5% of its weight

of water on prolong heating. 5 g of chalk on heating shows a loss in weight (due to evolution of 2CO and water) by 1.507 g. Calculate % of chalk in the sample.

Solution: Chalk has Clay + 3CaCO (Chalk) a g b g a + b = 5 … (1) On heating (1) Clay loses water (2) 3CaCO loses 2CO

Now, Weight loss of water by a g clay = 14.5 × a100

Weight loss of 2CO by b g 344 × bCaCO =

100

14.5 a 44 b 1.507100 100

… (2)

Solving eqs. (1) and (2), a = 2.349 g b = 2.651 g

% of chalk, i.e., 32.651CaCO 100 53%

5

5.1 AVERAGE ATOMIC WEIGHT For elements, which have atoms with different relative masses (isotopes) the atomic mass is taken as

weighted mean of the atomic masses. General formula for Average atomic weight = %of isotope×Atomic weight of Isotope

For example, chlorine contains two isotopes of atomic masses 35 amu and 37 amu. The relative abundance of these two is in the ratio of 3:1. Thus the atomic mass of chlorine is the average of

different relative masses. Therefore atomic mass of chlorine is equal to 4

137335 = 35.5 amu.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Illustration13: Calculate average atomic wt. of silicon if relative abundance is 92.23% Si28,

4.77%Si29, 3% Si30

Solution : Av at wt 92.23 28 4.77 29 3 30100

= 28.1 amu

Illustration14: Calculate % abundance of Ag109 if it is known that silver exist in only two isotopes

Ag107& Ag109 & average atomic weight = 108.5 Solution : let the % abundance of Ag109 be x

109 100 107108.5

100x x

10850 10700 2x 75%x Shortcut to calculate % abundance when an element X is having only two isotopes

XA & XB & average atomic wt is Xavg

% 100B

avgAA b

X wt of Xof X

wt of X wt of X

% obtained above is mole %. Try to obtain a similar expression for mass percent

5.2 AVERAGE MOLECULAR WEIGHT For homogenous mixture of several substances having number of moles, ni & molecular mass M(i)

for ith species, the average molecular weight is given as

Average Molecular wt. 1 1 2 2

1 2

n M n M ....n n .........

. .Total weight

Total no of moles i i

i

n Mn

Illustration 15: Dry air has a molar composition as 20% O2 & 80% N2. Calculate average molecular wt. of dry air

Solution :. Av. Molecular wt 20 32 80 28 28.8100

5.3 MINIMUM MOLECULAR WEIGHT It is the molecular weight of a compound shown by presence of minimum number of atoms [i.e. for monomer = 1 , Dimer = 2 , Trimer = 3 , Tetramer = 4 ] e.g. Insulin contains 3.4% S, find its MMW. 100g insulin contains 3.4g S (sulphur). Assuming it to be monomer [as nothing specified] one sulphur atom 32 amu (atomic mass) ,We can write , 3.4 g S is contained by 100g insulin

32g S is contained by

32

4.3100 = MMW

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

6. DETERMINATION OF MOLECULAR & EMPERICAL FORMULAE The molecular formula of a compound may be defined as the formula which gives the actual number of atoms of various elements present in the molecule of the compound. For example, the molecular formula of the compound glucose can be represented as C6H12O6. A molecule of glucose contains six atoms of carbon, twelve atoms of hydrogen and six atoms of oxygen.

In order to find out molecular formula of a compound, the first step is to determine its empirical formula from the percentage composition.

The empirical formula of a compound may be defined as the formula which gives the simplest whole number ratio of atoms of the various elements present in the molecule of the compound. The empirical formula of the compound glucose (C6H12O6) is CH2O which shows that C,H and O are present in the simplest ratio of 1:2:1.

Empirical formula mass of substance is equal to the sum of atomic masses of all the atoms in the empirical formula of the substance.

Molecular formula is a whole number multiple of empirical formula. Thus

Molecular formula = (Empirical formula) n where n = 1,2,3…

mass formula Empiricalmass Molecualr

formula Empiricalformula Molecularn

6.1 STEPS FOR WRITING THE EMPIRICAL FORMULA

The percentage of the elements in the compound is determined by suitable methods and from the data collected, the empirical formula is determined by the following steps:

i) Divide the percentage of each element by its atomic mass. This will give the relative number of moles of various elements present in the compound.

ii) Divide the quotients obtained in the above step by the smallest of them so as to get a simple ratio of moles of various elements.

iii) Multiply the figures, so obtained by a suitable integer if necessary in order to obtain a whole number ratio.

iv) Finally write down the symbols of the various elements side by side and put the above number as the subscripts to the lower right hand corner of each symbol. This will represent the empirical formula of the compound.

Steps for writing the molecular formula i) Calculate the empirical formula as described above. ii) Find out the empirical formula mass by adding the atomic masses of all the atoms present in the

empirical formula of the compound. iii) Divide the molecular mass (determined experimentally by some suitable method) by the empirical

formula mass and find out the value of n.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

6.2 DENSITY For liquid and solids :

Absolute density = massvolume

Relative density or specific gravity = o

densityof the substancedensity of water at 4 C

For Gases :

Absolute density ( mass / volume) = Molar mass PM

Molar volume RT

Where P is pressure of gas, M = mol. Wt. of gas , R is the gas constant , T is the temperature. Relative density or Vapour density : Vapour density is defined as the density of the gas with respect to hydrogen gas at the same

temperature and pressure.

Vapour density = 22

gas

HH

PMgasd RT

PMdRT

V. D. = 2

2gas gas

H

M MM

Mgas = 2 V.D

Relative density can be calculated w.r.t. to other gases also. Illustration16 : What is the V.D. of SO2 with respect to CH4

Solution:. V.D. =

2

4

SO

CH

MW

MW V.D. = 64 4

16

Illustration17 : 11.2 litre of the particular gas at S.T.P. weight 16 gram. What is the V. D. of gas. Solution wt.of 11.2 litre = 16 gram.

moles = 11.2 1622.4

M

32M gm mole V.D.= 32 162

Illustration 18: A substance, on analysis, gave the following percentage composition: Na = 43. 4%, C = 11.3%, O = 45.3%. Calculate its empirical formula. {Na = 23, C = 12, O = 16]

Solution:

Element SYMBOL

%

Atomic Mass Relative number of moles

Simple ratio of moles

Simplest whole no. ratio

Sodium Na 43.4 23 88.123

4.43 2

94.088.1

2

Carbon C 11.3 12 94.012

3.11 1

94.094.0

1

Oxygen O 45.3 16 83.216

3.45 3

94.083.2

3

Therefore, the empirical formula is Na2CO3.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Illustration 19:A compound has the following composition: Mg = 9.76%, S = 13.01%, O = 26.01%,

H2O = 51.22%. What is its empirical formula? [Mg = 24, S = 32, O = 16, H = 1] Solution

Element Symbol % age Atomic Mass

Relative number of moles

Simple ratioof moles

Simplest Wholeno. ratio

Magnesium Mg

9.76 24 406.02476.9

1406.0406.0

1

Sulphur S 13.01 32 406.032

01.13 1

406.0406.0

1

Oxygen O 26.01 16 625.116

01.26 4

406.0625.1

4

Water H2O 51.22 18 846.218

22.51 7

406.0846.2

7

Hence, the empirical formula is MgSO4. 7H2O.

Illustration 20:What is the simplest formula of the compound which has the following percentage composition. Carbon 80%, Hydrogen 20%. If the molecular mass is 30, calculate its molecular formula.

Solution:

Element % age At. Mass Relative

number of moles


Simples whole no.ratio

C 80 12 66.61280

166.666.6

1

H 20 1 201

20 3

66.620

3

Empirical formula is CH3

Empirical mass = 12 1 + 1 3 = 15 21530

mass formula Empricalmass Molecularn

Molecular formula = Empirical formula 2 = CH3 2 = C2H6

Illustration 21: A compound on analysis gave the following percentage composition:

C = 54.54%, H = 9.09%, O = 36.36%. The vapour density of the compound was found to be 44. Find out the molecular formula of the compound.

Solution: Calculation of empirical formula.

Element % age At. Mass Relative number of moles

Simple ratio of mole


C 54.54 12 53.412

54.54 2

27.253.4

2

H 9.09 1 09.9109.9

427.209.9

4

O 36.36 16 27.216

36.36 1

27.227.2

1

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Empirical formula is C2H4O. Calculation of molecular formula: Empirical formula mass = 12 2+14+161 = 44 Molecular mass = 2 Vapour density = 2 44 = 88

24488

mass formula Empiricalmass Molecularn

Molecular formula = Empirical formula n = C2H4O 2= C4H8O2.

Illustration22: An organic compound on analysis gave the following data: C = 57.82%, H = 3.6%, and the rest is oxygen. Its vapour density is 83. Find its empirical and molecular formula.

Solution: Calculation of empirical formula:

Element % age At. Mass Relative number of moles



C 57.82 12 80.412

82.57 2

4.28.4 4

H 3.60 1 60.3160.3

5.13.26.3 3

O 38.58 16 40.216

58.38 1

4.24.2 2

Empirical formula is C4H3O2. Empirical formula mass = 124+ 13 + 2 16 = 83 Molecular mass = 2 V.D. = 2 83 = 166

283

166 mass formula Emprical

mass Molecularn

Molecular formula = Empirical formula n = C4H3O2 2 = C8H6O4

Illustration 23: 2.746 gm of a compound gave on analysis 1.94 gm of silver, 0.268 gm of sulphur and 0.538 gm of oxygen. Find the empirical formula of the compound. (At masses : Ag = 108, S = 32, O = 16)

Solution: To calculate percentage composition. Percentage composition of the compound is calculated as under:

Silver = 6.70100746.294.1

%

Sulphur = 75.9100746.2268.0

% Oxygen = 0.538 1002.746

= 19.6%

To calculate empirical formula:

Element % At. Mass Relative number of moles

Simplest ratio of moles

Simplest whole no.ratio

Ag 70.6 108 654.0108

6.70 2

305.0654.0

2

S 9.75 32 305.03275.9

1305.0305.0

1

O 19.6 16 22.116

6.19 4

305.022.1

4

Empirical formula is Ag2SO4

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

7. STOICHIOMETRY The word ‘stoichiometry’ is derived from two Greek words - stoicheion (meaning element)

and metron (meaning measure). Stoichiometry, thus, deals with the calculation of masses (sometimes volumes also) of the reactants and the products involved in a chemical reaction. Before understanding how to calculate the amounts of reactants required or those produced in a chemical reaction, let us study what information is available from the balanced chemical equation of a given reaction. Let us consider the combustion of methane. A balanced equation for this reaction is as given below:

CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O(g) The above balance reaction gives the following information:

a) For every 1 mole of CH4, 2 mole of O2 will be required to produce 1 mole of CO2 and 2 moles of H2O. this signifies Mole – Mole relation

b) For every 16 gms of CH4 , 64 gms of O2 will be required to produce 44gms of CO2 and 36 gms of H2O this signifies Mass – Mass relation

c) Ratio of moles of CO2 : H2O at any time = 1 : 2 d) There will be no change in total mass of all reactants and products at any time for any chemical

reaction. e) For the above reaction only, there will be no change in total number of moles of all reactants and

products.

In order to solve the problems based on chemical calculations the following steps, in general,are quite helpful.

i) Write the balanced chemical equation.

ii) Write the atomic mass/molecular mass/moles/molar volumes of the species involved in calculations.

iii) Calculate the result by applying unitary method.

7.1 INTERPRETATION OF BALANCED CHEMICAL EQUATIONS Once we get a balanced chemical equation then we can interpret a chemical equation by following ways

Mass – mass analysis Mass – volume analysis Volume – volume analysis Now you can understand the above analysis by following example Mass – mass analysis Consider the reaction

3 22 2 3KClO KCl O According to stoichiometry of the reaction Mass – mass ratio : 2 x 122.5 : 2 x 74.5 : 3 x 32

Or 3 2 122.52 74.5

Mass of KClOMass of KCl

3

2

2 122.53 32

Mass of KClOMass of O

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Illustration 24 : 367.5 gram KClO3 ( M = 122.5) is heated. How many gram KCl and oxygen is produced. Solution : Balanced chemical equation for heating of KClO3 is 3 22 2 3KClO KCl O Mass – mass ratio: 2 x 122.5 gm 2 x 74.5 gm : 3 x 32 gm

3 2 122.52 74.5

Mass of KClOMass of KCl

367.5 122.5

74.5W

WKCl = 3 x 74.5 = 223.5 gm

3

2

2 122.53 32

Mass of KClOMass of O

367.5 2 122.5

3 32W

Woxygen = 144 gm

Mass – volume analysis : Now again consider decomposition of KClO3 3 22 2 3 KClO KCl O mass volume ratio 2 x 122.5 gm : 2 x 74.5 gm : 3 x 22.4 lt at S.T.P. we can use two relation for volume of oxygen.

3

2

2 122.53 22.4

Mass of KClOvolume of O at STP lt

…(i)

And 2

2 74.53 22.4

Mass of KClvolume of O at STP lt

…(ii)

Illustration 25: Calculate the volume of 2O and volume of air needed for combustion of 1 kg carbon

at STP. Solution: 2 2C + O CO

12 g C requires 2O = 22.4 litre of 2O = 1 mole of 2O = 32 g of 2O

1000 g C requires 222.4 1000O litre

12

= 1866.67 litre 2O

2air OV 5 V 5 1866.67 9333.35litre

Volume – Volume Relationship : It relates the volume of gaseous species ( reactants or product )

with the volume of another gaseous species ( reactant or product ) involved in a chemical reaction. Illustration26: What volume of oxygen gas at NTP is necessary for complete combustion of 20 litre

of propane measured at 0oC and 760 mm pressure. Solution : The balanced equation is

3 8 2 2 25 3 4

1 51 5

C H O CO H Ovol vollitre litre

1 litre of propane requires = 5 litre of oxygen 20 litre of propane will require = 5 x 20 = 100 litre of oxygen at 760 mm pressure and 0oC.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Illustration 27: The percentage by volume of 3 8C H in a mixture of 3 8C H , 4CH and CO is 36.5.

Calculate the volume of 2CO produced when 100 mL of the mixture is burnt in excess of 2O .

Solution : 3 8 2 2 2C H 5O 3CO 4H O l

4 2 2 2CH 2O CO 2H O l

2 21CO O CO2

Let a mL, b mL and c mL be volumes of 3 8 4C H ,CH and CO respectively in 100 mL given simple, then a + b + c = 100 and a = 36.5

Now 2CO is formed as a result of combustion of mixture.

Vol. of 2CO formed = 3 8 2

4 2

2

1 vol. C H gives 3 vol. CO3a + b + c 1vol. CH gives 1 vol. CO

1 vol. CO gives 1 vol. CO

3 36.5 100 36.5 = 173 mL

7.2 LIMITING REAGENT

In many situations one of the reactants is present in excess therefore some of this reactant is left over on completion of the reaction. For example, consider the combustion of hydrogen.

2H2(g) + O2(g) 2H2O(g)

Suppose that 2 moles of H2 and 2 moles of O2 are available for reaction. It follows from the equation that only 1 mole of O2 is required for complete combustion of 2 moles of H2 ; 1 mole of O2 will, therefore, be left over on completion of the reaction. The amount of the product obtained is determined by the amount of the reactant that is completely consumed in the reaction. This reactant is called the limiting reagent. Thus, limiting reagent may be defined as the reactant which is completely consumed during the reaction.

In the above example H2 is the limiting reagent. The amount of H2O formed will, therefore, be determined by the amount of H2. Since 2 moles of H2 are taken, it will form 2 moles of H2O on combustion.

The best method to identify limiting reagent is by dividing given moles of each reactant by their stoichiometric coefficient, the one with least ratio is limiting reagent. It is particularly useful when number of reactants are more than two.

Illustration 28: How much magnesium sulphide can be obtained from 2.00 g of magnesium and 2.00 g of sulphur by the reaction Mg + S MgS? Which is the limiting reagent? Calculate the amount of the reactants which remains unreacted.

Solution : First of all each of this masses are converted into moles:

2.00 g of Mg = 3.24

00.2 = 0.0824 moles of Mg

2.00 g of S = 1.32

00.2 = 0.0624 moles of S

From the equation, Mg + S MgS, it follows that one mole of Mg reacts with one mole of S. We are given more moles of Mg than of S. Therefore, Mg is in excess and some of it will remain unreacted when the reaction is over. S is the limiting reagent and

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

will control the amount of product. From the equation we note that one mole of S gives

one mole of MgS, so 0.0624 mole of S will react with 0.0624 mole of Mg to form 0.0624 mole of MgS.

Molar mass of MgS = 56.4 g Mass of MgS formed = 0.0624 56.4 g = 3.52 g of MgS Moles of Mg left unreacted = 0.0824 –0.0624 moles of Mg = 0.0200 moles of Mg Mass of Mg left unreacted = moles of Mg molar mass of Mg = 0.0200 24.3 g of Mg = 0.486 g of Mg Illustration 29: 4 mole of 3MgCO is reacted with 6 moles of HCl solution. Find the volume of 2CO

gas produced at STP. The reaction is 3 2 2 2MgCO 2HCl MgCl CO H O

Solution: From the reaction 3 2 2 2MgCO 2HCl MgCl CO H O

Given moles 4 mole 6 mole Given mole ratio 2 : 3 Stoichiometric Coefficient ratio 1 : 2 There should be one limiting reagent. To find the limiting reagent, divide the given moles by stoichiometric coefficient. 3MgCO HCl

4 41 6 3

2

HCl is limiting reagent.

2moles of CO producedmoles of HCl = 2 1

moles of 2CO produced = 3 moles

volumes of 2CO produced at S.T.P, = 3 22.4 67.2L

Illustration 30 : What weight of AgCl will be precipitated when a solution containing 4.77 g NaCl is

added to a solution of 5.77g of 3AgNO ?

Solution: 3 3AgNO + NaCl AgCl + NaNO

mmoles. mixed 5.77 1000170

4.77 100058.5

= 33.94 = 81.54 0 0 mmoles. left 0 47.60 33.94 33.94 mmoles of AgCl formed = 33.94

w 1000 33.94143.5

AgClW 4.87g

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

7.3 CALCULATION INVOLVING PERCENT YIELD

In general, when a reaction is carried out in the laboratory we do not obtain actually the theoretical amount of the product. The amount of the product that is actually obtained is called the actual yield. Knowing the actual yield and theoretical yield, the percentage yield can be calculated as

% yield = yieldlTheoretica

yieldActual 100

The actual amount of any limiting reagent consumed in such incomplete reactions is given by [% yield × given moles of limiting reagent] [For reversible reactions]. For irreversible reaction with % yield less than 100, the reactants are converted to product (desired) and waste.

7.4 PRINCIPLE OF ATOM CONSERVATION (POAC)

In chemical reaction atoms are conserved, so moles of atoms shall also be conserved. This is known as principle of atomic conservation. This principle is helpful in solving problems of nearly all stoichiometric calculations e.g.

KClO3(s) KCl(s) + O2(g)

Applying POAC for K atoms

Moles of K atoms in KClO3 = Moles of K atoms in KCl

Since one mole of KClO3 contains 1 mol of K atom. Similarly 1 mol of KCl contains one mole of K

atoms. 3

3

3

KClO KClKClO KCl

KClO KCl

W W1 n 1 n 1M M

(Mass-mass relationship)

Applying POAC for ‘O’ atoms Moles of O atom in KClO3 = Moles of O atom in O2

3 3KClOn = 2

2On 3

3

KClO 2

KClO

W Vol.of O atSTP3 × = 2 × M Standard Molar Volume

(Mass volume relationship of reactant and product)

In this way applying POAC we can break the chemical equation into a number of arithmetic equations without balancing the chemical equation. Moreover number of reactions and their sequence from reactants to products are not required. It is important to note that POAC can be applied for the atoms which remain conserved in chemical reaction.

Please note that the balanced reaction is essential in Mole method of solving .Only while using POAC (which would be applicable only in certain cases), balanced reaction is not required.

You can use POAC for all atoms in the compound only IF all the reactants and products are known.

Apart from the usual Mole method used for solving stoichieometric problems, sometimes POAC method are also used depending upon the type of problem. Students are advised to have an understanding of the utility of each of the method for solving stoichiometric problems.

Illustration 31: A sample of KClO3 on decomposition yeilded 448 mL of oxygen gas at NTP Calculate : (i) weight of oxygen produced , (ii) weight of KClO3 originaly taken (iii) weight of KCl produced ( K = 39 , Cl = 35.5 and O = 16 )

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Solution (i) Mole of oxygen = 448 0.0222400

Wt. of oxygen = 0.02 x 32 = 0.64gm (ii) KClO3 KCl + O2

Applying POAC for O atoms, Moles of O atoms in KClO3 = moles of O atoms in O2 3 (moles of KClO3 )= 2 (moles of O2) (1 mole of KClO3 contains 3 moles of O and 1 mole of O2 contains 2 moles of O)

3

3

. . ( )3 2. . 22.4

wt of KClO vol at NTP litremol wt of KClO

3. . ( )3 2122.5 22.4

wt of KClO vol at NTP litre

Wt. of KClO3 = 1.634 g (iii) Again applying POAC for K atoms, Moles of K atoms in KClO3 = 1 x moles of KCl ( 1 mole of KClO3 contains 1 mole of K and 1 mole of KCl contains 1 mole of K)

3

3

. .1 1. . . .

wt of KClO wt of KClmol wt of KClO mol wt of KCl

Wt. of KCl = 0.9937 g. Illustration 32: 27.6 g of K2CO3 was treated by a series of reagents so as to convert all of its carbon to

K2Zn3 [ Fe(CN)6]2. Calculate the weight of the product. Solution : 2 3 2 3 6 2[ ( ) ]Several

stepsK CO K Zn Fe CN

Since C atoms are conserved, applying POAC for C atoms, Moles of C in K2CO3 = moles of C in K2Zn3 [ Fe(CN)6 ]2

1 x mole of K2CO3 = 12 x moles of K2Zn3 [ Fe(CN)6]2

2 3 2 3 6 21 1 & 1 [ ( ) ]12

mole of K CO contains mole C mole of K Zn Fe CNcontains moleof C

2 3

2 3

wt. of K CO wt. of the product = 12 × mol. wt. of K CO mol. wt. of product

Wt. of 2 3 6 2

27.6 698K Zn Fe CN = × = 11.6 g.138 12

[mol. wt. of 2 3K CO 138 and mol. wt. of 2 3 6 2K Zn Fe CN = 698 ]

Illustration 33: In a gravimetric determination of P, an aqueous solution of dihydrogen phosphate ion 2 4H PO is treated with a mixture of ammonium and magnesium ions to precipitate magnesium ammonium phosphate, 4 4 2Mg NH PO . 6H O . This is heated and decomposed to magnesium pyrophosphate, 2 2 7Mg P O which is weighed. A solution of 2 4H PO yielded 1.054 g of 2 2 7Mg P O . What weight of 2 4NaH PO was present originally?

Solution : heated22 4 4 4 4 2 2 2 7NaH PO Mg NH Mg NH PO .6H O Mg P O

Since P atoms are conserved, applying POAC for P atoms, moles of P in 2 4NaH PO = moles of P in 2 2 7Mg P O

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

2 4 2 2 71 × moles of NaH PO = 2 × moles of Mg P O

( 1 mole of 2 4NaH PO contains 1 mole of P and 1 mole of 2 2 7Mg P O contains 2 moles of P)

2 2 72 4

2 4 2 2 7

wt. of Mg P Owt. of NaH PO = 2 × mol. wt. of NaH PO mol. wt. of Mg P O

2 4wt. of NaH PO 1.054 = 2 × 120 222

Wt. of 2 4NaH PO = 1.14 g.

Illustration 34: What weight of CO is required to form 2 10Re CO from 2.50 g of 2 7Re O according to the unbalanced reaction:

2 7 2 210Re O CO Re CO CO

(Re = 186.2, C = 12 and O = 16 )

Solution : Suppose the relative moles of each reactant and product are as follows (just for convenience)

2 7 2 210bmoles dmolesa moles cmoles

Re O CO Re CO CO

Applying POAC for Re atoms,

Moles of Re in 2 7Re O = moles of Re in 2 10Re CO

2 7 2 102 × moles of Re O = 2 × moles of Re CO

2a = 2c

or a = c … (i)

Applying POAC for C atoms,

Moles of C atoms in CO = moles of C in 2 10Re CO + moles of C in 2CO

2 2101 × moles of CO = 10 × moles of Re CO + 1 × moles of CO

Or b = 10c + d …. (ii)

Applying POAC for O atoms,

Moles of O in 2 7Re O + moles of O in CO

= moles of O in 2 10Re CO + moles of O in 2CO

2 77 × moles of Re O + 1 × moles of CO

2 10 2= 10 × moles of Re (CO) + 2 × moles of CO

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

or 7a + b = 10 c + 2d …. (iii)

From the eqns. (i), (ii), (iii), we get,

17a = b

i.e., 2 717 × moles of Re O = moles of CO

2 7mol. wt. of Re O = 484.42.50 wt. of CO in g17 × = .484.4 28 mol. wt. of CO = 28

Wt. of CO = 2.46 g.

Illustration 35: 1.84 g of a mixture of 3CaCO and 3MgCO was heated to a constant weight. The constant weight of the residue was found to be 0.96 g. Calculate the percentage composition of the mixture. (Ca = 40, Mg = 24, C = 12, O = 16)

Solution : On heating 3CaCO and 3MgCO , one of the products, 2CO , escapes out.

We have,

3 3 2yg 0.96 y gx g 1.84 x gCaCO MgCO CaO MgO CO

Applying POAC for Ca atoms,

Moles of Ca atoms in 3CaCO = moles of Ca atoms in CaO

31 × moles of CaCO = 1 × moles of CaO

x y100 56

3CaCO 100CaO 56

… (i)

Again applying POAC for Mg atoms,

Moles of Mg in 3MgCO = moles of Mg in MgO

31 × moles of MgCO = 1 × moles of MgO

3MgCO 841.84 x 0.96 y84 40 MgO 40

…. (ii)

From eqns. (i) and (ii), we get x = 1 g, y = 0.84 g

31% of CaCO = × 100 = 54.34 %

1.84

3% of MgCO = 45.66 %

Second Method Apply POAC for C atoms.

Illustration 36: How many grams of oxygen (O2) are required to completely react with 0.200 g of hydrogen (H2) to yield water (H2O)? Also calculate the amount of water formed. (At. Mass H = 1; O = 16).

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Solution : The balanced equation for the reaction is 2H2 + O2 2H2O 2 mol 1 mol 2 mol 4 g 32 g 36 g Now, 4g of H2 require oxygen = 32 g

0.200 g of H2 require oxygen = 4

32 0.200 = 1.6 g

Again, 4g of H2 produce H2O = 36 g

0.200 g of H2 produce H2O = 200.04

36 = 1.8 g.

Illustration 37: What volume of oxygen at N.T.P. can be produced by 6.125 g of potassium chlorate

according to the reaction 2KClO3 2KCl + 3O2. Solution: The given chemical equation is : 2KClO3 2KCl + 3O2 2 mol 2 mol 3 mol 2 122.5 g 3 22.4L at N.T.P Now 245 g of KClO3 produce oxygen at N.T.P. = 3 22.4 L

6.125 g of KClO3 produce oxygen = 245

4.223 6.125 = 1.68 L at N.T.P.

Ilustration 38: What mass of zinc is required to produce hydrogen by reaction with HCl which is enough to produce 4 mol of ammonia according to the reactions.

Zn + 2HCl ZnCl2 + H2 3H2 + N2 2NH3

Solution: The given equations are Zn + 2HCl ZnCl2 + H2 3H2 + N2 2NH3 From the equations it is clear that 2 mol of NH3 require = 3 mol of H2 ; 3 mol of H2 require = 3 mol of Zn Thus, 2 mol of NH3 require = 3 mol of Zn = 3 65 g of Zn

4 mol of NH3 require = 2653 4 = 390 g of Zn.

8. CONCENTRATION TERMS Many chemical reactions occur in the solution state and hence stoichiometric problems may

sometimes involve different concentration terms of solutions to give an idea of amount of solute and solvent present in the solution. There are various ways to represent concentration of the solution as mentioned below.

In a solution the designation of solute and solvent is often a matter of convenience, however many a times the one present in smaller quantity is termed as solute. Also, a solution may have more than one solute but solvent cannot be more than one.

All the concentration terms given below are applicable only when there is a homogenous solution or when solute completely dissolves in the solvent.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

The following concentrations terms are used to expressed the concentration of a solution. These are

a. Molarity (M)

b. Molality (m)

c. Mole fraction (x)

d. % calculation

e. Normality (N) (will be discussed in volumetric analysis)

f. ppm

Please remember that all these concentration terms are related to one another. By knowing one concentration term you can also find the other concentration terms. Let us discuss all of them one by one.

8.1 MOLARITY (M)

The number of moles of a solute dissolved in 1 L (1000 ml) of the solution is known as the molarity of the solution.

i.e., Molarity of solution = number of molesvolume of solution in litre

Let a solution is prepared by dissolving w gm of solute of mol. wt. M in V ml water.

Number of moles of solute dissolved = wM

V ml water have wM

mole of solute

1000 ml water have in ml

w 1000M V

Molarity (M) = in ml

w × 1000Mol. wt of solute × V

Some other relations may also useful.

Number of millimoles = in ml

mass of solute × 1000 = Molarity of solution × VMol. wt. of solute

Molarity of solution may given as :

Number of millimole of soluteTotal volume of solution in ml

Molarity is a unit that depends upon temperature. It varies inversely with temperature.

Mathematically : Molarity decreases as temperature increases.

1Molaritytemperature volume

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

If a particular solution having volume 1V and molarity = 1M is diluted to 2V mL the

1 1 2 2M V M V 2M : Resultant molarity

If a solution having volume 1V and molarity 1M is mixed with another solution of same solute having volume 2V mL & molarity 2M

Then 1 1 2 2 R 1 2M V M V M V V

RM = Resultant molarity = 1 1 2 2

1 2

M V M VV V

Illustration 39: 149 gm of potassium chloride (KCl) is dissolved in 10 Lt of an aqueous solution. Determine the molarity of the solution (K = 39, Cl = 35.5)

Solution : Molecular mass of KCl = 39 + 35.5 = 74.5 gm

Moles of KCl = 149gm 274.5gm

Molarity of the solution = 2 0.2M10

Illustration 40: What volume of a 3.0 M HCl solution be mixed with 500 mL of a 7 M HCl solution to prepare a HCl solution whose molarity will be 4.0?

Solution : Let V mL of 3.0 M HCl solution is taken, then

3 V + 500 × 7 = 4 × 500 + V V = 1500 mL

Illustration 41: Determine molarity of a solution obtained by mixing 50 mL of a 0.26 M 2 4H SO

solution with another 150 mL 0.48 M 4 4H SO solution. Solution : It is a case of mixing of two solutions of different molarities. Applying the mixing

formula:

1 1 2 23

3

M V M V 50 0.26 150 0.48MV 50 150

= 0.425 M

Illustration 42: What volume of a 5.00 M 2 4H SO solution should be added to a 150 mL 1.0 M

2 4H SO solution to obtain a solution of sulphuric acid of molarity 2.5? Solution : It is again a case of mixing of two solutions. Let us assume that V mL of the stock

solution of 2 4H SO is added.

5V + 150 × 1.0 = 2.5 × 150 + V

2.5 V = 225

225V 90 mL2.5

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Illustration 43: A 150 mL 0.25 M NaCl solution, 250 mL 0.45 M 2CaCl solution and a 100 mL 0.60

M 3AlCl solution are mixed together and diluted to a final volume of 750 mL by

adding enough water. Determine molarity of chloride ion Cl in solution assuming that all three salts are completely soluble as well as completely dissociated.

Solution : First we need to calculate total mmoles of Cl from the three salt solutions: mmoles of Cl ion from NaCl = 37.5 mmoles of Cl ion from 2 2CaCl = 2 × mmoles of CaCl = 2 × 112.5 = 225 mmoles of Cl ion from 3 3AlCl = 3 × mmoles of AlCl 3 60 180 Total mmoles of Cl in final solution = 37.5 + 225 + 180 = 442.5

Molarity of –

– mmoles of Cl 442.5Cl = = = 0.59 MmL of solution 750

8.2 MOLALITY (m) The molality is the number of moles of solute present in one Kg of solvent

1000.

solute

solvent

wmM wt w gm

Molality is independent of temperature changes.

Illustration 44 : 255 gm of an aqueous solution contains 5 gm of urea. What is the concentration of the solution in terms of molality. (Mol. wt. of urea = 60)

Solution : Mass of urea = 5 gm

Molecular mass of urea = 60

Number of moles of urea = 5 0.08360

Mass of solvent = (255 – 5) = 250 gm

Molality of the solution = Number of moles of solute × 1000Mass of solvent in gram

= 0.083 1000 0.332250

Illustration 45: The molarity and molality of a solution are M and m respectively. If the molecular weight of the solute is M , calculate the density of the solution in terms of M, m and M .

Solution : Let weight of solute be w g and weight of solvent is W g and volume of solution is

V mL.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

w 1000MM V

… (1)

w 1000mM W

… (2)

w WDV

… (3)

By Eq. (1) MM Vw1000

… (4)

By Eq. (2) w 1000 MM V 1000WM m 1000 M m

by Eq. (4)

MVWm

… (5)

By Eq. (3)

MV MM Vm 1000D

V

1 MD Mm 1000

Try deriving and get the relation between mole fraction and molarity

Molarity(M) and Molality(m) for Pure Substances: 1. Water : Let the sample of water has 1000 ml Mass of water = 1000 gm [density of water = 1gm/mL.]

Moles of water 100018

mol

100018 55.551

Molarity M

& molality =

100018 55.551

molm

kg

2. Pure ethanol : d gm/ml (density of ethanol) (C2H5OH) let volume of ethanol taken be 1000 ml. wt of ethanol in 1000 ml = 1000 d gm

Mol of ethanol =1000d46

Molarity = 100046

d

& molality of ethanol =

1000100046

1000 461000

d mol

d kg

Parts per million (ppm) Amount of solute ( in g ) with 106 g solvent Parts per billion ( ppb) amount of solute ( in g ) with 109 g solvent

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

8.3 MOLE FRACTION (X) The ratio of number of moles of the solute or solvent present in the solution and the total number of

moles present in the solution is known as the mole fraction of substance concerned. Let number of moles of solute in solution = n Number of moles of solvent in solution = N

Mole fraction of solute 1nX

n N

Mole fraction of solvent 2NX

n N

Also 1 2X X 1 Mole fraction is a pure number. It will remain independent of temperature changes.

8.4 PERCENTAGE CONCENTRATION The concentration of a solution may also be expressed in terms of percentage in the following

way. % weight by weight (w/w): it is given as mass of solute present in per 100 gm of solution.

i.e. mass of solute in gm% w/w = × 100mass of solution in gm

% weight by volume (w/v) : It is given as mass of solute present in per 100 ml of solution

i.e. mass of solute in gm% w/v = × 100volume of solution in ml

% volume by volume (V/V) : It is given as volume of solute present in per 100 ml solution.

i.e Volume of solute in ml% V/V = × 100Volume of solution in ml

Illustration 46: 0.5 g of a substance is dissolved in 25 g of a solvent. Calculate the percentage amount

of the substance in the solution. Solution : Mass of substance = 0.5 g Mass of solvent = 25 g

Percentage of the substance (w/w) 0.5 100 1.960.5 25

Illustration 47: 320cm of an alcohol is dissolved in 380cm of water. Calculate the percentage of

alcohol in solution. Solution : Volume of alcohol = 320cm Volume of water = 380cm

percentage of alcohol 20 100 2020 80

%

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Illustration 48 : Calculate the weight of iron which will be converted into its oxide by the action of 18 g

of steam by the reaction 2 3 4 23Fe 4H O Fe O 4H Solution : The reaction occurs as: 2 3 4 23Fe 4H O Fe O 4H Mole ratio of reaction suggests:

2

Mole of Fe 3 = Mole of H O 4

Mole of 18 3 3Fe18 4 4

weight of 3Fe 56 42g4

SOLVED OBJECTIVE PROBLEMS

Problem 1: Assuming that petrol is octane (C8H18) and has density 0.8 g/ml, 1.425 litre of petrol on complete combustion will consume

(A) 50 mole of O2 (B) 125 mole of O2 (C) 100 mole of O2 (D) 200 mole of O2 Solution: Mass of octane = 1.425 103 0.8 g

Moles of octane = 1425 0.8114 = 10 moles

C8H18 + 2

25 O2 8CO2 + 9H2O

From the equation it can be seen

For 1 mole octane oxygen required = 2

25 moles

for 10 mole octane oxygen required = 2

25 10 = 125 moles (B)

Problem 2: Weight of 1 atom of an element is 6.644 10–23 g. What is number of atoms of element in 40 kg of it.

