Sure Short Questions

UNIT 1 THE SOLID STATE

Q-1. What is the non- stoichiometry defect in the crystals? Ans. These defects occur when the ratio of the cations and anions in the resulting compound is different from that as indicated by the laws of the chemical combinations. Q-2. How many atoms are there in a simple or primitive unit cell body centered cubic unit cell face centered cubic unit cell Ans-.a) one b) two c) four Q-3.What type of semiconductors are when Si doped with – a) Phosphorous and b) Gallium. Ans. a) n-type and b) p-type Q-4. What is the coordination no. of an octahedral void and tetrahedral void? A-. Six(6) and Four(4) Q-5. Which of the two will show the Schottky defect when added to the AgCl crystal NaCl or CdCl2. Ans. CdCl2 . This is because divalent Cd2+ ions occupy the Ag+ sites and thus produce cationic vacancies in the crystal. Q-6 What type of compounds show the Schottky defect? Ans. Ionic compounds which have cations and anions of nearly the same size. Q-7.Name the compound in which both Schottky and Frenkel defects are found together? Ans- AgBr. Q-8.What are F- centers? Why are the solids containing the F- centers are paramagnetic? Ans. The free electrons trapped in the anion vacancies are termed as the F- centers Q-9 A unit cell consists of a cube in which there are anions at each corner and one at the center of the unit cell. The cations are the center of the each face . How many A) cations and B)anions make up the unit cell? C) What is the simplest formula of the compound? Ans. A)The cation at the center of each face is shared by two unit cells.

Hence no. of cations= 6 X ½ = 3 B) The anion at each corner is shared by 8 unit cells . The anion at the center is not shared by any other unit cell. Hence no. of anions= 8 X 1/8 + 1 = 2 C) Since there are 3 cations and 2 anions the simplest formula of the compound is A3B2. Q-10. Copper crystallises in face-centred cubic lattice and has a density of 8.930 g mol-3 at 293 K. Calculate the edge length of unit cell. [At. mass of Cu = 63.5 a.m.u, Avogadro's constant NA = 6.02 x 1023]. Ans. Density = Mass of unit cell/Volume of unit cell 8.93 = (4 x 63.5)/(a3 x 6.02 x 1023). a3 = 47.24 x 10-24 a = 3.6 x 10-8 cm. = 360 pm. Q.11 In corundum, oxide ions are arranged in hcp arrangement and the aluminium ions occupy 2/3 of the octahedral voids. What is the formula of corundum. Ans Number of oxide Ions = 8 x 1/8 = 1 per unit cell Number of Al ions = 2/3 formula Al2/3 O or Al2 O3

OR Rank of hcp = 6 = No. of octahedral voids.

No. of Al ions = 6 = 4

Formula Al4O6 or Al2O3.

Q12. What is the difference between Schottky and Frankel defect?

Ans.

Serial no.

Schottky defect Frankel defect

1. It decreases the density of the crystal.

It does not decrease the density of the crystal.

2. It occurs in compounds with high Co-ordination number.

It occurs in compounds with low number.

3.

It occurs in compounds in which cations and anions are of similar size. Examples: NaCl, KCl, KBr, CsCl.

It occurs in compounds in which cations and anions differ in their size to a large extent. Examples: ZnS, AgCl, AgBr, Agl.

3

2

UNIT 2: SOLUTIONS Q1. 0.1 mole of NaOH is present in 250 ml of solutions. Calculate the molarity(M). Ans: Molarity = No. of moles of solute. X 1000 Vol. Of Solution

= 0.1 X 1000 / 250 = 0.4 moles/litre. Q2. How many moles of urea are present in 5 litres a solution 0.02 M. Ans: Moles of Urea = Molarity X Vol. Of solution = 0.02 X 5 = 0.1 ml. Q3. What is effect of temperature on ? a) Molarity b) Molality. Ans: Molarity: It will decrease with the increase in temperature Molality: It does not change with temperature. Q4. Why molality does not depend upon temperature? Ans: Molality is not related to the volume of solution or solvent but it is related to the mass of solvent which does not change with temperature, hence the molality is independent of temperature. Q5. What is the relative lowering of the vapour pressure? Ans: Po

A - P is known as relative lowering of the vapour pressure where Po A

Po A and P are the vapour pressure of pure solvent and solution

respectively. Q6. What is the effect of the addition of non-volatile solution on the vapour pressure of pure liquid? Ans: It causes the lowering of vapour pressure i.e addition of non-volatile solute will decrease the vapour pressure of the liquid Q7. Define the Raoult’s law. Ans: The vapour pressure of the solution containing the non-volatile solute is directly proportional to the mole fraction of solvent. Q8. Define the molal elevation constant. Ans: Molal elevation constant is equal to the elevation of boiling points produce by one molal solution. Q9. What are the units of molal elevation constant? Ans: K kg mol –1 .

Q10. What is the relation between the elevation of boiling points and molality of the solutions? Ans: Elevation of boiling point is directly proportional to the molality of solutions. Tb ∞ m. Q11. What is the relation between osmotic pressure and relative lowering of vapour pressure of a solution? Ans: The relative lowering of vapour pressure is directly proportional to the osmotic pressure of the solution. P - P s ∞ л P Q12. A and B liquids on mixing produced a warm solution. Which type of deviation is there and why? Ans: Negative type of deviation is present. In the negative deviation the solute-solution (A-A) interaction and solvent-solvent (B-B) interaction will be weaker than solute-solvent(A-B ) interaction. Since the new forces are stronger therefore heat is evolved and solution becomes warm. Q13. Which type of deviation is shown by the solution formed by mixing cyclohexane and ethanol? Ans: Hydrogen bonds exist among the molecules of ethanol when cyclohexane is added to ethanol, the cyclohexane molecule will come in between the ethanol molecules and disturb the hydrogen bonding. The forces between ethanol and cyclohexane will be weaker than the previous forces (H-bonding) hence there will be positive deviation. Q.14 What mass of non - volatile solute (NH2CONH2 ) urea need to be dissolved in 50 gm of water in order to decrease the vapour pressure of water by 25 %. What will be the molality of the solution. Solution : Let the amt. Of urea dissolved in 50 g =x mol Vap. Pressure of solution= 100-25=75 mm Relative lowering in dil.solutions =P0-Ps/P0=XB=Nb/Na 100-75/100=X/50/18 AMT. OF UREA ADDED TO 50 g. =25/36 X60 G/MOL =41.67 g. Molality =no. of moles of solute/amt.of solvent in Kg =25/36/50/1000=13.9m Q.15 Which solution has higher concentration 1 Molar or 1 molar having same solute? Ans. (a) If density of solvent is 1 gm/ml then 1M solution is more concentrated then 1 molal. (b) If density of solvent is less than 1 gm/ml then 1 molar solution is less concentrated than 1 molal. (c) If density of solvent is more than 1 gm/ml then 1 molar solution is more concentrated than 1 molal.

Q16. Why osmotic pressure is a colligative property? Ans-- Osmotic pressure is a colligative property because it depends on the number of the particles. Q17. What happens when red blood cells are placed in 0.1 per cent NaCl solution? Ans—Water from NaCl solution passes into red blood cells (due to endosmosis), therefore they swell up and burst. Q18. Why molality is preferred over molarity? Ans. Molality is unaffected by temperature as it does not involve volume. Q19. Why solutions deviate from their ideal behavior? Ans. Due to the difference in their intermolecular interactions. Q20. What is value of van,t Hoff factor for (a) sodium sulphate (b) K4[Fe(CN)6]? Ans. (a) 3 (b) 5

UNIT -3 ELECTROCHEMISTRY

Q1. What is the effect of temperature on the electrical conduction of (i) Metallic conductor. (ii) Electrolyte conductor? Ans. With increase of temp. the electrical conduction of metals decreases whereas that of electrolyte increases. Q2. How is cell constant calculated from conductance values? Ans. Cell constant = specific conductance/observed conductance. Q3. What is the reference electrode in determining the standard electrode potential? Ans. Normal hydrogen electrode (NHE). Q-4. What are the units of specific conductance ? Ans-. ohm-1 cm-1 or Scm-1 . Q-5 What is the effect of the decreasing concentration on the molar conductivity of a weak electrolyte.? A-ns.Molar conductance of a weak electrolyte increase with decrease of the concentration. Q-6. Write an expression that relates the molar conductivity of a wek electrolyte to its degree of dissociation. Ans-. Degree of dissociation(αe )= Λem / Λωm

Q-7 How does an electrochemical cell help in predicting the feasibility of a redox reaction ? Ans-14.If E0 of the cell is +ve it will yield –ve ∆G0 Value which indicates the reaction is spontaneous. ∆G0 = -n F E0 Q.8 Why for acetic acid cannot be determined experimentally? Ans. Molar conductivity of weak electrolytes keeps on increasing with dilution and does not become constant even at very large dilutions. Q-9 The electrolysis of a metal salt solution was carried out by passing a current of 4 amp for 45 minutes. It resulted in deposition of 2.977 g of a metal. If atomic mass of the metal is 106.4g/mol, calculate the charge on the metal cation. Ans. Quantity of electricity passed = 4 x 45 x 60 C= 10800C. Let charge on metal ion be n+

Quantity of electricity required to deposit 106.4g of metal = n x96500C

Quantity of electricity required to deposit 2.977g of metal= =

nx2700C which is equal to 10800C, n = +4 Q.10. Calculate e.m.f of the cell at 298K.

