16 class 10 semester 2 chemistry

7/31/2019 16 Class 10 Semester 2 Chemistry

1/68

1B Panditya Road, Kolkata 29 www.edudigm.in 40034819

Chemical Bonding: Perhaps the chapter that will solve almost all doubts and concept related

questions in Chemistry (Trust me I am not exaggerating). It is of utmost importance to

understand this chapter and all the concepts in great details.

: There are a lot of elements but very few of them exist in their native form

(inert gases). Most of the elements exist as compounds, as the atoms (which are not

stable) combine to form more stable molecules. There is some attractive force in themolecules which is called chemical bond. But why and how do the atoms combine and

why different compounds have different properties?

The Octet Rule requires all atoms in a molecule to have 8 valence electrons--either

by sharing, losing or gaining electrons--to become stable. For Covalent bonds, atoms

tend to share their electrons with each other to satisfy the Octet Rule. It requires 8electrons because that is the amount of electrons needed to fill a shell(electron

configuration); also known as a noble gas configuration. Each atom wants to become

as stable as the noble gases that have their outer valence shell filled because noblegases have a charge of 0. Although it is important to remember the "magic number",

8, note that there are many Octet rule exceptions.

Example: As you can see from the picture below, Phosphorus has only 5 electrons in

its outer shell (bolded in red). Argon has a total of 8 electrons (bolded in red), whichsatisfies the Octet Rule. Phosphorus needs to gain 3 electrons to fullfill the Octet

Rule. It wants to be like Argon who has a full outer valence shell.
http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Electronic_Configurationshttp://chemwiki.ucdavis.edu/Inorganic_Chemistry/Electronic_Configurationshttp://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Lewis_Theory_of_Bonding/Violations_of_the_Octet_Rulehttp://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Lewis_Theory_of_Bonding/Violations_of_the_Octet_Rulehttp://chemwiki.ucdavis.edu/Inorganic_Chemistry/Electronic_Configurationshttp://chemwiki.ucdavis.edu/Inorganic_Chemistry/Electronic_Configurations


2/68


Lewis structures are also known as electron dot structures. The diagrams are named

for Gilbert N. Lewis, who described them in his 1916 article entitled The Atom and the

Molecule. Lewis structures depict the bonds between atoms of a molecule as well as any

unbonded electron pairs. You can draw a Lewis dot structure for any covalent molecule

or coordination compound.

Drawing Lewis structures can be a straightforward process if the proper steps are

followed. There are several different strategies to constructing Lewis structures. These

instructions outline the Kelter strategy to draw Lewis structures for molecules.

In this step, add up the total number of valence electrons from all the atoms in the

molecule.

An atom is considered "happy" if the atom's outer electron shell is filled.

Elements up to period four on the periodic table need eight electrons to fill their outer

electron shell. This property is often known as the "octet rule".
http://chemistry.about.com/od/famouschemists/p/lewisbio.htmhttp://chemistry.about.com/od/famouschemists/p/lewisbio.htm


3/68


4/68


The real ambition of the atoms in molecules is to maximize their bonding

interactions. There are limitations. One simply cannot keep drawing lines between

atoms without justification.

Each bond represents two electrons, there can only be so many bondsbased on the number of valence electrons present.

each atom has only a certain number of valence orbitals which it can usewhile making bonds. Hydrogen only has a 1s orbital available so it's total electron countis limited to 2 ( a single bond). The second period elements (with only 2s + 2pavailable) are limited to total electron counts of 8.

Geometric constraints arise because the atomic orbitals on atoms have

certain directional character (with the exception of "s" orbitals which arespherical). This allows them to bond effectively in only certain ways. This limitation

will be expounded upon later, but suffice it to say that two carbon atoms bonded in thegas phase (C2) cannot exhibit a bond order of 4 because all three 2p orbitals cannotlocalize electron density between the nuclei simultaneously.

*Atoms in molecules want to make bonds (within reason), not necessarily attain anoctet of electrons.

*The number of available orbitals and electrons gives rise to an "octet" rule for 2ndperiod elements which maximize their bonding interactions when the attain a total

electron count of 8.

We have already seen and are quite comfortable with the fact that hydrogen cannot

handle anything but a total electron count of 2. This should open your eyes to thereality underlying Lewis Structures! This is the first example of a violation to the so-called octet rule.