(A) 103 g atom (B) 102 g atom (C) 104 g atom (D) 10 g atom

Solution: Weight of Avogadro number ( AN ) of atoms of the element = 6.644 10–23 6.022 1023 = 40 g 40 g = weight of 1g atom 40 103 g = weight of 103 g atom (A)

Problem 3: 8 g of O2 has the same number of molecules as in (A) 7 g of CO (B) 14 g of CO (C) 28 g of CO (D) 11 g of CO2

Solution: 8 g O2 = 328 i.e.,

41 moles of O2 ( mol.wt. of O2 = 32)

Now, 7 g of CO = 287 mole of CO ( Mol.wt.of CO = 28) =

41

28 g of CO = 2828 = 1 mole of CO 14 g of CO =

2814 =

21 mole of CO

11 g of CO2 = 4411 =

41 mole CO2 2mol.wt.of CO 44

Same number of moles means same number of molecules . (A) and (D) Problem 4: A compound contains 3.2% of oxygen. The minimum mol wt. of the compound is (A) 300 (B) 440 (C) 350 (D) 500

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Solution: The compound must contain at least one oxygen atom So, a minimum of 1 g atom of oxygen

will be present in 1 g molecule i.e., 1 mole of the compound. If M is the mol. wt. of the compound then since 16 is the atomic mass of oxygen so

minimum of 16 g of oxygen will be present in M g of the compound

Thus, % of oxygen = M16 100 or 3.2 =

M10016 or M = 500 (D)

Problem 5: Arrange the following in order of increasing mass (i) 3.0115 1023 molecules of white phosphorus 4P (ii) 10 moles of H2 gas

(iii) 1 g molecule of anhydrous Na2CO3 (iv) 33.6 L of CO2 gas at S.T.P. (A) ii i iv iii (B) iii vi i ii (C) i ii iv iii (D) i iv iii ii

Solution: i) At. wt. of P = 31 and atomicity of P in white P is 4 Mol.wt. of white P = 31 4 = 124 6.023 1023 molecules of white P weigh 124 g

3.0115 1023 molecules of white P weigh 015.3023.6

124 = 62 g

ii) Wt. of 1 mole of H2 gas = 2 g wt. of 10 moles of H2 gas = 2 10 = 20 g iii) 1 g molecule of anhydrous Na2CO3 = Mol wt. of Na2CO3 in g = 106 g iv) At STP 22.4 lit CO2 weighs 44 g

33.6 lit CO2 weighs 4.22

44 33.6 g= 66 g

So, the correct choice , ii i iv iii (A)

Problem 6: Arrange the following in order of decreasing mass i. 1F atom ii. 1 N atom iii. 1 O atoms iv. 1 H atom (A) i iii iv ii (B) iv > ii > iii >i (C) i iii ii iv (D) iii i ii iv

Solution: Mass of 1F atom 2319

6.022 10

(since At.wt. of F = 19)

Mass of 1N atom 2314

6.022 10

Mass of 1 H atom 231

6.022 10

Mass of 1 O atom 2316

6.022 10

i iii ii iv (C)

Problem 7: The number of millimoles contained in 0.160 g of NaOH is : (A) 0.04 (B) 0.4 (C) 4 (D) 40

Solution: Number of millimoles wt in gm 0.16 10001000 4mol wt 40

(C)

Problem 8: The number of moles present in 1m3 of any gas at NTP is : (A) 32. 4 (B) 54.6 (C) 44.6 (D) 28.2

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Solution: 22.4 lit of any gas contains 1 mole or 3104.22 m3 of any gas contains 1 mole

1 m3 of any gas contains 4.22

103 moles = 44.6 (C)

Problem 9: The minimum quantity of H2S needed to precipitate 63.5 gm of Cu2+ will be nearly. (A) 63.5 gm (B) 31.75 gm (C) 34 gm (D) 32 gm

Solution: Since the ppt is of CuS, hence all the S atoms will be conserved

34

2SHw = moles of Sulphur = moles of Copper =

5.635.63

SH2w = 34 gm (C)

Problem 10: How many molecules are present in 12 L of liquid CCl4? The density of the liquid is 1.59 g cm–3

(A) 7.44 1026 (B) 0.744 1026 (C) 1.59 1026 (D) 15.9 1026

Solution: 1 cc of CCl4 contains 1.59 gms of it = 1.59 0.0103 moles154

12 L of liquid CCl4 will contain = 12 1000 0.0103 0.744 1026 molecules of it (B)

Problem 11: 13.4g of a sample of unstable hydrated salt: Na2SO4nH2O was found to contain 6.3g of

water. Determine the number of water of crystallisation. (A) 6 (B) 5 (C) 7 (D) 8

Solution: sampleof.wtwaterof.wt =

compoundtheof.wt.molcompoundtheinwaterof.wt

6.3 18 713.4 142 18

n nn

(C)

Problem 12: The mass of 3BaCO produced when excess of 2CO is bubbled through a solution of 0.205 mol 2Ba OH is

(A) 81 g (B) 40.5 g (C) 20.25 g (D) 162 g

Solution: 2 3 22Ba OH CO BaCO H O

Mol. wt. of 3BaCO 137 12 16 3 197

wt. of substanceNo. of mole=mol wt.

1 mole of 2Ba OH gives 1 mole of 3BaCO

0.205 mole of 2Ba OH will give 0.205 mole of 3BaCO

wt. of 0.205 mole of 3BaCO will be 0.205 197 40.385gm 40.5gm (B)

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Problem 13: What will be the volume of 2CO at NTP obtained on heating 10 grams of (90% pure)

limestone (a) 22.4 litre (B) 2.016 litre (C) 2.24 litre (D) 20.16 litre

Solution: 3 210 gm

CaCO CaO + CO

990% pure= 9 gm = mole100

Mole of 3 2CaCO = moles of CO = 0.09 mole

At NTP vol. 2CO 0.09 22.4 2.016 L (B)

Problem 14: The number of water molecules present in a drop of water (volume 0.0018 ml) at room

temperature is

(A) 196.022 10 (B) 181.084 10 (C) 174.84 10 (D) 235.023 10

Solution: MassDensity = Volume

Weight of 0.0018ml = 0.0018 gm (since 2H Od 1gm ml )

–4weight 0.0018No. of moles = = = 1 × 10Molecular weight 18

No. of water molecules = 23 –46.022 × 10 × 1 × 10 19= 6.023 × 10 A

Problem 15: From 160 g of 2SO g sample, 241.2046 10 molecules of 2SO are removed then find out

the volume of left over 2SO g at STP.

(A) 11.2L (B) 22.4 L (C) 44.82 L (D) 61.2 L

Solution: To calculate no. of moles in 160 gm sample of 2SO .

W 160gn 2.5moleM 64g / mole

To calculate no. of moles of 2SO removed.

24

23A

given no. of molecules 1.2046 × 10n = = = 2moleN 6.022 × 10

no. of mole of 2SO left = 2.5 – 2 = 0.5 mole

volume of 2SO left at STP.

V = n × 22.4 L = 11.2 L (A)

Problem16: 340 g 3NH (M = 17) when decompose how many litres of nitrogen gas is produced at STP.

(A) 224 Lt (B) 448 Lt (C) 226Lt (D) 322 Lt

Solution: 3 2 22NH N + 3H

Mole – mole ratio: 2 mole : 1 mole 3 mole

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Mass volume ratio: 2 17g : 22.4 L at STP

2

3

Volume of N at STP 22.4L = mass of NH 2 × 17

222.4LVol. of N at.STP ×340=224L.2 × 17

(A)

Problem17: 518 gm of an aqueous solution contain 18 gm of glucose (mol.wt. = 180). What is the molality of the solution.

(A) 0.2 (B) 0.4 (C) 0.6 (D) 0.8 Solution: Mass of solution = 518 g. Mass of solute = 18 g Mass of solvent = 518 – 18 = 500 g

Number of moles of glucose = 18 0.1180

Number of moles of soluteMolality of the solution = × 1000mass of solvent in gram

0.1 1000 0.2m500

(A)

Problem18: 0.25 g of a substance is dissolved in 6.25 g of a solvent. Calculate the percentage amount of the substance in the solution.

(A) 3.85% (B) 6.85% (C) 48.5 % (D) 91.2 % Solution: Mass of solute = 0.25 g Mass of solvent = 6.25 g Mass of solution = 0.25 +6.25 = 6.50 g.

Percentage amount of the solute = 0.25 100 3.85%6.50

(A)

Problem19: The vapour density of a mixture containing 2NO and 2 4N O is 38.3 at 027 C . Calculate the mole of 2NO in 100 mole mixture.

(A) 33.48 (B) 44.66 (C) 76.46 (D) 91.22

Solution: Mol. wt. of Mixture of 2NO and 2 4N O 38.3 2 76.6

Let a mole of 2NO are present in mixture

wt of 2NO + wt of 2 4N O = Total wt of mixture

a 46 100 a 92 100 76.6

a = 33.48 mole (A) Problem20: A compound contains 28% N and 72% of a metal by weight. Three atoms of metal

combine with two atoms of N. Find the atomic weight of metal. (A) 24 (B) 26 (C) 32 (D) 48

Solution: Given that, 3 23M + 2N M N

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Let a is at. wt, of metal (3a + 28)g 3 2M N has metal = 3a gm

100 g 3 2M N has metal = 3a 1003a 28

gm

3a 100 723a 28

a = 24 (A)

Problem21: On heating 1.763 g of hydrated 2BaCl to dryness, 1.505 g of anhydrous salt remained. What is the formula of hydrate?

(A) 2 2.2BaCl H O (B) 2 2.3BaCl H O (C) 2 2.4BaCl H O (D) 2 2.5BaCl H O

Solution: Δ2 2 2 2

Mol.wt(208+18n) 208BaCl . nH O BaCl + nH O

(208 + 18n)g 2 2 2BaCl .nH O gives = 208 g BaCl

1.763 g 2 2 2208 × 1.763BaCl . nH O = g BaCl208 + 18n

1.505

n 1.98 2

formula is 2 2BaCl .2H O (A)

SOLVED SUBJECTIVE PROBLEMS Problem 1: 0.05 gm of commercial sample of KClO3

on decomposition liberated just sufficient oxygen for complete oxidation of 20 ml CO at 27°C and 750 mm pressure. Calculate % of KClO3 in sample

Solution: Applying PV = nRT for CO:

Moles of CO (nCO) = 4PV 750 20 8.01 10RT 760 0.0821 300 1000

CO + 21 O2 CO2,

2

CO

O

n 1n 1 2

= 2

moles of O2 required 2

nn COO2

= 2

1001.8 4

2KClO3 2KCl + 3O2 As 3 moles of O2 are given by 2 mole of KClO3

2

1001.8 4 mol of O2 are given by = 2

1001.832 4 mol of KClO3

= 2.66 10–4 mol of KClO3 Weight of KClO3 = 2.66 10–4 122.5 gm = 3.26 10–2 gm

% of KClO3 in the mixture = 23.26 10 100 65%

0.05

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Problem 2: Upon mixing 45.0 ml of 0.25 M lead nitrate solution with 25 mL of 0.10 M chromic sulphate solution precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Also calculate the concentrations of the species left behind in final solution. Assume that lead sulphate is completely insoluble.

Solution: 3Pb(NO3)2 + Cr2(SO4)3 3PbSO4 + 2Cr(NO3)3 mmoles. (initial) 45 0.25 25 0.1 0 0

= 11.25 = 2.5 (limiting reagent) mmoles (final) 3.75 0 7.5 5

Hence, mmoles of PbSO4 precipitated = 7.5

Final Volume of Solution = 70 ml

[Pb2+] = 3.7570

= 5.36 10–2 M

[NO3–] = 3.75 2 5 3 22.5

70 70

= 0.32 M [Cr3+] = 2.5 2 570 70

= 7.14 10–2 M

Problem 3: A mixture of propane and methane is contained in a vessel of unknown volume V at a temperature T and exerts a pressure of 320 mm Hg. The gas is burnt in excess O2 and all the carbon is recovered as CO2. The CO2 is found to have a pressure of 448 mm Hg in a volume V, at the same temperature T. Calculate mole fraction of propane in mixture.

Solution: Let the moles of propane (C3H8) = n1 Moles of methane CH4 = n2 We know that PV = nRT 320 × V = (n1+n2) RT ...(1) after combustion

OH4CO3O5HC 2

n322

n83

11

OH2COO2CH 2

n22

n4

22

total moles of CO2 formed = (3n1+n2) Once again we have PV = nRT 448 ×V = 1 23n n RT ….(2)

Dividing eqn (2) by (1) we have

1 2 1

1 2 2

3448 0.25320

n n nn n n

Mol fraction of C3H8 = 1

1 2

nn n

= 2

2 2

0.25 0.20.25

nn n

Problem 4: An organic compound contains 69% C and 4.8% hydrogen the remainder being oxygen. Calculate the masses of Carbon dioxide and Water formed by the complete combustion of 0.20 g of the substance.

Solution.

2

2

substance

substance

12 100 % 44%44 12 100

COCO

W C WC WW

69 44 0.20 0.50612 100

g

2

2

substacne

substance

2 100 % 18%18 2 100

H O

H O

W H WH WW

4.8 18 0.2 0.08642 100

g

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Problem 5: K – 40 is a naturally occurring radioactive isotope having natural abundance 0.012% of

potassium isotopes. How many K – 40 atoms do you ingest by drinking one cup of whole milk containing 370 mg K ?

Solution: Amount of K – 40 in 370 mg K = 370 0.012 mg100

= 0.0444 mg 40 g K – 40 has atoms of K – 40 = 236.022 10

30.0444 10 g K – 40 has atoms = 23 36.022 10 0.0444 10

40

176.69 10 atoms Problem 6: P and Q are two elements which forms 2 3P Q and 2PQ . If 0.15 mole of 2 3P Q weights 15.9

g and 0.15 mole of 2PQ weights 9.3 g. what are atomic weights of P and Q? Solution: Let at. Wt. of P and Q are a and b respectively. Mol. wt. of 2 3P Q 2a 3b and Mol. wt. of 2PQ a 2b

now given that 0.15 mole of 2 3P Q weight 15.9 g

15.92a 3b0.15

wt. molemol.wt.

Similarly 9.3a 2b0.15

Solving these two equations b = 18 a = 26 Problem 7: From the following reaction sequence

2 2

3

3 4

Cl 2KOH KCl KClO H O

3KClO 2KCl KClO4KClO 3KClO KCl

Calculate the mass of chlorine needed to produce 100 g of 4KClO . Solution: Taking the given reactions

3 4

3

2 2

4KClO 3KClO KCl

12KClO 8KCl 4KClO

12Cl 24KOH 12KClO 12H O 12KCl

Adding 2 4 212Cl 24KOH 3KClO 21KCl 12H O

Mol. wt. of 4KClO 138.5

43 138.5gKClO is formed by = 212 71g Cl

100 g 4KClO will be formed = 12 71 1003 138.5

= 205.05 g Problem8: A hydrocarbon contains 10.5 g of carbon per g of H. One litre vapours of hydrocarbon

at 0127 C and 1 atm pressure weights 2.8g. find molecular formula of hydrocarbon.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Solution: Given, C : H10.5 1

Total = 11.5

Now from wPV RTM

for vapours of compound

2.81 1 0.0821 400M

Mol.wt. of compound = 92 11.5 g has 1 g H

92 g has 92 1 8g H 8g11.5

atom of H

And thus, 92 g has 84 g carbon = 7g atom carbon Molecular formula = 7 8C H Problem9: The reaction, 22C O 2CO is carried out by taking 24 g of carbon and 96 g of 2O ,

find out: (a) Which reactant is left in excess? (b) How much of it is left? (c) How many mole of CO are formed? (d) How many g of other reactant should be taken so that nothing is left at the end of

reaction? Solution: 22C + O 2CO

Mole before reaction 2412

9632

= 2 = 3 0 Mole ratio of Mole after reaction 0 2 2 2C : O : CO :: 2 :1: 2

(a) 2O is left in excess.

(b) 2 mole of 2O or 64 g 2O is left. (c) 2 mole of CO or 56 g CO is formed. (d) To use 2O completely total 6 mole of carbon or 72 g carbon is needed.

Problem10: A polystyrene, having formula 3 6 3 8 8 nBr C H C H was prepared by heating styrene with tribromobenzoyl peroxide in the absence of air. If it was found to contain 10.46% bromine by weight, find the value of n.

Solution: Let the weight of polystyrene prepared be 100 g.

No. of mole of Br in 100 g of polystyrene = 10.4680

= 0.1308 mole

From the formula of polystyrene, we have, No. of moles of Br = 3 6 3 8 8 n3 × mole of Br C H C H

wt. 3 1000.1308 3mol.wt. 315 104n

n = 19

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Problem11: A sample of gaseous hydrocarbon occupying 1.12 litre at NTP, when completely burnt

in air produced 2.2 g 2CO and 1.18 g 2H O . Calculate the weight of hydrocarbon taken and the volume of 2O at NTP required for its combustion.

Solution: Formula of hydrocarbon be a bC H

a b 2 2 2b bC H a O aCO H O l4 2

22.4 litre a bC H gives 44a g 2CO

1.12 litre a bC H gives = 244a 1.12 g CO

22.4

44a 1.12 2.222.4

a = 1

22.4 litre or 1 mole a bC H gives 2b 18g H O2

1.12 litre a bC H gives 2b 18 1.12 g H O2 22.4

b 18 1.12 1.82 22.4

4b

Hydrocarbon is 4CH

Wt. of 1.12 litre 4CH at NTP = 16 1.12 0.8g22.4

Also moles of 2O used in combustion for 22.4 litre 4CH

ba4

= 1 + 1 = 2 mole 22.4 litre 4CH requires 2 mole 2O or 2 22.4 litre 2O

1.12 litre 4CH requires = 2 22.4 1.1222.4

= 2.24 litre 2O

Problem12: The density of 3M solution of 2 2 3Na S O is 11.25g mL . Calculate,

(a) the % by weight of 2 2 3Na S O

(b) mole fraction of 2 2 3Na S O

(c) the molalities of Na and 22 3S O ions.

Solution: Given 2 2 3Na S O has molarity = 3 mole 1litre

Mole of 2 2 3Na S O 3

Wt. of 2 2 3Na S O 3 158 474g And V of solution = 1 litre = 1000 mL Wt. of solution = 1000 1.25 1250g Wt. of water = 1250 – 474 = 776 g

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

(a) % by wt. of 2 2 32 2 3

2 2 3 2

wt. of Na S ONa S O = wt. of Na S O + wt. of H O

474 100 37.921250

(b) Mole fraction of 2 2 32 2 3

2 2 3 2

Mole of Na S ONa S O = Mole of Na S O + Mole of H O

3 0.0653 776 /18

(c) Molality of +

+ Mole of NaNa = × 1000wt. of water in g

6 1000 7.732776

Molality of 22 3

3 1000S O 3.865776

Problem 13: Calculate molality of 1 litre solution of 93% 2 4H SO ( w v ). The density of solution is

1.84 g 1mL . Solution: Given 2 4H SO is 93% ( w v ).

Wt. of 2 4H SO 93g Volume of solution = 100 mL Wt. of solution = 100 1.84 184g Wt. of water = 184 – 93 = 91 g

MoleMolality=Wt. of water in kg

93 10.4291981000

Problem14: A solution of density 11.6g mL is 67% by weight. What will be the % by weight of the solution of same acid if it is diluted to density 11.2gmL ?

Solution: Consider 100 mL of solution 67% by weight Wt. of solution = 100 1.6 160g

Wt. of solute = 67 160 107.2g100

Now suppose X g of 2H O added in it 2H Od 1gm ml

Wt. of new solution = (160 + X) g Also volume of new solution = (100 + X) mL Wt. of new solution = 100 X 1.2g 120 1.2X g

160 +X = (120 + 1.2X) g X = 200 mL or 200g

% by weight of new solution =

107.2 100 29.78%160 200

Problem15: 10 mL of a solution of KCl containing NaCl gave on evaporation 0.93 g of the mixed salt

which gave 1.865 g of AgCl by the reaction with 3AgNO . Calculate the quantity of NaCl in 10 mL of solution.

Solution: 3AgNo

agNaCl AgCl

3AgNo

bgKCl AgCl

a + b = 0.93 … (1) 58.5 g NaCl gives = 143.5 g AgCl

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

a g NaCl gives = 143.5 a g AgCl58.5

Similarly, 74.5 g KCl gives = 143.5 g AgCl

b g KCl gives = 143.5 b74.5

143.5a 143.5 b 1.86558.5 74.5

... (2)

Solving Eqs. (1) and (2), a = 0.14 g , b = 0.79 g Problem16: A sample of metal chloride weighing 0.22 g required 0.51 g of 3AgNO to recipitate the

chloride completely. The specific heat of the metal is 0.057. find out the molecular formula of the chloride, if the symbol of the metal is ‘M’.

(Ag = 108, N = 14, O = 16, Cl = 35.5) Solution: Let the valency of the metal M be x x 3MCl AgNO AgCl As it is clear from the question that both Ag and Cl atoms are conserved, applying POAC for

Ag atoms, we get Moles of Ag atoms in 3AgNO = moles of Ag atoms in AgCl

31 × moles of AgNO = 1 × moles of AgCl

0.51 moles of AgCl170

…(1)

Again applying POAC for Cl atoms, we get Moles of Cl atoms in xMCl = moles of AgCl

xx × moles of MCl = 1 × moles of AgCl

x

0.22x × = moles of AgClmol. wt. of MCl

Now, at. Wt. of 6.4M 112.30.057

(Dulong and Petit;s law)

molecular weight of xMCl = (112.3 + 35.5x)

0.22x × = moles of AgCl112.3+35.5x

…(2)

From equations (1) and (2), x 3 Since valency is a whole number x = 3 and the formula of the metal chloride is 3MCl .

Problem17: Calculate the resultant molarity of following: (a) 200 ml 1 M HCl + 300 ml water (b) 1500 ml 1 M HCl + 18.25 g HCl (c) 200 ml 1M HCl + 100 ml 0.5 M HCl.

Solution: (a) 1 1R

R

M V 1 200M 0.4MV 500

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

(b) Milimoles of HCl = 18.25 1000 50036.5

Milimoles of HCl present in solution = 1500 1 1500 Total milimoles = 1500 + 500 = 2000. MV = 2000

2000 2000 4M MV 1500 3

(c) 1 1 2 2R

1 2

M V M V 1 200 0.5 100 5M MV V 300 6

FOUNDATION BUILDER (OBJECTIVE PROBLEM) MOLE CONCEPT 1. Maximum no. of hydrogen atoms are in: (A) 16g of CH4 (B) 31.2g of C6H4 (C) 34.2g of C12H22O11 (D) 36g of C6H12O6 2. 3 mol of ammonia contains: (A) 18 gm of hydrogen (B) 42 gm of nitrogen (C) both (D) None 3. Total no. of protons in 36 ml of water at 4°C (where of water = 1 g/ml) is (A) 20 (B) 16 (C) 20 NA (D) 16NA 4. In which of the following pairs both members have same no. of atoms (A) 1 gm O2, 1 gm O3 (B) 1 gm N2, 2 gm N (C) Both (D) None 5. The molecular wt. of green vitriol is M0. The wt. of 103 NA molecules of it is (A) M0 gm (B) M0 mg (C) 103 M0 gm (D) 103 M0mg 6. A sample contains 200 atoms of hydrogen, 0.05 gm atom of nitrogen, 1020 gm atom of oxygen. What

is the approximate no. of total atoms (A) 200 (B) 6223 (C) 31022 (D) none of these 7. The element A at wt.=75 and B at wt. =32 combine to form a compound X. If 3 mol of B combine with

2 mol of A to give 1 mol of X, the weight of 5 mol of X is (A) 246 gm (B) 1230 amu (C) 1.23 kg (D) None of these 8. Which of the following has the highest mass (A) 1g atom of C (B) ½ mole of CH4 (C) 10ml of H2O (D) 3.0111023 atom of oxygen.

9. Which one of the following samples contains the largest number of atoms. (A) 2.5 mole CH4 (B) 10 mole He (C) 4 mole SO2 (D) 1.8 mole S8

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

10. Which of the following has greatest no. of oxygen atoms (A) 10 mole water (B) 200gm of C12H22O11 (C) 144 gm O3 (D) 2.5 mole H2SO3

11. Which of the following substances contains the greatest mass of chlorine (A) 5 gm Cl2 (B) 60 gm NaClO3 (C) 0.10 mole of KCl (D) 0.5 mole of Cl2

12. Which of the following samples contains the smallest no. of atoms (A) 1 g of CO2 (g) (B) 1 g of C8H18 (l) (C) 1 g of C2H6 (g) (D) 1 g of LiF (s) 13. The wt. of NO having the same no. of oxygen atom present in 9.2 gm of 2NO is (A) 9.2 gm (B) 6 gm (C) 12 gm (D) 24 gm

14. The no. of atoms in 52 a.m.u. of He is (A) 131023 (B) 1.31023 (C) 13 (D) 103

15. The no. of electrons in 2 gm ion of nitrate ion (NO3-) is

(A) 64 (B) 64NA (C) 32 (D) 32NA

16. The mass of carbon present in 0.5 moles of K4[Fe(CN)6] is (A) 1.8 g (B) 18 g (C) 3.6 g (D) 36 g

17. The largest no. of molecules is in (A) 28 g of CO2 (B) 46 g of C2H5OH (C)36 g of H2O (D) 54 g of N2O5

18. How many electrons are present in 180 gm. of water (A) 1 mole (B) 10 moles (C) 18 moles (D) 100 moles

19. How many molecules of H2O are contained in 2.48 g of Na2S2O3.5H2O (at.wt. of Na=23,S=32) (A) 31020 (B) 31021 (C) 31022 (D) 31023

20. The no. of silver atoms present in a 90% pure silver wire weighing 10 g. is (at.wt. of Ag=108) (A) 8x1022 (B) 0.62x1023 (C) 5x1022 (D) 6.2x1029

21. The no. of moles of carbon dioxide which contain 8g of oxygen is (A) 0.50 mole (B) 0.20 mole (C) 0.40 moles (D) 0.25 moles

22. Which of the following weighs the least (A) 0.2 g atom of N (at. wt. N=14) (B) 3x1023 atoms of C (at. wt. C = 12) (C) 1 mole atoms of S (at. wt. S=32) (D) 7 g silver (at. wt. Ag=108) 23. Which one of the following statements is not correct? (A) One gm atom of silver equals 108 gms (B) One mole of CO2 and NH3 at NTP occupies same volume (C) One mole Ag weighs more than one mole of Zn (D) One gm molecule of CO2 is 44 times heavier than one molecule of CO2

24. A mixture contains n moles of H2 and 2n moles of CH4. The ratio of no. of C:H atoms in the mixture is : (A) 1/5 (B) 2/3 (C) 4/5 (D) 1/3

25. The charge on 1 gram ion of Al3+ is (e represents magnitude of charge on 1 electron) (A) 1/27 NAe coulomb (B) 1/3 NAe coulomb (C) 1/9 NAe coulomb (D) 3 NAe coulomb

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

26. The number of molecules of water in 333 g of Al2(SO4)3.18H2O is (A) 186.021023 (B) 96.021023 (C) 18 (D) 36 27. The number of water molecules present in a drop of water weighing 0.018 g is (A) 6.021026 (B) 6.021023 (C) 6.021020 (D) 6.021019 28. If NA is Avogadro’s number, then the number of valence electrons in 4.2 g of nitride ion (N3-) is

(Given One atom of N has 5 valence electrons) (A) 2.4 NA (B) 4.2 NA (C) 1.6 NA (D) 3.2 NA

29. A person adds 3.42 of sucrose (C12H22O11) in his cup of tea to sweeten it. How many atoms of carbon does he add?

(A) 132.44 1021 atoms (B) 66.22 1021 atoms (C) 0.1 atoms (D) 72.27 1021 atoms

30. The total number of protons in 8.4 g of MgCO3 is (NA = 6.021023) : (A) 2.521022 (B) 2.521024 (C) 3.011024 (D) 3.011022

31. 4.4g of 2CO and 2.24 litre of 2H at STP are mixed in a container. The total number of molecules present in the container will be

(A) 236.022 10 (B) 231.2044 10 (C) 2mole (D) 246.023 10

32. Which sample contains the largest number of atoms: (A) 4 101mg of C H (B) 21mg of N (C) 1mg of Na (D) 1mL of water

33. The atomic weight of a triatomic gas is a. The correct formula for the number of moles of gas in its w g is:

(A) 3wa

(B) w3a

(C) 3wa (D) a3w

34 Number of atoms in 558.5g Fe at.wt 55.85 is:

(A) Twice that in 60g carbon (B) 226.023 10

(C) Half in 8g He (D) 23558.5 6.023 10 35. How many moles of magnesium phosphate, 3 4 2Mg PO will contain 0.25mole of oxygen atoms?

(A) 0.02 (B) 23.125 10 (C) 21.25 10 (D) 22.5 10 PERCENTAGE COMPOSITION AND GRAVIMETRIC ANALYSIS

36. Cortisone is a molecular substance containing 21 atoms of carbon per molecule. The weight percentage of carbon in cortisone is 69.98%. What is the molecular weight of cortisone?

(A) 176.5 (B) 252.2 (C) 287.6 (D) 360.1

37. A partially dried clay mineral contains 8% water. The original sample contained 12% water and 45% silica. % of Silica in the partially dried sample is nearly

(A) 50% (B) 49% (C) 55% (D) 47%

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

38. A compound contains 28% nitrogen and 72% metal by mass 3 atoms of the metal combine with 2 atoms of nitrogen. The atomic mass of metal is

(A) 36 (B) 20 (C) 24 (D) 36

39. The hydrated salt Na2SO4.xH2O undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of x will be

(A) 5 (B) 3 (C) 7 (D) 10 EMPERRICAL AND MOLECULAR FORMULA

40. The simplest formula of a compound containing 50% by mass of element X (at. wt. 10) and 50% by mass of element Y (at. wt. 20) is ;

(A) XY (B) X2Y (C) XY2 (D) X2Y3

41. The hydrated salt Na2SO410H2O undergoes X% loss in weight on heating and becomes anhydrous. The value of X will be

(A) 10 (B) 45 (C) 56 (D) 70 42. An oxide of iodine (I =127) contains 25.4 g of iodine for 8 g of oxygen. Its formula could be (A) 2 3I O (B) 2I O (C) 2 5I O (D) 2 7I O

43. An organic compound on analysis was found to contain 0.014% of nitrogen. If its molecule contains two N atoms, then the molecular mass of the compound

(A) 200 (B) 2000 (C) 20,000 (D) 200000

44. The chloride of a metal contains 71% chlorine by weight and the vapour density of it is 50. The atomic weight of the metal will be

(a) 29 (b) 58 (c) 35.5 (d) 71

DENSITY AND VAPOUR DENSITY

45. Vapour density of air is (considering air as 80% 2N and 220%O ) (A) 0.001293 (B) 1.293 (C) 14.4 (D) 28.9 46. The density of chlorine relative to air is (A) 2.44 (B) 3 (C) 71 (D) 4 47. A gaseous oxide contains 30.4% of nitrogen, one molecule of which contains one nitrogen atom. The

density of the oxide relative to oxygen gas is (A) 0.9 (B) 1.44 (C) 1.50 (D) 3.0 STOICHIOMETRY 48. If two mole of methanol (CH3OH) completely burns to carbon dioxide and water, the weight of water

formed is about (A) 22 g (B) 18 g (C) 36 g (D) 72 g 49. How many g of KCl would have to be dissolved in 60 g 2H O to give 20% by wt. of solution (A) 15 g (B) 1.5 g (C) 11.5 g (D) 31.5 g

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

50. 2.7 g of Ag2CO3 on being heated strongly yields a residue weighing (A) 2.11 g (B) 2.48 g (C) 2.32 g (D) 2.64 g

51. If 1 mole of ethanol (C2H5OH) completely burns to CO2 and H2O, the weight of CO2 formed is about. (A) 22 g (B) 45 g (C) 66 g (D) 88 g 52. The percent loss in weight after heating a pure sample of KClO3 (molecular

weight = 122.5) will be (A) 12.25 (B) 24.50 (C) 39.18 (D) 49.0

53. Calculate the weight of iron which will be converted into its oxide by the action of 18g of steam on

it. From the reaction 2 2 3 22Fe 3H O Fe O 3H . (A) 37.3 gm (B) 3.73 gm (C) 56 gm (D) 5.6 gm 54. A 10.0 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to

precipitate the calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 gms. The % by mass of CaCl2 in the original mixture is :

(A) 15.2% (B) 32.1% (C) 21.8% (D) 11.07% 55. 8 g of sulphur is burnt to form SO2 which is oxidised by Cl2 water. The solution is treated with

BaCl2 solution. The amount of BaSO4 precipitated is (A) 1 mole (B) 0.5 mole (C) 0.24 mole (D) 0.25 mole 56. A mixture of KBr and NaBr weighing 0.560 gm was treated with aqueous Ag+ and all the bromide

ion was recovered as 0.970 gm of pure AgBr. The weight of KBr in the sample is (A) 0.25 gm (B) 0.212 (C) 0.36 (D) 0.285

57. In an experiment, it is found that 2.0769 g of Pure X produces 3.6769 g of pure X2O5. The number of moles of X is

(A) 0.04 (B) 0.06 (C) 0.40 (D) 0.02 58. 2.4 kg of carbon is made to react with 1.35 kg of aluminium to form Al4C3. The maximum amount in

kg of aluminium carbide formed is (A) 5.4 (B) 3.75 (C) 1.05 (D) 1.8

59. Consider the reaction 2A 2B, B 2C, 3C 4D. The no. of moles of D formed starting 4 moles of A, are

(A) 8 (B) 16 (C) 4 (D) 10.67

60. If 0.5 mole of BaCl2 is mixed with 0.2 mole of Na3PO4, the maximum number of moles of Ba3(PO4)2 that can be formed is

(A) 0.7 (B) 0.5 (C) 0.30 (D) 0.10

61. 0.5 mole of H2SO4 is mixed with 0.2 mole of Ca(OH)2. The maximum number of mole of CaSO4 formed is

(A) 0.2 (B) 0.5 (C) 0.4 (D) 1.5

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

62. For the reaction A + 2B C + 3D, 5 moles of A and 8 moles of B will produce

(A) 5 moles of C (B) 4 moles of C (C) 8 moles of C (D) None of these CONCENTRATION TERMS

63. An aqueous solution of urea containing 18 g urea in 1500 cc of solution has a density of 1.052 g/cc. If the mol.wt. of urea is 60, then the molality of solution is

(A) 0.2 (B) 0.192 (C) 0.064 (D) 1.2 64. Molarity of 1g H2SO4 solution in 1 lit. water is nearly (A) 0.1 (B) 0.20 (C) 0.05 (D) 0.01

65. 20 ml of 0.2 M Al2(SO4)3 is mixed with 20 ml of 0.6 M BaCl2. Concentration of Al3+ ion in the solution will be

(A) 0.2 M (B) 10.3 M (C) 0.1 M (D) 0.25 M 66. 50 ml of 0.01 M FeSO4 will react with what volume of 0.01 M KMnO4 solution in acid medium? (1 mole KMnO4 requires 5 mole of FeSO4 for complete reaction) (A) 50 ml (B) 25 ml (C) 100 ml (D) 10 ml

67. The number of H+ ions present in 100 ml of 0.001M H2SO4 solution will be (A) 120.4 1019 (B) 1.20 1020 (C) 6.023 1020 (D) 6.023 1021

68. 3.0 molal NaOH solution has a density of 1.11 g ml . The molarity of the solution is (A) 2.97 (B) 3.05 (C) 3.64 (D) 3.050 69. Element X reacts with oxygen to produce a pure sample of 2 3X O . In an experiment it is found that

1.00g of X produces 1.16g of 2 3X O . Calculate the atomic weight of X .