Cr Cr3+ Fe2+ Fe 0.1M 0.01M

Given: E = -0.75V, E = -0.45V

Ans.

= +0.2606V Q.11. Predict the products of electrolysis of each of the following: An aqueous solution of AgNO3 using silver electrode. An aqueous solution of AgNO3 using platinum electrode. A dilute solution of H2SO4 using platinum electrode. An aqueous solution of CuCl2 using platinum electrode. Ans At cathode At Anode A- Ag Ag+

At cathode At Anode B- Ag O2

At cathode At Anode C- H2 O2

At cathode At Anode

m

4.106

977.296500n

CrCr3

FeFe2

2

32

3

cellcell

Fe

Crlog

6

0591.0EE

D- Cu Cl2

Q-12. Describe the characteristics of variation of molar conductivity with dilution for (a) weak and (b) strong electrolytes. Ans. Strong Mol.con. weak Con 1/2

with increasing dilution the molar conductance increases rapidly because of greater extent of ionization at greater dilution. These are dissociated more or less completely at low dilutions. Their molar conductances therefore show slight increase with increasing dilution. At low conc, the interionic attractions are more and when dilution increases these attractions become weak and the conductance increases gradually. Q-13.Can nickel spatula to be used to stir a copper sulphate solution? Support your answer with a reason. E0Cu2+/Cu =0.34 , E0Ni 2+ /Ni = - 0.25 Ans. Cu2+ + 2e Cu(s) E0 =0.34

Ni(s) Ni 2+ + +2e E0 =-

0.25

Cu2+ (aq.) + Ni(s) Ni 2+ (aq.) + Cu(s)

E0 = 0.59 V Since the value of E0 cell is positive , nickel spatula will displace copper from copper sulphate solution, hence nickel spatula cannot be used for stirring. Q-14. Calculate the maximum possible electrical work that can be obtained from the following cell under standard conditions at 25 0 c. Zn(s) / Zn 2+ (aq.)// Cu2+ (aq) /Cu(s) At 25 0 C E0 Zn 2+ / Zn = -0.76 V , E0 Cu2+ (aq) /Cu(s) = +0.34 V. Ans. Maximum electrical work = + ∆G 0

- ∆G 0 = n F E0

Cu2+ (aq.) + Zn(s) Zn 2+ (aq.) + Cu(s) E0 = E R –E L = 0.34 –(- 0.76) =1.10 - ∆G 0 = 2 X 96500 X1.10 ∆G 0 = -212.3 Kj Q-15. Calculate the equilibrium constant for the reaction at 25 0 c Cu2+ (aq.) + Ni(s) Ni 2+ (aq.) + Cu(s) E0Cu2+/Cu =0.34 , E0Ni 2+ /Ni = - 0.25 R = 8.314 J/K/mol F= 96500 C. Ans. E CELL

E R - E L = 0.34 – (-0.25) = 0.59 V - ∆G 0 =2 X 96500X 0.59

∆G 0 = -113.87 k J = -113870 j ∆G 0 = - 2.303 RT log K -113870 = -2.303 X8.314 X298 log K log K= 113870 / 2.303 X 8.314 X 298 = 19.9597 K= 9.051 X 1019 . Q-16 Write the reaction taking place at cathode and anode during the operation of Daniel cell. Ans. A Zn - Cu cell is known as Daniel cell. The reactions taking place are described as under : Zn (s) Zn2+ + 2e- (Anode) Cu2+ + 2e- Cu (s) (Cathode) Zn (s) + Cu2+ Zn2+ + Cu (s) (Overall reaction) Q-17. A conductivity cell whose cell constant is 3.0/cm is filled with 0.1 M acetic acid solution , its resistance is 4000 ohms. find molar conductance of 0.1 M acetic acid. Degree of dissociation of acetic acid. Molar conductance of acetic acid at infinite dilution is 400 ohm –1 cm2 mol-1. Ans. Λm = 1000/M X 1/R X l/a =1000/.1 X 1/4000 X 3 = 7.5 S cm2 mol-1 α= (Λm)c/ (Λm)ω = 7.5 /400 =0.01875 Q- 18. What is the basic principle of fuel cell ? give two advantages of fuel cell. Ans. FUEL CELL is used to convert the chemical energy of fuel into electrical energy Chemical reactions: Reaction at Cathode 2 2( ) 2 ( ) 4 4 ( )O g H O l e OH aq

Anode 2 22 4 4 4H OH H O e

Overall Reaction 2 2 22 ( ) ( ) 2 ( )H g O g H O l

Advantages: 1-it does not create pollution 2- it has high efficiency. Q19. Describe normal hydrogen electrode and its application. Ans. The construction of normal hydrogen electrode is explained below :

Application : (i) The potential of normal hydrogen electrode has been arbitrarily taken as zero. Thus in combination with other electrodes, cell can be constructed and

the potentials of electrodes can be determined. (ii) Normal hydrogen electrode can be used to determine the pH of a solution. Q-20.What is Kohlrausch`s law ? calculate Λm

∞acetic acid Given Λm

∞ HCl=426 S cm2 mol-1 Λm

∞ NaCl= 126 S cm2 mol-1 Λm

∞ CH3COONa =91 S cm2 mol-1 Ans. : According to this law, molar conductivity of an electrolyte, at infinite dilution can be expressed as the sum of contributions from its individual ions Λm

∞ (CH3COOH) = Λm∞ CH3COO- + Λm

∞ H+ = Λm

∞ CH3COO- + Λm∞ Na+ - [Λm

∞ Na+ +Λm∞ Cl-] + [+Λm

∞ H+ +Λm∞ Cl-]

=Λm∞ (CH3COONa) –Λm

∞ (NaCl)+ Λm∞ (HCl)

=[91-126+426] Scm2/mol =391 S cm2 mol-1 Q-21. State and explain Faraday`s laws of electrolysis? What is the relationship between the chemical equivalent and electrochemical equivalent? Ans. Faraday – First law of electrolysis : The amount of substance deposited during electrolysis is directly proportional to quantity of electricity passed.

W zit Faraday – Second Law : The amount of different substances librated by the same quantity of electricity passing through the electrolytic solution is proportional to their chemical equivalent weights Z = EQUIVALENT WT. IN GM./ 96500 Q.22 Silver is electrodeposited on a metallic vessel of surface area 800cm2 by passing a current of 0.2 ampere for 3 hours. Calculate the thickness of the silver deposited. Given the density of silver as 10.47 gcm-3 (atomic mass of Ag = 107.92 U). Ans The quantity of electricity = 0.2amp x3 x60 x60sec= 2160C.

2160C of electricity deposit Ag= = 2.42g

Mass= area x thickness x density 2.42g= 800cm2 x thickness x 10.47gcm-3

Hence thickness = 2.89 x10-4 cm.

96500

216095.107

UNIT 4 CHEMICAL KINETICS

Q1. If K = 102 (rate constant), will the reaction go to completion? Ans. Yes. Q2. What is the order of a reaction when ethyl Acetate is hydrolysed? Ans2. CH3COOC2H5 + H2 3COOH + C2H5OH It is a Pseudo-unimolecular reaction. Q3. Why is a finely divided substance more efficient catalyst? Ans3. Because the surface area of a finely divided substance is very large. Q4. What are the units of rate constant for a zero order reaction? Ans4. Time-1 mole+1 litre-1. Q5. What are the units of rate constant for a second order reaction? Ans5. Litre+1 mole-1 sec-1.

Q6. The rate of a reaction at different times (- dx/dt) is found as Time (minutes) Rate (in mol s-1) 0 2.80 x 10-2 10 2.78 x 10-2 20 2.81 x 10-2 What is the order of a reaction?

Ans6. Since rate of reaction is constant, it is a zero order reaction.

Q7. A sample of a radio-active substance takes 80 days to lose half of its radio activity. What time will be required to lose 1/8 of its radio activity. Ans7. 80 + 80 + 80 = 240 days. Q8. If half life period of a first order reaction is 80 days, what is the value of rate constant? Ans8. K = 0.693/80 d-1 = 0.0693/8 = 0.00866 d-1. Q9. What is the order of a reaction? If rate = K|A|1/2.|B|-3/2.|C|2 Ans9. n = 1/2 - 3/2 + 2/1 = 2/2 = 1. Q.10 A gas decomposition of AB follows the rate law; rate = K 3/4 .Write

units of K. Ans Mol-1/4 L ¼ time-1

AB

Q.11 If half-life of a reaction is inversely proportional to initial concentration of the reactant, what is the order of the reaction? Ans. Second order Q.12 What is the order of the reaction represented in the following graph? Rate conc Ans. First order Q-13 What is the order of the reaction represented in the following graph? RATE C Ans ZERO ORDER. Q-14. For the assumed reaction : X2 + 2Y2 2XY2 ,write the rate equation in terms of rate of disappearance of Y2. Ans.-d[Y2]/dt= k[X2] [Y2]2

Q15. A first order reaction takes 69.3 minutes for 50% completion. How much time will be needed for 80% completion? Ans. t1/2 = 69.3 minutes k = 0.693/69.3 = 10-2 min-I Now t = 2.303/k log a/(a - x) = 2.303/10-2 x 0.6990 = 160.9 minutes. Q16. In a certain first order reaction, half the reaction was decomposed in 500 seconds. How long will it take for 90% completion? Ans. K = 0.693/(t1/2) = 0.693/500 = 0.001386 sec-1 Now, t = 2.303/0.001386 x log (100)/10 = 1661 sec. Q17. Find the value of rate constant for the reaction A + B AB if the rate of the reaction is 5 x 10-5 (mole/l)m-1 and |A| and |B| are 0.05 and 0.01 mol respectively.