What about a substance like BH3? The best Lewis structure is depicted below..


5/68


.

Each hydrogen exhibits it's maximum total electron count (2) and not surprisingly it'smaximum number of bonds. The boron atom in BH3, on the other hand, has only 6 total

electrons. This apparent problem is not a problem when one considers the principles

discussed above AND the fact that the real "rule" is that boron cannot EXCEED an octet

of electrons.

Consider BF3 as well..

.

I consider the best Lewis Structure to be the one shown on the left (I), while few wouldargue that the one on the right (II) is better. The presence of lone pair electrons on eachF atom potentially allows additional bonding interactions. The problem with II is the

formal charges which imply that electronegative fluorine is losing the electron densitytug-of-war with boron! (The electronegativity difference between these two elements issimilar to the difference between Na and Cl, a strong electrolyte!)

Another case where the "octet rule" must be violated are reactive species with an oddnumber of valence electrons. A perfect example is nitrogen monoxide (nitric oxide),

NO. This species has 11 valence electrons. There is absolutely NO way to have all theelectrons pair up and one of the atoms must have an odd total electron count aswell. The best Lewis structure maximizes the number of bonds but does not violate the

true octet rule: neither 2nd period atom exceeds an octet!

.

.

The last type of "apparent" exception to the octet rule occurs when an element of the

3rd period or higher is the central atom in a molecule or molecular ion AND it is bonded

to electronegative substituents such as O, F, Cl, Br. Examples: XeF4, SF6, PCl5, [SeBr6]2-,

IO4-, ClO4- etc...


6/68


The VSEPR theory, proposed by R.J.Gillespie and R.S. Nyholmm in 1957, is based on the

repulsions between the electron-pairs in the valence-shell of the atoms in the molecule.

According to VSEPR Theory Definition, it was developed to predict the shapes of themolecules in which the atoms are bonded together with single bonds only. VSEPR

Theory is abbreviated as Valence Shell Electron Pair Repulsion Theory.

In this article, we will learn the main postulates of VSEPR theory.

Given below are some of the postulates of VSEPR Theory.

The shape of the molecule is determined by both the total number of electronpairs (bonding and non-bonding) around the molecules central atom and the

orientation of these electron pairs in the space around the central atom.

In order to minimize the repulsion forces between them, electron pairs aroundthe molecules central atom, tend to stay as far away from each other as possible.

Electron pairs around the molecule's central atom can be shared or can be lonepairs. The 'shared pairs' of electrons are also called bond pairs of electrons. The

presence of lone pair(s) of electrons on the central atom causes some distortionsin the expected regular shape of the molecule.

The strength of repulsions between different electron pairs follows the order:Lone pair - Lone pair > Lone pair - Shared pair > Shared pair - Shared pair.

To use VSEPR theory for predicting the shapes of molecules, the number of electron

pairs (both, shared and lone pairs) is simply counted. This is illustrated by taking a

typical molecule of the type ABn. 'A' is the central atom, 'B' atoms are bonded to 'A' by

single covalent bonds (single electron pair bonds), and 'n' is the number of 'B' atoms

bonded to one atom of 'A'.

In a molecule having two bond pairs of electrons around its central atom, the bond pairs

are located on the opposite sides (at an angle of 180o) so that the repulsion between

them is minimum. Such molecules are therefore linear. Some molecules, which show

linear geometry are: BeF2 (beryllium fluoride), BeCl2 (beryllium chloride),

BeH2 (beryllium hydride), ZnCl2 (zinc chloride), and HgCl2 (mercuric chloride).


7/68


In a molecule having three bond pairs of electrons around its central atom, the electron

pairs form an equilateral triangular arrangement around the central atom. These

molecules have trigonal planar (or triangularplanar) shape and the three bond pairs areat 120C with respect of each other.

In a molecule of the type AB3, the three bond pairs of electrons are located around A in a

triangular arrangement and the molecule AB3, has a triangular planar geometry. Some

molecules that show triangular planar geometry are BCl3, BF3, etc.

Boron Trifluoride is a trigonal planar molecule

Molecule having four bond pairs of electrons around the central atom, arrange their

electrons tetrahedrally. These molecules have tetrahedral shapes and the four bond

pairs are at an angle of 109.