Given: atomic weight of oxygen, 116.0g mol . (A) 67 (B) 100.2 (C) 125 (D) 150 70. The mole fraction of NaCl in a solution containing 1 mole of NaCl in 1000g of water is: (A) 0.0177 (B) 0.001 (C) 0.5 (D) 1.5 71. If half mole of oxygen combine with Al to form 2 3Al O , the weight of Al used in the reaction is: (A) 27 g (B) 40.5 g (C) 54 g (D) 18 g MISCELLANEOUS PROBLEMS 72. One mole of potassium chlorate is thermally decomposed and excess of aluminium is burnt in the

gaseous product. How many mole of aluminium oxide are formed: (A) 1 (B) 1.5 (C) 2 (D) 3

73. The density of a 3.60M sulphuric acid solution that is 29% 2 4H SO (molar mass 198g mol ) by mass will be:

(A) 1.22 (B) 1.45 (C) 1.64 (D) 1.88

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

74. 10 moles 2N and 15moles of 2H were allowed to react over a suitable catalyst. 10 moles of 3NH

were formed. The remaining moles of 2N and 2H respectively are: (A) 5 moles, 0 mole (B) 0 moles, 5 mole (C) 9 moles, 12 mole (D) 0 moles, 0 mole 75. Reaction is: 2 4 2 4 33Fe SO BaCl BaSO FeCl

How many moles of 2BaCl is needed to produce 12 mole of 3FeCl ?

(A) 5 moles2

(B) 1 mole2

(C) 3 mole4

(D) 2moles

76. How many gms of copper at.wt 64 would be displaced from the copper sulphate solution by

adding 27 gm of aluminium at.wt 27 (A) 32 (B) 64 (C) 96 (D) 160 77. 5 moles of 4CH is burned with 8 moles of 2O than calculate mole of 2CO formed and remaining

moles of excess reagent (A) 4,1 (B) 1, 4 (C) 0,5 (D) 5,0 78. A hydrocarbon 10 xC H requires 40moles of 2O for combustion of 2.5moles . Calculate value of x ? (A) 24 (B) 34 (C) 12 (D) 22 79. Calcium carbonate reacts with aqueous HCl to give 2CaCl according to the reaction,

3 2 2 22 CaCO s HCl aq CaCl aq CO g H O l . The mass of 3CaCO required to react completely with 25mL of 0.75M HCl is

(A) 0.1 g (B) 0.84 g (C) 8.4 g (D) 0.94 g 80. 25.0 ml of HCl solution gave, on reaction with excess 3AgNO solution 2.125g of AgCl . The

molarity of HCl solution is (A) 0.25 (B) 0.6 (C) 1.0 (D) 0.75

FOUNDATION BUILDERS (SUBJECTIVE)

MOLE CONCEPT 1. The atomic wt. of two elements A and B are 20 & 40 resp. If X g of A contains Y atoms, how many

atoms are present in 2X g of B.

2. Calculate the total number of electrons present in 1.6g of CH4. 3. Calculate the total number of electrons present in 18 ml of water

4. Calculate the number of electrons, protons and neutrons in 1 mole of O2 ions 5. Mass of one atom of ‘X’ is 6.6421023 g. What is its atomic mass?

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

6. Mass of one atom of an element A is 3.98541023 g. How many atoms are contained in 1 g of the element of A?

7. From 200mg of CO2, 1021 molecules are removed. How many moles of CO2 are left?

8. Find the charge in coulomb on 1 g ion of N3

9. Calculate the ratio of no. of oxygen atoms present in 3.2 mg of SO2 to the no. of S atoms present in 5 millimoles of Na2S2O3.5H2O

10. Calculate the total no. of oxygen and nitrogen atoms present in a mixture containing 8.2 gm of

calcium 10 millimoles of sodium nitrate and NA/6 molecules of NO2. 11. How many year it would take to spend Avogadro number of rupees at the rate of 10 Lack rupees per

second?

12. One atom of an element X weighs 6.644 ×1023 g. Calculate the number of gram atoms in 40 kg of it.

13. A dot at the end of this sentence has a mass of about 1microgram.Assuming that the black stuff is carbon; calculate the approximate number of atoms of carbon needed to make such a dot.

14. How many iron atoms are present in a stainless steel ballbearing having a radius 0.1 inch? The stainless steel contains 85.6% Fe by weight and has a density of 7.75g/cc.

15. A sample of potato starch was ground in a ball mill to give a starch like molecule of lower molecular

weight. The product analyzed 0.086% phosphorus. If each molecule is assumed to contain one atom of phosphorus, what is the molecular weight of material?

16. Calculate volume of 3.4 g NH3 at STP. 17. Oxygen is present in a 1 litre flask at a pressure of 760 mm of Hg. Calculate the number of oxygen

molecules in the flask at 0°C.

18. 600 ml of a mixture of O3 and O2 weighs 1 g at NTP. Calculate the volume of ozone in the mixture.

EMPERICAL AND MOLECULAR FORMULA

19. Potassium manganate is a dark green crystalline substance whose composition is 40.2% K, 26.8%

Mn and rest P. What is empirical formula?

20. A drug marijuana owes its activity to tetrahydrocannabinol, which contains 70% as many carbon atoms as hydrogen atoms and 15 times as many hydrogen atoms as oxygen atoms. The number of mole in a gram of tetrahydrocannabinol is 0.00318. Determine the molecular formula.

21. A given sample of Xenon fluoride contains molecules of a single type XeFn where n is a whole number. If 9.03×1020 molecules has a mass of 0.311 g then find the value of n.

22. A poisonous compound cadaverine has 58.77%C,13.81%H , and 27.42% N . Its molar mass is

102g mol . Determine its molecular formula.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

23. What is the empirical formula of a compound 0.2801 gm of which gave on complete combustion

0.9482gm of carbon dioxide and 0.1939gm of water.

DENSITY AND VAPOUR DENSITY 24. An alloy has Fe, Co and Mo equal to 71%, 12% and 17% respectively. How many cobalt atoms are

there in a cylinder of radius 2.5 cm and a length of 10 cm? The density of alloy is 8.20 g/mL. Atomic weight of cobalt =58.9.

25. The vapour density (hydrogen = 1) of a mixture consisting of NO2 and N2O4 is 38.3 at 26.7°C.

Calculate the number of moles of NO2 in 100 g of the mixture. 26. An aqueous solution of ethanol has density 1.025 g/mL and it is 8.0 M. Determine molality m of this

solution.

PERCENTAGE COMPOSITION AND GRAVIMETRIC ANALYSIS 27. Determine the percentage composition of a mixture of anhydrous sodium carbonate and sodium

bicarbonate from the following data: wt. of the mixture taken = 2g

Loss in weight on heating = 0.124 g. 28. LSD is a complex compound whose mass is made up of 74.27% carbon, 7.79% hydrogen, 12.99%

nitrogen and 4.95% oxygen. What percent of the atoms in LSD are carbon atoms?

29. In a process for producing acetic acid, oxygen gas is bubbled into acetaldehyde, CH3CHO, under pressure at 60°C in presence of suitable catalyst.

2CH3CHO(l) + O2(g) 2CH3COOH(l) In a laboratory test of this reaction, 20 g CH3CHO and 10 g O2 were put into a reaction vessel. (A) How many gram of acetic acid can be produced by these amounts of reactants?

(B) How many gram of the excess reactant remain after the reaction is complete? (C) If actual yield is 23.8 g, calculate percentage yield.

30. Reaction Yield of reaction (i) 2A + B 3C + D 20% (ii) 2C + E 4F 40% (iii) 7H + 3F 8G 50% where B, E and H given in excess

Starting with 10 moles of A. Moles of G formed are (A) 6.4 (B) 9.6 (C) 3.2 (D) 1.6

31. A 5.0 g sample of a natural gas consisting of 4 2 4CH ,C H was burnt in excess of oxygen yielding

214.5g CO and some 2H O as product. What is weight percentage of 4CH and 2 4C H in mixture.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

STOICHIOMETRY

32. 27.6 g of K2CO3 was treated by a series of reagents so as to convert all of its carbon to K2Zn3[Fe(CN)6]2. Calculate the moles of the product.

33. Find the mass of Cu(NO3)2.3H2O produced by dissolving 10g of copper in nitric acid and then

evaporating the solution.

34. What weight of AgCl will be precipitated when solution containing 4.77 g of NaCl is added to a solution of 5.77 g of AgNO3? (Na = 23, Cl = 35.5, Ag = 108, N = 14, O =16)

35. 1.84 g of a mixture of CaCO3 and MgCO3 was heated to a constant weight. The constant weight of the residue was found to be 0.96g.Calculate % composition of the mixture.

36. Potassium chlorate (KClO4) is made in the following sequence of reactions:

OHKClOKClKOHgCl 22 )(

3KClOKClKClO

KClKClOKClO 43 What mass of Cl2 is needed to produce 1.385 kg of KClO4?

37. For the above set of reactions what will be amount of KCl produced when 142 g of Cl2 is taken 38. 1 gm sample of KClO3 was heated under such conditions that a part of it decomposed according

to the equation i) 2KClO3 2KCl + 3O2 ii) and the remaining underwent change according to the equation 4KClO3 3KClO4 + KCl

If the amount of O2 evolved was 146.8 ml at STP, calculate percentage by weight of KClO4 in the residue.

39. Based on the following information, determine value of x and y:

)()()( 3343 sAgClAlyClgCHxAlClCH AgNO

yx 0.643 g 0.222 g 0.996g

40. How much quantity of zinc will have to be reacted with excess of dilute HCl solution to produce sufficient hydrogen gas for completely reacting with the oxygen obtained by decomposing 6.125 g of KClO3?

41. (A) What is Limiting reagent? (B) Which is in excess? (C) Formed moles of C?

A + 2B 7C

(t=0) 4/3 moles 1 mol

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

42. By the reaction of carbon and oxygen, a mixture of CO and CO2 is obtained. What is the composition

of the mixture obtained in moles when 20 grams of O2 reacts with 12 grams of carbon ? CONCENTRATION TERMS 43. What is the strength in g/litre of a solution of H2SO4, 12 cc of which neutralizes 15 cc of M/10

NaOH? 44. 100 ml solution of KOH contains 10 millimoles of KOH. Calculate its strength in molarity and

gram/litre. 45. Calculate the number of sulphate ions in 100 ml of 0.001 M H2SO4 solution. 46. What weight of CuSO4 .5H2O must be taken to make 0.5 lt. of 0.01 M Cu2+ ions. 47. To 50 ml of 0.5 M H2SO4, 75 ml of 0.25 M H2SO4 is added. What is the concentration of the final

solution? 48. 3 g of a salt of molecular weight 30 is dissolved in 250 g of water. Calculate the molality of the

solution. 49. Mole fraction of I2 in C6H6 is 0.2. Calculate molality of I2 in C6H6. 50. The density of a 10% by mass KCl solution in water is 1.06 g/ml. Calculate molarity of KCl solution.

51. 105 ml of pure water (4°C) is saturated with NH3 gas producing a solution of density 0.9 g/ml. If this

solution contains 30% of NH3 by wt. Calculate the volume of the solution. 52. An aqueous solution of ethanol has density 1.025 g/mL and it is 8.0 M. Determine molality of this

solution. MISCELLANEOUS PROBLEMS 53. If 5 moles each of 2SO and 2O at STP reacts and form 3SO then calculate number of moles of

3SO gas produced at STP, 2 2 32SO O 2SO 54. Calculate the no. of moles of 2 3Fe O produced at STP when 600gm of 2FeS reacts with 800gm of 2O

2 2 2 3 24FeS 11O 2Fe O 8SO 55. Calculate the weight of 3NH required to neutralize 146gm of HCl

3 4NH HCl NH Cl

56. 6.00gm of 2H reacts with 29.00gm of 2O to yield 2H O (i) which is limiting reagent (ii) Calculate the maximum amount of 2H O that can be formed (iii) Calculate the amount of reactants which remains unreacted

2 2 22H O 2H O

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

57. What mass of NaCl would contain the same total number of ions as 245 g of 2MgCl ? 58. A mixture of pure AgCl and pure AgBr is found to contain 60.94% Ag by mass. What are mass

percentages of Cl and Br in the mixture 108, 35.5, 80Ag Cl Br

59. How much gas ( in litres) will be produced at 0 C and 760mm of pressure when 10 g of oxalic acid (H2C2O4) was heated with concentrated sulphuric acid?

60. 0.607 g of a silver salt (Ag3X type) of a tribasic organic acid (H3X type) was quantitatively reduced

to 0.370 g of pure silver. Calculate the molecular weight of the acid. 108Ag . GET EQUIPPED FOR JEE MAIN 1. Potassium combines with two isotopes of chlorine (35Cl and 37Cl) respectively to form two samples of

KCl. Their formation follows the law of: (A) constant proportions (B) multiple proportions (C) reciprocal proportions (D) none of these 2. H2S contains 5.88% hydrogen, H2O contains 11.11% hydrogen while SO2 contains 50% sulphur. These

figures illustrate the law of: (A) conservation of mass (B) constant proportions (C) multiple proportions (D) reciprocal proportions 3. The best standard of atomic mass is: (A) carbon-12 (B) oxygen-16 (C) hydrogen-1.008 (D) chlorine-35.5 4. The chemical formula of a particular compound represents: (A) the size of its molecule (B) the shape of its molecule (C) the total number of atoms in a molecule (D) the number of different types of atoms in a molecule 5. Two containers P and Q of equal volume (1 litre each) contain 6 g of O2 and SO2 respectively at 300 K

and 1 atmosphere. Then (A) Number of molecules in P is less than that in Q (B) Number of molecules in Q is less than that in P (C) Number of molecules in P and Q are same (D) Either (A) or (B) 6. The product of atomic mass and specific heat of any element is a constant, approximately 6.4. This is

known as: (A) Dalton’s law (B) Avogadro’s law (C) Gay-Lussac law (D) Dulong Petit’s law

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

7. 250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. If 10 ml of this solution is

diluted to one litre, what is the concentration of the resultant solution? (mol wt. of Na2CO3 = 106) (A) 0.1 M (B) 0.001M (C) 0.01 M (D) 104 M 8. 7.5 grams of a gas occupy 5.6 litres of volume at STP. The gas is (A) NO (B) N2O (C) CO (D) CO2 9. The weight of a molecule of the compound C60H122 is (A) 1.4 1021 g (B) 1.09 1021 g (C) 5.025 1023 g (D) 16.023 1023 g 10. 1.0 mole of CO2 contains: (A) 6.02 × 1023 atoms of C (B) 6.02 × 1023 atoms of O (C) 18.1 × 1023 molecules of CO2 (D) 3 g-atoms of CO2 11. The number of atoms in 1.4 g nitrogen gas is: (A) 6.02 × 1022 (B) 3.01 × 1022 (C) 1.20 × 1023 (D) 6.02 × 1023

12. Which of the following has the smallest number of molecules? (A) 22.4 × 103 ml of CO2 gas (B) 22 g of CO2 gas (C) 11.2 litre of CO2 gas (D) 0.1 mole of CO2 gas 13. The number of grams of H2SO4 present in 0.25 mole of H2SO4 is (A) 0.245 (B) 2.45 (C) 24.5 (D) 49.0 14. At NTP 1.0 g hydrogen has volume in litre: (A) 1.12 (B) 22.4 (C) 2.24 (D) 11.2 15. 19.7 kg of gold was recovered from a smuggler. The atoms of gold recovered are: (Au = 197) (A) 100 (B) 6.02 × 1023 (C) 6.02 × 1024 (D) 6.02 × 1025 16. The molecular mass of CO2 is 44 amu and Avogadro’s number is 6.02 × 1023. Therefore, the mass of one

molecule of CO2 is: (A) 7.31 × 10–23 (B) 3.65 × 10–23 (C) 1.01 × 10–23 (D) 2.01 × 10–23 17. The number of moles of H2 in 0.224 litre of hydrogen gas at NTP is: (A) 1 (B) 0.1 (C) 0.01 (D) 0.001

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

18. Irrespective of the source, pure sample of water always yields 88.89% mass of oxygen and 11.11% mass

of hydrogen. This is explained by the law of: (A) conservation of mass (B) constant composition (C) multiple proportion (D) constant volume 19. An element, X, have three isotopes X 20, X 21 and X 22. The percentage abundance of X 20is 90% and its

average atomic mass of the element is 20.11. The percentage abundance of X 21 should be (A) 9% (B) 8% (C) 10% (D) 0%

20. The O18/O16 ratio in some meteorites is greater than that used to calculate the average atomic mass of

oxygen on earth. The average mass of an atom of oxygen in these meteorites is ………. that of a terrestrial oxygen atom.

(A) equal to (B) greater than (C) less than (D) None of these 21. 6.023 ×1023 molecules of Ca (OH)2 react with 3.01×1022 molecules of HCl, number of moles of CaCl2

obtained are (A) 0.05 (B) 0.10 (C) 0.025 (D) 3.01 22. A copper sulphate solution contains 1.595% of CuSO4 by weight. Its density is 1.2 g / ml, Its molarity

will be (A) 0.12 (B) 0.06 (C) 1.20 (D) 1.595 23. Which of the following samples contains 2.0 1023 atoms? (A) 8.0 g O2 (B) 3.0 g Be (C) 8.0 g C (D) 19.0 g F2 24. Choose the wrong statement:- (A) 1 Mole means 6.02 1023 particles (B) Molar mass is mass of one molecule (C) Molar mass is mass of one mole of a substance (D) Molar mass is molecular mass expressed in grams 25. What quantity of limestone (CaCO3) on heating will give 56 Kg of CaO? (A) 1000 Kg (B) 44 Kg (C) 56 Kg (D) 100 Kg 26. Simplest formulae of a compound containing 20% of element X (atomic weight 10) and 80% of element

Y (atomic weight 20) is (A) XY (B) X2Y (C) XY2 (D) X2Y3

27. At room temperature and pressure two flask of equal volumes are filled with H2 and SO2 respectively.

Particles which are equal in number in two flasks are (A) Atoms (B) Electrons (C) Molecules (D) Neutrons

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

28. Chlorophyll contains 2.68% of magnesium by mass. Calculate the number of magnesium atoms in 3.00 gms of chlorophyll.

(A) 2.01 1021 atoms (B) 6.023 1023 atoms (C) 1.7 1020 atoms (D) 2.8 1022 atoms 29. What is the total number of atoms present in 25.0 mg of camphor C10H16O? (A) 9.89 1019 (B) 6.02 1020 (C) 9.89 1020 (D) 2.67 1021

30. 2 mol of H2S and 11.2 L SO2 at N.T.P. reacts to form x mol of sulphur; x is SO2 + 2H2S 3S + 2H2O (A) 1.5 (B) 3 (C) 11.2 (D) 6 31. How many grams of phosphoric acid (H3PO4) would be needed to neutralise 100 g of magnesium

hydroxide (Mg(OH)2). (A) 66.7 g (B) 252 (C) 112.6 g (D) 168 g 32. If 0.5 mol of BaCl2 is mixed with 0.2 mol of Na3PO4, the maximum number of mol of Ba3(PO4)2 that

can be formed is (A) 0.7 (B) 0.5 (C) 0.2 (D) 0.1 33. The number of electron in the telluride ion. 52Te2– is (A) 50 (B) 52 (C) 53 (D) 54 34. An ore contains 1.34% of the mineral argentite, Ag2S, by weight. How many grams of this ore would

have to be processed in order to obtain 1.00 g of pure solid silver, Ag? (A) 74.6 g (B) 85.7 g (C) 134.0 g (D) 171.4 g 35. Hydrogen evolved at NTP on complete reaction of 27 gm of Al with excess of aq NaOH would be (Chemical reaction: 2Al + 2NaOH + 2H2O 2NaAlO2 + 3H2) (A) 22.4 lit (B) 44.8 lit (C) 67.2 lit (D) 33.6 lit

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

GET EQUIPPED FOR JEE ADVANCE

(ONLY ONE OPTION IS CORRECT) 1. N2 and H2 are mixed in 14:3 mass ratio. After certain time ammonia was found to be 40% by mol. The

mole fraction of N2 at that time in the mixture of N2, H2 and NH3 is (A) 0.15 (B) 0.3 (C) 0.45 (D) none of these 2. Two flasks of equal volumes are evacuated, then one is filled with gas A and other with gas B at the

same temperature and pressure. The weight of B was found to be 0.80 gm while the weight of gas A is found to be 1.40 gm. What is the weight of one molecule of B in compared to one molecule of A

(A) 1.40 times as heavy as A (B) 0.40 times as heavy as A (C) 0.57 times as heavy as A (D) 0.80 times as heavy as A 3. On reduction with hydrogen, 3.6 g of an oxide of metal left 3.2 g of the metal. If the atomic weight of

the metal is 64, the simplest formula of the oxide would be (A) M2O3 (B) M2O (C) MO (D) MO2

4. A certain compound has the molecular formula X4O6. if 10 g of X4O6 has 5.72 g of X, atomic mass of

X is : (A) 32 amu (B) 37 amu (C) 42amu (D) 98 amu 5. If 224 ml of a triatomic gas has a mass of 1 g at 273 K and 1 atm pressure, then the mass of one atom

is (A) 55.31023 g (B) 0.5531023 g (C) 5.531023 g (D) 5531023 g

6. The weight of 350mL of a diatomic gas at 00C and 2 atm pressure is 1g. The wt of one atom is

(A)16 NA (B) 32 NA (C) A

16N

(D) A

32N

7. 25.4 g of iodine and 14.2g of chlorine are made to react completely to yield a mixture of ICl and ICl3. Calcualte the ratio of moles of ICl and ICl3.

(A) 1:1 (B) 1:2 (C) 1:3 (D) 2:3

8. A mixture contains 4FeSO and 2 4 3Fe SO . If both 4FeSO and 2 4 3Fe SO provide equal number of

sulphate ions then, the ratio of 2Fe and 3Fe ions in mixture is (A) 1 : 2 (B) 2 :3 (C) 2 : 1 (D) 3 : 2 9. In what volume ratio a 0.36 M 3HNO solution should be mixed with another 0.15 M 3HNO solution

to obtain a 0.24 M 3HNO solution? (A) 4 : 3 (B) 2 : 3 (C) 4 : 9 (D) 3 : 4 10. One atom of an element weighs 3.981023g. Its atomic mass is (A) 18 (B) 29.9 (C) 24 (D) 108 11. Weight ratio of Fe:C in Fe2(Fe(CN)6) is: (A) 3/7 (B) 7/5 (C) 7/3 (D) 5/7

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

12. Which of the following has the highest mass:

(A) 12g of C atom (B)13 g of H atom (C)9 moles of NH3 (D)80g molecule of H2SO4 13. Ratio of carbon to hydrogen atom is maximum in: (A) 16 g CH4 (B) 31.2 g C6H4 (C) 34.2 g C12H22O11 (D)36 g C6H12O6 14. How many grams are contained in 1 g atom of Nitrogen (A) 14 g (B) 7 g (C) 1 g (D) Data Insufficient

15. The volume of 3M HCl required to completely react with 1.431g of sodium carbonate

(A) 10 ml (B) 9 ml (C) 8 ml (D) 4.5 ml

16. In which of the following pairs do 1 g of each have an equal number of molecules? (A) N2O and CO (B) N2 and C3O2 (C) N2 and CO (D) NO2 and CO2

17. How many spherical colloidal oil particles 0

20 A in radius can be made from a spherical oil drop whose radius is 2 microns?

(A) 910 (B) 610 (C) 410 (D) 210 18. A 10 g sample of 3KClO , gave on complete decomposition , 2.24 L of oxygen at NTP. What is the

percentage purity of the sample of potassium chlorate? (A) 61.2 (B) 81.6 (C) 96.6 (D) 24.6 19. A 0.65 M 2BaCl solution is prepared by dissolving pure solid 2 2BaCl .2H O in water. Determine the

mass of hydrated salt dissolved per milliliter of solution and mass of anhydrous 2BaCl present per milliliter of solution. Molar masses are : Ba = 137, Cl = 35.5.

(A) 0.158 g, 0.135 g (B) 0.226 g, 0.135 g (C) 0.248 g, 0.163 g (D) 1.1 g, 2.2 g 20. What volume of a 1.36 M HCl solution should be added to a 200 mL 2.4 M HCl solution and finally

diluted to 500 mL so that molarity of final HCl solution becomes 1.24 M. (A) 29.2 mL (B) 102.94 mL (C) 46.34 mL (D) 9.4 mL 21. Potassium salt of benzonic acid 6 5C H COOK can be made by the action of potassium permanganate

on toluene as follows: 6 5 3 4 6 5 2 2C H CH KMnO C H COOK MnO KOH H O If the yield of potassium benzonate can’t realistically be expected to be more than 71%, what is the

minimum number of grams of toluene needed to achieve this yield while producing 11.5 g of 6 5C H COOK ?

(A) 6.23 (B) 9.3 (C) 4.23 (D) 5.63 22. A mixture of 4 2CuSO .5H O and 4 2MgSO .7H O is heated until all the water is driven-off. If 5.0 gm of

a mixture gives 3g of anhydrous salts, what is the percentage by mass of 4 2CuSO .5H O in the original mixture?

(A) 44% (B) 64% (C) 74% (D) 94%

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

23. Aspirin 9 8 4C H O , is prepared by heating salicylic acid, 7 6 3C H O with acetic

anhydride, 4 6 3C H O . The other product is acetic acid, 2 4 2C H O

7 6 3 4 6 3 9 8 4 2 4 2C H O C H O C H O C H O when 2.00 g of salicylic acid is heated with 4.00 g of acetic anhydride? If the actual yield of aspirin is

2.1 g, what is the percentage yield? (A) 80.7% (B) 40.7% (C) 25.2 % (D) 43.9% 24. An element X forms an iodide 3XI and a chloride 3XCl . The iodide is quantitatively converted to the

chloride when it is heated in a stream of chlorine 3 2 3 22XI 3Cl 2XCl 3l

If 0.5000 g of 3Xl is treated, 0.2360 g of 3XCl is obtained. Calculate the atomic weight of the element X. (A) 246 (B) 139 (C) 180 (D) 196

25. A solution of palmitic acid in benzene contains 4.24 g of acid per litre. When this solution is dropped on a surface, benzene gets evaporated and palmitic acid forms a unimolecular film on the surface. If we wish to cover an area of 2500cm with unimolecular film, what volume of solution should be used? The area covered by one palmitic acid molecule may be taken as 20.21nm .

Mol. Wt. of palmitic acid = 256. (A) 54.38 10 (B) 52.4 10 (C) 414 10 (D) 55.6 10 26. 6.0 g of a sample containing 2CuCl and 2CuBr is dissolved in 100 – mL water. A 10 – mL portion of

this solution on treatment with 3AgNO solution results in complete precipitation of Cl and Br giving 0.9065 gram of precipitate. The precipitate thus obtained was shaken with dilute solution of NaBr where all AgCl gets convererted into AgBr. Mass of the new precipitate was found to be 1.005 g. Determine % mass of 2CuCl and 2CuBr in the original sample.

(A) 25 %, 58% (B) 50%, 50% (C) 75%, 25% (D) 20%, 80% 27. An element (X) reacts with hydrogen leading to formation of a class of compounds that is analogous

to hydrocarbons. 5.00 g of X forms 5.628 g of a mixture of two compounds of 4 2 6X XH and X H in the molar ratio of 2:1. Determine the molar mass of X.

(A) 28 (B) 58 (C) 72 (D) 83 28. A 2.00 g portion of a sample containing NaBr and 2 4Na SO was dissolved and diluted to 250 mL.

One fifth aliquot portions were titrated by silver nitrate, an average of 42.5 mL solution being required for the aliquot portion. In standardization 1.00 mL 3AgNO is found to be equimolar to 0.0125 g KBr. Calculate percentage of Br in sample.

(A) 42% (B) 52% (C) 33% (D) 12% 29. The molecular mass of an organic acid was determined by the study of its barium salt. 4.290 g of salt

was quantitatively converted to free acid by the reaction with 21.64 mL of 2 40.477 M H SO . The barium salt was found to have two mole of water of hydration per 2Ba ion and acid is mono basic. What is molecular weight of anhydrous acid?

(A) 122 (B) 142 (C) 108 (D) 110

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

30. What volume of 0.010 M NaOH aq is required to react completely with 30 g of an aqueous acetic acid solution in which mole fraction of acetic acid is 0.15?

(A) 108.55 lt (B) 18.55 lt (C) 34.66 lt (D) 42 lt 31. A mixture of ethane 2 6C H and ethene 2 4C H occupies 40 litre at 1.00 atm and at 400 K. The

mixture reacts completely with 130 g of 2O to produce 2CO and 2H O . Assuming ideal gas behavior, calculate the mole fraction of 2 4C H and 2 6C H in the mixture.

(A) 0.34, 0.66 (B) 0.66, 0.34 (C) 0.50, 0.50 (D) 0.20, 0.80 32. A crystalline hydrated salt on being rendered anhydrous, looses 45.6% of its weight. The percentage

composition of anhydrous salt is: Al = 10.5% , K = 15.1% , S = 24.8% and O = 49.6%. The empirical formula of the crystalline salt.

(A) 2 8 2KAlS O .12H O (B) 2 2 2 8 2K Al S O .12H O

(C) 2 2 8 2KAl S O .12H O (D) None of these

33. A crystalline polymer molecule is uniform prismatic in shape with dimensions as shown in below,

If density of this polymer is 1.2 g/cm2, the molar mass is. (A) 1000 (B) 3939 10 (C) 939 (D) 32300 10

MORE THAN ONE CHOICE

1. The density of 3M sodium thiosulphate 2 2 3Na S O is 11.25g mL . Identify the correct statements among the following:

(A) % by weight of sodium thiosulphate is 37.92 (B) The mole fraction of sodium thisoulphate is 0.065 (C) The molarity of Na is 2.53 and 2

2 3S O is 1.25.

(D) The molality of Na is 7.732 and 22 3S O is 3.866

2. The density of air is 30.001293g / cm at STP. Identify which of the following statement is correct. (A) Vapour density is 14.48 (B) Molecular weight is 28.96 (C) Vapour density is 30.001293g / cm (D) Vapour density and molecular weight cannot be determind 3. Which of the following has same mass (A) 1.0 moles of 2O (B) 233.011 10 molecules of 2SO

(C) 0.5 moles of 2CO (D) 1 g atom of sulphur

A100

A300

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

4. 100 mL of 0.06 M 3 2Ca NO is added to 50 mL of 0.06 M 2 2 4Na C O . After the reaction is complete.

(A) 0.003 moles of calcium oxalate will get precipitated (B) 0.003 M of excess of 2Ca will remain in excess (C) 2 2 4Na C O is limiting reagent.

(D) 3 2Ca NO is excess reagent.

5. A sample of mixture of 2CaCl and NaCl weighing 4.44 gm was treated to precipitate all the Ca as

3CaCO , which was then heated and quantitatively converted to 1.12g of CaO . (At. Wt. Ca = 40, Na = 23, Cl = 35.5) (A) Mixture contains 50% NaCl (B) Mixture contains 60% 2CaCl

(C) Mass of 2CaCl is 2.22 gm (D) Mass of 2CaCl is 1.11 gm 6. 1 M 100 ml NaCl is mixed with 3 M 100 ml HCl solution and 1 M 200 ml 2CaCl solution. (A) The ratio of concentration of cation and anion = 3/4 (B) The ratio of concentration of cation and anion = 2 (C) Cl 2M

(D) Cl 3/ 2M

7. Number of hydrogen atoms are equal in (A) 16 gm CH4 (B) 52 gm of C6H6 (C) 34.2 gm C12H22O11 (D) 36 gm water 8. 3 mole of ammonia contains (A) 9 gm hydrogen (B) 42 gm nitrogen (C) Total 18.06 x 1023 molecules (D) Total 7.226 x 1024 atoms 9. 1g molecule of 2 5V O contains: (A) 5 mole of oxygen atom (B) 2 mole of V atom

(C) 1mole of oxygen atom (D) 2.5 mole of oxygen atom 10. Which of the following quantities are independent of temperature (A) Molarity (B) mole fraction (C) molality (D) % (w/v) 11. Which of following will be present in the solution formed when 50mL of 0.1M HCl is mixed with

50mL of 0.1M NaOH ?

(A) 4.5m molof H (B) 0.05m molof OH

(C) 0.05M NaCl (D) 710 M of H ion

12. Which of the following expressions is correct ( n no. of moles of the gas, AN Avgadro constant, m mass of molecule of the gas, N no. of molecules of the gas, M = Molar Mass)

(A) An mN (B) Am M / N (C) AN nN (D) Am mn N

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

13. Among the following, which solutions contain equal numbers of millimoles? (A) 2 4100mLof 0.05M H SO (B) 200mLof 0.02M NaOH

(C) 2 2 4100mLof 0.10M Na C O (D) 200mLof 0.025MKOH

14. 11.2 L of gas at STP weighs 14.0g . The gas could be: (A) 2N O (B) 2NO (C) 2N (D) CO

MATCH THE FOLLOWING 1. Hexachlorophere, C13H6Cl6O2 (M wt = 407) is a germicide is soap & helps to clear germs. Match the

two columns regarding composition of the germicide. Column I Column II

(I) wt % of C (A) 1.47% (II) wt % of H (C) 1: 35.5 (III) ratio of wt% of H: Cl (E) 6.5 : 1 (IV) ratio of mol of C:O (P) 38.33%

2.

Column – I Column – II (a) Vapour density of 2SO with respect to 2O (P) 22

(b) Specific gravity of the solid with mass 10 gm and volume 5 cc

(Q) 32

(c) Molar mass of the compound having V.D. 16 (R) 9 (d) Number of atoms in 132 amu 2CO (S) 2

3. Column – I Column – II

(a) 100 ml of 0.2 M 3AlCl solution + 400 ml of 0.1 M HCl solution

(P) concentration of cation = 0.12 M

(b) 50 ml of 0.4 M KCl + 50 ml 2H O (Q) 24SO 0.06M

(c) 30 ml of 0.2 M 2 4K SO + 70 ml 2H O (R) 24SO 2.5M

(d) 200 ml 24.5% (w/v) 2 4H SO (S) Cl 0.2M

4. Column-I Column-II (A) 20.5molof SO g (p) occupy 11.2LatSTP

(B) 1g of 2H g (q) weights 24g

(C) 30.5mole O g (r) total no. of atoms A1.5 N

(D) 1g molecule of 2O g (s) weight 32 gm

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

COMPREHENSIVE TYPE PASSAGE-1 The terms mole first used by Ostwald in 1896 refers to the ratio of molecular weight of

molecule to mass of one molecule of substance in gm. Also, 1 mole of gaseous compound occupies 22.4 litre at NTP and contains 6.022 x 1023 molecules of gas.

1. Weight of 1 atom of hydrogen is: (A) 1.66 x 10 -24 amu (B) 3.32 x 10 -24 g (C) 1.66 x 10 -24 g (D) 3.32 x 10 -24 amu

2. The amount of sulphur required to produce 100 mole of H2SO4 is : (A) 3.2 x 103 g (B) 32.65 g (C) 32 g (D) 3.2 g 3. A substance contains 3.4% sulphur. If it contains two atoms of sulphur per molecule the molecular

weight of substance will be: (A) 941 (B) 1882 (C) 470.5 (D) 1411.5 4. The volume of air at STP required for burning 12 g carbon completely is: (Assuming air contains

20% oxygen). (A) 22.4 litre (B) 112 litre (C) 44.8 litre (D) 50 litre

PASSAGE-II The concentration of solutions can be expressed in number of ways such as Molarity, Molality, Mole

fraction, % weight, % by volume % by strength and many others. All these are inter-convertible if certain data like density of solution, molecular mass of solute and solvent are known. Also, addition of water to a solution changes all these terms, though a change in temperature does not change molality, mole fraction and % by weight terms.

1. A 6.90 M KOH solution in water has 30% by weight of KOH. The density of KOH solution is: (A) 1.288 g/mL (B) 12.88 g/mL (C) 0.1288 g/ml (D) None of these 2. Two litres of NH3 at 30o C and 0.20 atm is neutralized by 134 mL of acid H2SO4. The molarity of

H2SO4 is: (A) 0.12 (B) 0.24 (C) 0.06 (D) 0.03 3. The volume of water required to make 0.20M solution from 1600 ml of 0.2050 M solution (A) 40 ml (B) 80 ml (C) 120 ml (D) 180 ml 4. What volume of 0.2 M 2 4H SO is required to produce 34.0 g of 2H S by the reaction?