Ans. A + B AB R = K[A][B] K = (5 x 10-5)/(0.05 x 0.01) = 10-1 = 0.1 (mol/l)-1min-1 Q.18 Following reaction takes place in one step 2NO (g) + O2 (g) 2 NO2 (g) How will the rate of above reaction change if the volume of the reaction vessel is diminished to one third of its original volume? Will there be any change in order of reaction with the reduced volume? Ans. Rate of the reaction will become 27 times, order will remain unchanged Q-19 The rate of decomposition of a substance A becomes 8 times when its concentration is doubled. What is the order of the reaction? Ans. Rate= k [A]n

r = k [A]n i

8r= k [2A]n ii

dividing ii by I 8r/r = k [2A]n / k [A]n 8 =[2]n n=3 order is 3 Q.20 The kinetics of the reaction: A + 2B Products; obeys the rate equation Rate = k X Y. For it, find

Order of the reaction Apparent molecularity of reaction Order of reaction when B is in large excess. Ans a) X+Y b) 3 c) X Q.21 In the start of summer, a given sample of milk turns sour at room temperature (300K) in 48 hours. In a refrigerator at 275K, milk can be stored three hours longer before it sours. Calculate activation energy of the souring of milk.

Ans. = 3, Ea = 30.146 kJ mol –1

Q.22. Decomposition of SOCl2 at 600K in gas phase is a first order reaction

with rate constant of 2.3 10-5 sec –1 .Calculate the % of SOCl2 that would remain undecomposed after 200 minutes of reaction time.

Ans. log =0.12

A B

275

300

k

k

A

A

=1.318

% of SOCl2 that remains undecomposed is = 75.8%

Q-23. Consider the following data for the reaction A + B Products Determine the order of the reaction w.r.t. A and B and the overall order of reaction.

s.n. Initial con.[A] mol/l

Initial con.[B] Mol/l

INITIAL RATE(mol/l/sec)

1. 0 2. 3.

0.10 0.20 0.20

1.0 1.0 2.0

2.1 x 10-3

8.4 x 10-3 8.4x10-3

A-6) The rate law may be written as Rate = k[A]p [B]q

R1 = k[0.10]p [1]q 2.1 x 10-3= k[0.10]p [1]q (i)

R2 = k[0.20]p [1]q

8.4 x 10-3 = k[0.20]p [1]q (2) R3= k[0.20]p [2]q

8.4x10-3 = k[0.20]p [2]q (3)

dividing eq. (2) by (1) 8.4 x 10-3 /2.1 x 10-3= k[0.20]p [1]q /[0.10]p [1]q or 4 = 2 p

p =2 i.e. order w.r.t A =2 SIMILARLY by dividing (3) by(2) we get Q=0 i.e. w.r.t B =0 Overall order = 2+0= 2

A

A

318.1

100

UNIT 5 SURFACE CHEMISTRY

Q1. What is adsorption? Ans. The phenomenon of attracting and retaining the molecules of a substance on the surface of a solid or a liquid is called adsorption. Q2. What is adsorbate and adsorbent? Ans. The substance adsorbed on the surface is called adsorbate and the substance on which it is adsorbed is called adsorbent. Q3. What is desorption? Ans. The reverse process i.e., removal of adsorbed substance from the surface is called desorption. Q4. What is occlusion? Ans. The adsorption of gases on the surface of metals is called occlusion Q5. How is the physically adsorbed mass of a gas on a unit mass of an adsorbent related to pressure of the gas? Ans. x/m = kp Q6. Out of NH3 and N2 which gas will be adsorbed more readily on the surface of charcoal and why? Ans. NH3 as it is easily liquefiable gas. Q7. 0.1 M AlCl3 solution is more effective than 0.1 M NaCl solution in coagulating As2S3 sol while 0.1 M AlCl3 is less effective than0 .1 M Na3PO4 solutions in coagulating ferric hydroxide sol. Ans. AlCl3 – contains Al+3 ion and NaCl – contains Na+ ions Greater the charge on the opposite ion greater will be coagulation.(Hardy Schulze Rule) Similarly PO4

-3 ion is more effective than Cl- ion in coagulating Fe(OH)3 solution. Q8. Why delta is formed at the point where river enters the sea? Ans Due to coagulation of colloidal clay. Q9. Preparation of colloid such as protein can be made more stable if the colloid is dialysed? Why Ans. This is because dialysis helps in removing undesirable ions from colloidal preparation in order to stabilize the colloid. Q10. Give reasons for the development of electrical charge on colloidal particles? Ans : Due to preferential adsorption of common ion

Q-11. Write Freundlich adsorption isotherm equation. Ans. Log x/m =log k+ 1/n log p. Q-12. What is Kraft` temperature? Ans. Micelles form ionic surfactants only above a certain temperature called as Kraft`s temperature. Q-13.What is the difference between a sol and a gel? Ans. Both are colloidal solutions. A sol has a solid dispersed phase in liquid as dispersion medium. A gel has liquid as dispersed phase and solid as dispersion medium. Q14. Explain electrophoresis. Ans. Movement of colloidal particles under the influence of electric fields is termed as electrophoresis. When electric current is passed through a colloidal solution, colloidal particles are discharged at an electrode depending upon the type of charge on them. In this way nature of the charge on colloidal particle can be ascertained.

Q15. What is Brownian movement? What is its advantage? Ans. The movement of colloidal particles in Zig-zag way, when viewed through ultra-microscope is called Brownian movement. This is due to unequal bombardment of colloidal particles by the particles of dispersion medium.

As a result of Brownian movement colloidal solution is stable and colloidal particles do not settle at the bottom of vessel or it is not coagulated Q16. What is peptization? How a colloidal solution can be prepared with its help? Ans. Process of conversion of precipitate to colloid is termed as peptization. By adding Fe+++ in a suspension of ferric hydroxide colloidal solution of hydrated ferric oxide can be prepared.

+ Fe+++

Q-17. What are macromolecular colloids? Ans. When certain substances having big size molecules, called macromolecules having large molecular masses are dissolved in a suitable liquid, they form a solution in which the molecules of the substance. i.e., the dispersed phase particles have size in the colloidal range. Such substances are called macromolecular colloids. For example, starch, cellulose, proteins and gelatin. Q-18. What is positive and negative adsorption? Ans. Positive adsorption: When there is a more concentration of adsorbate at the surface of the adsorbent than that in the bulk, it is called positive adsorption. Negative adsorption: When there is a less concentration of a solute (Adsorbate) on the surface layer than in the bulk part of the solution it is called negative adsorption. Q-19. How the following sols are produced? Sulphur sol collodion Ans. By bubbling H2S gas through an oxidizing agent like bromine water. H2S + Br2 2HBr + S cellulose nitrate colloid can be prepared by dispersing it in a mixture of ethanol and ether. This is commercially called as collodion. Q20- Differentiate between oil in water and water in oil type of emulsions. Ans-OIL IN WATER- 1) when diluted with water no separate layer is formed. 2) Less viscous and more conductive. WATER IN OIL – 1) When diluted with water show separate layer 2) More viscous low conductivity. Q21 Name the dispersion medium in hydrosol, alcosol, benzosols and aerosol. Ans- WATER, ALCOHOL, BENZENE AND GASES. Q-22. Describe Bredig`s arc method for the preparation of colloidal solutions. Ans. Electric arc is struck between the electrodes of the metal immersed in the dispersion medium. The intense heat produced vaporizes some of the metal, which then condenses to form particles of the colloidal size. Q-23. Give some importance of adsorption in daily life. Ans. 1) Gas masks 2) decolorizers 3) volumetric analysis 4) removes hardness of water. 5) Chromatography 6) heterogeneous catalysis. 7) De-humidizers. Q24. Explain Tyndall effect. What is its significance? Ans. (1) If a beam of light is passed through a sol, the beam is visible due to scattering of light by the colloidal particles. It is called Tyndall effect and the illuminated path as Tyndall cone or beam. The intensity of scattered light

depends upon the difference of the refractive index of the dispersed phase and that of dispersion medium. Lyophobic sols have appreciable difference hence show the Tyndall effect, while in the lyophilic sols this difference is very small. So the Tyndall effect is very small. Q25. What is a colloidal state? Give differences between a true solution, colloidal solution and a suspension. Ans. (a) Colloids: Property True solution Colloidal state Suspension (i) Nature Homogenous Heterogeneous Heterogeneous

(ii) Particle size Less than 1 nm = 10-9 m

1 nm to 100 nm Greater than 100 nm.

(iii) Diffusion Diffuse rapidly Diffuse slowly Do not diffuse

(iv) Visibility Completely invisible and do not scatter light

Invisible to naked eye but scatter light

Visible to naked eye

(v) Tyndall effect Do not exhibit Exhibit Do not exhibit

(vi) Coagulation

These can be precipitated by adding suitable electrolyte

These can also be coagulated by adding suitable electrolyte

These can or cannot coagulated

Q-26. What are lyophilic and lyophobic colloids? Give differences between them. Ans. Lyophilic Colloids Lyophobic Colloids (i) These are prepared by shaking or warming the dispersed phase in dispersion medium and do not require any electrolyte for their Stabilizations

These are difficult to prepare. Special methods are used. Addition of stabilizer is a must for their stabilization.