2 4 2 4 2 2 28KI 5H SO 4K SO 4I H S 4H O (A) 25 litre (B) 50 litre (C) 75 litre (D) 100 litre

PASSAGE-3 The term mole was introduced by Ostwald in 1896. In Latin word ‘moles’ meaning heap or pile. A

mole is defined as the number of atoms in 12.00 g of carbon – 12. The number of atoms in 12 g of

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

carbon -12 has been found experimentally to be 236.022 10 . This number is also known as Avogadro’s number named in honour of Amedeo Avogadro (1776 - 1856).

The mass of one mole atoms of any element is exactly equal to the atomic mass in gram (gram atomic mass) of that element. One mole of a gas occupies 22.4 litre at S.T.P. Atomic mass unit is the unit of atomic mass.

241amu 1.66 10 g Answer the following questions: 1. Mass of one molecule of water is approximately: (A) 1 g (B) 0.5 g (C) 241.66 10 g (D) 233 10 g

2. x L 2N gas at S.T.P. contains 233 10 molecules. The number of molecules in x L ozone at S.T.P. will be:

(A) 233 10 (B) 234 10 (C) 236.023 10 (D) 243 10

3. If Avogadro’s number is 23 11 10 mol then the mass of one atom of oxygen would be:

(A) 16 amu6.02

(B) 16 6.02amu (C) 16amu (D) 2316 10 amu

4. The maximum number of atoms presents are in: (A) 4 g He (B) 4 g O2 (C) 4 g O3 (D) 4 g H2O2

PASSAGE-4 The concentration of solutions can be expressed in number of ways such that Normality,

Molarity, Molality, Mole fractions, Strength , % by weight , % by volume and % by strength. The molarity of ionic compound is usually expressed as formality because we use formula weight of ionic compound. Addition of water to a solution changes all these terms, however increase in temperature does not change molality, mole fraction and % by weight terms.

1. The weight of AgCl precipitated by adding 5.77 g 3AgNO to 4.77g NaCl in a solution is: (A) 4.88g (B) 5.77 g (C) 4.77 g (D) None of these 2. The weight of 2 4H SO in 1200 mL of 0.1M solution is: (A) 11.76g (B) 5.83g (C) 16.42g (D) 2.92 INTEGER TYPE 1. Number of moles of electrons in 0.5 mole of 3N will be ________. 2. How many gm atoms of carbon are there in 132 gm CO2. 3. A sample of metal chloride weighing 0.22 g required 0.51g of 3AgNO to precipitate the chloride

completely. The specific heat of the metal is 0.057. Find out the valency of metal, if the symbol of the metal is ‘M’. (Ag = 108, N = 14, O = 16, Cl = 35.5)

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

4. A complex compound of iron has molar mass = 2800 and it contain 8 % iron by weight. The number

of iron atoms in one formula unit of complex compound is 5. If x mL 5 M HCl is mixed with 20 mL, 2 M HCl, a 2.6 M HCl is produced. The x is 6. A gaseous alkane n 2n 2C H on explosion with 2O gives 2CO , the volume of 2O required for

complete combustion of alkane to 2CO formed is in the ratio 7 : 4 the value of n is 7. The number of moles of fullerene 60C (a soccer ball structure molecule discovered in 1980) in

1.44 kg Fullerene are 8. Weight of a 5% by weight of a NaCl solution to yield 0.3 g of NaCl is 9. Haemoglobin contains 0.25% iron by weight, the molecular weight of Haemoglobin is 89600.

calculate the number of Iron atoms per molecule of Haemoglobin.

10. 1g of dry green algae absorbs 34.7 10 mole of 2CO per hour by photosynthesis. If the fixed carbon atoms were all stored after photosynthesis as starch, 6 10 5 nC H O , how long would it take for the algae to double their own weight assuming photosynthesis takes place at a constant rate?

Answer should be reported to the nearest integer value. EXPERTISE ATTAINERS 1. Cobalt is precipitated with - nitroso - - naphtol as 6 10 3

Co C H O NO and ignited in a stream of

2O to 3 4Co O . Alternatively, the precipitate is ignited in stream of hydrogen gas and weighed as Co. what weight of cobalt would have been obtained from the same weight of sample that produced 0.2125 g 3 4Co O ? What was the weight of precipitate that gave 0.2125 g 3 4Co O ?

2. 0.5 g 3Fe is precipitated as hydrated ferric oxide. During ignition, 90% of iron is converted into

2 3Fe O and the remainder is present as 3 4Fe O (a) What does the ignited precipitate weigh? (b) What should it would have weighed if all the iron were in 2 3Fe O form? 3. A mixture contains NaCl and an unknown chloride MCl. (i) 1 g of this is dissolved in water. Excess of acidified 3AgNO solution is added to it. 2.567 g of a

white precipitate is formed. (ii) 1.0 g of the original mixture is heated to 0300 C . Some vapours come out which are absorbed

in acidified 3AgNO solution. 1.341 g of a white precipitate is obtained. Find the molecular weight of the unknown chloride.

4. 0.220 g of a sample of a volatile compound, containing carbon, hydrogen and chlorine yielded on

combustion in oxygen 0.195 g of 2CO , 0.0804 g of 2H O . 0.120 g of the compound occupied a volume of 37.24 mL at 0105 C and 768 mmHg pressure. Calculate the molecular formula of the compound.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

5. One mole of a mixture of N2, NO2 and N2O4 has a mean molar mass of 55.4. On heating to a

temperature at which N2O4 is completely dissociated : N2O4 2NO2, the mean molar mass tends to the lower value of 39.57. What is the mole ratio of N2 : NO2 : N2O4 in the original

mixture?

6. Chile salt peter a source of NaNO3 also contains NaIO3. The NaIO3 can be used as a source of Iodine produced in the following reactions:

IO3– + 3HSO3

– I– + 3H+ + 3SO42–

5I– + IO3– + 6H+ 3I2 + 3H2O

One litre of chile salt peter solution containing 5.80 gm NaIO3, is treated with stoichiometric quantity of NaHSO3. Now additional amount of same solution is added to the reaction mixture to bring about the second reaction. How many grams of NaHSO3 are required in step 1 and what additional volume of chile salt peter must be added in step II to bring in complete conversion of I– to I2?

7. 1 gm sample of AgNO3 is dissolved in 50 mL of water. It is reacted with 50 mL of KI solution. The AgI precipitated is filtered off. Excess of KI in filtrate is titrated with M/10 KIO3 in presence of 6 M HCl till all I– converted into ICl. It requires 50 mL of M/10 KIO3 solution. 20 mL of the same stock solution of KI requires 30 mL of M/10 KIO3 under similar condition. Calculate % of AgNO3 in sample. The reactions involved are

AgNO3 + KI KNO3 + AgI(ppt), KIO3 + 2KI + 6HCl 3ICl + 3KCl + 3H2O

WINDOWS TO COMPETITIVE EXAMS Objective Questions

1. A gaseous mixture contains oxygen and nitrogen in the ratio of 1:4 by weight. Therefore, the ratio of their number of molecules is (JEE 1979) (A) 1:4 (B) 1:8 (C) 7:32 (D) 3:16

2. The total number of electrons in one molecule of carbon dioxide is (JEE 1979)

(A) 22 (B) 44 (C) 66 (D) 88

3. The largest number of molecules is in (JEE 1979) (A) 36 g of water (B) 28 g CO (C) 46 g of ethyl alcohol (D) 54g of nitrogen pentaoxide (N2O5)

4. When the same amount of zinc is treated separately with excess of sulphuric acid and excess of sodium

hydroxide, the ratio of volume of hydrogen evolved is (JEE 1979) (A) 1:1 (B) 1:2 (C) 2:1 (D) 9:4

5. 2.76g of silver carbonate on being strongly heated yields a residue weighing (JEE 1979)

(A) 2.16g (B) 2.48 g (C) 2.32 g (D) 2.64 g

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

6. If 0.50 mole of BaCl2 is mixed with 0.20 mole of Na3PO4, the maximum number of moles of Ba3(PO4)2

that can be formed is (JEE 1981) (A) 0.70 (B) 0.50 (C) 0.20 (D) 0.10

7. A molal solution is one that contains one mole of solute in (JEE 1986)

(A) 1000g of solvent (B) 1.0 L of solvent (C) 1.0 L of solution (D) 22.4 L of solution

8. In which mode of expression, the concentration of a solution remains independent of temperature?

(JEE 1988) (A) Molarity (B) Normality (C) Formality (D) Molality

9. How many moles of electron weights one kilogram? (JEE 2002)

(A) 236.023 10 (B) 311 10

9.108

(C) (D) 81 10

9.108 6.023

10. Which has maximum number of atoms? (JEE 2003)

(A) 24 g of C(12) (B) 56 g of Fe(56) (C) 27 g of Al(27) (D) 108 g of Ag(108)

11. Which of the following concentration factor is affected by change in temperature ? [AIEEE 2002]

(A) Molarity (B) Molality (C) Mole fraction (D) Weight

12. Number of atoms in 560g of Fe (atomic mass 56g/mol) is : [AIEEE 2002] (A) Twice that of 70g N (B) Half that of 20g H (C) Both (A) and (B) (D) None of these

13. In an organic compound of molar mass 108 g/mol C, H and N atoms are present in 9 : 1 : 3.5 by weight. Molecular formula can be : [AIEEE 2002] (A) C6H8N2 (B) C7H10N (C) C5H6N3 (D) C4H18N3

14. What volume of hydrogen gas at 273 K and 1 atm pressure will be consumed in obtaining 21.6 gm of elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen-

[AIEEE 2003] (A) 44.8 lit. (B) 22.4 lit. (C) 89.6 lit. (D) 67.2 lit.

15. 6.02 ×1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is [AIEEE 2004] (A) 0.001 M (B) 0.01 M (C) 0.02 M (D) 0.1 M

16. If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will [AIEEE 2005]

(A) decrease twice (B) increase two fold (C) remain unchanged (D) be a function of the molecular mass of the substance

546.023 109.108

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

17. Density of a 2.05M solution of acetic acid in water is 1.02 g/ml. The molality of the solution is : [AIEEE-2006] (A) 1.14 mol/kg (B) 3.28 mol/kg (C) 2.28 mol/kg (D) 0.44 mol/kg 18. The density (in g/mL) of a 3.60 M sulphuric acid solution that is 29% by mass will be

[AIEEE-2007] (A) 1.22 (B) 1.45 (C) 1.64 (D) 1.88

19. Given that the abundances of isotopes 54Fe, 56Fe and 57Fe are 5%, 90% and 5% respectively, the atomic

mass of Fe is (JEE 2009) (A) 55.85 (B) 55.95 (C) 55.75 (D) 56.05

20. Dissolving 120g of urea (mol.wt. 60) in 1000g of water gave a solution of density 1.15 g/mL. The molarity of the solution is (JEE 2011) (A) 1.78 M (B) 2.00M (C) 2.05M (D) 2.22M

Fill in the blanks 1. The modern atomic mass unit is based on the mass of _______ . (JEE 1980) 2. The total number of electrons present in 18 mL of water is ______. (JEE 1980) 3. 3.0 g of a salt of molecular weight 30 is dissolved in 250 g water. The molality of the solution is _____.

(JEE 1983) 4. The weight of 1× 1022 molecules of CuSO4.5H2O is _______. (JEE 1991)

Subjective Questions

1. Naturally occurring boron consists of two isotopes whose atomic weights are 10.01 and 11.01. The atomic weight of natural boron is 10.81. Calculate the percentage of each isotope in natural boron. (JEE 1978)

2. Account for the following. Limit your answer to two sentences, “Atomic weights of most of the elements are fractional”. (JEE 1979)

3. The vapour density (hydrogen = 1) of a mixture consisting of NO2 and N2O4 is 38.3 at 26.7°C. Calculate the number of moles of NO2 in 100g of the mixture. (JEE 1979)

4. In the analysis of 0.5 g sample of feldspar, a mixture of chlorides of sodium and potassium is obtained, which weight 0.1180g. Subsequent treatment of the mixed chlorides with silver nitrate gives 0.2451g of silver chloride. What is the percentage of sodium oxide and potassium oxide in the sample? (JEE 1979)

5. 5.00mL of a gas containing only carbon and hydrogen were mixed with an excess of oxygen (30mL) and the mixture exploded by means of electric spark. After explosion, the volume of the mixed gases remaining was 25mL. On adding a concentrated solution of KOH, the volume further diminished to 15 mL, the residual gas being pure oxygen. All volumes have been reduced to NTP. Calculate the molecular formula of the hydrocarbon gas. (JEE 1979)

6. (a) 1.0L of a mixture of CO and CO2 is taken. This mixture is passed through a tube containing red hot charcoal. The volume now becomes 1.6 L. The volumes are measured under the same conditions. Find the composition of mixture by volume.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

(b) A Compound contains 28 % of nitrogen and 72 % of a metal by weight. 3 atoms of metal combine with 2 atoms of nitrogen. Find the atomic weight of metal. (JEE 1980)

7. A sugar syrup of weight 214.2 g contains 34.2 g of sugar (C12H22O11). Calculate

(i) molal concentration and (ii) mole fraction of sugar in syrup. (JEE 1988)

8. n-butane is produced by monobromination of ethane followed by Wurtz’s reaction. Calculate volume of ethane at NTP required to produce 55g n-butane, if the bromination takes place with 90% yield and the Wurtz’s reaction with 85% yield. (JEE 1989)

9. A solid mixture (5.0g) consisting of lead nitrate and sodium nitrate was heated below 600°C until the weight of the residue was constant. If the loss in weight is 28 %, find the amount of lead nitrate and sodium nitrate in the mixture. (JEE 1990)

10. Calculate the molality of 1.0L solution of 93% H2SO4, (weight/volume). The density of the solution is 1.84 g/mL. (JEE 1990)

11. ‘A’ is a binary compound of a univalent metal. 1.422g of A reacts completely with 0.321 g of sulphur in an evacuated and sealed tube to give 1.734 g of white crystalline solid B, that forms a hydrated double salt C with Al2(SO4)3. Identify A,B and C. (JEE 1994)

12. 8.0575 ×10-2 kg of Glauber’s salt is dissolved in water to obtain 1 dm3 of solution of density 1077.2 kg m-3. Calculate the molality, molarity and mole fraction of Na2SO4 in solution. (JEE 1994)

13. A plant virus is found to consist of uniform cylindrical particles of 150 A

in diameter and 5000 A

long. The specific volume of the virus is 0.75cm3/g. If the virus is considered to be a single particle, find its molar mass. (JEE 1999)

14. Find the molarity of water. Given density of water = 1000kg/m3 (JEE 2003)

15. In a solution of 100mL 0.5M acetic acid, one gram of active charcoal is added, which adsorbs acetic acid. It is found that the concentration of acetic acid becomes 0.49 M. If surface area of charcoal is 3.01 × 102m2. Calculate the area occupied by single acetic acid molecule on surface of charcoal. (JEE 2003)

16. 20% surface sites have adsorbed N2. On heating N2 gas evolved from sites and was collected at 0.001 atm and 298 K in a container of volume is 2.46cm3. Density of surface sites is 6.023×1014/cm2 and surface area is 1000 cm2; find out the number of surface sites occupied per molecule of N2. (JEE 2005)

ANSWER KEY 1. A 2. B 3. C 4. A 5. B 6. C 7. C 8. A 9. D 10. A 11. D 12. A 13. C 14. C 15. B 16. D 17. C 18. D 19. C 20. C 21. D 22. A 23. D 24. A 25. D 26. B 27. C 28. A 29. D 30. B 31. B 32. D 33. B 34.A 35.B 36. D 37. D 38. C 39. D 40. B 41. C 42. C

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

43. D 44. A 45. C 46. A 47. B 48. D 49. A 50. A 51. D 52. C 53. A 54. B 55. D 56. B 57. A 58. D 59. D 60. D 61. A 62. B 63. B 64. D 65. A 66. D 67. B 68. A 69. D 70. A 71. D 72. A 73. A 74. A 75. C 76. C 77. A 78. A 79. D 80. B FOUNDATION BUILDER SUBJECTIVE 1. [Y atoms] 2.[6.0224 1023] 3.[10 × 6.022 ×1023]

4. 10 6.022 1023, 8 6.022 1023, 86.022 1023 5. 40 6. 2.50921022 atom 7. 0.00288 8. [2.894 105 C] 9. [0.01] 10. [0.363N, 0.176N] 11.[1.90991010 year] 12.[1000] 13. [5 1016] 14. [4.911021] 15. [3.6104 g] 16. [4.48 L] 17. [2.651022]

18. [200 ml] 19. [K2 MnP2] 20. [C21H30O2] 21. [n=4] 22. 5 14 2C H N 23. CH 24. [19.81023] 25. [0.437]

26. 12.18 27. %NaHCO3 = 16.8, % Na2CO3= 83.2 28. 40.8% 29. [27.3 g, 2.736 g O2, 87.2%] 30. (C)

31. 42 4

60%40%

CHC H

32. [0.0166] 33. [38 g]

34. [4.87 g] 35.[%CaCO3=54.34%] 36. 2840 gm 37.. 216

moles

38. 49.8 % 39. x = 2 y = 1 40. 9.795 gm 41. (A) – B, (B) - [A: left moles = (4/31/2)], (C) - [7/2] 42. CO : CO2 = 3 : 1 43. [6.125g/lt.] 44. [0.1, 5.6] 45. [6.0221019] 46. [1.2475] 47. [0.35 M] 48. [0.4 m] 49. [3.205] 50. [1.4228] 51. [166.67 ml] 52. 12.18 53. ( 5 mole) 54. (2.5 mole) 55. 68g 56. A – (O2), B – (32.625), C – (1.18) 57. 225 g 58. 4.85%Cl , 34.19%Br 59. 4.97 Lt 60. 210 GET EQUIPPED FOR JEE-MAIN 1. D 2. D 3. A 4. D 5. B 6. D 7. B 8. A 9. A 10. A 11. A 12. D 13. C 14. D 15. D 16. A 17. C 18. B 19. A 20. B 21. C 22. A 23.B 24. B 25. D 26. C 27. C 28. A 29. A 30. A 31. C 32. D 33. D 34. B 35. D

GET EQUIPPED FOR JEE ADVANCE(SINGLE ANSWER CORRECT) 1. A 2. C 3. B 4. A 5. C 6. C 7. A 8. D 9. D 10. C 11. C 12. D 13. B 14. A 15. B 16. C 17. A 18. B 19. A 20. B 21. B 22. C 23. A 24. B 25. B 26. a 27. a 28. b 29. A 30. B 31. A 32. A 33. B

MORE THAN ONE CHOICE 1. A, B, D 2. A, B 3. A, B, D 4. A, C, D 5. A, C 6. A, C 7. A, B, D 8. A, B, C, D 9. A,B 10. B,C 11. C,D 12. B,C 13. A,D 14. C,D

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

MATCH THE FOLLOWING 1. I – P, II – A, III - C, IV – E 2. a – S, b – S, c- Q, d – R 3. a – PS, b – S, c – PQ, d – R 4. A- p ,r,s B-p, C- p,q,r, D-s COMPREHENSIVE TYPE PASSAGE – I 1. C 2. A 3. B 4. B PASSAGE – II 1. A 2. C 3. A 4. A PASSAGE – III 1. D 2. A 3. C 4. A PASSAGE-IV 1. A 2. A INTEGER TYPE 1. (5) 2. (3) 3. (3) 4. (4) 5. (5) 6. (2) 7.(2) 8. (6) 9. (4) 10. (8) EXPERTISE ATTAINERS 1. 0.156g ,(1.52 g) 2. a) (0.712), b) (0.714) 3. (53.5) 4. 2 4 2C H Cl 5. 0.5 : 0.1 : 0.4,

6. 9.14 gm, 200mL 7. 85%

WINDOWS TO COMPETITIVE EXAMS OBJECTIVE QUESTIONS

1. (C) 2. (A) 3. (A) 4. (A) 5. (A) 6. (D) 7. (A) 8. (D) 9. (D) 10.(A) 11. (A) 12. (C) 13. (A) 14. (D) 15. (B) 15. (C) 17. (C) 18. (A) 19. (B) 20. (C) Fill in blanks: 1. C-12 isotope 2. 6.023 × 1024 3. 0.4 m 4. 4.14g Subjective Answers

1. Therefore, natural boron contain 20% (10.01) first isotope and 80% other isotope. 2. Most of the elements found in nature exist as a mixture of isotopes whose atomic weights are different.

The atomic weight of an element is the average of atomic weights of all its naturally occurring isotopes.

3. 0.437

4. 10.6%

5. C2H4. 6. (a) 40% CO and 60% CO2 by volume. (b) 24

7. (i) 0.55

(ii) 39.9 10

8. 55.50L. 9. 1.7 g 10. 10.43

11. A= KO2, B=K2SO4, C= K2[Al2(SO4)4] 12. Molarity of solution = 0.25 M

molality = 0.24 m Mole fraction of Na2SO4= -34.3×10

13. 70.92 × 106 g 14. -155.56mol L

15. 0.5 ×10-18 m2 16. 2

JEE CHEMISTRY

The IITian’s Hub I IIT/MEDICAL/FOUNDATION/OLYMPIAD

IITIAN'S HUBTH

ET

HE

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

IIT–JEE Syllabus Development of structure of Atom and historical Models; Spectrum of hydrogen atom; Bohr model; de Broglie relation, Uncertainty principle, Quantum model; Electronic configuration of elements (upto to atomic number 36); Aufbau principle, Pauli’s exclusion principle and Hund’s rule; shapes of s, p, and d orbitals; Probability Density Curves (Radial and Angular)

CONTENTS

1. THEORY 74-109 2. SOLVED OBJECTIVE PROBLEMS 110- 112 3. SOLVED SUBJECTIVE PROBLEMS 113- 116 4. FOUNDATION BUILDER (OBJECTIVE) 117-126 5. FOUNDATION BUILDER (SUBJECTIVE) 127-130 6. GET EQUIPPED FOR IIT-JEE 130- 140 7. EXPERTISE ATTAINER 141-143 8. WINDOWS TO IIT-JEE 143-151

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

1. INTRODUCTION The existence of atoms has been proposed since the time of early Indian and Greek philosophers (400 B.C.) who were of the view that atoms are the fundamental building blocks of matter. According to them, the continued subdivisions of matter would ultimately yield atoms which would not be further divisible. The word ‘atom’ has been derived from the Greek word ‘a-tomio’ which means ‘uncutable’ or ‘non-divisible’. These earlier ideas were mere speculations and there was no way to test them experimentally. These ideas remained dormant for a very long time and were revived again by scientists in the nineteenth century. In this unit we start with the experimental observations made by scientists towards the end of nineteenth and beginning of twentieth century.

1.1 DALTON’S ATOMIC THEORY

All the objects around you, this book, your pen or pencil and things of nature such as rocks, water and plant constitute the matter of the universe. Matter is any substance which occupies space and has mass.

Dalton, in 1808, proposed that matter was made up of extremely small, indivisible particles called atoms. (In Greek atom means which cannot be cut). This concept was accepted for many years.

The main postulates of Dalton’s atomic theory are

Matter is made up of small indivisible particles, called atoms.

Atoms can neither be created nor destroyed. This means that a chemical reaction is just a simple rearrangement of atoms and the same number of atoms must be present before and after the reaction.

Atom is the smallest particle of an element which takes part in a chemical reaction.

Atoms of the same element are identical in all respects especially, size, shape and mass.

Atoms of different elements have different mass, shape and size.

Atoms of different elements combine in a fixed ratio of small whole numbers to form compound atoms, called molecules.

Drawbacks: 1. The discovery of isotopes and isobars showed that atom of same element may have different

atomic mass (isotopes) and atom of different kinds may have same atomic masses (isobars). 2. Atoms can be split into more fundamental particles: electrons, protons and neutrons.

2. DISCOVERY OF SUBATOMIC PARTICLES

Dalton’s atomic theory was able to explain the law of conservation of mass, law of constant composition and law of multiple proportions very successfully. However, it failed to explain the results of many experiments; for example, it was known that substances like glass or ebonite when rubbed with silk or fur generate electricity. Many different kinds of sub-atomic particles were discovered in the twentieth century. However, in this section we will talk about three particles, namely electron, proton and neutron.

2.1 CATHODE RAYS AND DISCOVERY OF ELECTRON Michael Faraday showed that chemical changes occur when electricity is passed though an

electrolyte. He stated that electricity is made up of particles called atoms of electricity. G.J Stoney suggested the name of electron for the atoms of electricity. However, the real credit for the discovery of electrons goes to J.J. Thomson. In mid 1850s many scientists mainly Faraday began to study electrical discharge in partially evacuated tubes, known as cathode ray discharge tubes. A cathode ray tube is made of glass containing two thin pieces of metal, called electrodes, sealed in it. The electrical discharge

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

through the gases could be observed only at very low pressures and at very high voltages. The pressure of different gases could be adjusted by evacuation.

When sufficiently high voltage is applied across the electrodes, current starts flowing as a stream of

particles moving in the tube from the negative electrode (cathode) to the positive electrode (anode). These were called cathode rays or cathode ray particles. The flow of current from cathode to anode was further checked by making a hole in the anode and coating the tube behind anode with phosphorescent material zinc sulphide. When these rays, after passing through anode, strike the zinc sulphide coating, a bright spot on the coating is developed (same thing happens in a television set) The results of these experiments are summarised below. (i) The cathode rays start from cathode and move towards the anode. (ii) These rays themselves are not visible but their behaviour can be observed with the help of certain

kind of materials (fluorescent or phosphorescent) which glow when hit by them. Television picture tubes are cathode ray tubes and television pictures result due to fluorescence on the television screen coated with certain fluorescent or phosphorescent materials.

(iii) In the absence of electrical or magnetic field, these rays travel in straight lines (iv) In the presence of electric or magnetic field, the behaviour of cathode rays is similar to that

expected from negatively charged particles, suggesting that the cathode rays consist of negatively charged particles, called electrons.

(v) The characteristics of cathode rays (electrons) do not depend upon the material of electrodes and

the nature of the gas present in the cathode ray tube. Thus, we can conclude that electrons are basic constituent of all the atoms.

(vi) Cathode rays produce heating effect. When these rays are made to strike on a metal foil, the latter

gets heated. (vii) Cathode rays produce X-rays when they strike on surface of hard metals such as tungsten, copper

molybdenum etc. (viii) Cathode rays can pass through thin foils of metals like aluminium. However, these are stopped if

the foil is quite thick. (ix) Cathode rays ionize the gas through which they pass.

J. J. Thomson (1856–1940):

Sir J.J. Thomson confirmed these findings in 1897. Thomson performed a series of

experiments in which he was able to determine the charge/mass em

ratio of the particles that

make up the cathode ray by measuring the deflection of the rays with varying magnetic and electric

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

fields. Thomson performed the same experiments using different metals for the cathode and anode as well as different gases inside the tube.

This value em

is found to be 111.76 10 coulomb per kg.

The e/m ratio for electron was found to be same irrespective of the nature of cathode and nature of gas taken in discharge tube. This shows that the electrons are universal constituent of all matter.

2.2 Quantization Of Charge

R.A. Millikan (1868-1953) devised a method known as oil drop experiment (1906-14), to determine the charge on the electrons. Millikan concluded that the magnitude of electrical charge, q, on the droplets is always an integral multiple of the electrical charge, e, That is, q = n e, where n = 1, 2, 3............... That means charge is quantised. He found that the value of e is –1.6 ×10–19 C. The present accepted value of electrical charge is – 1.6022 ×10–19 C. The mass of the electron (me) was determined by combining these results with Thomson’s value of e/m ratio.

Calculation of mass of an electron: With the help of the experiments discussed above, it is possible to calculate the values of e/m ratio and also the charge (e) on the electron. The mass (m) of the electron can be calculated as follows.

Mass of electron (m) 19

8 1(1.60 10 )

/ (1.76 10 )e C

e m Cg

28 319.10 10 9.10 10g kg .

An electron may be defined as: A fundamental particle present in an atom, which carries one unit negative charge 19(1.60 10 )C and negligible mass 28(9.1 10 )g which is 1/1837 of the mass of an atom of hydrogen.

2.3 DISCOVERY OF PROTONS (GOLDSTEIN)

We know that an atom is electrically neutral, if it contains negatively charged electrons it must also contain some positively charged particles. This was confirmed by Goldstein in his discharge tube experiment with perforated cathode. On passing high voltage between the electrodes of a discharge tube it was found that some rays were coming from the side of the anode which passed through the holes in the cathode. These anode rays (canal rays) consisted of positively charged particles formed by ionization of gas molecules by the cathode rays. The charge to mass ratio (e/m value) of positively charge particles was found to be maximum when the discharge tube was filled with hydrogen gas as hydrogen is the lightest element. e/m varies with the nature of gas taken in the discharge tube. The positive particles are positive residues of the gas left when the gas is ionized.

2.4 DISCOVERY OF NEUTRONS (CHADWICK)

The neutral charge particle, neutron was discovered by James Chadwick by bombarding boron or beryllium with –particles.

9 4 12 14 2 6 0Be C n

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Characteristics of the three fundamental particles are:

The electron and proton have equal, but opposite, electric charges while the neutron is uncharged.

3. ATOMIC MODELS We know the fundamental particles of the atom. Now let us see, how these particles are arranged in an atom to suggest a model of the atom.

3.1 THOMSON’S MODEL

J.J. Thomson, in 1904, proposed that there was an equal and opposite positive charge enveloping the electrons in a matrix. This model is called the plum – pudding model after a type of Victorian dessert in which bits of plums were surrounded by matrix of pudding.

electron

Positive sphere

This model could not satisfactorily explain the results of scattering experiment carried out by Rutherford who worked with Thomson.

3.2 RUTHERFORD’S MODEL

– Particles emitted by radioactive substance were shown to be dipositive Helium ions (He++) having a mass of 4 units and 2 units of positive charge.

Rutherford allowed a narrow beam of –particles to fall on a very thin gold foil of thickness of the order of 0.0004 cm and determined the subsequent path of these particles with the help of a zinc sulphide fluorescent screen. The zinc sulphide screen gives off a visible flash of light when struck by an particle, as ZnS has the remarkable property of converting kinetic energy of particle into visible light.

Electron Proton Neutron

Approximate relative mass 1/1836 1 1

Approximate relative charge –1 +1 No charge

Mass in kg 9.10910–31 1.67310–27 1.67510–27

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

OBSERVATIONS :

i) Majority of the –particles pass straight through the gold strip with little or no deflection.

ii) Some –particles are deflected from their path and diverge.

iii) Very few –particles are deflected backwards through angles greater than 90.

iv) Some were even scattered in the opposite direction at an angle of 180 [ Rutherford was very much surprised by it and remarked that “It was as incredible as if you fired a 15–inch shell at a piece of tissue paper and it came back and hit you”]. There is far less difference between air and bullet than there is between gold atoms and -particles assuming of course that density of a gold atom is evenly distributed.

CONCLUSIONS: 1. The fact that most of the - particles passed straight through the metal foil indicates the most part

of the atom is empty. 2. The fact that few - particles are deflected at large angles indicates the presence of a heavy

positively charge body , for such large deflections to occur - particles must have come closer to or collided with a massive positively charged body.

3. The fact that one in 20,000 have deflected at 180° backwards indicates that volume occupied by this heavy positively charged body is very small in comparison to total volume of the atom.

POSTULATES OF RUTHERFORD ATOMIC MODEL :

On the basis of the above observation, and having realized that the rebounding -particles had met something even more massive than themselves inside the gold atom, Rutherford proposed an atomic model as follows.

i) All the protons (+ve charge) and the neutrons (neutral charge) i.e nearly the total mass of an atom is present in a very small region at the centre of the atom. The atom’s central core is called nucleus.

ii) The size of the nucleus is very small in comparison to the size of the atom. Diameter of the nucleus is about 10–13cm while the atom has a diameter of the order of 10–8 cm. So, the size of atom is 105 times more than that of nucleus.

iii) Most of the space outside the nucleus is empty. iv) The electrons, equal in number to the net nuclear positive charge, revolve around the nucleus with

fast speed in various circular orbits. v) The centrifugal force arising due to the fast speed of an electron balances the coulombic force of

attraction of the nucleus and the electron remains stable in its path. Thus according to him atom consists of two parts (A) nucleus and (B) extra nuclear part.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

DEFECTS OF RUTHERFORD’S ATOMIC MODEL :

1. Position of electrons: The exact positions of the electrons from the nucleus are not mentioned.

2. Stability of the atom: Neils Bohr pointed out that Rutherford’s atom should be highly unstable. According to the law of electro–dynamics, any charged particle under acceleration should continuously lose energy. The electron should therefore, continuously emit radiation and lose energy. As a result of this a moving electron will come closer and closer to the nucleus and after passing through a spiral path, it should ultimately fall into the nucleus.

It was calculated that the electron should fall into the nucleus in less than 10–8 sec. But it is known that electrons keep moving outside the nucleus. To solve this problem Neils Bohr proposed an improved form of Rutherford’s atomic model. Before going into the details of Neils Bohr model we would like to introduce you some important atomic terms.

4. ATOMIC TERMS (a) Atomic Number (Z): The atomic number of an element is the number of protons contained in the

nucleus of the atom of that element. (b) Nucleons: Protons and neutrons are present in a nucleus. So, these fundamental particles are

collectively known as nucleons. (c) Mass Number (A): The total number of protons and neutrons i.e, the number of nucleons present

in the nucleus is called the mass number of the element. (d) Nuclide: Various species of atoms in general. A nuclide has specific value of atomic number and

mass number. (e) Isotopes: Atoms of the element with same atomic number but different mass number e.g. 1H1, 1H2,

1H3. There are three isotopes of hydrogen. (f) Isobars: Atoms having the same mass number but different atomic numbers, e.g. 15P32 and 16S32

are called isobars. (g) Isotones: Atoms having the same number of neutrons but different number of protons or mass

number, e.g. 6C14, 8O16, 7N15 are called isotones. (h) Isoelectronic: Atoms, molecules or ions having same number of electrons are isoelectronic e.g.

N2,CO, CN–. (i) Isosters : Molecules having same number of atoms and also same number of electrons are called

isosters. e.g., (i) N2 and CO ii) CO2 and N2O iii) HCl and F2

(j) Atomic mass unit: Exactly equal to 1

12

of the mass of 6C12 atom

1 amu = 1.66 10–27 kg, If it is converted to energy then E= 931.5 MeV (k) Isodiapheres: Atoms having same difference between neutrons & protons are called isodiapheres.

They have same value of N - Z or A - 2Z

Example: i) 238 23492 90&U Th ii) 3 7

1 3&H Li

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

5. CHARACTERISTICS OF WAVES

A wave is a sort of disturbance which originates from some vibrating source and travels outward as a continuous sequence of alternating crests and troughs. Every wave has five important characteristics, namely, wavelength (), frequency (), velocity (C), wave number and amplitude (A).

a

CrestCrest

TroughTrough A sinusoidal wave

Ordinary light rays, X–rays,–rays, etc. are called electromagnetic radiations because similar waves can be produced by moving a charged body in a magnetic field or a magnet in an electric field. These radiations have wave characteristics and do not require any medium for their propagation.

i) Wave length (): The distance between two neighbouring troughs or crests is known as wavelength. It is denoted by and is expressed in cm, m, nanometers (1nm=10–9m) or Angstrom (1Å=10–10m); 1 micron 6( ) 10 ,m 1 milli micron 9( ) 10m m 121 10 .pm m

ii) Frequency (): The frequency of a wave is the number of times a wave passes through a given point in a medium in one second. It is denoted by (nu) and is expressed in cycles per second (cps) or hertz (Hz) 1Hz = 1cps.

The frequency of a wave is inversely proportional to its wave length () 1 or =

c

iii) Velocity: The distance travelled by the wave in one second is called its velocity. It is denoted by c

and is expressed in cm sec–1. c = or = c

iv) Wave number : It is defined as number of wavelengths per cm. It is denoted by and is

expressed in cm–1. = 1 (or) =

c

v) Amplitude: It is the height of the crest or depth of the trough of a wave and is denoted by a. It determines the intensity or brightness of the beam of light & is also expressed in the unit of length

Ex.1 A source of sound 330velocity m s produces waves in the frequency range 500 Hz to 6600 Hz . Find the wavelength range of sound produced

Ans: . 330 V m s

maxmin

330 0.66500

V m

min

max

330 0.056600

V m

Ex.2 Find the distance traveled by a wave of frequency 20 KHz in 2 minutes if its wavelength is 1.5 cm

Ans: .V 2 31.5 10 20 10 m Hz 300 m s distance 300 120v t s 36 km

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

6. PLANCK’S QUANTUM THEORY When a black body is heated, it emits thermal radiations of different wavelengths or frequency. To explain these radiations, Max Planck put forward a theory known as Planck’s quantum theory. The main points of quantum theory are

i) Substances radiate or absorb energy discontinuously in the form of small packets or bundles of energy.

ii) The smallest packet of energy is called quantum. In case of light the quantum is known as photon.

iii) The energy of a quantum is directly proportional to the frequency of the radiation .E (or) E = h were is the frequency of radiation and h is Planck’s constant having the value 6.626 10–27

erg – sec or 6.626 10–34 J–sec. iv) A body can radiate or absorb energy in whole number multiples of a quantum. Hence it can be h,

2h,3h………..nh. where n is a positive integer.

v) Energy of photons can also be represented by nhc Joules

. Where,

n : number of photons

h : 346.626 10 J s .

8C 3 10 m s

: Wavelength in 'm ' .

In eV units,

photon12400E eV

in A

where, 191eV 1.6 10 Joules

7. ELECTROMAGNETIC WAVES Wave theory was given by C. Huygens. In 1856, James Clark Maxwell stated that light, X-rays,

-rays and heat etc. emit energy continuously in the form of radiations or waves and the energy is called radiant energy. These waves are associated with electric and magnetic fields and are, therefore, known as electromagnetic waves (or radiations). A few important characteristics of these waves are listed:

(i) They emit energy continuously in the form of radiations or waves. (ii) The radiations consist of electric and magnetic fields which oscillate perpendicular to each other and also perpendicular to the direction in which the radiations propagate. (iii) All the electromagnetic waves travel with the velocity of light 8 1(3.0 10 )ms . (iv) These rays do not require any medium for propagation. (v) They are not affected by electric or magnetic fields.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

WAVELENGTHS OF ELECTROMAGNETIC RADIATIONS: The complete range of the electromagnetic waves is known as electromagnetic spectrum. It may be defined as: The arrangement of different electromagnetic radiations in order of increasing wavelength or decreasing frequency. Ex.3 Find the energy range of photons belonging to the visible region Given:

4000violet A

and Re 8000d A

Sol: Maximum energy (minimum )

max12400 3.14000

E eV eV

Minimum energy (maximum ) min12400 1.55

8000E eV

eV

Ex.4 Two waves differ in frequency by 1510 Hz . If one wave has 2000 A

. Find possible values of other wave.

Sol: Given, 1 2000 A

151 1.5 10

c

Possible 2 values 15 150.5 10 , 2.5 10 Hz 2 values are:

(i) 15 60000.5 10

C A

(ii) 15 12002.5 10

C A

Electromagnetic radiations Wave length (Å)

Radio waves 31014 to 3 107

Micro waves 3109 to 3 106

Infrared (IR) 6106 to 7600

Visible 7600 to 3800

Ultra violet (UV) 3800 to 150

X–rays 150 to 0.1

Gamma rays 0.1 to 0.01

Cosmic rays 0.01 to zero

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Ex.5 Find the number of photons required to raise the temperature of 5 g of water from 20 C to 22 C ,

given that water has a specific heat capacity of 4.2 J g C and that 5000 A

. It is also given that photon energy is 80% utilized to heat the water

Ans: Heat required . .m s T

5 4.2 2g J g C C 42 J

i.e ,8042

100 photon totalJ E

80 42

100

hcie n

10

34 8

42 100 5000 10

80 6.626 10 3 10n

201.32 10 photons

8. BOHR’S ATOMIC MODEL Bohr developed a model for hydrogen atom and hydrogen like one–electron species (hydrogenic species). He applied quantum theory in considering the energy of an electron bound to the nucleus.

Important postulates

An atom consists of a dense stationary nucleus situated at the centre with the electron revolving around it in circular orbits without emitting any energy. The force of attraction between the nucleus and an electron is equal to the centrifugal force on the moving electron.

Out of many circular orbits around the nucleus, an electron can revolve only in those orbits whose angular momentum (mvr) is an integral multiple of factor 2h

mvr = 2

nh ; where, m = mass of the electron

v = velocity of the electron ; n = orbit number in which electron is present ; r = radius of the orbit

As long as an electron is revolving in such an orbit it neither loses nor gains energy. Hence these orbits are called stationary states. Each stationary state is associated with a definite amount of energy and it is also known as energy levels. The greater the distance of the energy level from the nucleus, the more is the energy associated with it. The different energy levels are numbered as 1,2,3,4, (from nucleus onwards) or K,L,M,N etc.

Ordinarily an electron continues to move in a particular least possible energy stationary state without losing energy. Such a stable state of the atom is called as ground state or normal state.

If energy is supplied to an electron, it may jump (excite) instantaneously from lower energy (say 1) to higher energy level (say 2,3,4, etc.) by absorbing one photon. This new state of electron is called as excited state. The quantum of energy absorbed is equal to the difference in energies of the two concerned levels.

Since the excited state is less stable, atom will lose its energy and come back to the ground state.

Energy absorbed or released in an electron jump, (E) is given by E = E2 – E1 = h

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Where E2 and E1 are the energies of the electron in the first and second energy levels, and is the frequency of radiation absorbed or emitted.

[Note: If the energy supplied to hydrogen atom is less than 13.6 eV, it will accept or absorb only those quanta which can take it to a certain higher energy level i.e., all those photons having energy different from what is required for a particular transition will not be absorbed by hydrogen atom. But if energy supplied to hydrogen atom is more than 13.6 eV then all photons are absorbed and excess energy appears as kinetic energy of emitted photo electron].

Radius and Energy levels of hydrogen atom

Consider an electron of mass ‘m’ and charge ‘e’ revolving around a nucleus of charge Ze (where, Z = atomic number and e is the charge of the proton) with a tangential velocity v. r is the radius of the orbit in which electron is revolving.

By Coulomb’s Law, the electrostatic force of attraction between the moving electron and nucleus is

Coulombic force = 2

2

rKZe

K = o4

1 (where o is permittivity of free space) ; K = 9 109 Nm2 C–2

In C.G.S. units, value of K = 1 dyne cm2 (esu)–2

The centrifugal force acting on the electron is r

mv 2

Since the electrostatic force balance the centrifugal force, for the stable electron orbit.

r

mv 2 =

2

2

rKZe … (1) (or) v2 =

mrKZe2

… (2)

According to Bohr’s postulate of angular momentum quantization, we have

mvr = 2

nh ; v = mr 2

nh

; v2 = 222

22

4 rmhn

… (3)

Equating (2) and (3) 222

222

4 rmhn

mrKZe

solving for r we get r = 22

22

mKZe4πhn , where n = 1,2,3 - - - - -

Hence only certain orbits whose radii are given by the above equation are available for the electron. The greater the value of n, i.e., farther the energy level from the nucleus, the greater is the radius.

The radius of the smallest orbit (n=1) for hydrogen atom (Z=1) is ro.

ro = Kme

hn22

22

4 =

9219312

2342

109106.110914.34

10626.61

= 5.29 10–11 m = 0.529 Å

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Radius of nth orbit for an atom with atomic number Z is simply written as

rn = 0.529 Zn2

Å

Calculation of velocity :

We know that ; mvr = 2

nh ; v = mr2

nh

By substituting for r we are getting ; v = nhKZe2 2

Where excepting n and z all are constants ; v = 2.18 108 nZ cm/sec.

Calculation of energy of an electron:

The total energy, E of the electron is the sum of kinetic energy and potential energy. Kinetic energy of the electron = ½ mv2

Potential energy = r

eK 2Z

Total energy = ½ mv2 – reK 2Z … (4)

From equation (1) we know that r

mv2

= 2

2

rKZe ; ½ mv2 =

rKZe

2

2

Substituting this in equation (4)

Total energy (E) = r2

KZe2

– r

KZe2

= r2

KZe2

Substituting for r, gives us E = 22

2422

hnKemZ2π where n = 1,2,3……….

This expression shows that only certain energies are allowed to the electron. Since this energy expression consist of so many fundamental constant, we are giving you the following simplified expressions.

E = –21.8 10–12 2

2

nZ erg per atom = –21.8 10–19

2

2

nZ J per atom = –13.6

2

2

nZ eV per atom

(1eV = 3.83 10–23 kcal, 1eV = 1.602 10–12 erg, 1eV = 1.602 10–19J)

E = –313.6 2

2

nZ kcal / mole (1 cal = 4.18 J)

The energies are negative since the energy of the electron in the atom is less than the energy of a free electron (i.e., the electron is at infinite distance from the nucleus) which is taken as zero. The lowest energy level of the atom corresponds to n=1, and as the quantum number increases, E becomes less negative.

When n = , E = 0, which corresponds to an ionized atom i.e., the electron and nucleus are infinitely separated.

H H++ e– (ionisation).

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Ex.6 Find the distance between 2nd and 4th orbit in He

Ans: Required answer 4, 2,He Her r . Now, radius 2

0.529 n AZ

Distance 2 20.529 4 2 3.1742

A A

Ex.7 Find the time taken for half a revolution by electron in 3rd orbit of 2Li

distancetime = 1 2speed

circumferencevelocity

2

32

3

2 ,12 ,

r LiV Li

2

23

0.529 3, 1.5873

r Li A

2 6 63, 2.18 10 2.18 10ZV Li m s

n

Time 166

1.587 2.287 102.18 10

A sm s

Ex.8 Find the wavelength of photon absorbed upon transition of an electron from 4th orbit to 6th orbit in

He Ans: 6, 4,photon He HeE E E

But, 2

, 213.6n ZZE eVn

2 21 113.6 4 1.894 6photonE eV

12400 65601.89

A A

MERITS AND DRAWBACKS OF BOHR’S MODEL Bohr’s model of the hydrogen atom was no doubt an improvement over Rutherford’s nuclear model, as it could account for the stability and line spectra of hydrogen atom and hydrogen like ions (for example, He+, Li2+, Be3+ and so on). Merits of Bohr’s theory i) The experimental value of radii and energies in hydrogen atom are in good agreement with that

calculated on the basis of Bohr’s theory. ii) Bohr’s concept of stationary state of electron explains the emission and absorption spectra of

hydrogen like atoms. iii) The experimental values of the spectral lines of the hydrogen spectrum are in close agreement with

that calculated by Bohr’s theory.

Limitations of Bohr’s theory

i) It does not explain the spectra of atoms having more than one electron.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

ii) Bohr’s atomic model failed to account for the effect of magnetic field (Zeeman Effect) or electric field (Stark effect) on the spectra of atoms or ions. It was observed that when the source of a spectrum is placed in a strong magnetic or electric field, each spectral line further splits into a number of lines. This observation could not be explained on the basis of Bohr’s model.

iii) De Broglie suggested that electrons like light have dual character. It has particle and wave character. Bohr treated the electron only as particle.

iv) Another objection to Bohr’s theory came from Heisenberg’s Uncertainty Principle. According to this principle “It is impossible to determine simultaneously the exact position and momentum of a small moving particle like an electron”. The postulate of Bohr, that electrons revolve in well defined orbits around the nucleus with well defined velocities is thus not tenable.

9. SPECTRUM

When light coming from a source is dispersed by a prism, light of different wavelength are deviated through different angles and get separated. This phenomenon is called dispersion and such a dispersed light may be received on a photo graphic plate or it may be viewed directly by eye. A collection of dispersed light giving its wave length composition is called a spectrum.

9.1 KINDS OF SPECTRUM (I) Emission Spectra: The spectrum of radiation emitted by a substance that has absorbed energy is called an emission

spectrum. Atoms, molecules or ions that have absorbed radiation are said to be “excited”. To produce an emission spectrum, energy is supplied to a sample by heating it or irradiating it and the wavelength (or frequency) of the radiation emitted, as the sample gives up the absorbed energy, is recorded. If the atom gains energy the electron passes from a lower energy level to a higher energy level, (energy is absorbed). That means a specific wave length is absorbed. Consequently, a dark line will appear in the spectrum. This dark line constitutes the absorption spectrum.

If the atom loses energy, the electron passes from higher to a lower energy level, energy is

released and a spectral line of specific wavelength is emitted. This line constitutes the emission spectrum. There are two types of emission spectrum.

(i) Continuous Spectrum: When white light is dispersed a bright spectrum continuously distributed

on the dark back ground is obtained. The colours are continuous during change and there are no sharp boundaries in between various colours. Colours appear to be merging into each other. Such a spectrum is known as a continuous spectrum.

Red orange

IndigoViolet

VIBGYOR

White light

Energy

Frequency

1Wavelength

Red

Violet

VIBGYORAngle of dispersion

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

(ii) Discontinuous Spectrum: (a) Line Spectrum (For atoms) : When an electron in excited state makes a transition to lower energy

states, light of certain fixed wave lengths are emitted. When such a light is dispersed we get sharp bright lines in dark back ground, such a spectrum is line emission spectrum:

+ Energy (Heat)

Electron in ground state

Electron in excited state

Transition

+ Photon of wavelength Specturm

(b) Band Spectrum (In molecules):

Band

(II) Absorption spectrum: When white light (composed of all visible photon frequencies) is passed through atomic hydrogen

gas, certain wave lengths are absent. The resulting spectrum consists of bright background with some dark lines. The pattern of the dark lines is called an absorption spectrum.

The missing wavelengths are same as the ones observed in the corresponding emission

spectrum.

9.2 ATOMIC SPECTRUM We have seen earlier that when electromagnetic radiation interacts with matter, atoms and molecules may absorb energy and reach to a higher energy state. With higher energy, these are in an unstable state. For returning to their normal state (more stable, lower energy states), the atoms and molecules emit radiations in various regions of the electromagnetic spectrum. These lines constitute the atomic spectrum of the elements. The atomic spectrum of the elements is a “characteristic property” of the elements and is often termed as “finger prints” of the elements.

Gas

absorbed wave length

Transmittedwave length

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

9.3 HYDROGEN SPECTRUM

If an electric discharge is passed through hydrogen gas taken in a discharge tube under low pressure, and the emitted radiation is analysed with the help of spectrograph, it is found to consist of a series of sharp lines in the UV, visible and IR regions. This series of lines is known as line or atomic spectrum of hydrogen. The lines in the visible region can be directly seen on the photographic film.

Each line of the spectrum corresponds to a light of definite wavelength. The entire spectrum consists of six series of lines, each series known after their discoverer as Lyman, Balmer, Paschen, Brackett, Pfund and Humphrey series. The wavelength of all these series can be expressed by a single formula which is attributed to Rydberg.

1 = R

2

221

11nn

Where, = wave number ; = wave length

R = Rydberg constant (109678 cm–1) = 110967820m

n1 and n2 have integral values as follows

Series n1 n2 Main spectral lines

Lyman 1 2,3,4, etc Ultra – violet

Balmer 2 3,4,5 etc Visible

Paschen 3 4,5,6 etc Infra – red

Brackett 4 5,6,7 etc Infra – red

Pfund 5 6,7,8, etc Infra – red

[Note: All lines in the visible region are of Balmer series but reverse is not true. i.e., all Balmer lines will not fall in visible region]

Total possible transitions for jump from 2 1n n

2 11

n n 1n to n i

2

, where 2 1n n n . This

also gives us the number of spectral lines observed under the given circumstances As discussed earlier, the above pattern of lines in atomic spectrum is characteristic of hydrogen.

n=1

n=2

n=3

n=4

n=5

n=6

n=

10.2

eV

3.6

eV

1.9

eV

3.4

eV0.

65 e

V 1.5

eV

Lyman(ultraviolet region)

Balmer(visible)

Paschen(Infrared)

Bracket(Infrared)

Pfund(Infrared)

Humphery(Infrared)

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Explanation for hydrogen spectrum by Bohr’s theory According to the Bohr’s theory electron neither emits nor absorbs energy as long as it stays in a particular orbit. However, when an atom is subjected to electric discharge or high temperature, and electron in the atom may jump from the normal energy level, i.e., ground state to some higher energy level i.e. excited state. Since the life time of the electron in excited state is short, it returns to the ground state in one or more jumps.

During each jump, energy is emitted in the form of a photon of light of definite wavelength or frequency. The frequency of the photon of light thus emitted depends upon the energy difference of the two energy levels concerned (n1, n2) and is given by

h = 2 1n nE E =

2

24222h

KemZ

21

22

11nn

; = 3

24222h

KemZ

22

21

11nn

The frequencies of the spectral lines calculated with the help of above equation are found to be in good agreement with the experimental values. Thus, Bohr’s theory elegantly explains the line spectrum of hydrogen and hydrogenic species.

Bohr had calculated Rydberg constant from the above equation.

= c = 3

2422

hKemZ2

22

21

11nn

; ch

KemZ213

2422

22

21

11nn

where ch

Kme23

242 = 1.097 10–7m–1 or 109678 cm–1 i.e., Rydberg constant (R)

Further application of Bohr’s work was made, to other one electron species (Hydrogenic ion) such as He+ and Li2+. In each case of this kind, Bohr’s prediction of the spectrum was correct.

Now after obtaining the explanation of Rydberg’s equation from Bohr’s theory, can you derive what could be the equation for other uni-electronic species? What would be the value of Rydberg’s constant for He+1 , Li+2 ?

Ex.9 Find out the longest wavelength of absorption line for hydrogen gas containing atoms in ground state.

Solution:

22

21

2

n1

n1RZ1

for longest wavelength E should be smallest i.e. transition occurs from n = 1 to n = 2

n = 7 n = 6

n = 5

n = 4

n = 3

n = 2

n = 1

i.e. 1 = 109678 cm–1 12

221

11

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

1 = 109678

43 cm–1

= 1

43 109678cm

= 1.2157 10–5 cm = 121.6 nm

Ex.10: The series limit for the Paschen series of hydrogen spectrum occurs at 8205.8Å. Calculate. (A) Ionization energy of hydrogen atom (B) Wave length of the photon that would remove the electron in the ground state of the

hydrogen atom.

Solution: (A) Energy corresponding to 8205.8 Ao = 10

834

108.820510310626.6

= 2.422 × 10-19 J = 1240820.58

=1.512 eV ; 21, 2 2

1 2

1 1HE E Z

n n

1.512 eV = E1,H × (1)2 ×

221

31 ; 1.512 eV =

9E H1

E1, H = 13.608 eV Ionisation energy of hydrogen atom = 13.6 eV

(B) 19

834

10602.16.1310310626.6

Ehc

= 1240 in nm13.6

= 91.6 nm

Ex.11 Calculate frequency of the spectral line when an electron from 5th Bohr orbit jumps to

the second Bohr orbit in a hydrogen atom

Solution: 1 =

22

21 n

1n1R = 109673

22 51

21 = 2.304 106 m–1

= C = 2.304 106 m–1 2.998 108 m/s = 6.906 1014 Hz

Ex.12: Calculate the energy of an electron in 3rd Bohr orbit.

Solution: En = 2n6.13

eV = 236.13

= – 1.51 eV = – 2.42 10–19 J

Ex.13: Calculate the energy in kJ per mole of electronic charge accelerated by a potential of 1 volt.

Solution: Energy in joules = charge in coulombs potential difference in volt

= 1.6 10–19 6.02 1023 1 ; = 9.632 104 J or 96.32 kJ

Ex.14: What is highest frequency photon that can be emitted from hydrogen atom? What is wavelength of this photon?

Solution: Highest frequency photon is emitted when electron comes from infinity to 1st energy level.

E = 2

2

1Z6.13

= – 13.6eV

or 13.6 1.6 10–19 Joule = 2.176 10–18 Joule

E = h = hE =

18

342.176 10 J6.626 10 Js

= 0.328 1016 Hz ; =

C

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

= 16

8

10328.0103

= 9.146 10–8 m

Ex.15: Calculate the longest wavelength transition in the Paschen series of He+.

Solution:

22

21

2H n

1n1ZR ; For He; Z = 2; For Paschen series n1 = 3

For longest wavelength n2 = 4

222

41

3121096781 = 109678×4×

161

91 = 109678 × 4 ×

1447

= 4689 Ao

Ex.16: Calculate the ratio of the wavelength of first and the ultimate line of Balmer series of Li2+?

Solution: wave number of first line of Balmer, 1 = 4

536

9531

21

222 RRRZ

wave length of first line of Balmer = R54

wave number of ultimate line of Balmer, 2 =

1

212

2RZ = 4

9R

wave length of ultimate line of Balmer = R94 Ratio =

59

10. PHOTO ELECTRIC EFFECT

Sir J.J. Thomson observed that when a light of certain frequency strikes the surface of a metal, electrons are ejected from the metal. This phenomenon is known as photoelectric effect and the ejected electrons are called photoelectrons. A few metals, which are having low ionisation energy like Cesium, show this effect under the action of visible light but many more show it under the action of more energetic ultraviolet light.

V

A

–+ electrons

Evacuated quartz tube

Light

An evacuated tube contains two electrodes connected to a source of variable voltage, with the metal plate whose surface is irradiated as the anode. Some of the photoelectrons that emerge from this surface have enough energy to reach the cathode despite its negative polarity, and they constitute the measured current. The slower photoelectrons are repelled before they get to the cathode. When the voltage is increased to a certain value Vo, of the order of several volts, no more photoelectrons arrive, as indicated by the current dropping to zero. This extinction voltage is also referred as stopping potential corresponds to the maximum photoelectron kinetic energy i.e.,

eVo = ½ mv2

The experimental findings are summarized as below:

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

i) Electrons come out as soon as the light (of sufficient frequency) strikes the metal surface.

ii) Light of any random frequency will not be able to cause ejection of electrons from a metal surface. There is a minimum frequency, called the threshold (or critical) frequency, which can just cause the ejection. This frequency varies with the nature of the metal. The higher the frequency of the light, the more energy the photoelectrons have. Blue light results in faster electrons than red light.

iii) Photoelectric current is increased with increase in intensity of light of same frequency, if emission is permitted i.e., a bright light yields more photoelectrons than a dim one of the same frequency, but the electron energies remain the same.

Light must have stream of energy particles or quanta of energy (h). Suppose, the threshold frequency of light required to eject electrons from a metal is o, when a photon of light of this frequency strikes a metal it imparts its entire energy (ho) to the electron.

E = ho E > ho

K.Emax = h – hoK. E = 0

Metal

“This energy enables the electron to break away from the atom by overcoming the attractive influence of the nucleus”. Thus each photon can eject one electron. If the frequency of light is less than o there is no ejection of electron. If the frequency of light is higher than o (let it be ), the photon of this light having higher energy (h), will impart some energy to the electron that is needed to remove it away from the atom. The excess energy would give a certain velocity (i.e, kinetic energy) to the electron. h = ho + (K.E)max ; h = ho + ½ mv2 ; ½ mv2 = h–ho Where, = frequency of the incident light o = threshold frequency

ho is the threshold energy (or) the work function denoted by = ho (minimum energy of the photon to liberate electron). It is constant for particular metal and is also equal to the ionization potential of gaseous atoms.

The kinetic energy of the photoelectrons increases linearly with the frequency of incident light. Thus, if the energy of the ejected electrons is plotted as a function of frequency, it result in a straight line whose slope is equal to Planck’s constant ‘h’ and whose intercept is ho.

It is important to note that the expression involves (KE)max. In reality e- have K.E lesser than this.

K.Emax of Photoelectrons Threshold frequency

tan h

0

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Total energy falling on plate = n h v photon intensity = nAt

Intensity of energy falling on the metal surface= n h vAt

Variation of P.E current vs voltage for different photon intensities I1, I2, I3

Ex.17: A photon of wavelength 5000 A

strikes a metal surface, the work function of the metal being 2.20 eV. Calculate (i) the energy of the photon in eV (ii) the kinetic energy of the emitted photo electron and (iii) the velocity of the photo electron.

Solution: i) Energy of the photon

E = h = hc =

m105ms103Js106.6

7

1834

= 3.96 10–19 J (1 eV = 1.6 10–19 J)

Therefore E = 19

19

3.96 101.6 10 /

JJ eV

= 2.475 eV

ii) Kinetic energy of the emitted photo electron Work function = 2.20 eV Therefore, KE = 2.475 – 2.20 = 0.275 eV = 4.4 10–20 J

iii) Velocity of the photo electron ; KE = 2mv21 = 4.4 10–20 J

Therefore, velocity (v) = 31

20

101.9104.42

= 3.11 105 ms–1

Ex.18:

When photons of wavelength 1240 A

strikes a metal surface, the photoelectrons of maximum KE

has thrice as much KE as the fastest electrons ejected by a source of wavelength 2480 A

from the same metal surface.

(a) Find work function of the metal (b) Find threshold wavelength of the metal

(c) Find stopping potential when light of wavelength 3100 A

is incident on this meal. Solution:

(a) 1240 ,1240A photon A

KE E work function

2480 ,2480A photon A

KE E

Given, 1240 2480

3A A

KE KE

P.E current

1I

2I

3I1 2 3I I I

V

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

ie 12400 1240031240 2480

or 10 3 5eV eV

or 2 15 10 2.5eV eV

(b) 012400 4960 A

(c) max12400 1.53100

KE eV

Stopping potential 1.5volts

11. Particle-Wave duality of Matter and Radiation In case of light some phenomenon like diffraction and interference can be explained on the basis of its wave character. However, the certain other phenomenon such as black body radiation and photoelectric effect can be explained only on the basis of its particle nature. Thus, light is said to have a dual character. Such studies on light were made by Einstein in 1905.

Louis de Broglie, in 1924 extended the idea of photons to material particles such as electrons and he proposed that matter also has a dual character-as wave and as particle.

11.1 DE-BROGLIE EQUATION

The wavelength of the wave associated with any material particle was calculated by analogy with photon.

In case of photon, if it is assumed to have wave character, its energy is given by

E = h ------------------ (i) (according to the Planck’s quantum theory)

Where is the frequency of the wave and ‘h’ is is is Planck’s constant

If the photon is supposed to have particle character, its energy is given by

E = mc2 ------------------ (ii) (according to Einstein’s equation)

where ‘m’ is the mass of photon, ‘c’ is the velocity of light.

By equating (i) and (ii) h = mc2 ; But = c/ ; h c = mc2 ; (or) = h /mc

The above equation is applicable to material particle if the mass and velocity of photon is replaced by the mass and velocity of material particle. Thus for any material particle like electron

= h/mv (or) = ph ; Where mv = p is the momentum of the particle.

What can you say about wave nature of the fast train from Churchgate to Virar?

Although the de Broglie equation is applicable to all material objects but it has significance only in case of microscopic particles. Since, we come across macroscopic objects in our everyday life, de Broglie relationship has no significance in everyday life.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Ex.19: Two particles A & B are in motion. If the wavelength associated with particle A is 5 10–8 m,

calculate the wavelength associated with particle B if its momentum is half of A.

Solution: According to de Broglie equation

A = Ap

h and B = Bp

h B

A

=

A

B

pp

But pB = ½ pA (given)

B

A

=

A

A

pp2/1

=1/2 B = 2A = 2 510–8 m = 10–7 m

Ex.20: Calculate the de Broglie wavelength of a ball of mass 0.1kg moving with a speed of 60ms–1.

Solution: mvh

= 601.0

106.6 34

= 1.1 10–34 m.

11.2 Explanation of quantization of Angular Momentum

According to Bohr’s model, the electron revolves around the nucleus in circular orbits. According to de Broglie concept, the electron is not only a particle but has a wave character also.

If the wave is completely in phase, the circumference of the orbit must be equal to an integral multiple of wave length () .Therefore 2r =n where ‘n’ is an integer and ‘r’ is the radius of the orbit But = h/mv 2r = nh /mv (or) mvr = n h/2

Which is Bohr’s postulate of angular momentum, where ‘n’ is the principal quantum number.

“Thus, the number of waves an electron makes in a particular Bohr orbit in one complete revolution is equal to the principal quantum number of the orbit”.

Alternatively

Number of waves ‘n’ = r2 =

mvhr2 =

hmvr2

Where v and r are the velocity of electron and radius of that particular Bohr orbit in which number of waves are to be calculated, respectively.

The electron is revolving around the nucleus in a circular orbit. Can you guess how many revolutions it can make in one second?

11.3 HEISENBERG’S UNCERTAINTY PRINCIPLE

All moving objects that we see around us e.g., a car, a ball thrown in the air etc., move along definite paths. Hence their position and velocity can be measured accurately at any instant of time. Is it possible for subatomic particle also?

As a consequence of dual nature of matter, Heisenberg, in 1927 gave a principle about the uncertainties in simultaneous measurement of position and momentum (mass velocity) of small particles.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

This principle states that “It is impossible to measure simultaneously the position and momentum of a small microscopic moving particle with absolute accuracy or certainty” i.e., if an attempt is made to measure any one of these two quantities with higher accuracy, the other becomes less accurate. The product of the uncertainty in position (x) and the uncertainty in the momentum (p = m.v where m is the mass of the particle and v is the uncertainty in velocity) is equal to or greater than h/4 where h is the Planck’s constant.

Thus, the mathematical expression for the Heisenberg’s uncertainty principle is simply written as x . p h/4

Explanation of Heisenberg’s uncertainty principle Suppose we attempt to measure both the position and momentum of an electron, to pinpoint the position of the electron we have to use light so that the photon of light strikes the electron and the reflected photon is seen in the microscope. As a result of the hitting, the position as well as the velocity of the electron is disturbed. The accuracy with which the position of the particle can be measured depends upon the wavelength of the light used. The uncertainty in position is . The shorter the wavelength, the greater is the accuracy. But shorter wavelength means higher frequency and hence higher energy. This high energy photon on striking the electron changes its speed as well as direction. But this is not true for macroscopic moving particle. Hence Heisenberg’s uncertainty principle is not applicable to macroscopic particles.

Ex.21: If the uncertainty in the position of an electron is 0.33 pm, what will be uncertainty in its velocity?

Solution: 120.33 10x m

hx p4

12 31 h0.33 10 9.1 10 v4

348

12 316.6 10v 1.75 10

4 3.14 .33 10 9.1 10m/sec.

Ex.22 Calculate the uncertainty in velocity of a ball of mass

132 gms if the uncertainty in

position is of the order of 1 Ao.

Solution:

hx.m. v4

3423

3 106.6 10v 2.5 10 m / sec

4 132 10 1 10

12. QUANTUM MECHANICAL MODEL OF ATOM The atomic model which is based on the particle and wave nature of the electron is known as wave or quantum mechanical model of the atom. This was developed by Ervin Schrödinger in 1926. This model describes the electron as a three dimensional wave in the electronic field of positively charged nucleus. Schrödinger derived an equation which describes wave motion of an electron. The differential equation is

0)VE(h

m8dzd

dyd

dxd

2

2

2

2

2

2

2

2

(r, , )

(0,0,0)

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

where x, y, z are certain coordinates of the electron, m = mass of the electron E = total energy of the electron. V = potential energy of the electron; h = Planck’s constant and (psi) = wave function of the electron. When Schrödinger equation is solved for hydrogen atom, the solution gives the possible energy levels the electron can occupy and the corresponding wave function () of the electron associated with each energy level. Actual view and properties of quantum mechanical model can be better understood in polar coordinates instead of normal Cartesian coordinate.In a polar coordinate any point in space can be represented in terms of r, and ; where r = distance from the origin; and = angles from any of two Cartesian coordinates Amplitude of electronic wave inside the atom. It can change if any of r, and changes. So has to be the function of r, and . i.e. = f(r, , ) Fortunately this whole function can be written as multiplication of two different functions as

(r, , ) = R(r) × A (, ) Where R(r), Radial Wave function is dependent only on distance from the nucleus and A(, ), Angular Wave function depends only on the two angles. Significance of The wave function may be regarded as the amplitude function expressed in terms of coordinates x, y and z. The wave function may have positive or negative values depending upon the value of coordinates. The main aim of Schrödinger equation is to give solution for probability approach. When the equation is solved, it is observed that for some regions of space the value of is negative. But the probability must be always positive and cannot be negative, it is thus, proper to use 2 instead of .

Significance of 2: 2 gives us probability density. It describes the probability of finding an electron within a small space. The space in which there is maximum probability of finding an electron is termed as orbital. The important point of the solution of the wave equation is that it provides a set of numbers called quantum numbers which describe energies of the electron in atoms, information about the shapes and orientations of the most probable distribution of electrons around nucleus.

12.1 QUANTUM NUMBERS

An atom contains large number of shells and subshells. These are distinguished from one another on the basis of their size, shape and orientation (direction) in space. The parameters are expressed in terms of different numbers called quantum numbers.

Quantum numbers may be defined as a set of four numbers with the help of which we can get complete information about all the electrons in an atom. It tells us the address of the electron i.e., location, energy, the type of orbital occupied and orientation of that orbital.

i) Principal quantum number (n): It tells the main shell in which the electron resides and the approximate distance of the electron from the nucleus. This value determines to a large extent energy of the orbital. It also tells the maximum number of electrons a shell can accommodate is 2n2, where n is the principal quantum number.

Shell K L M N Principal quantum number (n) 1 2 3 4 Maximum number of electrons 2 8 18 32

Radial Wave function

Total Wave function

Angular Wave function

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Permissible values of n: all positive integers.

ii) Azimuthal or angular momentum quantum number ( ): This represents the number of subshells present in the main shell. These subsidiary orbits within a shell will be denoted as s,p,d,f… This tells the shape of the sub shells. The orbital angular momentum of the electron is given as:

1 2h (or) 1 for a particular value of ' '

2h where .

For a given value of n, possible values of vary from 0 to n – 1. This means that there are 'n ' possible shapes in the thn shell.

iii) The magnetic quantum number (m): An electron due to its angular motion around the nucleus generates an electric field. This electric field is expected to produce a magnetic field. Under the influence of external magnetic field, the electrons of a subshell can orient themselves in certain preferred regions of space around the nucleus called orbitals. The magnetic quantum number determines the number of preferred orientations of the electron present in a subshell. The values allowed depends on the value of , the angular momentum quantum number, m can, assume all integral values between – to + including zero. Thus m can be –1, 0, +1 for = 1. Total values of m associated with a particular value of are given by 2 + 1.

iv) The spin quantum number (s): Just like earth not only revolves around the sun but also spins about its own axis, an electron in an atom not only revolves around the nucleus but also spins about its own axis. Since an electron can spin either in clockwise direction or in anticlockwise direction, therefore, for any particular value of magnetic quantum number, spin quantum number can have two values, i.e., +1/2 and –1/2 or these are represented by two arrows pointing in the opposite directions, i.e., and . When an electron goes to a vacant orbital, it can have a clockwise or anticlockwise spin. This quantum number helps to explain the magnetic properties of the substances.

Spin angular momentum s

hs(s 1)2

where s = ½.

Another term, defined as multiplicity is given as 2|S|+1 where |S| is total spin

= no. of unpaired electrons × 1/2.

Can you derive the following expressions?

1. no. of orbital = 2n or (2 1) for ‘n’ ' ' respectively

2. no. of e- = 2n2 or 2(2 1) for ‘n’ ' ' respectively Nodes : The region where the probability of finding an electron is zero or the probability density function

reduces to zero is called a nodal surface or simply nodes. Nodes are classified as radial nodes and angular nodes. In general, an orbital with principal quantum number = n and azimuthal quantum number = l, has

Total nodes = n – 1 ; Radial nodes = n – – 1 ;Angular nodes = ; At Nodes, 2=0, =0

12.2 SHAPES AND SIZE OF ORBITALS

An orbital is the region of space around the nucleus within which the probability of finding an electron of given energy is maximum (B > 90%). The shape of this region (electron cloud) gives the shape of the orbital. It is basically determined by the azimuthal quantum number , while the orientation of orbital depends on the magnetic quantum number (m). Let us now see the shapes of orbitals in the various subshells.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

s–orbitals: These orbitals are spherical and symmetrical about the nucleus. The probability of finding the electron is maximum near the nucleus and keeps on decreasing as the distance from the nucleus increases. There is vacant space between two successive s–orbitals known as radial node. But there is no radial node for 1s orbital since it is starting from the nucleus.