(ii) These are reversible (ii) These are irreversible

(iii) These are very stable and are not easily coagulated

(iii) These are generally unstable and are coagulated by the addition of an electrolyte

(iv) Particles carry no or very little charge depending on PH of the solution

(iv) Colloidal particles carry +ive or -ive charge

(v) Viscosity is much higher than medium

(v) Viscosity is same as that of the medium.

(vi) Surface tension is usually less than medium

(vi) Surface tension is usually the same as that of medium

(vii) Migration in electric field : Particles may or may not migrate in an electric field

(vii) Colloidal particles migrate towards cathode or anode under the influence of an electric field depending upon charge on them

(viii) Visibility : Particles are seen under an ultra-microscope

(viii) Particles are invisible, but can be detected by ultra-microscope

(ix) Tyndall effect : Less distinct (ix) More distinct (x) Cataphoresis : Measurement is difficult

(x) Cataphoresis can be measured easily.

(xi) A Large amount of electrolyte is needed for their coagulation.

(xi) A small amount of electrolyte is needed for their coagulation.

(xii) Examples : Mostly organic substances e.g. starch gums, gelatin etc.

(xii) Generally in organic substances e.g. metal sols, sulphides and oxides sols.

(xiii) These show high conductivity which can be measured

(xiii) Conductivity can rarely be measured over a considerable range of conc.

UNIT 6 General Principal and isolation of elements Q.1- What is the role of cryolite in electrometallurgy of aluminium? Ans- alumina cannot be fused easily because of high melting point. Dissolving of alumina in cryolite furnishes Al3+ ions, which can be electrolyzed easily. Q.2 Describe the method of refining of nickel. Ans- Mond Process, Ni is heated in a stream of CO forming a volatile complex, which then decomposes at higher temperature to give Ni. At 330-350K: - Ni + 4CO → Ni (CO) 4 At 450-470K Ni (CO)4 → Ni + 4 CO Q.3- What is Zone Refining? Explain with example. Ans- It is based on the principal that impurities are more soluble in molten state of metal than solidified state. In this method, a rod of impure metal is moved slowly over circular heater. The portion of the metal being heated melts & forms the molten zone. As this portion of the rod moves out of heater, it solidified while the impurities pass into molten zone. The process is repeated to obtain ultrapure metal and end of rod containing impure metal cutoff. Q.4- Describe the method of refining of Zirconium and Titanium. Ans- Van Arkel process – it is used for obtaining ultrapure metal. The impure metal is converted into volatile compound, which then decomposes electrically to get pure metal. At 850K: - Zr impure) + 2 I2 → ZrI4 At 2075K:- ZrI4 → Zr (pure) + 2 I2 Q.5. Name one each of a) Acidic flux b) Basic flux Ans: a) Silica b) Lime Q.6. What is basic difference between Calcination and Roasting? Ans: Calcination the concentrated ore is heated either in the absence or in the limited supply of air on the othar hand roasting is carred out in excess of air. Q.7. What is the significance of Leaching in the extraction of Aluminium? Ans: Leaching is significant as it helps like SiO2, Fe2O3 etc from the Bauxite ore.

Q.8. State the role of

I. Depressant in froth flotation process. II. Silica in the metallurgy of copper. III. Graphite rod in the electrolytic reduction of alumina.

Ans.

i. It prevent certain sulphides like ZnS to enter the froth in presence of Pbs, Therefore, Helps in their separation. Sodium Cyanide is used as depressant in the separation of ZnS and PbS.

ii. It acts as flux. iii. Graphite rod acts as anode in the extraction of aluminium.

UNIT 7 p-Block Elements

1. N2 is less reactive at room temperature? because of high bond

dissociation enthalpy

2. PCl5 more covalent than PCl3? because of pentavalent cation has more

polarizing power than PCl3

3. BiH3 strogest reducing agent ammomgest all hydrides of grp 15 element

because of lowest bond dissociation enthalpy due to larger size of Bi

4. Bond angle in PH4 + is higher than that in PH3. Because PH4+ has sp3

hybridization with no lone pair, whereas PH3 has pyramidal shape with

one lone pair.

5. NH3 form H – bond but PH3 does not. because more electronegativity

and small size of N

6. PCl5 is ionic in nature in the solid state. Because it exist as [PCl4]+ [PCl

7. P4 is more reactive than N2 due to larger size of phosphorous than

nitrogen.

8. All the P-Cl bonds in PCl5 are not equivalent.Due to BP – BP electron

repulsion.

9. Bi (v) strong oxidising agent than Sb(v) Due to inert pair effect

10. BiH3 strongest reducing agent amongst all the hydride of group 15

elements Because of its lowest bond dissociation enthalpy

11. R3P= O exist but R3N= O does not. Because absence of d – orbital

12. Nitrogen show catenation properties less than phosphorus because of N –

N bond is weaker than P – P bond due to e – e repulsion due to small size.

13. H3PO3 is dibasic and H3PO4 is tribasic. Because of presence of ionisable

two P – OH bond in H3PO3 whereas presence of ionisable three P – OH

bond in H3PO4

14. Nitrogen does not form pentahelides. Because of absence of d – orbital

and small size

15. NO2 dimerise Because of presence of unpaired or odd electron on N

16. Stability of +5 oxidation State decreases and +3 oxidation state stability

increases down the group of 15 group elements. Because of inert pair

effect.

17. Which neutral molecule would be isoelectronic with ClO- ? OF2 and ClF

18. H2O liquid but H2S gas because of presence of H – Bonding in water

whereas H2S is not

19. O3 act as powerful oxidizing agent because of libration of nascent oxygen.

20. H2S is less acidic than H2Te. Because of S atomic size less than Te. Therefore S cannot

release H+ ion easily.

21. SF6 is kinetically an inert. Because of steric effect

22. H2S is more acidic than H2O. Because of S atomic size more than O. Therefore S can

release H+ ion easily.

23. All the bonds in SF4 are not equivalent Because of all bonds are in different planes.

24. Interhalogen compound are more reactive than halogens Because of interhalogen

bond are weaker or low bond dissociation enthalpy than halogen.

25. ICl more reactive than I2 Because weaker bond in ICl than I2

26. Halogens are strong oxidizing agents Because high electronegativity, therefore can

gain electrons easily.

27. F2 is most reactive of all the other halogens. Because inter electronic repulsion due

to small size

28. F2 is stronger oxidizing agent than Chlorine. Because inter electronic repulsion due

to small size of fluorine

29. F2 has low bond enthalpy than Cl2. Because inter electronic repulsion due to small

size of fluorine

30. Negative value of electron gain enthalpy of Fluorine less than Chlorine. because

inter electronic repulsion due to small size of fluorine

31. Noble gas Xenon forms known Compounds. Because of larger atomic size and high

polarizing power of Xenon.

32. Noble gases have comparatively larger atomic size. Because inter electronic repulsion

and low ionization enthalpy

33. Noble gas has very low boiling points. Because of presence of weak vander waal force

of attraction and larger size.

34. Draw the structure of XeOF4, XeO3, XeF4, XeF6, BrF3, H3PO3, H2SO5, H2S2O7

UNIT 8 d- and f- Block elements

Q.1. Why d-block elements or their compounds are good catalytic agents?

Ans. Since a good catalyst must provide a large surface to adsorb or to form activated

complex with reactants. d-block elements or their compounds have this unique property and

are thus good catalytic agents.

Q.2. Give comparison between lanthanides and actinides. Ans.

Lanthanides Actinides

1. These show lanthanide contraction. 1. These show actinide contraction.

2. These are separated by ion-exchange

behaviour. 2. These also show ion exchange behaviour.

3. Most of lanthanides are colourless. 3. Most of the ions of actinides are coloured.

4. These have lesser tendency to form

complexes. 4. These have greater tendency to form

complexes.

5. Their magnetic behaviour can be easily

explained. 5. Their magnetic behaviour cannot be easily

explained.

6. These show mostly +3 oxidation state. 6. These show even higher oxidation state +3

to +7.

Q.3. (a) Why transition elements exhibit:-

(i) variable oxidation states (ii) form complexes (iii) form coloured compounds (iv)

and show catalytic properties. Ans. (a) (i) incomplete d-sub shell and participation of ns and (n - 1)d electrons in bond

formation

(ii) small size, high nuclear charge, presence of valence vacant d-orbitals, d-block elements

or their ions accommodate lone pairs of electrons from electron pair donor species and thus

complexes are formed for example

(iii) The Transition metal ions have unpaired d-electrons and d-d transition

(iv) the presence of unpaired electrons in d-orbitals.

Q.4.Most of the transition elements are paramagnetic?

Ans. due to presence of one or more unpaired electrons in atomic orbitals. Q5.. What is the lanthanoid contraction? What are its causes and consequences? Ans. Lanthanoid contractions – The cumulative effect of the regular decrease in size or radii of Lanthanoid with increase in atomic number is called Lanthanoid contraction. Causes - The shape of f orbitals is diffused. They have poor shielding effect due to which the effective nuclear charge increase with increase in atomic number. This causes a decrease in atomic radii Consequences – Due to Lanthanoid contraction- 1. Radii of the members of the third transition series is similar to those of second transition series. 2. It becomes difficult to separate Lanthanoids.