The size of the orbital depends upon the value of principal quantum number (n). Greater the value of n, larger is the size of the orbital. Therefore, 2s–orbital is larger than 1s orbital but both of them are non- directional and spherically symmetrical in shape.

p–orbitals ( =1): The probability of finding the p–electron is maximum in two lobes on the opposite sides of the nucleus. This gives rise to a dumb–bell shape for the p– orbital. For p–orbital l = 1. Hence, m = –1, 0, +1. Thus, p–orbital have three different orientations. These are designated as px , py & pz depending upon whether the density of electron is maximum along the x, y and z axis respectively. As they are not spherically symmetrical, they have directional character. The two lobes of p–orbitals are separated by a nodal plane, where the probability of finding electron is zero.

The three p-orbitals belonging to a particular energy shell have equal energies and are called degenerate orbitals.

d–orbitals ( =2): For d–orbitals, l =2. Hence m= – 2,–1,0,+1,+2. Thus there are 5 d orbitals. They have relatively complex geometry. Out of the five orbitals, the three (dxy, dyz,dzx) project in between the axis and the other two 2zd and 22 yd

x lie along the axis.

Y

XZ

d x y2 2d z 2

Dough–nut shape or Babysoother shape

Clover leaf shape

ZY

dxy

X X Y

Z

dyzdxz

Y

pz py

px Z

X

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

12.3 PROBABILITY DISTRIBUTION CURVE

It is found that wave functions ( ) can be expressed as the product of two functions, one of which the radial part R(r) depends only on the distance from the nucleus, the other being the angular part f( , ) depends only on the angles & . (r, , ) R(r) f( , )

Radial wave function angular wave function

Hence, Probability distribution curves which give the variation of probability of finding the electron can also be classified into two types (i) one which give the variation of probability of finding the electron with radial distance (r), termed as radial probability distribution curves, and (ii) and one which give the variation of probability of finding the electron with angle keeping the radial distance same (θ & ). Let us understand each of them separately.

RADIAL PROBABILITY DISTRIBUTION CURVES:

We had earlier studied that ψr2 gives the radial probability density of finding an electron at a point. It

refers to the radial probability of finding an electron in a unit volume in an atom at a radial distance of r from the nucleus. Hence, total radial probability in a spherical shell of thickness dr at a radial distance of r from the nucleus (which will have a volume of 4πr2dr) is given by 4πr2ψr

2dr. Sometimes, ψr2 is often represented as R2.

Calculation of Radial probability distribution function As stated above, the radial probability density at a radial distance r is R2(r). Therefore radial

probability of finding the election in a volume dv will be R2(r) dv. The radial probability is the probability of finding the election in a radial shell between spheres of

radii r and r + dr, where dr is small radial distance.

Volume of spherical shell of thickness 3 34 4dr (r dr) r3 3

3 3 34 r (dr) 3rdr(r dr) r

3

3 2 24 (dr) 3r dr 3r(dr)3

Since dr is very small, so the terms can be neglected.

2 24dv 3r dr 4 r dr3

Radial probability of finding an e- in a shell of thickness dr at a distance ‘r’ = 2 2R (r) 4 r dr ; = 2 24 r R (r)dr Summary of Radial Probability Density and Radial Probability Distribution Curves

R(r)

r

R

r

node

R

r

1 s 2 s 2 p

1 s 2 s 2 p

r

d r

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

R2

r

R2

r

node

r

R2

1 s 2 s 2 p

rr

node

0.529 A

Bohr rdius= 0.529 A

2 2 24 r R (r) 2 24 r R 2 2 24 r R

r

(3 s) (3 s) (3 s)

r

2 2 24 r RR(r)

nodesR2(r)

r r

(3 p) (3 p) (3 p)

r

2 2 24 r R (r)R(r) R2(r)

r r

(3 d) (3 d) (3 d)

r

R(r) R(r)

r r

R(r)2R (r) 2 2 24 r R (r)

Radial Probability Density curves are between ψr

2 vs r whereas Radial Probability Distribution curves which would be more useful are between 4πr2ψr

2 vs r.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Comparison between 3 s, 3 p and 3d orbitals:

r

3d (There is no node)

rmax

rmax

rmax

3 s

3 p

3 d

2 24 r R

1. For the same value of n, the distance of max probability, rmax of various orbitals is inversely dependent upon the value of .

max, r

max 3s max 3p max 3d(r ) (r ) (r )

2. Penetration power of an orbital is a measure of its closeness to the nucleus. Due to the additional maximas in 3 s curve, electron in 3 s spends some of its time near the nucleus making it to be more penetrating than 3 p which in turn more penetrating than 3 d.

3s 3p 3dDecreasing order of penetration power

Important points to note about the 4r2R2 vs. r plots: i) Radial probability is ALWAYS SMALL near the nucleus (4r2 small near the nucleus). ii) The maximum in the 4r2R2

1s vs. r plot occurs at 0.53Å - just the radius of the n=1 orbit of the Bohr model.

iii) On average a 2s electron spends its most time at a greater distance from the nucleus than the 1s electron which is consistent with the observation that (r1s) max < ( r2s) max

iv) The position of the principal (i.e. largest) maximum depends on n and l. For fixed l, as n increases the position of the principal maximum moves to larger r values. (For fixed n, the position of the

principal maximum moves to shorter r values as .) i.e., (r2p)max < (r2s)max ; (r3d) max < (r3p)max < (r3s)max ; r2p max < ( r3p) max Plots of Angular wave functions Angular probability density 2f ( , ) determines the shape of orbitals and its orientation in space f( , ) for s-orbitals does not depend upon angles & , for all other orbitals f( , ) will be a

function in terms of & . The angular probability density curves are same as the shape and orientation of the orbital in space.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

z

Elengatedtwo lobes

2f ( , )

Sphericaltwo lobes

2f ( , ) Angular nodes (Nodal Planes): Total no. of angular nodes for any orbital = For s orbital there will be no angular node.

z

x

px

For px orbital, = 1 no. of angular nodal = 1 (yz plane) For py orbital, nodal plane (xz plane) For pz orbital, nodal plane (xy plane) For dxy orbital, = 2 therefore 2 nodal planes, nodal planes: xz & yz planes

y

x

xyd

For dyz orbital : xy & xz planes are nodal planes.

Z

Y

For dzx orbital : xy & yz planes are nodal planes.

Z

X

x

y

nodal planes

2 2x yd

Can you guess where will be the two angular nodes for dz2 located ?

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

no. of radial nodes = n - - 1 no. of angular nodes = total no. of nodes = n - - 1 + = (n – 1) No. of peaks in 2

r vs r curve is (n - )

13. ELECTRONIC CONFIGURATION The electron configuration of an atom is the particular distribution of electrons among available shells. It is described by a notation that lists the subshell symbols, one after another. Each symbol has a superscript on the right giving the number of electrons in that subshell. For example, a configuration of the lithium atom (atomic number 3) with two electrons in the 1s subshell and one electron in the 2s subshell is written 1s22s1. The notation for electron configuration gives the number of electrons in each subshell.

13.1 RULES FOR FILLING OF ELECTRONS IN VARIOUS ORBITALS

The atom is built up by filling electrons in various orbitals according to the following rules.

Aufbau Principle: This principle states that the electrons are added one by one to the various orbitals in order of their increasing energy starting with the orbital of lowest energy. The increasing order of energy of various orbital is

1s,2s,2p,3s,3p,4s,3d,4p,5s,4d,5p,6s,4f,5d,6p,5f,6d,7p……………………

How to remember such a big sequence? To make it simple we are giving you the method to write the increasing order of the orbitals. Starting from the top, the direction of the arrows gives the order of filling of orbitals.

1s

2s

3s

4s

5s

6s

7s

2p

3p

4p

5p

6p

3d

4d

5d

4f

Alternatively, the order of increasing energies of the various orbitals can be calculated on the basis of (n+l) rule.

The energy of an orbital depends upon the sum of values of the principal quantum number (n) and the azimuthal quantum number (l). This is called (n+ l) rule. According to this rule,

“In neutral isolated atom, the lower the value of (n+ l) for an orbital, lower is its energy. However, if the two different types of orbitals have the same value of (n+ l), the orbitals with lower value of n has lower energy’’.

Illustration of (n + l) rule

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Type of orbitals Value of n Values of l Values of (n+ l) Relative energy

1s 1 0 1+0=1 Lowest energy 2s 2 0 2+0=2 Higher energy than 1s orbital 2p 2 1 2+1=3 2p orbital (n=2) have lower

energy than 3s orbital (n=3) 3s 3 0 3+1=3 (n + ) rule is applicable for multi electronic systems only. For uni-electronic system like

H, order of energy of orbitals is not “significantly influenced” by . Now can you write the order of energy of orbitals for uni-electronic system?

Pauli’s Exclusion principle

According to this principle, an orbital can contain a maximum number of two electrons and these two electrons must be of opposite spin.

Two electrons in an orbital can be represented by or

Hund’s rule of maximum multiplicity

This rule deals with the filling of electrons in the equal energy (degenerate) orbitals of the same sub shell (p,d and f). According to this rule,

“Electron pairing in p,d and f orbitals cannot occur until each orbital of a given subshell contains one electron each or is singly occupied & that too with the same spin.

This is due to the fact that electrons being identical in charge, repel each other when present in the same orbital. This repulsion can, however, be minimized if two electrons move as far apart as possible by occupying different degenerate orbitals. All the electrons in a degenerate set of orbitals will have same spin. Multiplicity is given by 2|S| + 1. Can you now comment why the rule is called Hund’s rule

of maximum multiplicity.

13.2 ELECTRONIC CONFIGURATION OF ELEMENTS

Electronic configuration is the distribution of electrons into different shells, subshells and orbitals of an atom.

n l xNumber of electrons in thesubshell

Symbol of subshell ororbitals (s,p,d,f)

Principal quantum number

Alternatively

Orbital can be represented by a box and an electron with its direction of spin by arrow. To write the electronic configuration, we need to know (i) the atomic number (ii) the order in which orbitals are to be filled (iii) maximum number of electrons in a shell, sub–shell or orbital.

(A) Each orbital can accommodate two electrons

(B) The number of electrons to be accomodated in a subshell is 2 number of degenerate orbitals.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Subshell Maximum number of electrons s 2 p 6 d 10 f 14

(C) The maximum number of electron in each shell (K,L,M,N…) is given by 2n2. Where n is the principal quantum number.

(D) The maximum number of orbitals in a shell is given by n2 where n is the principal quantum number.

Ex. 23 Write the electronic configuration of nitrogen (atomic number= 7)

Solution:

1s2 2s2 2p3

Exceptional Configurations (Extra stability of half-filled and fully-filled sub shell) : The electronic configuration of most of the atoms follow the Aufbau’s rule. However, in certain elements such a Cr, Cu etc. electron fills in 3d in preference to 4s provided the subshell become either half-filled or fully filled. 24Cr [Ar] 3d5, 4s1 and not [Ar] 3d4, 4s2 ; 29Cu [Ar] 3d10 4s1 and not [Ar] 3d9, 4s2 It has been found that there is extra stability associated with these electronic configurations. This stabilization is due to the following two factors 1. Symmetrical distribution of electron:

It is well known that symmetry leads to stability. The completely filled or half-filled subshells have symmetrical distribution of electrons in them and are therefore more stable. This effect is more dominant in d and f-orbitals.

This means three or six electrons in p-subshell, 5 or 10 electrons in d-subshell, and 7 or 14 electrons in f-subshell forms a stable arrangement.

2. Exchange Energy:

This stabilizing effect arises whenever two or more electrons with the same spin are present in the degenerate orbitals of a subshell. These electrons tend to exchange their positions and the energy released due to this exchange is called exchange energy. The number of exchanges that can take place is maximum when the subshell is either half-filled or full filled. As a result the exchange energy is maximum and so is the stability.

(1)

4-exchanges by electron 1

(2)


(3)


(4)


Total exchanges = 10 If n is the number of electron with parallel spins then can you calculate total number of

possible exchanges?

e.g. Total exchanges possible are = 6

e.g. 24Cr : Total exchanges possible are = 6

Total exchanges possible are = 10

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

The stabilation due to exchange energy will compensate for the energy required for

excitation from 4s to 3d. However in case of carbon (6C): 1s2, 2s2 2p



The stabilation due to exchange energy will not be able to compensate for the energy required for excitation from 2s to 2p.

Electronic configuration of ions : Note that while filling electrons in various orbitals they are filled according to the three laws –

Aufbau, Pauli and Hund’s. For removing electrons to form cations, electrons are removed from outermost shell as they are bound to the nucleus by lesser forces of attraction because of shielding effect.

For example for iron, 26Fe 1s2, 2s2 2p6, 3s2 3p6 3d6, 4s2 & the configuration of ions would be Fe2+ 1s2, 2s2 2p6, 3s2 3p6 3d6 & Fe3+ 1s2, 2s2 2p6, 3s2 3p6 3d5 Similarly for copper 29Cu 1s2, 2s2 2p6, 3s2 3p6 3d10, 4s1 & for its ions Cu+ 1s2, 2s2 2p6, 3s2 3p6 3d10 & Cu2+ 1s2, 2s2 2p6, 3s2 3p6 3d9

The anions are formed by adding electrons to the vacant orbital of lowest energy [follow (n +l) rule] For example 9F 1s2, 2s2 2p5 & that of its ion F- 1s2, 2s2 2p6 Similarly for Chlorine 17Cl 1s2, 2s2 2p6, 3s2 3p5 & that of its ion Cl- 1s2, 2s2 2p6, 3s2 3p6

Some Exceptional electronic configuration : 24Cr: 1s2, 2s2 2p6, 3s2 3p6 3d5, 4s1 47Ag : 1s2, 2s2 2p6, 3s2 3p6 3d10, 4s2 4p6 5s1 29Cu: 1s2, 2s2 2p6, 3s2 3p6 3d10, 4s1 42Mo: 1s2, 2s2 2p6, 3s2 3p6 3d10, 4s2 4p6 4d5, 4s1 Applications of electronic configuration: (1) Calculation of Magnetic Moment (µ) : 2n n Bohr Magneton

Where n no. of unpaired electrons & 1 BM (Bohr Magneton) = e

e.h4 m

When = 0 (Diamagnetic repelled, by magnetic field) If 0 , paramagnetic attracted by magnetic field. (2) Colour: It has been observed that the species having unpaired e- generally impart colour.

SOLVED PROBLEMS (OBJECTIVE)

Problem 1: For a p-electron, orbital angular moment is (A) 2 (B) (C) 6 (D) 2

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Solution: Orbital angular momentum L = )1( where

2h

L for p electron = 2)11(1 (A)

Problem 2: For which of the following species, Bohr theory doesn’t apply (A) H (B) He+ (C) Li2+ (D) Na+

Solution: Bohr theory is not applicable to multi electron species (D)

Problem 3: If the radius of 2nd Bohr orbit of hydrogen atom is r2. The radius of third Bohr orbit will be

(A) 2r94 (B) 4r2 (C) 2r4

9 (D) 9r2

Solution: 22

22

mZe4hnr

2

2

3

2

32

rr

r3 =

2r49 (C)

Problem 4: Number of waves made by a Bohr electron in one complete revolution in 3rd orbit is (A) 2 (B) 3 (C) 4 (D) 1

Solution: Circumference of 3rd orbit = 2r3 According to Bohr, angular momentum of electron in 3rd orbit is

mvr3 = 2

h3 or 3r2

mvh 3

by De-Broglie equation

= mvh =

3r2 3

2r3 = 3

i.e. circumference of 3rd orbit is three times the wavelength of electron or number of waves made by Bohr electron in one complete revolution in 3rd orbit is three. (B)

Problem 5: The degeneracy of the level of hydrogen atom that has energy 16RH is

(A) 16 (B) 4 (C) 2 (D) 1

Solution: En = 2

H

nR

16R

nR H

2

H i.e. for 4th sub-shell

- 1 0 +1 - 2 –1 +1 0 +2 – 3 – 2 – 1 0 +1 +2 +3

3 2 1

three p five d Seven f

n = 4 l = 0 m = 0 one s

i.e. 1 + 3 + 5 + 7 = 16 degeneracy is 16

Problem 6: An electron is moving with a kinetic energy of 4.55 10–25 J. What will be de Broglie wave length for this electron?

(A) 5.28 10–7 m (B) 7.28 10–7 m (C) 2 10–10 m (D) 3 10–5 m

Solution: KE = 21 mv2 = 4.55 10–25 ; v2 = 6

31

25

101101.9

1055.42

; v = 103 m/s

De Broglie wave length = mvh = 331

34

10101.910626.6

= 7.28 10–7 m

(B)

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Problem 7: Suppose 10–17J of energy is needed by the interior of human eye to see an object. How many photons of green light (=550 nm) are needed to generate this minimum amount of energy?

(A) 14 (B) 28 (C) 39 (D) 42

Solution: Let the number of photons required = n

n 1710hc ; n =

hc1710 = 834

917

10310626.61055010

= 27.6 = 28 photons (B)

Problem 8: Photoelectric emission is observed from a surface for frequencies 1 and 2 of the incident radiation (1>2). If the maximum kinetic energies of the photoelectrons in two cases are in ratio 1:K then the threshold frequency 0 is given by

(A) 1K12

(B)

1K21K

(C)

1KK 12

(D)

K12

Solution: KE1 = h1 – ho ; KE2 = h2–ho It is given that ; 2

1

KEKE =

K1

K1

hhhh

o2

o1

; 1KK o21 ; o = 1K

K 21

(B)

Problem 9: The velocity of electron in the ground state hydrogen atom is 2.18 106 ms–1. Its velocity in the second orbit would be

(A) 1.09 106 ms–1 (B) 4.38 106 ms–1 (C) 5.5 105 ms–1 (D) 8.76 106 ms–1-

Solution: We know that velocity of electron in nth Bohr’s orbit is given by

= 2.18 106 nZ m/s ; for H, Z = 1 ; v1 =

11018.2 6 m/s

v2 = 2

1018.2 6 m/s = 1.09 106 m/s ; (A)

Problem 10: The ionization energy of the ground state hydrogen atom is 2.1810–18J. The energy of an electron in its second orbit would be

(A)–1.09 10–18 J (B) –2.18 10–18J (C) –4.36 10–18J (D) –5.45 10–19J

Solution: Energy of electron in first Bohr’s orbit of H–atom

E = Jn 2

181018.2 ( ionization energy of H = 2.18 10–18J) ; E2 = 2

18

21018.2 J

= –5.45 10–19J ; (D)

Problem 11: Magnetic moments of V (Z = 23), Cr (Z = 24) and Mn (Z = 25) are x, y, z. Hence (A) z y x (B) x = y = z (C) x z y (D) x y z

Solution: Magnetic moments = .M.B)2n(n where n is the number of unpaired electron V ( Z = 23) (Ar) 3d34s2 n = 3, 15 BM = x Cr (Z = 24) (Ar) 3d5 4s1 n = 6, 48 BM = y Mn (Z = 25) (Ar) 3d5 4s2 n = 5, 35 BM = z C

Problem 12: The speed of a proton is one hundredth of the speed of light in vacuum. What is the de Broglie wavelength? Assume that one mole of protons has a mass equal to one gram. h = 6.626 10–27 erg sec

(A) 3.31 10–3Å (B) 1.33 10–3 Å (C) 3.13 10–2 Å (D) 1.31 10–2Å

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Solution: m = g10023.6

123

; = mvh = 18

27

seccm103110626.6

6.023 1023

= 1.33 10–11 cm ; B

Problem 13: The wave number of first line of Balmer series of hydrogen atom is 15200 cm–1. What is the wave number of first line of Balmer series of Li2+ ion.

(A) 15200 cm–1 (B) 6080 cm–1 (C) 76000 cm–1 (D) 1,36800 cm–1 Solution: For Li+2 vv for H Z2 = 15200 9 = 1, 36,800 D

Problem 14: The speed of the electron in the 1st orbit of the hydrogen atom in the ground state is

[C is the velocity of light]

(A) 1.37

C (B) 1370

C (C) 13.7

C (D) 137C

Solution: V = mr2h

= 2.189 108 cm sec–1 ; C = 3 1010 cm/s, vC = 137 D

Problem 15: The quantum number not obtained from the Schrödinger’s wave equation is (A) n (B) l (C) m (D) s

Solution: n, l and m quantum numbers can be obtained from Schrodinger equation. s is obtained from spectral evidence. D

Problem 16: Which of the following sets of quantum numbers is not allowed (A) n = 3, l = 1, m = +2 (B) n = 3, l = 1, m = +1 (C) n = 3, l = 0, m = 0 (D) n = 3, l = 2, m = 2

Solution: If n = 3 l = 0,1,2

for l = 0, m = 0 l = 1, m = –1, 0, +1 l = 2, m = –2, –1, 0, +1, +2 If l = 1, the value of m can not be 2 (A)

Problem 17: Assuming that a 25 watt bulb emits monochromatic yellow light of wave length

0.57 m.The rate of emission of quanta per sec. will be (A) 5.89 1015 sec–1 (B) 7.28 1017 sec–1 (C) 5 1010 sec–1 (D) 7.18 1019 sec–1

Solution: Let n quanta are evolved per sec.

hcn = 25 J sec–1

n 6

834

1057.010310626.6

= 25

n = 7.18 1019 sec–1 (D) Problem 18: The binding energy of an electron in the ground state of the He atom is equal to 24.6 eV. The energy required to remove both the electrons from the atom will be (A) 59 eV (B) 81 eV (C) 79 eV (D) None of these

Solution: Ionization energy of He+

= 6.132

2

nz = 6.13

12

2

2 = 54.4 eV

Energy required to remove both the electrons

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

= binding energy + ionization energy = 24. 6 + 54.4 = 79 eV (C)

Problem 19: The wave number of the shortest wave length transition in Balmer series of atomic hydrogen will be :

(A) 4215Å (B) 1437 Å (C) 3942 Å (D) 3647 Å

Solution: shortest

1 = RZ2

22

21

11nn

= 109678 12

221

21 = 3.647 10–5 cm

= 3647 Å (D)

SOLVED PROBLEMS (SUBJECTIVE)

Problem 1: Which electronic transition in Balmer series of hydrogen atom has same frequency as that of n = 6 to n = 4 transition in He+.

Solution: 22 2He

1 1RZ4 6

=

16361636R4 =

36R5

H = R 12

22 n1

21

HHe

36R5

= 2n

R4R

On solving above equation n2 = 9 n = 3

Or corresponding transition from 3 2 in Balmer series of hydrogen atom has same frequency as that of 6 4 transition in He+.

Problem 2: Calculate ionisation potential in electron volts of (A) He+ and (B) Li2+

Solution: I.E. = 2

2

nZ6.13

= 2

2

1Z6.13 [Z =2 for He+] = 13.6 4 = 54.4 eV

Similarly for Li2+ = 2

2

136.13 = 13.6 9 = 122.4 eV

Problem 3: What fraction of the volume of an atom of radius 10–8 cm is occupied by its nucleus if nuclear radius is 10–12 cm?

Solution: Assuming atom to be spherical having definite boundary its volume can be given by

3r34 (where r is atomic radius). Similarly volume of nucleus can be given by 3r

34 where r

is radius of nucleus.

3

3

r34

r34

atomofVolumenucleusofVolume

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

= 38

312

3

3

)10(10

rr

= 24

36

1010

= 10–12

Problem4: Calculate the ratio of K.E and P.E of an electron in the Bohr orbit?

Solution: K.E. = r

Ze2

2

P.E. = rZe2 P.E. = –2K.E

21

P.EK.E

Problem 5: How many spectral lines are emitted by atomic hydrogen excited to nth energy level? Solution:

n-1 n-2n-1

n-2

n-3

n-4

nn-1

Thus the total number of lines will be equal to Number of lines coming to 1st level + number of lines coming to 2nd level + number of lines

coming to 3rd level + …….. number of lines to n-1 level. = (n -1) + (n - 2) + (n – 3) + ( n – 4) +……. +3 + 2 + 1 = sum of first ( n – 1) natural numbers.

= 2

111 nn = 21 nn

Number of spectral lines that appear in hydrogen spectrum when an electron de excites

from nth energy level = 2

1nn

These total lines can be observed only if there is sufficiently large number of Hydrogen

atoms in the sample so that all the possible transitions can take place.

Can you drive the expression for total number of lines when an electron makes a transition from a higher level n2 to a lower level n1?

Problem 6: Calculate (A) the de Broglie wavelength of an electron moving with a velocity of 5.0 105 ms–1 and (B) ratio of de Broglie wavelength of an atom of hydrogen and atom of oxygen moving with the same velocity (h = 6.63 10–34 kg m2 s–1)

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Solution: (A) = mvh =

34 2 1

31 5 1

6.63 10 kgm s9.11 10 kg 5.0 10 ms

Wavelength = 1.46 10–9m

(B) An atom of oxygen has approximately 16 times the mass of an atom of hydrogen. In

the formulamvh

, h is constant while the conditions of problem make v, also

constant. This means that and m are variables and varies inversely with m. Therefore, for the hydrogen atom would be 16 times greater than for oxygen atom.

Problem 7: In photoelectric effect, an absorbed quantum of light results in the ejection of an

electron from the absorber. The kinetic energy of the electron is equal to the energy of the absorbed photon minus the energy of the longest wavelength that causes the effect. Calculate the kinetic energy of an electron produced in cesium by 400 nm light. The critical (maximum) wavelength for the photoelectric effect in cesium is 660 nm.

Solution: Kinetic energy of electron = h – hcritical =

hc - critical

hc

KE = nm660

eVnm1240nm400

eVnm1240 = 1.22 eV

Problem 8: It has been found that gaseous Iodine molecules dissociate into separated atoms after absorption of light at wavelengths less than 4995Å. If each quantum is absorbed by one molecule of I2, what is the minimum input in kcal/mole, needed to dissociate I2 by this photo chemical process?

Solution: E (per mole) = NAh

E = NA

23 1 34 8 1

10

6.022 10 6.626 10 3 10

4995 10

mol Js mshcm

= 239.5 kJ/mol

kJ184.4

kcal1 = 57.1 kcal/mole

Problem 9: What is the wavelength associated with 150 eV electron

Solution: = .E.Km2

h

=

34

31 19

6.626 10

2 9.1 10 150 1.6 10

Js

kg J =

50

34

10436810626.6

= 10–10 m = 1Å

Alternatively we can also use the expression DB(for electron)150

E (ev) to obtain

wavelength in Å directly.

Problem 10: The energy of electron in the second and third Bohr orbit of the hydrogen atom is – 5.42 10–12 erg and –2.41 10–12 erg, respectively. Calculate the wavelengths of emitted radiation when the electron drops from third to second orbit.

Solution: E3 – E2 = h =

hc ; – 2.41 10–12 – (– 5.42 10–12) =

1027 10310626.6

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

= 12

1027

1001.310310626.6

= 6.604 10–5 cm = 6604 Å

Problem 11: Show that frequencies of emitted photons are additive but their wavelengths are not.

Solution:

n = 3

n = 2

n = 1

E2 3

E1 2

X

E1 3

E

E1 3 = E1 2 + E2 3 ; h1 3 = h1 2 + h2 3

1 3 = 1 2 + 2 3 ; i.e. frequencies like energies are additive, on the other hand E1 3 = E1 2 + E2 3

322131

hchchc

; 322131

111

3221

2132

31

1

; 3221

322131

i.e. wavelengths are not additive.

Problem 12: O2 undergoes photochemical dissociation into one normal oxygen and one excited oxygen atom, 1.967 eV more energetic than normal. The dissociation of O2 into two normal atoms of oxygen atoms requires 498 KJ mole–1. What is the maximum wavelength effective for photochemical dissociation of O2?.

Solution: O2 ON + excitedO ;

O2 ON + ON

E = 498 103 J / mole = J23

3

10023.610498

per molecule = 8.268 10–19 J

Energy required for excitation = 1.967 eV = 3.146 10–19J Total energy required for photochemical dissociation of O2 = 8.268 10–19 + 3.146 10–19

= 11.414 10–19 J ; hc = 11.414 10–19 J

= 19

834

10414.1110310626.6

= 1.7415 10–7 m = 1741.5 Å

Problem 13: Compare the wavelengths for the first three lines in the Balmer series with those which arise from similar transition in Be3+ ion.

Solution: 2H 2 2

1 1R 12 n

; 32

2 2Be

1 1R 42 n

3

3

Be H

H Be

= 16

So we can conclude that all transitions in Be3+ will occur at wavelengths 161 times the

hydrogen wavelengths.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

FOUNDATION BUILDER (OBJECTIVE)

ATOMIC TERMS 1. In the following reaction 3Li6 + ? 2He4 + 1H3, the missing particles is (A) Neutron (B) Proton (C) Electron (D) Deuterium 2. The increasing order (lowest first) for the magnitude of e/m (charge/mass) for electron (e), proton

(p), neutron (n) and alpha particle () is (A) e,p,n, (B) n,p,e, (C) n,p,,e (D) n,,p,e, 3. X2― has 56 electrons, the atomic number X is (A) 56 (B) 58 (C) 28 (D) 54 4. 11Na23 and 12Mg24 are

(A) Isotopes (B) Isobars (C) Isodiaphers (D) Isotones 5. Particle in cathode ray has same charge/ mass ratio as (A) α particle (B) β particle (C) γ particle (D) proton 6. During Muliken’s oil drop experiment, out of the following, which is not a possible charge on oil

droplet? (A) 1.6×10−19 C (B) 2.4×10−19 C (C) 3.2×10−19 C (D) 4.8×10−19 C ELECTROMAGNETIC WAVES & PLANCK’S QUANTUM THEORY 7. In a time period of 2 sec., the following wave pattern is observed:

Then the frequency of wave in Hz is (A) 1 (B) 2 (C) 3 (D) 4 8. Yellow light is more energetic than (A) Violet (B) blue (C) Indigo (D) Red 9. Rutherford’s experiment on scattering of -particles showed for the first time that the atom has (A) electrons (B) protons (C) nucleus (D) neutrons 10. The wave number of radiation of wavelength 500 nm is (A) 5 10–7 m–1 (B) 2 107 m–1 (C) 2 106 m–1 (D) 500 10–9 m–1 11. The ratio of the energy of a photon of 2000Å wavelength radiation to that of 4000Å radiation is (A) ¼ (B) ½ (C) 2 (D)4 12. Radio city broadcasts on a frequency of 5,090 KHz. What is the wavelength of electromagnetic

radiation emitted by the transmitter? (A)10.3 m (B) 58.9 m (C) 60.5 m (D) 75.5 m

13. A 1000 watt radio transmitter operates at a frequency of 880 kilocycle/sec. How many photons per

sec does it emit? (A) 2.01×1029 (B) 1.72×1030 (C) 1.51×1029 (D) 1.77×1031

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

14. The eyes of certain members of reptile family pass a visual signal to the brain when the visual

receptors are struck by photons of wavelength 890 nm. If a total energy of 3.15×10−14 J is required to trip signal, what is the minimum number of photons that must strike the receptor?

(A) 3.05×1019 (B) 1.72×109 (C) 1.41×105 (D) 2.75×1010 15. A certain dye absorbs light of wavelength 4500 Ao and then fluorescence light of 5000 Ao. Assuming

that, under given conditions 50% of the absorbed energy is re-emitted out as fluorescence. Calculate the ratio of quanta emitted to the number of quanta absorbed?

(A) 0.55 (B) 2.1 (C) 1.8 (D) 0.75 BOHR MODEL

16. An electron in an atom jumps in such a way that its kinetic energy changes from x to .4x

The change in potential energy will be: (A) x

23

(B) x83

(C) x43

(D) x43

17. The potential energy of an electron in the Hydrogen atom is – 6.8 eV. Indicate in which excited

state, the electron is present? (A) first (B) second (C) third (D) fourth 18. What is the potential energy of an electron present in N-shell of the Be3+ ion? (A) – 3.4 eV (B) – 6.8 eV (C) – 13.6 eV (D) – 27.2 eV 19. The kinetic and potential energy (in eV) of electron present in third Bohr’s orbit of hydrogen atom

are respectively: (A) – 1.51, – 3.02 (B) 1.51, – 3.02 (C) – 3.02, 1.51 (D) 1.51, – 1.51 20. The distance between 4th and 3rd Bohr orbits of He+ is: (A) m1010645.2 (B) m1010322.1 (C) m1010851.1 (D) None 21. What is the atomic number (Z) correspond to which 4th orbit would fit inside the 1st Bohr orbit of H-

atom? (A) 3 (B) 4 (C) 16 (D) 25 22. The ratio of velocity of the electron in the third and fifth orbit of Li2+ would be: (A) 3 : 5 (B) 5 : 3 (C) 25 : 9 (D) 9 : 25 23. If radius of second stationary orbit (in Bohr’s atom) is R. Then radius of third orbit will be: (A) R/3 (B) 9R (C) R/9 (D) 2.25 R 24. Which state of Be3+ has the same orbit radius as that of the ground state of hydrogen atom? (A) 3 (B) 2 (C) 4 (D) 5

25. Select the incorrect graph for velocity of e in an orbit Vs. n

Z 1, and n:

(A) (B) (C) (D) 26. What is the frequency of revolution of electron present in 2nd Bohr’s orbit of H-atom? (A) 11610016.1 s (B) 11610065.4 s

(C) 11510626.1 s (D) 14 18.13 10 s 27. The number of photons of light having wave number ‘x’ in 10 J of energy source is:

v

n

v

1/n

v

Z

v

n

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

(A) 10hcx (B) x

hc10

(C) hcx10 (D) None of these

28. The ionization potential for an electron in ground state of the hydrogen atom is 13.6 eV. What would

be the ionization potential for the electron in the first excited state of H atom? (A) 13.6 eV (B) 6.8 eV (C) 3.4eV (D) 27.2 eV 29. According to Bohr’s theory, angular momentum of electron in any orbit of Hydrogen is directly

proportional to

(A) 1

nr (B) 1

nr (C) 2

nr (D) nr

30. If the revolutions per second by the electron in 3rd orbit of H is , then the revolutions per second by

the electron in 2nd orbit of He+ is (A) (B) 13.5 (C) 1.5 (D) 0.07 31. If the kinetic energy of electron moving in 4th orbit of hydrogen is €, then the total energy in 1st orbit

of Li2+ is (A) –144 € (B) –0.0069 € (C) –(27/9) € (D) – €

32. If same energy is supplied to electron in ground state of Hydrogen as well as He+, electron jump to

5th main shell in Hydrogen, then final orbit of electron in He+ is (A) 2nd (B) 1st (C) 3rd (D) 4th 33. If force of attraction between the electron and nucleus in 2nd orbit of Li2+ is, force of attraction if

electron present in 1st orbit of H is (A) 6

49 (B) 12

25 (C) 8

81 (D) 16

27

34. If acceleration of electron in 1st orbit of He+ is, acceleration of electron 2rd orbit of Be3+ is (A) (B) 2 (C) 1

2 (D) 1

4

35. The spacing between the orbits in terms of distance is maximum in the case of (A) 1st and 2nd (B) 2nd and 3rd (C) 3rd and 4th (D) 4th and 5th 36. The spacing between the orbits in terms of energy is maximum in the case of (A) 1st and 2nd (B) 2nd and 3rd (C) 3rd and 4th (D) 4th and 5th HYDROGEN SPECTRUM 37. Transition of an electron from n = 2 to n = 1 level results (for a H - atom) in (A) IR spectrum (B) UV spectrum

(C) Visible spectrum (D) X - ray spectrum 38 The emission spectrum of He+ ion is the consequence of transition of electron from orbit n2 to orbit

n1. Given that 2n2 + 3n1 = 18 and 2n2 – 3n1 = 6, then what will be the maximum number of spectral lines in atomic spectrum when electron transits from n2 to orbit n1?

(A) 10 (B) 15 (C) 20 (D) 21 39. Find the value of wave number )(v in terms of Rydberg’s constant, when transition of electron takes

place between two levels of He+ ion whose sum is 4 and difference is 2. (A)

98R (B)

932R (C)

43R (D) None of these

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

40. When an electron makes a transition from (n + 1) state to nth state, the frequency of emitted

radiations is related to n according to (n >> 1):

(A) 3

22n

cRZv (B) 4

2

ncRZv (C)

2

2

ncRZv (D)

2

22n

cRZv

41. What is the shortest wavelength line in the Paschen series of Li2+ ion? (A)

9R (B)

R9 (C)

R1 (D)

49R

42. What is the longest wavelength line in the Lyman series of He+ ion? (A) 3R (B)

R31 (C) 1

4R (D) None of these

43 An excited state of H atom emits a photon of wavelength and returns to the ground state, the

principal quantum number of excited state is given by: (A) )1( RR (B)

)1( RR

(C) R( R 1) (D)

RR

)1(

44. A dye absorbs a photon of wavelength and re-emits the same energy into two photons of

wavelength 1 and 2 respectively. The wavelength is related with 1 and 2 as:

(A) 21

21

(B) 21

21

(C) 21

22

21

(D) 2

21

21

)(

45. Which electronic transition in a hydrogen atom, starting from the orbit n = 7, will produce infrared

light of wavelength 2170 nm? (A) n = 7 to n = 6 (B) n = 7 to n = 5

(C) n = 7 to n = 4 (D) n = 7 to n = 3 46. A hydrogen atom in the ground state is excited by monochromatic radiation of wavelength .A The

resulting spectrum consists of maximum 15 different lines. What is the wavelength ? (A) 937.3 Ao (B) 1025 Ao (C) 1236 Ao (D) None of these 47. The emission spectra are observed by the consequence of transition of electron from higher energy

state to ground state in Li2+ ion. Six different types of photons are observed during the emission spectra, and then what is the excitation state of Li2+ ion?

(A) 3rd (B) 4th (C) 2nd (D) 5th 48 If 1, 2, and 3, is wave length of photon corresponding to 1st, 2nd Lyman series and 1st Balmer

series respectively, which of the following statement is correct? (A) 2 = 1 + 3 (B) 2 = 13/(1+3)

(C) 1 + 2 +3 =0 (D) 22 = 1

2 + 32

49. If the shortest wavelength of H atom in Lyman series is x, then longest wavelength in Balmer series

of He+ is (A)

5x9 (B)

5x36 (C)

4x (D)

9x5

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

HEISENBERG UNCERTAINTY PRINCIPLE AND DE BROGLIE 50. Uncertainty in position and momentum are equal. Uncertainty in velocity is:

(A) h (B)

2h (C)

h

m21 (D) none

51. Wavelength associated with Virar-F local train having mass 100 × 103 Kg moving with the speed of

23.76 kms/hr is: (plank’s constant =6.6× 10-34 Js) (A) 10−31 Ao (B) 10−35 Ao (C) 10−29 Ao (D) 10−40 Ao 52. If E1, E2 and E3 are kinetic energy of electron, alpha particle and proton having same De-Broglie

wave length, then (A) E1 E2 E3 (B) E1< E2< E3 (C) E2< E3< E1 (D) E1 =E2 =E3

53. If the radius of first Bohr orbit is x, then de Broglie wavelength of electron in 3rd orbit is nearly.

(A) 2x (B) 6x (C) 9x (D) 3x

54. A ball of mass 200 g is moving with a velocity of 10ms–1. If the error in measurement of velocity is 0.1%, the uncertainty in its position is:

(A) 3.310–3 m (B) 3.310–27 m (C) 5.310–25 m (D) 2.64 10–32 m

55. If the uncertainty x in the position is along X-axis, then uncertainty in the momentum is along (A) X-axis (B) Y-axis (C) Z-axis (D) Any axis 56. When applied, the uncertainty principle has significance in case of (A) Moving train (B) Spinning cricket ball (C) Moving α- particle (D) All the above PHOTOELECTRIC EFFECT 57. The kinetic energy of the electron emitted when light of frequency 3.51015 Hz falls on a metal

surface having threshold frequency, 1.51015 Hz is (h = 6.610-34 Js): (A) 1.3210-18 J (B) 3.310-18 J (C) 6.610-19 J (D) 1.9810-19J 58. 0 is the threshold wavelength of a metal for photoelectron emission. If the metal is exposed to the

light of wavelength , then the velocity of ejected electron will be 02h Km

. The value of K

is:

(A) speed of light (B) 1 (C) 0

C

(D) 0

1

59. When photons of energy 4.25 eV strike the surface of a metal A. The ejected photoelectrons have

maximum kinetic energy (T(A) (expressed in eV) and de-Broglie wavelength ((A). The max kinetic energy of photoelectrons liberated from another Metal B by photons of enegy 4.2 eV is TB.Where TB= (TA1.5). If De-Broglie wave length of these photoelectrons B (B =2A), then which of the following is not correct

(A) The work function of A is 2.25 eV. (B) The work function of B is 3.7 eV (C) TA = 2.0 eV, (D) TB =0 .75 eV 60. Work function WA for a photoelectric material A is 2 eV & WB for another photoelectric material B is

4 eV. If the photons of energy EA strike with surface of A the ejected photoelectrons have minimum

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

de-Broglie wavelength A and photons of energy EB strike the surface of B, the ejected photoelectrons have minimum de-Broglie wavelength B. Where EB= (EA+0.5) eV and B =2A, VA and VB are respective stopping potentials then which of the following is not correct

(A) EA = 4 eV (B) EA = 4.5 eV (C) VA = 2 Volts (D) VB =0 .5 Volts QUANTUM WAVE MECHANICAL MODEL AND ORBITALS 61. The number of nodal planes in a px orbital is (A) one (B) two (C) three (D) zero 62. For a 4d electron the orbital angular momentum is: ( 2

h )

(A) 6 (B) 12 (C) 2 (D) zero 63. If travelling at equal speeds, the longest wavelength of the following matter waves is that of (A) electron (B) proton (C) neutron (D) alpha particle 64. Which subshell doesn’t exist? (A) 7s (B) 3d (C) 3f (D) 5d 65. The quantum number not obtained from the Schrodinger’s wave equation is (A) n (B) l (C) m (D) s 66. The orbital angular momentum of an electron in 2s orbital is

(A) + 21

.2

h (B) zero (C) 2

h (D) 2

h.2

67. Which quantum number determines shape of the orbital? (A) Principal (B) Angular (C) Magnetic (D) Spin 68. Radial nodes are maximum in: (A) 4s (B) 4p (C) 3d (D) 5f 69. Total number of electrons in any orbit is:

(A)n

12(2 1)

l =

ll + (B)

n 1

12(2 1)

l =

ll + (C)

n 1

02(2 1)

l =

ll + (D)

n 1

02(2 1)

l =

ll +

70. Which of the following relates to photons both as wave motion and as a stream of particles? (A) Interference (B) E = mc2 (C) diffraction (D) E = h 71. Probability of finding the electron in the orbital is? (A) 100% (B) 5-10% (C) 90-95% (D) 50-60% 72. The Wave mechanical model of atom is based upon (A) De Broglie concept of dual nature (B) Heisenberg’s uncertainty principle (C) Schrodinger wave equation (D) All of these 73. Select the correct statement from following: A: Splitting of spectra line occurs when placed in a magnetic field or in an electric field. B: In case of 1s- orbital, the density of the electron cloud is the greatest near the nucleus and falls

off with the distance. C: Electron – density is concentrated along a particular direction in case of 2p-orbital D: A p-orbital can take maximum of six electrons

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

(A) A, B, D (B) A, B, C (C) B, C, D (D) A, C, D 74. The Magnetic quantum number signifies (A) Size of the orbital (B) Shape of the orbital (C) Orientation of orbital in space (D) Nuclear stability 75. The quantum number which specifies the location as well as energy is (A) n (B) l (C) m (D) s 76. Which of the following sets of quantum numbers is not allowed? (A) n = 3, l = 1, m = +2 (B) n = 3, l = 1, m = +1 (C) n = 3, l = 0, m = 0 (D) n = 3, l = 2, m = 2 77. Which set of quantum numbers is not possible for electron in 3rd shell? (A) n = 3, l = 2, m = – 1, s= +1/2 (B) n = 3, l = 2, m = – 1, s= –1/2 (C) n = 3, l = 2, m =0, s= +1/2 (D) n = 3, l = 3, m = 0, s= –1/2 78. Out of the following which sub shell has maximum energy? (A) 3d (B) 4s (C) 5s (D) 4p 79. Which two orbitals are located along the axes, and not between the axes?

(A) 2, zxy dd (B) zxy pd , (C) xyz pd , (D) 22,yxz dp

80. Which series of subshells is arranged in the order of increasing energy for multi-electron atoms? (A) 6s, 4f, 5d, 6p (B) 4f, 6s, 5d, 6p (C) 5d, 4f, 6s, 6p (D) 4f, 5d, 6s, 6p 81. The subshell that arises after f is called the g subshell. How many electrons may occupy the g

subshell? (A) 9 (B) 7 (C) 5 (D) 18 PROBABILITY DISTRIBUTION CURVES AND WAVE FUNCTION 82. Miss Ritika has two correct information from Mr. Gupta and Mr. Agarwal about a particular orbital of hydrogen atom. Identify the orbital

Mr. Gupta: (angular) of orbital is 1 21

4

Mr. Agarwal: The orbital has two radial nodes.

(A) 's' orbital with any principal quantum number (B) any orbital with principal quantum number 3 (C) 3s orbital (D) Information from Mr. gupta & Mr. Agarwal cannot predict the orbital.

83. Which of the following is the correct representation of plot radial probability (4r2R2) in Y-axis vs distance from the nucleus in X-axis for 1-elelectron of 4d-atomic orbital?

(A)

r

Y

X

4r2R

2

(B)

r

Y

X

4r2R

2

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

(C)

r

Y

X

4r2R

2

(D)

r

Y

X

4r2R

2

84. If the nodes at infinity are not neglected, then what is the total number of radial and angular nodes

of 5f-orbitals? (A) 4 (B) 3 (C) 5 (D) infinity

85. The distance of maximum probability for 3s, 3p, 3d are given in the order (A) spd rrr 3max3max3max )()()( (B) psd rrr 3max3max3max )()()(

(C) dps rrr 3max3max3max )()()( (D) None of these 86. The number of local maxima in the Radial distribution curve of 5d orbital is (A) Zero (B) 1 (C) 2 (D) 3 87. Out of the following, which is the correct match for radial probability of finding the electron of 2s

orbital?

(A) A-H, B-He+, C-Li2+ (B) A- He+, B-H, C-Li2+ (C) A- Li2+, B-He+, C-H (D) Can’t say

88. For a 3s-orbital, 3/2

2 /2

0 0

1 1 2 .(3 ) (6 6 ) ;39 3r Zs e where

a a

What is the maximum radial distance of node from nucleus?

(A) Z

a0)33( (B) Za0 (C)

Za0)33(

23 (D)

Za02

89. The Schrödinger wave equation for hydrogen atom is

2/22/3

0

)]128)(1[(416

1(radial)

e

aZ

Where a0 and Z are the constant in which answer can be expressed and 0

2aZr

. Minimum and

maximum positions of radial nodes from nucleus are……respectively. (A)

Za

Za 00 3, (B)

Za

Za 00 ,2

(C) Za

Za 00 3

,2

(D) Za

Za 00 4

,2

ELECTRONIC CONFIGURATION AND APPLICATION 90. Which of the following electronic configurations is correct for Iron, (atomic number 26)? (A) 61 34][ dsKr (B) 71 34][ dsKr (C) 62 34][ dsAr (D) 62 34][ dsKr 91. Which of the following representation of excited states of atoms is impossible? (A) 11 21 ss (B) 132 433][ spsNe (C) 6162 3433][ dspsNe (D) 2722 3221 spss

r

Y

X

4pr 2R22

A B

C

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

92. Which of the following has the maximum number of unpaired electrons? (A) Mn (B) Ti (C) V (D) Al 93. The magnetic moment of isolated Fe2+ ion is

(A) 2 6 BM (B) 15 BM (C) 3 BM (D) 35 BM 94. Total spin resulting from a d5 configuration is (A) 1 (B) ½ (C) 5/2 (D) 3/2 95. Electronic configuration of Ni is [Ar]3d8,4s2. The electronic configuration of next element is (A) [Ar]3d104s1 (B) [Ar]3d9 4s2 (C) 3d84s24p1 (D) None 96. Which of the following is having the maximum number of unpaired electrons? (A) Mg2+ (B) Ti3+ (C) V3+ (D) Fe2+

97. Which of the following violates the Pauli Exclusion Principle? (A) (B) (C) (D) 98. The value of the magnetic moment of a particular ion is 2.83 Bohr magneton. The ion is (A) Fe2+ (B) Ni2+ (C) Mn2+ (D) Co3+ 99. If an ion of 25Mn has a magnetic moment of 3.873 B.M. Then Mn is in which state. (A) + 2 (B) + 3 (C) + 4 (D) + 5 100. A compound of vanadium has a magnetic moment )( of 1.73 BM. If the vanadium ion in the

compound is present as Vx+, then, the value of x is? (A) 1 (B) 2 (C) 3 (D) 4 101. The ratio of magnetic moments of Fe (III) and Co (II) is: (A) 7:5 (B) 15:35 (C) 7 : 3 (D) 15:24

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Answers to Foundation Builder (Objective) 1. A 2. D 3. D 4. D 5. B 6. B 7. B 8. D 9. C 10. C 11. C 12. B 13. B 14. C 15. A 16. A 17. A 18. D 19. B 20. C 21. D 22. B 23. D 24. B 25. D 26. D 27. C 28. C 29. D 30. B 31. A 32. B 33. D 34. C 35. D 36. A 37. B 38. A 39. B 40. A 41. C 42. B 43. B 44. B 45. C 46. A 47. A 48. B 49. A 50. C 51. C 52. C 53. B 54. D 55. D 56. C 57. A 58. C 59. D 60. B 61. A 62. A 63. A 64. C 65. D 66. B 67. B 68. A 69. D 70. D 71. C 72. C 73. B 74. C 75. A 76. A 77. D 78. C 79. D 80. A 81. D 82. C 83. D 84. C 85. C 86. D 87. C 88. C 89. C 90. C 91. D 92. A 93. A 94. C 95. A 96. D 97. C 98. B 99. C 100. D 101. B

FOUNDATION BUILDER (SUBJECTIVE) ELECTROMAGNETIC WAVES, LIGHT AND PLANCKS QUANTUM THEORY 1. A monochromatic source of light operating at 600 watt emits 2 1022 photons per second. Find

the wavelength of the light.

2. The dissociation of I2 h 2I utilizes one photon per iodine molecule dissociated. The

maximum which can cause this dissociation is 4995 Å. Calculate number of moles of I2 dissociated per KJ of photon energy.

3. The quantum yield for decomposition of HI is 2. In an experiment 0.01 moles of HI are

decomposed. Find the number of photons absorbed.

moles dissociatedQ.Emoles of photonsabsorbed

4. Iodine molecule dissociates into atoms after absorbing light of 45000A . If one quantum of radiation

is absorbed by each molecule, calculate the kinetic energy of iodine atoms. (Bond energy of I2 = 240 kJ mol-1).

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

HYDROGEN SPECTRUM/ ATOMIC SPECTRUM 5. Find the wavelength of the radiation emitted by hydrogen in the following transitions. (a) n= 3 to n=2, (b) n = 5 to n = 4 and (c) n = 10 to n = 9. What is the order of the wavelengths obtained? 6. Calculate the wavelength of first line for hydrogen atom of the (i) Lyman series and (ii) Balmer

series (iii) Paschen series assuming the Rydberg constant as 109, 678 cm–1 7. Find the wavelength of the radiation required to excite the electron in Li + 2 from the first to the

third Bohr orbit. How many spectral lines are observed in the emission spectrum of the above excited system?

8. What electron transition in a hydrogen atom, starting from the orbit n=7, will produce infrared

light of wavelength 2170nm? Given: RH=1.09677×107/m. 9. Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in

Lyman series. What wavelength does this latter photon correspond to? 10. A sample of hydrogen atoms is exposed to electromagnetic radiation of 1028 Å which causes

emission of induced radiations. Calculate of induced radiations. 11. Calculate the energy emitted when electrons of 1.0 g atom of hydrogen undergo transition giving the

spectral line of lowest energy in the visible region of its atomic spectrum. 12. The hydrogen atom in the ground state is excited by means of monochromatic

radiation of wavelength x Å. The resulting spectrum consists of 15 different lines. Calculate the value of x.

13. The threshold wavelength for photoelectric effect for a metal is 230nm. Determine the

K.E. of the photoelectron ejected from the surface by U.V. radiation emitted from the 2nd longest wavelength transition (downward) of electron in Lyman series of atomic spectrum of hydrogen.

14. One of the series of the hydrogen spectrum can be represented by the equation in

terms of wave number

= 7 2 11.09603 10 (1 n ) m (where n = 2,3,…..) i) Calculate the maximum and minimum wavelength of lines in this series. ii) In what portion of the electromagnetic spectrum will this series be found?

15. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition

n = 4 to n = 2 of He+ spectrum? 16. Calculate the wave number for the shortest wavelength transition in the Balmer series of atomic

hydrogen. BOHR’S MODEL

17. Show that the speed of the electron revolving in a Bohr`s orbit of the hydrogen atom is Inversely

proportional to the quantum number of the orbit. Calculate this speed for n=2 and the time taken by the electron to complete 107 revolutions in the orbit.

18. The electron in a hydrogen atom revolves in the third orbit. Calculate (i) the energy of the electron in

this orbit (ii) the radius of the third orbit and (iii) the frequency and wavelength of the spectral line emitted when the electron jumps from the third orbit to the ground state.

19. Calculate the radii of the I, II and III permitted electron Bohr orbits in a hydrogen atom. What are

the corresponding values in the case of a singly ionized helium atom?

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

20. Calculate energy in kcal/mole necessary to remove an electron in a hydrogen atom from fourth Bohr’s orbit

21. An electron collides with a hydrogen atom in its ground state and excites it to a state of n = 3.

How much energy was given to the hydrogen atom in this inelastic collision? 22. The electron energy in hydrogen atom is given by E = – 21.7 10–12/n2 erg. Calculate the

energy required to remove an electron completely from n = 2 orbit. What is the longest wavelength (in cm) of light that can be used to cause this transition?

23. Ionisation energy of hydrogen atom is 13.6 eV. Calculate the energy of electron for Li2+ and

Be3+ in the first excited state. 24. A bulb emits light of wave length 4500Å. The bulb is rated as 150 watt and 8% of the energy is

emitted as light. How many photons of lights are emitted by bulb per second? 25. Calculate of the radiations when the electron jumps from III to II orbit of hydrogen atom. The

electronic energy in II and III Bohr orbit of hydrogen atoms are – 5.42 10–12 and –2.41 10–12 ergs respectively.

26. Calculate velocity of an electron placed in the third orbit of the hydrogen atom. Also calculate the

number of revolutions per second that this electron makes around the nucleus. 27. Average lifetime of a hydrogen atom excited to n = 2 state is

810s. find the number of

revolutions made by the electron on the average before it jumps to the ground state. 28. Estimate the difference in energy between Ist and 2nd Bohr orbit for a hydrogen atom. At what

minimum atomic number, a transition from n = 2 to n = 1 energy level would result in the emission of X – rays with = 3.0 10-8 m ? Which hydrogen atom – like species does this atomic number correspond to?

29. Find out the number of waves made by a Bohr electron in one complete revolution in its 3rd orbit. PHOTO ELECTRIC EFFECT

30. The photoelectric effect consists of the emission of electron from the surface of the metal when

the metal is irradiated with light. A photon with a minimum energy of 3.97 10–19J is necessary to eject an electron from barium.

(A) What is frequency of the radiation corresponding to this value? (B) Will the blue light with wave length 450 nm be able to eject the electron? 31. Find the threshold wavelengths for photoelectric effect from a copper surface, a sodium surface

and a cesium surface. The work function of these metals are 4.5 eV, 2.3 eV and 1.9 eV respectively.

32. Energy required to stop the ejection of electrons from Cu plate is 0.24 eV. when radiation of =

253.7 nm strikes the plate Calculate the work function. 33. Calculate the threshold frequency of metal if the binding energy is 180.69 KJ /mole 34. Calculate the minimum & maximum kinetic energy in eV of photoelectrons produced in cesium by

400 nm light, when the potential difference is 2V is applied as accelerating voltage .The critical (maximum) wavelength for the photoelectric effect in cesium is 660 nm,

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

35. A stationary He+ ion emitted a photon corresponding to a first line of the Lyman series. The photon liberated a photoelectron from a stationary H atom in ground state. What is the velocity of photoelectron?

36. The minimum energy necessary to overcome the attractive force between the electron and the

surface of silver metal is 7.52 10–19 J. What will be the maximum kinetic energy of the electrons ejected from silver which is being irradiated with ultraviolet light having a wavelength 360Å?

DUAL NATURE AND HEISENBERG UNCERTAINITY PRINCIPLE

37. Find wavelength for 100 g particle moving with velocity 100 ms–1. 38. A moving electron has 4.55 10-25 joules of kinetic energy. Calculate its wavelength 39. Calculate the de-Broglie wavelength of electron accelerated through 100 volt.

40. Calculate de-Broglie wavelength of an electron moving with a speed of nearly 201 th that of light

41. What is uncertainty in velocity of an electron if uncertainty in its position is 1Å? 42. A proton is accelerated to one- tenth of the velocity of light. If its velocity can be measured with a

precision + 1%. What must be its uncertainty in position. 43. An electron beam can undergo diffraction by crystals. Through what potential should a beam of

electron accelerated so that its wavelength becomes equal to 1.54 0A .

QUANTUM NUMBERS AND ELECTRONIC CONFIGURATION 44. Give reasons why the ground state outermost electronic configuration of silicon is:

3s 3p and not - 45. What is the maximum number of electrons that may be present in all the atomic orbitals with

principle quantum number 3 and azimuthal quantum number 2? 46. Predict the magnetic moment for S2–, Co3+ WAVE MECHANICAL MODEL & PROBABLITY DISTRIBUTION CURVES 47. The wave function of 2s electron is given by

0

r1 22a

2s0 0

1 1 r. 2 ea a2 2

If it has a node at r = r0, find relation between r0 and a0.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Answers to Foundation Builder (Subjective)

1. 6630 nm 2. 4.17 10–3 3. 3.01 × 1021 4. 2.165 2010 J 5. (a) 6564Å (b) 40253Å (c) 388701Å 6. 1215 Å, 6564 Å, 18756 Å 7. 11.39nm, 3 8. n2 = 7 to n1=4 9. 121.56 nm or 1215.6Å 10. 1028Å, 1216 Å, 6568 Å 11. 1.82 × 105 J/mol 12. 937.8 Å 13. 1.072 x 10 -18 J 14. 1.216 x 10 -7m, 0.912 x 10 -7m, Ultraviolet region. 15. n2 = 2 to n1 = 1 16. 27419.25 cm – 1

17. 1.09 x 106 m/sec, 1.219 x 10-8 sec 18. – 1.51eV, 4.761 Å, 2.905 x 1015 s-1, 1.03211 x 10-7 m 19. 0.529 Å, 2.11 Å, 4.76 Å, 0.2645 Å, 1.058 Å, 2.38 Å. 20. 19.6 Kcal/mol 21. 12.09 eV 22. 5.425 x 10 – 12 erg, 3.36 x 10-5 cm 23. 30.6 ev, 54.4 ev 24. 2.717 x 1019 25. 6603 Å 26. 0.73 x 106 m/sec, 0.0242 x 106 27. 8.2 x 106

28. He+. 29. 3 30. 0.599 x 1015, Yes 31. 2.76 x 10-7m, 5.4 x 10-7m, 6.54 x 10-7m 32. 4.65 eV 33. 4.5 ×1014 s–1 34. [2 eV, 3.22 eV] 35. 3.09 × 108 cm/sec 36. 4.7 59 x 10-18 J 37. 6.626 × 10–35 m 38. 7.27 10–7m 39. 1.227Å 40. 4.854 x 10-11 m 41. 5.8 x 105 m/sec 42. 0.525×10-13 m 43. 63.3 volt 45. 10 46. Zero, 4.89 B.M. 47. (i) r0 = 2a0,

GET EQUIPPED FOR IIT-JEE

( ONLY ONE OPTION CORRECT ) 1. What is the series limit for Paschen series of the He+ ion in Å? (A) 2502 (B) 2520 (C) 2250 (D) 2052 2. The time taken by an electron for one complete revolution in the nth Bohr orbit of the hydrogen atom

is (A) inversely proportional to n 2 (B)directly proportional to n3 (C) directly proportional to n (D) directly proportional to (n/h)2 3. The ionization potential difference is 2.55eV for two of the Bohr orbits of the atomic hydrogen of

quantum numbers n1 and n2, [n1<n2]. What are the values of n1 and n2? (A) 2, 3 (B) 3, 4 (C) 2, 4 (D) 3, 5 4. Which of the following statements is correct? (A) The 4p orbital has three nodes (B) 3s orbital has no node (C) Orbital angular momentum of the electron in the 1s orbital of the hydrogen atom is h/2π (D) The Rydberg constant has the same value for all hydrogen like species 5. An electron in a hydrogen atom in its ground state absorbs 1.5 times as much energy as the

minimum required escape from the atom. What is the wavelength of the emitted electron? (A) 4.7 Å (B) 4.7nm (C) 9.4 Å (D) 9.4nm

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

6. A one dimensional electron trap has an extension of 1nm. What is the uncertainty in its momentum

in kgm/s? (A) 1.551×10−24 (B) 1.055×10−25 (C) 1.115×10−23 (D) 5.2×10−26 7. Calculate the mass of the photon with a wavelength corresponding to the series limit of Balmer

transitions of the He+ ion in kg? (A) 4.22×10−36 (B) 2.24×10−34 (C) 2.42×10-35 (D) 4.22×10−36 8. The wavelength of a certain electron transition in the hydrogen spectrum is 4864Å. Identify the

Transition; (A) 3rd line Balmer (B) 1st line Lyman (C) 1st line Paschen (D) 2nd line Balmer 9. What according to the Bohr model would be the radius of the electron orbit in the first excited state

of the Li+2 ion?? (A) 0.751Å (B) 0.705Å (C) 0.925Å (D) 0.952Å 10. Photo-electrons are ejected from a metal surface using photons of energy 4×10-20J .The de Broglie

wavelength of the electron emitted with maximum K.E. =59Å . What is the photoelectric threshold in joules.

(A) 3.313×10−20 (B) 1.131×10−20 (C) 1.331×10−20 (D) 1.673×10−20 11. The series limit values are 8208Å and 22800 Å for the quantum numbers n1 and n2 in the atomic

spectrum of the hydrogen .What is the wavelength for the radiation emitted for the transition n2→n1 in Å?

(A) 15282 (B) 12258 (C) 12825 (D) 15822 12. An electron, practically at rest is initially accelerated through a potential difference of 100V .It then

has a de Broglie wavelength = λ1 Å .It then get retarded through 19 V and then has a wavelength λ2 Å. A further retardation through 32 V changes the wavelength to λ3Å.What is

(λ3― λ2 ) / λ1? (A) 20/41 (B) 10/63 (C) 20/63 (D) 10/41

13. Wave function vs distance from nucleus graph of an orbital is shown.

The number of nodal sphere of this orbital is

(A) 1 (B) 2 (C) 3 (D) 4 14. For an electron in a hydrogen atom the wave function ψ is proportional to e− (r/(a) where a is Bohr`s

radius . What is the ratio of probability of finding the electron at the nucleus to the probability of finding the electron at a

(A) e (B) e2 (C) 1/ e2 (D) Zero 15. What transition in He+ ion shall have the same wave number as the first line in Balmer series of H-

atom? (A) 3→2 (B) 6→4 (C) 5→2 (D) 7→5 16. The potential energy of the electron in an orbit of H-atom would be (A) −mv2 (B) −e2/r (C) –mv2/2 (D) −e2/2r

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

17. An electron is moving with a KE of 4.55×10−25J .What will be the de Broglie wavelength for this electron

(A) 5.28×10−7m (B) 7.27×10−7m (C) 2×10−10m (D) 3×10−5m 18. If each orbital can hold a maximum of 3 electrons. The number of elements in 4th period of periodic

table is (A) 48 (B) 57 (C) 27 (D) 36 19. The number of orbitals in a sub shell is equal to (A) n2 (B) 2l (C) 2l +1 (D) m 20. The ratio of the energy of the electron in ground state of hydrogen atom to that of the electron in the

first excited state of Be3+ is (A) 1:4 (B) 1:8 (C) 1:16 (D) 16:1 21. The electronic transition from n=2 to n=1 will produce the shortest wavelength in (A) H-atom (B) D-atom (C) He+ ion (D) Be3+ ion 22. The wave number of the first line of Balmer series of Hydrogen is 152000/cm. The wave number of

the first Balmer line of Li2+ ion is? (A) 15200/cm (B) 60800/cm (C) 76000/cm (D) 1368000/cm 23. Photoelectric emission is observed from a surface for frequencies v1 & v2 of the incident radiation

(where v1>v2) . If the maximum KE of the photoelectrons in two cases are in the ratio of 1:k ,then the threshold frequency v0 is given by

(A) (v1−v2)/k-1 (B) (kv1−v2)/k-1 (C) (kv2−v1)/k-1 (D) (v1−v2)/k 24. The difference of nth and (n+1)th Bohr`s radius of H-atom is equal to its (n -1)th Bohr`s radius. The

value of n is (A) 1 (B) 2 (C) 3 (D) 4 25. The electron in a hydrogen atom makes transition from M shell to L, the ratio of magnitude of initial

to final acceleration is (A) 9:4 (B) 81:16 (C) 4:9 (D) 16:81 26. The nucleus of an atom is located at x = y = z = 0 .If the probability of finding an s-orbital electron in

a tiny volume assumed x = a, y = 0 = z is 10−5, what is the probability of finding the electron in the same sized volume around x = z = 0, y = a?

(A) 10−5 (B) 10−5a (C) 10−5a2 (D) 10−5a−1 27. Suppose it is taken as a working hypothesis that an electron finds itself inside a typical nucleus with

an uncertainty in momentum, Δp≈10-20kgm/s, what is an approximate estimate of the size of the nucleus?

(A) 1.058×10−13m (B) 8.53×10−15 m (C) 3.58×10−16m (D) 5.27×10−15m 28. Energy will be absorbed for the process of separating (A) a proton from a proton (B) an electron from proton (C) an electron from an electron (D) a neutron from a neutron 29. Number of waves in third Bohr orbit of Hydrogen will be (A) 3 (B) 6 (C)9 (D)12 30. Principal, azimuthal and magnetic quantum numbers are respectively related to (A) size, orientation and shape (B) size, shape and orientation (C) shape, size and orientation (D) none of these 31. In the hydrogen atoms, the electrons are excited to the 5th energy level. The number of the lines

that may appear in the emission spectrum will be (A) 4 (B) 8 (C) 10 (D) 12

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

32. Light of wavelength λ shines on a metal surface with intensity x and the metal emits y electrons per second of average energy Z , what will happen to y and Z if x is doubled

(A) y will be doubled and Z will become half (B) y will remain same and Z will be doubled (C) Both y and Z will be doubled (D) y will be increased but Z will remain same 33. The ratio of potential energy and total energy of an electron in a Bohr orbit of hydrogen like species (A) 2 (B) −2 (C) 1 (D) −1 34. If λ1 and λ2 denotes the de Broglie wavelength of two particles with same masses but charges in the

ratio of 1:2 after they are accelerated from rest through the same potential difference then (A) λ1 > λ2 (B) λ1 = λ2 (C) λ1 < λ2 (D) None

MORE THAN ONE OPTION CORRECT 1. When α-particle are sent through a thin metal foil, most of them go straight through the foil,

because: (A) α-particle are much heavier than electrons (B) α-particle are positively charged (C) most part of the atom is empty space (D) α-particle move with high velocity 2. Rutherford`s α-scattering experiment led to the following conclusions (A) Atom has largely empty space (B) The centre of the atom has positively charged nucleus (C) The size of the nucleus is very small as compared to the size of the atom (D) The electrons revolve around the nucleus 3. The probability of finding the electron in Px orbital is (A) Maximum at two opposite sides of the nucleus along the x-axis (B) Zero at nucleus (C) Maximum at nucleus (D) Along the x-axis 4. Which of the statements is(are) correct? (A) Electrons in motion behave as if they are waves (B) s-orbital is non directional (C) An orbital can accommodate a maximum of the two electrons with parallel spins (D) The energies of the various sub-levels in the same shell are in order s>p>d>f 5. Which concerning Bohr`s model are true? (A) Predicts that probability of electron near the nucleus is more (B) Angular momentum of electron in H-atom =nh/2π (C) Introduces the idea of stationary states (D) Explains line spectrum of hydrogen 6. Which sets of quantum no. are consistent with the theory n l m s (A) 2 1 0 -1/2 (B) 4 3 -2 -1/2 (C) 3 2 -3 1/2 (D) 4 3 -3 1/2 7. Heisenberg uncertainty principal is not significant for (A) Moving electrons (B) Motor car (C) Stationary particles (D) All of the above 8. Consider the electronic configuration for neutral atoms 1) 1s22s22p63s1 2) 1s22s22p64s1. Which of the following statements is/are false? (A) Energy is required to change 1 to 2 (B) 1 represents Na atom

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

(C) 1 and 2 represent different elements (D) More energy is required to remove one electron from 1 than 2 9. For the energy levels in an atom which one of the following statements is/are correct? (A) 4f subshell has maximum 14 electrons. (B) The 2nd principal energy level can have 4 sub energy levels and contains a maximum of 8

electrons (C) The M energy level can have a maximum of 32 electrons (D) The 4s sub energy level is at lower energy than the 3d sub energy level 10. Which of the following statements are correct for an electron that has n = 4 and m =−2? (A) The electron may be in a d orbital (B) The electron is in the fourth principal electronic shell (C) The electron may be in a p-orbital (D) The electron may have the spin quantum number =0.5 11. Which of the following statement is/are correct?? (A) There is no probability of finding a p –electron exactly at the nucleus. (B) The orbital d2

z has two lobes of electron density directed along the z-axis and a ring of electron density (called dough nut) centered in the xy-plane

(C) The orientation of p-orbitals and d-orbitals minimize electron-electron repulsion in many electron atoms

(D) None is correct 12. Which of the following is/are correct?? (A) For all values of n the p-orbitals have the same shape but the overall size increases as n

increases, for a given atom (B) The fact that there is a particular direction, along which each p-orbital has maximum electron

density, plays an important role in determining molecular geometries (C) The charge cloud of a single electron in 2px atomic orbital consists of two lobes of electron

density (D) None is correct 13. Which of the following is the correct set of four quantum numbers for each of the electrons in the 3d

sub shell when it is fully occupied? (A) n=3; =2; m=0,1,─1,2, ─2 and s = ±0.5, ±0.5, ±0.5, ±0.5 and ±0.5 (B) n=3; =1; m=0,1,─1,2,─2 and s = ±0.5, ±0.5, ±0.5, ±0.5 and ±0.5 (C) n=4; =2; m=0,1,─1,2,─2 and s=±0.5, ±0.5, ±0.5, ±0.5 and ±0.5 (D) None is correct. 14. Which of the following statements is/are correct?? (A) The energy of an electron in a many electron atom generally increases with an increase in the

value of n, but for a given (n+l), the lower the value of l, the lower is the energy. (B) An electron close to the nucleus experiences a large electrostatic attraction (C) For a given value of n, an s-electron penetrates the nucleus more than a p-electron which

penetrates more than a d-electron and so on (D) None is correct 15. Which of the following statements is/are correct?? (A) All substances have magnetic properties (B) There are substances in which all electron spins are paired (C) There are substances in which one or more electrons have unpaired spins (D) The greater the number of unpaired electrons, the greater is Paramagnetism per mole of

substance 16. The mathematical expression for the uncertainty principal is (A) Δ x Δp ≥ h/4π (B) Δ EΔt ≥ h/4π (C) Δ x Δp ≥ h/p (D) ΔEΔt ≥ h/p

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

17. Choose the correct relations on the basis of Bohr`s theory (A) Velocity of electron α1/n (B) Frequency of revolution α 1/n3 (C) Radius of orbit α n2Z (D) Force on electron α 1/n4 18. Which of the following statements is/are correct? (A) A photon is a positively charged nuclear particle (B) A photon is a particle of light energy (C) A photon is a quantum of light (D) A photon is a bundle of energy of definite magnitude. 19. Which of the following statements is/are correct? (A) Stark effect is the splitting lines when source is placed in electric field (B) Beyond a certain limit a spectrum of an atom there is continuum (C) The intensities of spectral line in line spectrum decreases with increase in the value of n (D) Shielding effect is possible in H-atom 20. Which statements is/are true regarding white lights? (A) It is a form of energy (B) It can be deflected by a magnet (C) It consists of photons of same energy (D) It is part of electromagnetic spectrum 21. An electron jumps nth level to 1st level , the fact which is/are correct about H atoms is/are (A) Number of spectral lines =n(n-1)/2 (B) Number of spectral lines =∑(n-1) (C) If n=4, the number of spectral lines =6 (D) Number of spectral lines =n(n-1) 22. Photoelectric effect supports quantum nature of light because (A) There is a minimum frequency below which no emission of photoelectron is possible (B) Maximum KE of photo electrons depends only on frequency of light and not on intensity (C) Even when the metal surface is faintly illuminated the photoelectron leave the surface

immediately (D) Electric charge of the photoelectrons is quantized 23. Which of the following statements is/are correct?? (A) If the value of l=0, the electron distribution is spherical (B) The shape of the orbital is given by magnetic quantum number (C) Orbital angular momentum of 1s, 2s, 3s electrons are equal (D) In an atom, all electrons travel with the same velocity COMPREHENSION TYPE Comprehension – 1: Rutherford proposed the atomic model after his most striking experiment on scattering

leading to discovery of nucleus. Bohr later on modified the atomic model on the basis of Planck’s quantum theory of light and proposed the concept of stationery circular orbit of quantized angular

momentum2nh

. The collection of the line spectrum led Sommerfeld to give the idea of elliptical

orbits. He successfully explained the existence of subshells and their number in a shell. The orbital

angular momentum of subshells was proposed as 1 . .2hl l

the emission of a spectral line in

atomic spectra was supposed to be due to the jump of electron from one energy level to other.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

1. The orbital angular momentum of electron in 2p orbital is:

(A)2h

(B)2h

(C) 2h

(D) 2

2. The volume occupied by the nucleus is about …… times of volume of atom: (A) 10 – 15 (B) 1015 (C) 10 ─12 (D) 10 – 10

3. Non – directional orbital is (A) 3s (B) 4f (C) 4d (D) 4p 4. The total number of fundamental particles in 14

6 C is (A) 6 (B) 8 (C) 14 (D) 20 5. The minimum energy is given out when an electron jumps from one orbit to other from: (A) 2 to 1 (B) 3 to 2 (C) 4 to 3 (D) 5 to 4 6. An oil drop has ─6.39 x 10─19 coulomb charge. The number of electrons in this oil drop is: (A) 4 (B) 3 (C) 2 (D) 1 Comprehension – 2 : Bohr proposed his atomic model based on Planck’s quantum theory and derived following relations

for one electron system in C.G.S. units :

For H – atom eVEcmuArnu

uZEErnr nnn 6.13.;sec/1019.2;529.0;,, 18

1112

112

For 1 electron systems , other than H

;12

Zrn

r Hn

;2

21

nZE

E Hn

nZu

u Hn

1

Later on de Broglie proposed the dual nature of electron and put forward his wave concept. The wavelength of electron in an orbit was given by muh / .