Q.6. Assign reasons for the following:

(i) The enthalpies of atomisation of transition elements are high.

(ii) The transition metals and many of their compounds act as good catalysts.

(iii) Scandium (Z = 21) does not exhibit variable oxidation states and yet it is regarded

as a transition element.

Ans. (i) This is because transition metals have strong metallic bonds as they have a large

number of unpaired electrons.

(ii) The catalytic activity of transition metals is attributed to the following reasons:

(a) Because of their variable oxidation states transition metals form unstable intermediate

compounds and provide a new path with lower activation energy for the reaction.

(b) In some cases, the transition metal provides a suitable large surface area with free

valencies on which reactants are adsorbed.

(iii)This is because scandium has partially filled d orbitals in the ground state

(3d1 4s

2).

Q.7.Zinc, cadmium and mercury are not considered as transition metals. Why?

Ans. Zinc, cadmium and mercury have fully filled d10

configuration. Therefore they are not

considered as transition metal.

UNIT 9 CO ORDINATION COMPOUNDS

Q.1.Why Coordination compounds are coloured? Ans. Due to d-d transition. Q.2What is Homoleptic complexes? The complexes in which a metal is bound to only one kind of donor groups. E.g., [Co(NH3)6]

3+ Q.3.Write IUPAC name of the following [CO(NH3)6Cl3 Ans. Hexaaminocobalt(III)chloride

Q.4. Draw figure to show splitting of d- orbitals in an octahedral crystal field.

Q.5. Explain structure of[ [CO(NH3)6

3+ ] on the basis of valence bond theory.Structure on the basis of valence bond theory

In the diamagnetic octahedral complex,[Co(NH3)6]3+ , the cobalt ion is in +3 oxidation state and has the electronic configuration 3d6. The hybridisation scheme is as shown in diagram.

Six pairs of electrons, one from each NH3 molecule, occupy the six hybrid orbitals. Thus, the

complex has octahedral geometry and is diamagnetic because of the absence of unpaired

electron. In the formation of this complex, since the inner d orbital (3d) is used in

hybridisation, the complex, [Co(NH3)6]3+ is called an inner orbital or low spin or spin paired

complexes.

Q.6. [NiCl4]2− is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why?

Answer Though both [NiCl4]2− and [Ni(CO)4] are tetrahedral, their magnetic characters are different. This is

due to a difference in the nature of ligands. Cl− is a weak field ligand and it does not cause the pairing of

unpaired 3d electrons. Hence, [NiCl4]2− is paramagnetic.

In Ni(CO)4, Ni is in the zero oxidation state i.e., it has a configuration of 3d8 4s2.

But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also, it causes the

4s electrons to shift to the 3d orbital, thereby giving rise to sp3 hybridization. Since no unpaired electrons

are present in this case, [Ni(CO)4] is diamagnetic.

Q.7.Explain [Co(NH3)6]

3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex.

Ans.

[Co(NH3)6]3+

[Ni(NH3)6]2+

Oxidation state of cobalt = +3 Oxidation state of Ni = +2

Electronic configuration of cobalt = d6 Electronic configuration of nickel = d

8

NH3 being a strong field ligand causes the

pairing. Therefore, Ni can undergo d2sp

3

hybridization.

Hence, it is an inner orbital complex.

If NH3 causes the pairing, then only one 3d

orbital is empty. Thus, it cannot undergo d2sp

3

hybridization. Therefore, it undergoes sp3d

2

hybridization.

Hence, it forms an outer orbital complex.

Q.8. Draw the structures of optical isomers of:

(i) [Cr(C2O4)3]3− (ii) [PtCl2(en)2]2+ (iii) [Cr(NH3)2Cl2(en)]+

Answer

(i) [Cr(C2O4)3]3−

(ii) [PtCl2(en)2]2+

(iii) [Cr(NH3)2Cl2(en)]+

Q.9. Draw all the isomers (geometrical and optical) of:

(i) [CoCl2(en)2]+ (ii) [Co(NH3)Cl(en)2]2+ (iii) [Co(NH3)2Cl2(en)]+

Answer (i) [CoCl2(en)2]+

In total, three isomers are possible.

(ii) [Co(NH3)Cl(en)2]2+

Trans-isomers are optically inactive.

Cis-isomers are optically active.

(iii) [Co(NH3)2Cl2(en)]+

Q.10. What is spectrochemical series? Explain the difference between a weak field ligand and a

strong field ligand.

Answer. A spectrochemical series is the arrangement of common ligands in the increasing order

of their crystal-field splitting energy (CFSE) values. strong field ligands cause higher splitting in the

d orbitals than weak field ligands.

I− < Br−< S2−< SCN−<Cl−< N3< F−< OH−< C2O42−∼ H2O < NCS−∼ H−< CN− < NH3< en ∼ SO3

2−<

NO2−<phen< CO

CHEMISTRY MIND MAP

ORGANIC CHEMISTRY

NAME REACTIONS

1. Sandmeyer’s reaction

Benzene diazonium salt with cuprous chloride or cuprous bromide

results in the replacement of the diazonium group by –Cl or –Br.

2. Wurtz reaction

Alkyl halides react with sodium in dry ether to give hydrocarbons

containing double the number of carbon atoms present in the halide.

2 RX+ 2 Na R-R +2 NaX

Wurtz-Fittig reaction

A mixture of an alkyl halide and aryl halide gives an alkylarene when treated with sodium in dry ether.

3. Fittig reaction

Aryl halides when treated with sodium in dry ether give diphenyl.

4. Finkelstein reaction

R-X + NaI → R-I + NaX X=Cl, Br

5. Swarts reaction

H3C-Br +AgF → H3C-F + AgBr

6. Kolbe’s reaction

When phenol reacts with sodium hydroxide, phenoxide ion

generated undergoes electrophilic substitution with carbon dioxide

and forms salicylic acid.

8. Reimer-Tiemann reaction

On treating phenol with chloroform in the presence of sodium hydroxide,

a –CHO group is introduced at ortho position of benzene ring.

9. Williamson synthesis

An alkyl halide is treated with sodium alkoxide to form ether.

CH3Br + C2H5ONa CH3 – O - C2H5

10. Rosenmund reduction.

Acyl chloride (acid chloride) is hydrogenated over catalyst, palladium on

barium sulphate to give aldehyde..

11. Clemmensen reduction

Aldehydes and ketones are reduced to alkanes on treatment with zinc- amalgam

and concentrated hydrochloric acid.

CH3 –CHO CH3 - CH3

12. Wolff-Kishner reduction

Aldehydes and ketones are reduced to alkanes on reaction with hydrazine

followed by heating with potassium hydroxide in ethylene glycol.

CH3 –CO - CH3 CH3 –CH2 - CH3

13. Aldol condensation

Aldehydes and ketones having at least one α-hydrogen undergo a reaction in

the presence of dilute alkali form β-hydroxy aldehydes (aldol) or β-hydroxy

ketones

(ketol), respectively.

14. Hell-Volhard-Zelinsky reaction

Carboxylic acids having an α-hydrogen are halogenated at the α-position on

treatment with chlorine or bromine in the presence of small amount of red

phosphorus to give α-halocarboxylic acids.

15. Hoffmann bromamide degradation reaction

Primary amines can be prepared by treating an amide with bromine in an

aqueous or ethanolic solution of sodium hydroxide.The amine so formed contains

one carbon less than that present in the amide.

Important reactions

i) Chloroform is slowly oxidised by air in the presence of light to an extremely

poisonous gas, carbonyl chloride, also known as phosgene. It is therefore

stored in closed dark coloured bottles completely filled so that air is kept out.

ii) Phenol is manufactured from the hydrocarbon, cumene (isopropylbenzene)

iii) Carbylamine reaction

Aliphatic and aromatic primary amines on heating with chloroform and ethanolic

potassium hydroxide form isocyanides or carbylamines which are foul smelling

substances. Secondary and tertiary amines do not show this reaction.

iv) Coupling reaction

Benzene diazonium chloride reacts with phenol or aniline to form azo

dyes.

S

NO

NAME OF

THE

COMPOUND

CHEMICAL

REAGENT

OBSERVATION PAIRS OF COMPOUNDS THAT CAN BE

DISTINGUISHED

1 R - X Aqueous NaOH &

AgNO3 solution

Precipitate

AgCl – White

AgBr – Pale

yellow

AgI – Dark

yellow

1) Ethyl chloride & chlorobenzene

Ethyl chloride gives white ppt

Chlorobenzene does not give ppt.

2) Cyclohexylbromide & bromobenzene

Cyclohexylbromiide gives pale yellow ppt

Bromobenzene does not give the test.

3) Benzyl chloride & chlorobenzene

Benzyl chloride gives white ppt

Chlorobenzene does not give ppt.

4) Vinyl iodide & Allyl iodide

Vinyl iodide gives pale yellow ppt.

Allyl iodide does not give the test.

2 R - OH Lucas test – conc

HCl & anh ZnCl2

Iodoform test –

NaOH &I2

Note: When both

the alcohols given

are primary / sec

apply this test

10 – Does not

react.

20 – Forms

turbidity after

few min

30 - Forms

turbidity

immediately

Yellow ppt of

iodoform (CHI3)

To distinguish primary , sec & ter alcohols

Primary – Methanol, Ethanol, n – alkyl

alcohols / alkan – 1-ol, benzyl alcohol

Secondary – Isopropyl alcohol, alkan – 2 –

ol

Tertiary – ter- buytl alcohol (2- methyl –

propan – 2- ol)

Alcohols with CH3 – CH – group give this

test

OH

1) Ethyl alcohol & methyl alcohol (both are

10)

Ethyl alcohol gives yellow ppt

Methyl alcohol does not give this test.

2) Pentan – 2- ol & Pentan – 3- ol (both are

20)

DISTINCTION TESTS

3 Phenol Neutral ferric

chloride

Violet colour 1) Phenol & ethyl alcohol

Phenol gives violet colour

Ethyl alcohol does not give this test

4 Aldehydes Tollens test –

ammoniacal

AgNO3 , warm

Iodoform test –

NaOH &I2

Note: When both

the compounds

given are

aldehydes/

ketones apply this

test

Silver mirror

Yellow ppt of

iodoform (CHI3)

Aldehydes give positive test.

1. Acetaldehyde (Propanal) & Acetone

(Propanone)

Acetaldehyde (Propanal) gives Ag mirror

Acetone (Propanone) does not give this

test.

Ald. & ket. with CH3 – C – group give

this test

O

1) Acetaldehyde & benzaldehyde

Acetaldehyde gives yellow ppt.

Benzaldehyde does not give this test.

1) Acetaldehyde & formaldehyde

Acetaldehyde gives yellow ppt.

Formaldehyde does not give this test.

3) 2 – Pentanone & 3 – Pentanone

2 – Pentanone gives yellow ppt.

3 – Pentanone does not give this test.

4) Acetophenone & benzophenone

Acetophenone gives yellow ppt.

Pentan – 2- ol gives yellow ppt

Pentan – 3- ol does not give this test.

Benzophenone does not give this test.

5. Carboxylic

acids

Formic acid

( Methanoic

acid)

Sodium

bicarbonate test

Tollens test

Effervescence

due to evolution

of CO2 gas

Effervescence

due to evolution

of CO2 gas

All carboxylic acids give this test

HCOOH is the only carboxylic acid that

gives this test

HCOOH and CH3COOH

HCOOH gives this test . CH3COOH does not

give this test

6 Amines

Primary

amines

Aniline

Hinsberg test –

Benzene sulphonyl

chloride + NaOH

Carbylamine test –

Alcoholic KOH +

CHCl3

Azodye test –

NaNO2 + HCl +

phenol ( 00 to 5 0

C)

10 –Soluble

20 – Insoluble

30 – does not

react with

Hinsberg

reagent

Very unpleasant

smelling

isocyanide gas

evolves

Reddish orange

dye

To distinguish primary , sec & ter

amines

Alkyl amine – 10

Dialkylamine -20 (N- alkyl alkanamine)

Trialkylamine( N,N – dialkyl alkanamine - 30

To distinguish primary amines from

other amines

1)Ethylamine & Diethylamine

Ethylamine gives this test.

Diethylamine does not give this test

2) Aniline & Dimethylamine

Aniline gives this test.

Diethylamine does not give this test

Aniline & Benzylamine

Aniline gives this test

Benzylamine does not give this test.

MECHANISM FOR PREPARATION OF ALCOHOL FROM ALKENE

MECHANISM FOR PREPARATION OF ETHER FROM ALCOHOL

MECHANISM FOR DEHYDRATION OF ALCOHOL TO ALKENE

ACIDITY OF PHENOL OR o-p DIRECTING NATURE

STABILITY OF PHENOXIDE ION

UNIT 10 HALOALKANES & HALOARENES

1.What are ambident nucleophiles ? Give two eg.

Nucleophiles that possess two nucleophilic centres. Eg CN- & NO2 -

2. Why is chloroform stored in closed dark colour bottle ?

In the presence light & air it forms poisonous gas , phosgene.

2CHCl3 +O2 2COCl2 + 2HCl

3. Alkly halides though polar are immiscible with water. Why ?

Because they do not form intermolecular H bond with water.

4. Arrange the following set of compounds in order of increasing boiling

points.

1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.

Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane

NOTE : SN1 : 10 < 20 < 30 Tertiary cabocation is more stable

SN2 : 30 < 20 < 10 Steric hindrance is less in primary

5. Which one in the following pairs would undergo SN2 reaction faster?

Because iodine is a better leaving group.

6. Which one in the following pairs would undergo SN1 reaction faster?

7. An alkyl halide having molecular formula C4H9Cl is optically active

.What is its name?

Ans.= 2-chlorobutane

8. Explane why haloarenes are less reactive towards nucleophilic

substation reactions than Haloalkanes.

(a) Due to resonance in benzene ring there is partial double bond in C-X bond which is difficult to break.

(b) It is difficult for electron rich nucleophiles to approach electron rich

benzene ring in haloarenes.

9. How do the products differ with ethyl bromide react separately with ;

(a) Alc. KCN and Alc. AgCN. B) Aq. KOH and Alc. KOH.

Ans. (a)

Alc.KCN C2 H5 CN+ KBr

C 2H5Br

Alc.AgCN C2 H5 NC+ AgBr

(b)

Aq.KOH C2H5OH + KBr

C2H5 Br

Alc.KOH CH2=CH2 + KBr +H2O

10(a) Give the IUPAC name of the following ;

H3C CH3

H H CH3 Br

(b)Write Wurtz fittig reaction .

Ans. (a) 4-Bromo-3-methyl –pent-2-ene.

(b) 2Na + X-R R + 2NaX

Dry ether

11. Arrange the following compounds in order of reactivity towards SN2

displacement:

2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane

ANS:- 2-Bromo-2-methylbutane (ter) < 2-Bromopentane(sec) < 1-

Bromopentane (pri)

Because of less steric hindrance in 1-Bromopentane

12. Convert the following:-

(a) ethanol to propanenitrile (b) propene to 1-propanol

(c) 1-Bromopropane to 2-Bromopropane

Ans a) PCl5 KCN

CH3CH2OH CH3CH2Cl CH3CH2CN

HBr/peroxide KOH

b)CH3CH=CH2 CH3CH2CH2Br CH3CH2CH2OH

alc. KOH HBr

c) CH3CH2CH2Br CH3CH=CH2 CH3CHBrCH3

13.(a)Convert Nitrobenzene to m-Bromoiodobenzene.

X

(b)Draw the structure of the product of the Bromination of

following compounds in presence of FeBr3.

(b)Because –NO2 deactivates the ring but OCH3 activates.

14. Convert the following:-

(a) ethanol to propanenitrile

(b) propene to 1-propanol

(c) 1-Bromopropane to 2-Bromopropane

Ans a) PCl5 KCN

CH3CH2OH CH3CH2Cl CH3CH2CN

HBr/peroxide KOH

b)CH3CH=CH2 CH3CH2CH2Br CH3CH2CH2OH

alc KOH HBr

c) CH3CH2CH2Br CH3CH=CH2 CH3CHBrCH3

UNIT 11 ALCOHOLS PHENOLS & ETHERS

1. Phenol is acidic in nature. Why?

Ans. Phenol is acidic in nature because:

a) phenol , due to resonance, the positive charge rests on oxygen making

the shared pair of electrons more towards oxygen and hydrogen as H+

b) The carbon attached to OH is SP2 hybridize and is more

electronegative, this decreases the electron density on oxygen, increasing

the polarity of O-H bond and ionization of phenol.

c) The phenoxide ion formed by loss of H+ is more resonance stabilized

than phenol itself.

2. o- nitrophenol has lower boiling point ( is more volatile ) than p –

nitrophenol.

Ans. P- nitrophenol has intermolecular hydrogen bonding which increases the

boiling point while in o- nitro phenol due to presence of intra molecular

hydrogen bonding, there is a decrease in boiling point and increase in

volatility.

3. Methanol is miscible with water while iodomethane is not.

Ans.

Methanol can form intermolecular hydrogen bonding with water but there

is no hydrogen bonding in iodomethane and water. Therefore methanol in miscible

in water.

4 Alcohols have higher boiling points than isomeric ethers.

Ans. Alcohols can form intermolecular hydrogen bonds due to their high

polarity whereas, ether cannot. Therefore alcohols have higher boiling

points than isomeric ethers.

5. Ethers are soluble in water alkanes are not.

Ans. Ethers can form H- bonding with water molecule whereas alkenes cannot.

Therefore ethers are soluble in water and alkanes are not.

6. The order of acidic strength in alcohols is R CH2OH > R2 CHOH

> R3 COH

Ans. In alcohols, the acidic strength is due to polar nature of O-H bond. An

electron releasing group e.g., alkyl groups, increases electron density on

oxygen tending to decrease the polarity of O-H bond. This decreases the

acid strength. Therefore the order of acid strength is

7. During preparation of ester from alcohol and acid, water has to be

removed as soon as it is formed.

Ans.

8. Ethers cannot be prepared by dehydration of secondary or tertiary

alcohols.

Ans. For secondary and tertiary alcohols, elimination competes over

substitution and alkenes are formed on acidic dehydration as the reaction

follows Sn1 mechanism. Therefore the acidic dehydration of secondary or

tertiary alcohols does not give ethers.