1. The number of waves made by a Bohr electron in H – atom for one complete revolution in its 3rd

orbit are: (A) 1 (B) 2 (C) 3 (D) 4 2. The circumference (in m) of third Bohr orbit in H – atom is : (A) 3.0 710 (B) 3.0 810 (C) 3.0 610 (D) 3.0 910 3. The wavelength (in m ) of moving electron in 3rd orbit of H – atom is : (A) 1.0 910 (B) 2.0 910 (C) 1.0 710 (D) 1.0 810 4. The potential energy of electron in 3rd Bohr orbit of H – atom is: (A) - 2.42 erg1210 (B) -4.84 erg1210 (C) +4.84 erg1210 (D) 2.42 erg1210 5. The momentum of electron in 3rd Bohr orbit of H – atom is : (A) 6.65 125 sec10 mkg (B) 6.65 128 sec10 mkg

(C) 6.65 129 sec10 mkg (D) 6.65 120 sec10 mkg

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Comprehension – 3: De Broglie proposed dual nature for electron by putting his famous equation .

muh

Later on

Heisenberg proposed uncertainly principle as hx. p

2 2 . On the contrary particle nature

of electron was established on the basis of photoelectric effect. When a photon strikes the metal surface, it gives up its energy to the electron. Part of this energy (say W) is used by the electrons to escape from the metal and the remaining imparts the kinetic energy (1/2 mu2) to the photoelectron. The potential applied on the surface to reduce the velocity of photoelectron to zero is known as stopping potential.

1. With what velocity must an electron travel so that its momentum is equal to that of photon of

wavelength is A5200 : (A) 800 m s- 1 (B) 1400 m s- 1 (C) 400 m s- 1 (D) 200 m s-1 2. With what potential should a beam of electron be accelerated so that its wavelength becomes equal

to A54.1 . (A) 63.3 V (B) 6.33 V (C) 633 V (D) None of these 3. The binding energy of electron in a metal is 250 kJ mol – 1 . The threshold frequency of metal is: (A) 112 sec106 (B) 114 sec106 (C) 110 sec106 (D) 112 sec106 4. If uncertainties in position and momentum of an electron are same, then uncertainty in its velocity

can be given by:

(A) 24 mh

(B) m

u

4 (C) 22m

(D) Either of these

5. The metal most commonly used in photoelectric cell is (A) Na (B) Ba (C) Cs (D) Ni 6. The wavelength of a golf ball weighing 200g and moving at a speed of 5 meter /hr is of the order: (A) 10 – 10 m (B) 10 – 20 m (C) 10 – 30 m (D) 10 – 40 m Comprehension – 4: The electrons in a poly – electronic atom are filled one by one in order of increasing energy level.

The energy of subshells and orientation of orbitals depends upon the values of three quantum numbers (i.e. n , l and m respectively) derived from Schrodinger wave equation. The different orbitals of a subshells however posses same energy level and are called degenerate orbitals but their energy level changes in presence of magnetic field and the orbitals are non – degenerate. A spectral line is noticed if an electron jumps from one level to other. The paramagnetic nature of element is due to the presence of unpaired electron.

1. P3- is isoelectronic with: (A) N 3- (B) As3+ (C) Cl- (D) F- 2. The number of unpaired electrons in Cr atom is: (A) 2 (B) 3 (C) 5 (D) 6 3. The element which has as many ‘s’ electrons as ‘p’ electrons but belong to III period is : (A) O (B) Mg (C) Al (D) C 4. The total magnetic moment of Ni2+ ion is :

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

(A) BM6 (B) BM8 (C) BM15 (D) BM12 5. The number of spherical and angular nodes in 2p orbitals are: (A) 1, 1 (B) 2, 1 (C) 1, 0 (D) 0, 1 6. The possible number of spectral lines when an electron can jump from 5th shell to 2nd shell is: (A) 4 (B) 2 (C) 3 (D) 6 7. The correct order for energy levels in H – atom is: (A) 3s = 3p = 3d > 2s (B) 3d > 3p > 3s > 2s (C) 3d > 3p = 3s > 2s (D) 3d > 3p > 3s = 2s 8. Total number of valence electrons in

4NH is: (A) 9 (B) 8 (C) 6 (D) 11

MATRIX MATCH Column-I and Column-II contains four entries each. Entries of Column-I are to be matched with

some entries of Column-II. One or more than one entries of Column-I may have the matching with the same entries of Column-II.

1. If in Bohr’s model, for unielectronic atom following symbols are used znr , Radius of thn orbit with

atomic number Z; znU , Potential energy of e ;

znK , Kinetic energy of e : znV , Velocity of e ;

znT , Time period of revolution Column-I Column-II (A) U1,2 : K1,1 (P) 1 : 8 (B) r2,1 : r1,2 (Q) – 8 : 1 (C) V1,2 : V2,4 (R) 1 : 1 (D) T1,2 : T2,2 (S) 8 : 1 2. Column-I Column-II (A) The number of radial node of 5s atomic orbital is (P) 1 (B) The number of angular node of yzd3 atomic orbital is (Q) 4 (C) The sum of number of angular node and radial (R) 2 node of xyd4 atomic orbital (D) The number of angular node of 3p atomic orbital is (S) 3 3. Column-I Column-II

(A) Orbital angular momentum of an electron (P) 2

)1( hss

(B) Angular momentum of an electron in an orbit (Q) )2( nn

(C) Spin angular momentum of an electron (R) 2

nh

(D) Magnetic moment of atom (in B.M.) (S) 2

)1( hll

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

4. Column-I Column-II (A) No. of orbitals in the nth shell (P) 2(2l + 1) (B) Max. no. of electrons in a subshell (Q) n (C) No. of subshells in nth shell (R) 2l + 1 (D) No. of orbitals in a subshell (S) n2 5. Column-I Column-II (A) (P) 4s (B) (Q) 5px (C) Angular probability is dependent on and (R) 3s (D) No angular node is present (S) 6dxy 6. Column - I Column - II (A) m= 1 (P) p- subshell (B) l = 2 (Q) f- subshell (C) n = 3 (R) d- subshell (D) Angular nodes (S) s- subshell 7. Column - I Column - II (Bohr’s model) (A) Total Energy of e− (P) 2Z (B) Revolution frequency of e− (Q) 3Z (C) Acceleration of e− (R) Decreases as orbit number increases (D) Radius of orbit (S) Increases as orbit number increases

1

r

2 24 r

Distance from nucleus

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Answers to Get Equipped for IIT-JEE

ONLY ONE OPTION CORRECT

1. D 2. B 3. C 4. A 5. A 6. D 7. C 8. D 9. B 10. A 11. C 12. C 13. A 14. D 15. B 16. A 17. B 18. C 19. C 20. A 21. D 22. D 23. B 24. D 25. D 26. A 27. D 28. B

29. A 30. B 31. C 32. D 33. A 34. A

MORE THAN ONE OPTION CORRECT 1. A, C, D 2. A, B, C, D 3. A, B, D 4. A, B 5. B, C, D 6. A, B, D 7. B, C 8. C 9. A, D 10. A, B, D 11. A, B, C 12. A, B, C 13. A 14. B, C 15. A, B, C, D 16. A, B 17. A, B, D 18. B, C, D 19. A, B, C 20. A, D 21. A, B, C 22. A, B, C 23. A, C

ANSWERS (COMPREHENSION TYPE; ASSERTION & REASONS) COMPREHENSION -1 1. (B) 2.(A) 3.(A) 4.(D) 5. (D) 6. (A) COMPREHENSION -2 1. (C) 2. (D) 3. (A) 4. (B) 5. (A) COMPREHENSION -3 1. (B) 2. (A) 3. (B) 4. (D) 5. (C) 6. (C) COMPREHENSION -4 1. (C) 2. (D) 3. (B) 4. (B) 5. (D) 6. (D) 7. (A) 8.(B)

MATRIX MATCH 1. A Q; B S; C R; D P 2. A Q; B R; C S; D P 3. A S; B R; C P; D Q 4. A S; B P; C Q; D R 5. A P; B P, Q, S; C Q, S; D P, R 6. A P, Q, R; B R; C P, R, S; D P, Q, R 7. A P, S; B P, R; C Q, R; D S

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

EXPERTISE ATTAINERS 1. Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium.

The kinetic energy of the fastest photo electrons emitted from sodium is 0.73 eV. The work function for sodium is 1.82 eV. Find

i) The energy of the photons causing the photo electric emission. ii) The quantum numbers of the two levels involved in the emission of these photons. 2. An electron in order to have a wavelength of 500Å, through what potential difference it must

pass? 3. The photo electric emission requires a threshold frequency 0 for a certain metal. 1 = 2200Å

and 2 = 1900 Å, produce electrons with a maximum kinetic energy KE1 & KE2. If KE2 = 2KE1 calculate 0 and corresponding 0.

4. The absorption of energy by an atom of hydrogen in ground state ,results in the ejection of the

electron with the de Broglie wavelength λ= 4.7×10-10m. Given that the ionization energy is 13.6 ev, Calculate the energy of the photon which caused the ejection of electron.

5. What lines of atomic hydrogen spectrum fall within the wavelength range from 94.5 to

130.0nm?? Given: RH=1.1×107/m 6. Light of the prominent mercury line 2357Å ejects from a metal surface, electrons that have

stopping potential of 2.6 volts. The stopping potential of electrons ejected from the same metal by the other prominent mercury line 1849Å is 4.04 volts.

a) Calculate the value of Planck`s constant b) Calculate the value of threshold frequency using these data.

7. If the kinetic energy of an e- is given as “K” Joules with an uncertainty of 66 10 , then calculate the expression of uncertainty in velocity, uncertainty in position. Express your answers in terms of “K” only.

8. An electron already accelerated through a potential drop of V1 volts is further accelerated through a potential drop of V2 volts .Its de Broglie wavelength λ1 before the second acceleration and λ2 after it. The change Δλ is 47.48% of λ1. Calculate the ratio V1:V2.

9. It is known that (radial) 1, 0 0

3/ 2(zr / a )

0

z2 ea

for hydrogen like species. What will be the distance

(from the nucleus) where there is maximum probability of finding a 1 s electron in terms of a0 & z.

10. 1.8 g hydrogen atoms are excited to radiations. The study of spectra indicates that 27% of the

atoms are in 3rd energy level and 15% of atoms in 2nd energy level and the rest in ground state. If I.P. of H is 13.6 eV . Calculate

(i) No. of atoms present in III & II energy level. (ii) Total energy evolved when all the atoms return to ground state. 11. The energy of an excited H-atom is –3.4 eV. Calculate angular momentum of e–. 12. The eyes of certain member of the reptile family pass a single visual signal to the brain when the

visual receptors are struck by photons of wavelength 850 nm. If a total energy of 3.15 10-14 J is required to trip the signal, what is the minimum number of photons that must strike the receptor.

13. To what series does the spectral lines of atomic hydrogen belong if its wave number is equal to the

difference between the wave numbers of the following two lines of the Balmer series 486.1 and 410.2 nm. What is the wavelength of this?

14. The dye uriflarine, when dissolved in water has its maximum light absorption at 4530Å and its

maximum fluorescence emission at 5080Å. The number of fluorescence quanta is, on the

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

average 53% of the number of quanta absorbed. Using the wavelengths of maximum absorption and emission, what percentage of absorbed energy is emitted as fluorescence?

15. The ionisation energy of a H like Bohr atoms in terms Rydberg’s constant is 4RH.

[ 18H1R 2.18 10 J ]

(A)Calculate the wavelength radiated when electron jumps from the first excited state to ground state. (B) What is radius of 1st orbit of this atom? 16. A hydrogen like atom (atomic no. Z) is in a higher excited state of quantum number ‘n’. This

excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.2 eV and 17.00 eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energy 4.25 eV and 5.95 eV respectively. Determine the values of n and Z.

7. A single electron species has nuclear charge +Ze where Z is atomic number and e is electronic

charge. It requires 17 eV to excite the electron from the second Bohr orbit to third Bohr orbit. Find :

i) The atomic no. of element ii) The energy required for transition of electron from first to third orbit. iii) Wavelength required to remove electron from first Bohr orbit to infinity. iv) The kinetic energy of electron in first Bohr orbit. 18. A light beam of wavelength 400 nm is incident on a metal plate of work function 2.2 eV (a) A particular electron absorbs a photon and makes two collisions before coming out of the

metal. Assuming that 10% of the initial extra energy is lost to the metal in each collision, find the kinetic energy of this electron as it comes out of the metal.

(b) Under the same assumptions, find the maximum number of collisions the electron can suffer before it becomes unable to come out of the metal.

19. A certain series in the atomic spectrum of hydrogen like ion is observed to have a series limit of

2052 Å. Suggest the simplest possible solution identifying the nature of the ion and name of the series.

20. An unknown orbital is having the following equation for the radial component of its wave function.

0

3 / 2 2 2zr / 3a

20 0 0

2 z 2zr 2z r3 e3 3a a 9a

(a) identity the type of orbital. Justify (b) Number, location & shape of angular node (c) Number of radial nodes, the distance of the radial nodes & shapes of radial nodes. 21. Show that for large value of principal quantum number the frequency of an electron rotating in

adjacent energy levels of H-atom and the radiant frequency for transition between these all approach the same value.

22. He atom can be excited to 1s1 2p1 by = 58.44 nm. If lowest excited state for He lies 4857cm–1

below the above. Calculate the energy for the lowest excitation state. 23. The energy needed for the reaction is 20000 KJ/ mole. If the first ionization energy of lithium is 520KJ/ mole, calculate its second & third ionization energy

3gLi g Li 3e

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

Answers to EXPERTISE ATTAINERS

1. (i) 2.55 eV (ii) n = 2, 4 2. 610-4 V

3. 15 10 02613A 1.148 10 sec 4. 20.4 eV

5. 51, 41, 31, 21 6. (a) 6.610-34 (SI units) (b) 6.421014 sec

7. 21

h 2Kx4 5.73 10

96.29 10V2K

8. 0.381

9. 0az

10. 292.68 ×1021 atoms, 162.60 ×1021 atoms, 832 KJ 11. h/ 12. 1.35×105 13. Brackett; 2.63 ×10–4 cm 14. 47.26% 15. (a) 303.89 A (b) 0.2645 A 16. Z = 3, n = 6 17. (i) Z = 3 (ii) 108.8 eV (iii) 101.3 A (iv) 122.4 eV 18. (a) 0.72 eV (b) 10

19. n = 3, Z = 2 20. (a) 3s (b) 0 (c) 2 , Spherical. 0a 9 11

z 2 3

0a 9 11z 2 3

22. 3.3 × 10–18 J 23. [7690, 11790 KJ]

WINDOW TO IIT-JEE

OBJECTIVE QUESTIONS I 1. Rutherford’s experiment on scattering of α-particles showed for the first time that the atom has

(JEE 1981) (a) electrons (b) protons (c) nucleus (d) neutrons

2. Any p-orbital can accommodate up to (JEE 1983)

(a) Four electrons (b) six electrons (c) two electrons with parallel spins (d) two electrons with opposite spins

3. The principal quantum number of an atom is related to the (JEE 1983)

(a) size of the orbital (b) spin angular momentum (c) Orientation of the orbital in space (d) Orbital angular momentum

4. Rutherford’s scattering experiment is related to the size of the (JEE 1983) (a) nucleus (b) atom (c) electron (d) neutron

5. The increasing order (lowest first) for the values of e/m (charge/mass) for electron (e), proton(p),

neutron (n) and alpha particle (α) is (JEE 1984) (a) e,p,n,a (b) n,p,e,α (c) n,p,α,e (d) n,α,p,e

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

6. Correct set of four quantum numbers for the valence (outermost) electron of rubidium (Z=37) is (JEE 1984)

(a) 5,0,0,+12

(b) 5,1,0,+ 12

(c) 5,1,1,+ 12

(d) 6,0,0,+ 12

7. Which electronic level would allow the hydrogen atom to absorb a photon but not to emit a photon?

(JEE 1984) (a) 3s (b) 2p (c) 2s (d) 1s

8. Bohr’s model can explain (JEE 1985)

(a) the spectrum of hydrogen atom only (b) spectrum of an atom or ion containing one electron only (c) the spectrum of hydrogen molecule (d) the solar spectrum

9. The radius of an atomic nucleus is of the order of (JEE 1985)

(a) 10-10cm (b) 10-13cm (c) 10-15cm (d) 10-8cm

10. Electromagnetic radiation with maximum wavelength is (JEE 1985) (a) ultraviolet (b) radio wave (c) X ray (d) infrared

11. Rutherford’s alpha particle scattering experiment eventually let to the (JEE 1986)

(a) Mass and energy are related (b) electrons occupy space around the nucleus (c) neutrons are buried deep in the nucleus (d) the point of impact with matter can be precisely determined.

12. Which one of the following sets of quantum numbers represents an impossible arrangement?

(JEE 1986) n l m s

(a) 3 2 -2 12

(b) 4 0 0 12

(c) 3 2 -3 12

(d) 5 3 0 12

13. The ratio of the energy of a photon of 2000Åwavelength radiation to that of 4000 Å radiation is

(a) 14

(b) 4 (c) 12

(d) 2 (JEE 1986)

14. The wavelength of a spectral line for electronic transition is inversely related to (JEE 1988) (a) the number of electrons undergoing the transition. (b) the nuclear charge of the atom (c) the difference in the energy of the energy levels involved in the transition (d) the velocity of the electron undergoing the transition.

15. The orbital diagram in which the aufbau principle is violated (JEE 1988)

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

(a) (b)

(c) (d)

16. The outermost electronic configuration of the most electronegative elements is (JEE 1988,90) (a) ns2 np3 (b) ns2 np4 (c) ns2 np5 (d) ns2 np6

17. The correct ground state electronic configuration of chromium atom is (JEE 1989)

(a) [Ar]3d5 4s1 (b) [Ar]3d4 4s2 (c) [Ar]3d5 4s0 (d) [Ar]4d5 4s1

18. The correct set of quantum numbers for the unpaired electron of chlorine atom is (JEE 1989) n l m

(a) 2 1 0 (b) 2 1 1 (c) 3 1 1 (d) 3 0 0

19. Which of the following does not characterise X – rays? (JEE 1992) (a) The radiation can ionise gases (b) It causes ZnS to fluorescence (c) Deflected by electric and magnetic fields (d) Have wavelengths shorter than ultraviolet rays

20. Which of the following relates to photons both as wave motion and as a stream of particles?

(JEE 1992 ) (a) Interference (b) E = mc2 (c) Diffraction (d) E = hv

21. The orbital angulr momentum of an electron in 2s orbital is (JEE 1996)

(a) +1 .2 2

h

(b) zero

(c) 2h

(d) 2.2h

22. Which of the following has the maximum number of unpaired electrons? (JEE 1996)

(a) Mg2+ (b) Ti3+ (c) V3+ (d) Fe2+

23. The first use of quantum theory to explain the structure of atom was made by (JEE 1997)

(a) Heisenberg (b) Bohr (c) Planck (d) Einstein

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

24. For a d-electron, the orbital angular momentum is (JEE 1997)

(a) 62h

(b) 22h

(c) 2h

(d) 22h

25. The energy of an electron in the first Bohr orbit of H-atom is -13.6 eV. The possible energy value (s) of

the excited state (s) for electrons in Bohr orbits of hydrogen is (are) (JEE 1998) (a) -3.4eV (b) -4.2eV (c) -6.8 eV (d) + 6.8 eV

26. The electrons, identified by quantum numbers n and l, (i) n = 4, l = 1 (ii) n = 4, l = 0 (iii) n=3, l = 2 (iv)

n = 3, l = 1 can be placed in order of increasing energy, from the lowest to highest ,as (JEE 1999) (a) (iv)<(ii)<(iii)<(i) (b) (ii)<(iv)<(i)<(iii) (c) (i)<(iii)<(ii)<(iv) (d) (iii)<(i)<(iv)<(ii)

27. The electronic configuration of an element is 1s2, 2s22p6, 3s23p63d5, 4s1. This represents its.

(JEE 2000) (a) excited state (b) ground state (c) cationic form (d) anionic form

28. The number of nodal planes in a px orbital is (JEE 2000)

(a) one (b) two (c)three (d) zero.

29. The wavelength associated with a golf ball weighing 200g and moving at a speed of 5m/h is of the order (JEE 2001) (a) 10-10m (b) 10-20m (c) 10-30m (d) 10-40m

30. The quantum numbers +1/2 and -1/2 for the electron spin represent (JEE 2002)

(a) rotation of the electron in clockwise and anticlockwise direction respectively. (b) rotation of electron in anticlockwise and clockwise direction respectively (c) magnetic moment of electron pointing up and down respectively (d) two quantum mechanical spin states which have no classical analogue

31. If the nitrogen atom had configuration 1s7, it would have energy lower than that of the normal ground

state configuration 1s2 2s2 2p3, because the electrons would be closer to the nucleus, yet 1s7 is not observed because it violates (JEE 2002)

(a) Heisenberg uncertainty principle (b) Hund’s rule (c) Pauli exclusion principle (d) Bohr postulate of stationary orbits

32. Rutherford’s experiment , which established the nuclear model of the atom, used a beam of(JEE 2002)

(a) β-particles , which impinged on the metal foil and got absorbed (b) γ–rays which impinged on a metal foil and got scattered (c) helium atom which impinged on metal foil and got scattered (d) helium nuclei which impinged on metal foil and got scattered

33. Which hydrogen like atom have similar radius as that of Bohr orbit of hydrogen atom? (JEE 2004)

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

(a) n=2, Li2+ (b) n=2, Be3+ (c) n=2, He+ (d) n=3, Li2+

34. The number of radial nodes in 3sand 2p respectively are (JEE 2005) (a) 2 and 0 (b) 0 and 2 (c) 2 and 1 (d) 1 and 2 Objective Questions II [One or more than one correct option]

1. An isotone of 7632Ge is (JEE 1984)

(a) 7732Ge (b) 77

33 As (c) 7734 Se (d) 78

34 Se

2. Many elements have non-integaral atomic masses because (JEE 1984) (a) they have isotopes (b) their isotopes have non-integral masses (c) their isotopes have different masses (d) the constituents, neutrons, protons and electrons, combine to give fractional masses

3. When alpha particles are sent through a thin metal foil, most of them go straight through the foil because

(a) alpha particles are much heavier than electrons (JEE 1984) (b) alpha particles are positively charged (c) most part of the atom is empty space (d) alpha particles move with high velocity

4. The sum of the number of neutrons and proton in the isotope of hydrogen (JEE 1986)

(a) 6 (b) 5 (c) 4 (d) 3

5. The atomic nucleus contains (JEE 1986) (a) protons (b) neutrons (c) electrons (d) photons

6. The energy of an electron in the first Bohr orbit of H-atom is -13.6 eV. The possible energy value (s) of

the excited state (s) for electrons in Bohr orbits of hydrogen is(are) (JEE 1988)

(a) -3.4eV (b) -4.2eV (c) -6.8eV (d) +6.8eV

7. Which of the following statement (s) is (are) correct? (JEE 1998) (a) The electronic configuration of Cr is [Ar] 3d5 4s1. (Atomic number of Cr=24). (b) The magnetic quantum number may have a negative value (c) In silver atom, 23 electrons have a spin of one type and 24 of the opposite type. (Atomic number of

Ag = 47) (d) The oxidation state of nitrogen in HN3 is -3.

8. The ground state electronic configuration of nitrogen atom can be represented by

(JEE 1999) (a) (b) (c) (d)

Assertion and Reason

Read the following questions and answer as per the direction given below: (a) Statement I is true; Statement II is true, Statement II is the correct explanations of Statement I.

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

(b) Statement I is true; Statement II is true ,Statement II is NOT the correct explanations of Statement I. (c) Statement I is true; Statement II is false. (d) Statement I is false; Statement II is true.

1. Statement I: The first ionization energy of Be is greater than that of B. Statement II: 2p orbital is lower in energy than 2s. (JEE 2000)

Comprehension Based Questions The hydrogen – like species Li2+ is in a spherically symmetric state S1 with one radial node. Upon absorbing light the ion undergoes transition to a state S2. The state S2 has one radial node and its energy is equal to the ground state energy of the hydrogen atom. (JEE 2010) 1. The state S1 is

(a) 1s (b) 2s (c) 2p (d) 3s

2. Energy of the state S1 in units of the hydrogen atom ground state energy is (a) 0.75 (b) 1.50 (c) 2.25 (d) 4.50

3. The orbital angular momentum quantum number of the state S2 is

(a) 0 (b) 1 (c) 2 (d) 3 Match the Columns 1. According to Bohr’s theory

En = Total energy Kn= Kinetic energy Vn= Potential energy rn = Radius of nth orbit

Match the following: (JEE 2006)

Column I Column II (A) Vn/Kn= ? (B) If radius of nth orbit ∝ , ?x

n xE (C) Angular momentum in lowest orbital (D) 1/rn∝zy, y= ?

(p) 0 (q) -1 (r) -2 (s) 1

Match the entries in Column I with the correctly related quantum number (s) in column II. (JEE 2008)

Column I Column II (A) Orbital angular momentum of the

electron in a hydrogen – like atomic orbital

(B) A hydrogen – like one-electron wave function obeying Pauli’s principle

(C) Shape, size and orientation of hydrogen – like atomic orbital

(D) Probability density of electron at the nucleus in hydrogen – like atom

(p) Principal quantum number (q) Azimuthal quantum number (r) Magnetic quantum number (s) Electron spin quantum number

Fill in the Blanks 1. The mass of hydrogen is _____kg. (JEE 1982) 2. Isotopes of an element differ in the number of ____ in their nuclei (JEE 1982) 3. When there are two electrons in the same orbital, they have _____spins (JEE 1983) 4. Elements of the same mass number but of different atomic number are known as _____.(JEE 1983)

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

5. The uncertainty principle and the concept of wave nature of matter were proposed by ____ and ____

respectively (JEE 1988) 6. The light radiations with discrete quantities of energy are called _____. (JEE 1993) 7. Wave functions of electrons in atoms and molecules are called _____. (JEE 1993) 8. The 2px, 2py and 2Pz orbitals of atom have identical shapes but differ in their____. (JEE 1993) 9. The outermost electronic configuration of Cr is ______. (JEE 1994)

True/False 1. The outer electronic configuration of the ground state chromium atom is 3d44s2. (JEE 1982) 2. Gamma rays are electromagnetic radiations of wavelengths of 10-6cm to 10-5 cm. (JEE 1983) 3. The energy of the electron in the 3d-orbital is less than that in the 4s-orbital in the hydrogen atom.

(JEE 1983) 4. The electron density in the XY plane in 2 23

x yd

orbital is zero. (JEE 1986)

5. In a given electric field, β-particles are deflected more than α – particles in spite of α-particles having larger charge. (JEE 1993)

Integer Types Question 1. The maximum number of electrons that can have principal quantum number, n = 3 and apin quantum

number, ms = -12

, (JEE 2011)

2. The work function ( ) of some metals is listed below. The number of metals which will show photoelectric effect when light of 300nm wavelength falls on the metal is (JEE 2011) Metal Li Na K Mg Cu Ag Fe Pt W (eV) 2.4 2.3 2.2 3.7 4.8 4.3 4.7 6.3 4.75 Subjective Question:

1. The energy of the electron in the second and third Bohr’s orbits of the hydrogen atom is 125.42 10 erg and -2.41×10-12 erg respectively. Calculate the wavelength of the emitted light when the electron drops from the third to the second orbit. (JEE 1981)

2. Calculate the wavelength in Angstroms of the photon that is emitted when an electron in the Bohr’s orbit, n = 2 returns to the orbit, n = 1 in the hydrogen atom. The ionisation potential of the ground state hydrogen atom is 2.17 10-11 erg per atom. (JEE 1982)

3. The electron energy in hydrogen atom is given by En = 12

221.7 10

n

erg. Calculate the energy required

to remove an electron completely from the n = 2 orbit. What is the longest wavelength (in cm) of light that can be used to cause this transition? (JEE 1984)

4. Give reason why the ground state outermost electronic configuration of silicon is : (JEE 1985)

3s 3p 3s 3p and not

5. What is the maximum number of electrons that may be present in all the atomic orbital with principal quantum number 3 and azimuthally quantum number 2? (JEE 1985)

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

6. According to Bohr’s theory, the electronic energy of hydrogen atom in the nth Bohr’s orbit is given by: (JEE 1990)

19

221.7 10

nE Jn

What is the longest wavelength for He+ (n=3). 7. Estimate the difference in energy between 1st and 2nd Bohr’s orbit for a hydrogen atom. At what minimum

atomic number, a transition from n= 2 to n = 1 energy level would result in the emission of X-rays with l = 3.0×10-8m? Which hydrogen atom-like species does this atomic number correspond to?(JEE 1993)

8. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n=4 to n = 2 of He+ spectrum? (JEE 1993)

9. Find out the number of waves made by a Bohr’s electron in one complete revolution in its 3rd orbit. (JEE 1994)

10. Iodine molecule dissociates into atoms after absorbing light to 4500Å. If one quantum of radiation is absorbed by each molecule, calculate the kinetic energy of iodine atoms. (Bond energy of I2 = 240kJ mol-1). (JEE 1995)

11. Calculate the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen. (JEE 1996)

12. Consider the hydrogen atom to be proton embedded in a cavity of radius a (Bohr’s radius) whose charge is neutralised by the addition of an electron to the cavity in vacuum, infinitely slowly. Estimate the average total energy of an electron in its ground state in a hydrogen atom as the work done in the above neutralisation process. Also, if the magnitude of the average kinetic energy is half the magnitude of the average potential energy, find the average potential energy. (JEE 1996)

13. With what velocity should anα-particle travel towards the nucleus of a copper atom so as to arrive at a

distance 10-13m from the nucleus of the copper atom? (JEE 1997)

14. An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to 1.54 Å.

(JEE 1997)

15. Calculate the energy required to excite 1L of hydrogen gas at 1 atm and 298 k to the first excited sate of atomic hydrogen. The energy for the dissociation of H – H bond is 436kJ mol-1. (JEE 2000)

16. The wavelength corresponding to maximum energy for hydrogen is 91.2nm. Find the corresponding wavelength for He+ion. (JEE 2003)

17. (a) The Schrodinger wave equation for hydrogen atom is;

Ψ1s=

0

32

21

0 02

1 1 24 2

rar e

a a

Where, 0a is Bohr’s radius. Let the radial node in 2s be at 0r . Then r0 in terms of 0a (b) A baseball having mass 100g moves with velocity 100 m/s. Find out the value of wavelength of

baseball. (JEE 2004)

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

18. (a) Calculate velocity of electron in first Bohr orbit of hydrogen atom (Given, r = 0a )

(b) Find de-Broglie wavelength of the electron in first Bohr orbit. (c) Find the orbital angular momentum of 2p orbital in terms of h/2π units. (JEE 2005)

ANSWER KEY Objective Questions - I 1. (c) 2. (d) 3. (a) 4. (a) 5. (d) 6. (a) 7. (d) 8. (b) 9. (b) 10. (b) 11. (b) 12. (c) 13. (d) 14. (c) 15. (b) 16. (c) 17. (a) 18. (c) 19. (c) 20. (a) 21. (b) 22. (d) 23. (b) 24. (a) 25. (a) 26. (a) 27. (b) 28. (a) 29. (c) 30. (d) 31. (c) 32. (d) 33. (b) 34. (a) Objective Questions - II

1. (b,d) 2. (a,c) 3. (a,c) 4. (d) 5. (a,b) 6. (a) 7. (a,b,c) 8. (a,d)

Assertion and Reason 1. (c)

Comprehension Based Questions

1. (b) 2. (c) 3. (b)

Match the Columns

A → r; B→q; C→p; D→s

A→q; B→p,q,r,s C→p,q,r D→p,q,r Fill in the Blanks

1. 1.66×10-27 2. neutrons 3. Opposite 4.isobars 5. Heisenberg, de-Broglie 6.photons 7. orbital 8. Orientation in space 9. Cr = [Ar]3d5, 4s1 True/False

1. F 2. F 3. T 4. F 5. T Integer Type Questions 1. (9) 2. (4)

Subjective Questions:

1. 660nm 2. 1220Å 3. 3.66× 10-5cm , 5.425 x 10-12 erg 4. The 2nd configuration is against Hund’s rule of maximum multiplicity which states that the singly occupied degenerate atomic orbitals must have electrons of like spins. 5. Ten

JEE CHEMISTRY


IITIAN'S HUBTH

ET

HE

6. 471 nm 7. 1.635 x 10 -18 J , Z=2 (He+) 8. n1= 1 and n2 = 2 9. Number of complete waves formed in one complete revolution of electron in any Bohr orbit is equal to orbit number, hence three. 10. 2.165× 10-20 J/atom 11. 6 12.725 10 m

12. 2

22

eV Er

13. 8.96 × 106 ms-1

14. 63.56 V.

15. 98.39kJ. 16. 22.8nm 17.

0

25

( ) 2

( ) 6.625 10

a r a

b

Å(negligible small)

18. (a)2.18 × 106 ms-1 (b)3.3Å

(c) 2 12h for p orbital

chemistry...mole concept one mole is an amount of substance containing avogadro's number of...

Documents