9. Reaction of anisole with HI gives methyl iodide and phenol.

Ans.

10. o-nitrophenol is more acidic than o-methoxyphenol. Why ?

Ans. Because of electron withdrawing nitro group in o-nitrophenol there is

effective delocalisation negative charge.

11. Phenol is more acidic than alcohols. Why?

Ans. Phenoxide is more stabilised by resonance than alkoxide ion.

12. What is denaturation of alcohols ?

Ans. Commercial alcohol is made unfit for drinking by adding copper sulphate/

pyridine.

13. Arrange the following compounds in increasing order of their

acid strength: Propan-1-ol, 2,4,6-trinitrophenol, 3-nitrophenol, phenol, 4-

methylphenol Ans. Propan-1-ol < 4-methylphenol < phenol < 3-nitrophenol < 2,4, 6-

trinitrophenol.

14.

I) Butanol to Butanoic acid

ii). Ethanol to propanone

iii). Phenol to salicylic acid

Ans.

IV). Methanol to Ethanol

V. Ethanol to propanol

vi. Phenol to Benzyl Alcohol

Ans.

VII. Ethanal to propan -2- ol

Ans.

UNIT 12 ALDEHYDES KETONES & CARBOXYLIC ACIDS

1. Arrange the following set of compounds in order of

increasing boiling points. CH3CH2CH2CHO, CH3CH2CH2CH2OH, CH3CH2 OCH2 CH3

,CH3CH2CH2 CH3

Ans. CH3 ,CH3CH2CH2 CH3 < CH3CH2 OCH2 CH3 < CH3CH2CH2CHO <

CH3CH2CH2CH2OH 2. Ketones are less reactive than aldehydes towards nuclephilic

addition reaction. Why? Ans. Two electron releasing alkyl groups in ketones reduce the

electrophilicity of carbonyl. Also due to steric hindrance nucleophilic

attack is not easier.

3. Benzaldehyde is less reactive than propanal towards nucleophilic addition reaction. Why ?

Ans. Resonance in benzaldehyde reduces the electrophilicity of carbonyl carbon

4. (i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6- trimethylcyvlohexanone does not.

Ans. The three methyl groups reduce the elecrophilicity of carbonyl and also offer steric hindrance to the nucleophilic attack

(ii) In semicarbazide ,the NH2 group directly attached to the -C = O group is not involved in the formation of semicarbazones.Give

reason. The lone pair of electrons of the – NH2 group directly attached to –C

= O group is involved in resonance. Hence it does not act

nucleophile

5. Arrange the following compounds in increasing order of their

reactivity towards nuclephilic addition reaction. Methanal, propanone, ethanol

Ans. Propanone < Ethanal < Methanal

6. Acetaldehyde undergoes aldol condensation. Why ? Ans. Because it contains alpha hydrogen.

7. Formaldehyde & benzaldehyde do not undergo aldol

condensation. Why ? Ans. Because they do not contain alpha hydrogen.

8. Formaldehyde & benzaldehyde show Cannizaro reaction. Why ?

Ans. Because they do not contain alpha hydrogen 9. Carboxylic acids are higher boiling liquids than alcohols. Why

?

Ans. Carboxylic acids form more extensive hydrogen bond than alcohols and exisi as dimer.

10. Carboxylic acids are more acidic than phenol. Why ? OR

Ka of carboxylic acid is more than that of phenol.Why ? OR

pKa of carboxylic acid (acetic acid) is more than that of phenol. Why?

Ans. Carboxylate ion is more stabilised than phenoxide ion due to resonance with negative charge on more electronegative oxygen

atom. 11. Carboxylic acids do not show the reactions of carbonyl

group. Why ? Ans. Due to resonance the double bond character is reduced in

carboxylic acids.

12. Compare the acid strength of the following acids-

HCOOH , CH3COOH , C6H5COOH

Ans. HCOOH > C6H5COOH > CH3COOH

13. Why chloroacetic acid stronger than acetic acid.

Ans- Cl is electron withdrawing group thus chloroacetate ion is more stable than

acetate ion.

14. Carry out the following conversion—

a. Acetylene to Acetaldehyde

b. Toluene to Benzaldehyde

Ans—a. CHΞCH + H2O H2SO4/HgSO4 CH2=CHOH CH3CHO

b.

[O] CrO2Cl2 + H2O

UNIT 13 AMINES

1. Ammonolysis is not preferred for preparation for primary

amines.Why ? Ans. Because it gives mixture of amines.

2. Alkyl amines have higher b.pts than alkanes of comparable

molecular mass. Why? Ans. Because alkyl amines form intermolecular hydrogen bond.

3. Alcohols have higher b.pts than alkyl amines of comparable

molecular mass. Why?

Ans. Because in alcohols the strength of H – bond is more than that in amines since O is more electronegative than N.

4. Arrange the following set of compounds in order of

increasing boiling points. n-Butylamine(10), Diethylamine(20),Ethyl dimethylamine(30)

[isomeric amines] Ans. Ethyl dimethylamine(30) < Diethylamine(20) < n-Butylamine(10)

Reason - 30 amine – No H bond In primary amine 2 hydrogen atoms are available for H- bond. In

sec amine only one H atom available for H- bond.

5. (i) Ethylamine (aliphatic amines) is a stronger base than aniline (aromatic amine). Why ?

(ii) pKb of aniline is more than that of methyl amine. Why?

Ans. (i) In ethyl amine the electron releasing alkyl group increases the

electron density on nitrogen. In aniline the electrons on N is involved in resonance.

(ii) In methyl amine the electron releasing alkyl group increases

the electron density on nitrogen. In aniline the electrons on N is involved in resonance.

6. Why cannot primary amines be prepared by Gabriel phthalimide ?

Ans. Aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

CH3 CHO

7. Methylamine in water reacts with ferric chloride to

precipitate hydrated ferric oxide. Why ? Ans. Methylamine is more basic than water. So it forms OH- with water.

The OH- ions react with ferric chloride and gives ppt. 8. Aniline does not undergo Friedel- Crafts reaction. Why ?

Ans. In Friedel- Crafts reaction AlCl3 which is used a catalyst is a Lewis acid. It reacts with aniline and forms salt.

9. Nitration of aniline gives a substantial amount of m-nitroaniline besides ortho and para derivatives. Why ?

Ans. During nitration, in presence of strong acidic medium aniline is protonated to form the anilinium ion which is meta directing.

10. Arrange the following In decreasing order of basic strength:

C6H5NH2,C6H5(CH3)2,(C2H5)2NH and CH3NH2

Ans- (C2H5)3N > CH3NH2 > C6H5NH2 > C6H5N(CH3)2 .

11. How will you convert Aniline to Benzonitrile?

Ans-

NaNO2+HCl KCN

0 – 5 0C CuCN

Aniline Benzonitrile

12. Account for the following :-

(i) Pkb of aniline is morethan that of methylamine.

(ii) Ethylamine is soluble in water whreas aniline is not.

(iii)Why has aniline a weak basic nature than aliphatic amines.

Ans- (i) aniline has an electron withdrawing phenyl group so it is the weaker

base than ammonia. Methyl group in methylamine is electron donating

group so it is stronger base than ammonia.

(ii) ethyl group in ethylamine is comparatively a small group and causes

no hindrance in formation of hydrogen bonding and hence it is soluble in

water.

(i) Due to resonance in aniline.

CΞN

N2+Cl- NH2

UNIT 14 BIOMOLECULES

(Q.1) What is difference between Reducing and non-reducing sugars or

carbohydrates?(1 Mark)

(Ans) All those carbohydrates which contain aldehydic and ketonic group in the

hemiacetal or hemiketal form and reduce Tollen’s reagent or Fehling’s solution are called

reducing carbohydrates while others which do not reduce these reagents are called non-

reducing reagents.

(Q.2) Define glycosidic linkage? (1

Mark)

(Ans) The two monosaccharide units are joined together through an ethereal or oxide

linkage formed by the loss of a molecule of H2O. Such a linkage between two

monosaccharide units through oxygen atoms is called glycosidic linkage.

(Q.3) What is Milk sugar? Give its characteristics. (2

Marks)

(Ans) Lactose occurs in milk so, it is called milk sugar. Lactose on hydrolysis with dilute

acids yields an equimolar mixture of D-glucose and D-galactose. It is a reducing sugar

since it forms an osazone. It undergoes mutarotation and also reduces Tollen’s or

Fehling’s solution.

(Q.4) What do you understand by denaturation of proteins? (2

Marks)

(Ans) When a protein in its native form, is subjected to physical change like in

temperature or chemical change like change in pH, the hydrogen bonds are disturbed.

Due to this, globules unfold and helix get uncoiled and protein loses its biological

activity. This is called denaturation of protein.

(Q.5) Give a broad classification of vitamins? (2

Marks)

(Ans) Vitamins are complex organic molecules.They can be broadly classified as:

(i) Water soluble vitamins: These include vitamin B-complex and vitamin C.

(ii) Fat soluble vitamins: These are oily substances that are not readily soluble in

water. However, they are soluble in fat. These include vitamins A,D,E and K. Nucleic

acids are bipolar (i.e. polymers present in the living system). They are also called

polynucleotides since the repeating structural unit of nucleic acids is a nucleotide

(Q.6) Give a short note on Zwitter ion? (3

Marks)

(Ans) Amino acids are usually colourless, crystalline solids. These are water soluble ,

high melting solids and behave like salts rather than simple amines or carboxylic acids.

This behaviour is due to the presence of both acidic (carboxylic group) and basic (amino

group) groups in the same molecule. In aqueous solution, the carboxyl group can lose a

proton and amino group can accept a proton, giving rise to a dipolar ion known as

zwitter ion.

Q7. What is the primary structure of proteins?

Ans:- The sequence of amino acids in proteins.

Q. 8Define globular & Fibrous proteins

Ans: The protein with spherical shape and soluble in water are globular proteis eg Albumin and

Insulin

The protein having thread like structure and insoluble in water are fibrous proteins eg keratin and

myosin.

Q.9 Differentiate between DNA & RNA

DNA RNA

thymine uracil

Deoxy ribose sugar Ribose sugar

Genetic information Protein synthesis

Double helical Single helical

Q.10 What happens when glucose reacts with

(i) Br2 water

(ii) HI

(iii) Conc. HNO3

Ans:- (i) Gluconic acid

(ii) n-hexane

(iii) Saccharic acid

UNIT 15 POLYMERS

1. Define the term homopolymerisation giving an example.

Ans. Polymerisation process involving only single monomeric species is

known as homopolymerisation. For example formation of polyethene from

ethane.

2. Write the name and structure of the monomer of polymer:

PVC.

Ans. The monomer of PVC is vinyl chloride.

CH2=CH-Cl

3. Give an example of elastomers.

Ans. Natural rubber of Buna-S.

4. What are the monomer units of Bakelite?

Ans. Phenol and Formaldehyde.

5. Write the name and structure of monomer of natural rubber.

Ans. Isoprene or 2-Methylbuta-1,3-diene.

6. What are the monomer units of terylene?

Ans. Ethylene glycol and terepthalic acid.

7. Differentiate between thermoplastic and thermosetting polymers.

Give one example of each.

Ans.

Thermoplastic polymer Thermosetting polymer

1. They are linear polymer without cross-linking.

1. They have three dimensional network of

covalent bonds with cross

links.

2. Upon heating these

polymers become soft and becomes hard again on

cooling.

2.Upon heating they retain

their strength and does not become soft.

3. These plastics can be recycled. They can be

melted and reused.

3.These plastics cannot be recycled.

4.Example: PVC, Nylon,

Polystrene.

4.Example: Phenol formaldehyde resin,

urea formaldehyde resin.

8. Explain each of the following giving suitable example of each:

(i) Elastomers

(ii) Condensation polymers

(iii)Addition polymers

Ans. (i) Elastomers: Polymers which are rubber like with elastic

properties are called Elastomers.

(ii) Condensation polymers: The polymer formed by the repeated

condensation reaction between two same or different bifunctional monomeric

units with the elimination of simple molecules like water, ammonia etc., are

called as condensation polymers.

(iii)Addition polymers: The polymers formed by the repeated addition of

monomer molecules possessing double or triple bond are called addition

polymers.

8. Write the name and structures of the monomers of the following

polymers:

Buna-S, Dacron and Neoprene

Ans.

Polymer Monomer

Buna-S CH2=CH-CH=CH2 (Butadiene), CH2=CH-C6H5 (Styrene)

Dacron HOCH2-CH2OH ( Ethylene Glycol), HOOC-C6H4-COOH (Terepthalic acid)

Neoprene CH2=C(Cl)-CH-CH3 (Chloroprene)

9. What are biodegradable polymers? Give two examples.

Ans. Polymers which can be easily degraded by micro-organisms present in

nature are called biodegradable polymers. Common examples are poly-β-

hydroxybutyrate-co-β-hydroxyvalerate (PHBV), Nylon-2-nylon-6.

10. Give the differences between natural rubber and

vulcanised rubber.

OR

How does vulcanisation change the character of the natural

rubber?

Ans.

Natural rubber Vulcanised rubber

It becomes soft at temperature above

335K and brittle at temperature below

283K

It has a much wider useful temperature

range.

It shows high water absorption capacity. It absorbs water to a much lesser extent.

It is highly soluble in non-polar solvents. It is less soluble in non-polar solvents.

It is readily attacked by oxidising agents. It is not readily attacked by oxidising

agents.

UNIT 16 CHEMISTRY IN EVERYDAY LIFE

1. What is meant by narrow spectrum antibiotics?

Ans. Antibiotics which are mainly effective against gram positive of gram

negative bacteria are called as narrow spectrum antibiotics. For example

Penicillin G is a narrow spectrum antibiotic.

2. What is meant by broad spectrum antibiotics?

Ans. These are medicines which are effective against several different

types of micro-organisms. For example tetracycline, chloramphenicol etc.

3. What do you understand by the term ‘Chemotherapy’?

Ans. Use of chemicals for therapeutic effect is called as Chemotherapy.

4. Name a drug used in case of mental depression?

Ans. Equanil, barbituric acid derivatives such as seconal, neuronal are used

as antidepressents.

5. Give an example of Narcotic which is used as analgesic.

Ans. Morphine is a narcotic which is used as an analgesic i.e. to reduce pain.

6. What is the role of Bithional in toilet soaps?

Ans. Bithional acts as antiseptic when added to soap. Thus, it increases the

antiseptic properties of the soap.

7. Define antipyretic. Give two examples.

Ans. These are the chemicals which are used to bring down the body

temperature during high temperature. For example: Asprin, Paracetamol.

8. What is tincture of iodine? For what purpose it is used?

Ans. Alcohol water solution containing 2-3% iodine is known as tincture of

iodine. It is used as antiseptic.

9. Give an example of non-ionic detergent.

Ans. Lauryl alcohol ethoxylate.

10. Name the chemical responsible for antiseptic properties

of Dettol.

Ans. Chloroxylenol and terpineol in a suitable solvent.

11. To which class of drugs cimetidine and ranitidine

belongs?

Ans. Anti-histamines.

12. Name a food preservative which is most commonly

used by food producers.

Ans. The most common food preservative used by food producers is sodium

benzoate.

13. Describe antiseptics giving suitable examples.

Ans. Antideptics are the chemical substances which prevent the growth of

micro-organism and may even kill them. Boric acid, hydrogen peroxide,

bithional, is common antiseptics.

14. How are transparent soaps manufactured?

Ans. Transparent soaps are manufactured by dissolving soap in ethanol and

evaporating the excess solvent.

15. What are artificial sweetening agents? Give two

examples.

Ans. These are the chemical substances which are sweet in taste but do not

add to calorie intake. Some common examples are aspartame saccharin,

alitame etc.

16. Why the use of aspartame is is limited to cold food and

drinks?

Ans. This is because aspartame decomposes at baking and cooking

temperatures. Hence aspartame can be used only with cold drinks and foods.

17 What are biodegradable and non-biodegradable detergents? Give

one example of each.

Ans. Detergents which are easily degraded by micro-organisms present in

the rivers, ponds etc., are called as biodegradable polymers. For example

sodium-dodecylbenzene sulphonate.

Non-biodegradable detergents are not easily degraded by mico-

organisms. For example sodium-1,1-dimethyldecylbenzene sulphonate.

18. Explain the following giving one example of each type:

a. Antacids b. Disinfectants

Ans. (i) Antacids: The chemicals which control the release of acid in gastric

juices or neutralises the excess acid in gastric juices are called antacids.

(ii) Disinfectants: The chemical substances which are used to kill

microorganisms but they cannot be applied on living tissues are called as

disinfectants. For example 1% solution of phenol in water is a disinfectant.

19. Explain the following giving one example of each type:

a. Tranquilizers

Ans. a. Tranquilizers: These are chemicals used for the treatment of stress

and mild or even severe mental diseases. For example, veronal, serotonin,

derivatives of barbituric acid etc.

20. Describe the following giving one example:

a. Antifertility drugs b. Antioxidants

Ans. Antifertility drugs: These drugs are used to prevent unwanted

pregnancies. For example Norethindrone

Antioxidants: Chemicals which prevent the oxidation of food stuff during

storage etc., are called as antioxidants. Vitamin C and E are natural oxidants.

BHA ( Butylated Hydroxy Anisole) is a synthetic antioxidant.

21. Why are Cimitidine and ranitidine better antacid than sodium

hydrogen carbonate?

Ans. Excessive intake of sodium hydrogen carbonate can make stomach alkaline

and trigger the production of even more acid. Also these treatments controls

only the symptoms of hyperacidity and not the cause. Cimetidine, ranitidine are

better antacids because they control the cause of hyperacidity. It is observed

that hormone Histamine stimulates the production of pepsin and hydrochloric

acid in the stomach. Drugs like cimetidine and ranitidine prevent the interaction

of histamine with the receptors present in the stomach wall. This results in

production of lesser amount of acids and controlling hyperacidity.

22. Account for the following:

(i) Asprin drug helps in preventing the heart attack.

(ii) Detergents are non-biodegradable while soaps are biodegradable.

Ans. (i) Asprin is found to help in prevention of heart attack because of its anti-

blood clotting action. The blood clots if formed, are mainly responsible for heart

attack.

(ii) Common detergents which contain branched chain hydrocarbon part

in their molecules are non-biodegradable. This is because micro-organisms

present in the waste water cannot metabolise branched chain hydrocarbons. On

the other hand soaps contain linear chain hydrocarbons, which can be

biodegraded more easily and water pollution is prevented.

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