Five bond pairs orient themselves around the central atom in a trigonal bipyramidal

way. A molecule having five bond pairs around its central atom has a triangular

bipyramidal shape. Three bond pairs are arranged in an equatorial triangular plane and

are oriented at an angle of 120 with respect to each other. The other two bond pairs

are opposite to each other, and at right angles to the triangular plane formed by the

three bond pairs. Some other molecules, which show trigonal bipyramidal geometry

are; PCl5, PF5, SbCl5.

For example in a molecule of the type AB5, the five bond pairs are distributed in a

trigonal bipyramidal around the central atom 'A'. Therefore, the molecules of the type

AB5 are trigonal bipyramidal in shape.


8/68


AB5, five bond pairs are oriented around A in a trigonal bipyramidal

shape.Similarly,PCl5 also has a trigonal bipyramidal shape.

Six bond pairs in a molecule are distributed octahedrally around the central atom. A

molecule having six bond pairs around its central atom has an octahedral shape. In amolecule of the type AB6, the six 'B' atoms are placed octahedrally around 'A'. Thus, the

molecules of the type AB6 are octahedral. The molecule SF6 has an octahedral geometry.

AB6 type molecules are octahedral in shape.


9/68


2 linear

(180o)

0 AX2

linear

BeH2

3 trigonal planar

(120o)

0 AX3

trigonal planar

BF3

1 AX2E

bent

SO2

4 tetrahedral

(109.5o)

0 AX4

tetrahedral

CH4

1 AX3E

trigonal pyramidal

NH3

2 AX2E2

bent

H2O

5 trigonal bipyramidal

(90o, 120o)

0 AX5 PCl5


10/68


trigonal bipyramidal

1 AX4E

see-saw

SF4

2 AX3E2

T-shaped

BrF3

3 AX2E2

linear

XeF2

6 octahedral(90o)

0 AX6

octahedral

SF6

1 AX5E

square pyramidal

BrF5

2 AX4E2

square planar

XeF4


11/68


Given below is an example based on VSEPR Theory.

: Predict the shapes of the following molecules using the valence shell electron

pair repulsion (VSEPR) theory. AsF5, HgBr2.

According to the VSEPR theory, the electron pairs present is the valence-shell

of the central atom/ion arrange themselves in the space around it so as to keep them as

far as possible from each other, so as to minimize the electrostatic repulsions.

As has five electrons in its outermost orbit. Due to sharing of 5 electrons from 5F-atoms,

there are in all 5 electron pairs. These are distributed in space to form a trigonal

bipyramid as shown here.

Hg has only two electrons in its outermost orbit and sharing these electrons with two Br

gives 2 pairs of electrons around Hg. This gives a linear structure to (electron

pairs are positioned at 180)


12/68


The VSEPR theory is simple yet powerful. Nevertheless, like any simplified model, it has

its limitations. First, although it predicts that the bond angle in H2O is less than thetetrahedral angle, it does not make any attempt to predict the magnitude of the

decrease. Second, the theory makes no predictions about the lengths of the bonds,

which is another aspect of the shape of a molecule. Third, it ascribes the entire criterion

of shape to electrostatic repulsions between bonding pairs, when in fact there are

numerous contributions to the total energy of a molecule, and electrostatic effects are

not necessarily the dominant ones. Fourth, the theory relies on some vague concepts,

such as the difference in repelling effects of lone pairs and bonding pairs. There also are

some species for which VSEPR theory fails. Nevertheless, despite these limitations and

uncertainties, VSEPR theory is a useful rule of thumb and can be used with reasonable

confidence for numerous species.

To Sum Up:

(i) It does not explain the shapes of the molecules having very polare.g. should have same structure but is linear while is

angular

(ii) It does not explain the shapes of the molecules or ions which are extensive bydelocalized -electron system

(iii) It does not explain the shapes of some molecules which have an inert pair ofelectrons.

(iv) It does not explain the shapes of certain compounds of transitional metals e.g.the shape of the compound electronic configuration of the central atom, is

square planar and not tetrahedral as predicted b this theory


13/68


The valence bond theory was proposed by Heitler and London to explain the

formation of covalent bond quantitatively using quantum mechanics. Later on, LinusPauling improved this theory by introducing the concept ofhybridization.

The main postulates of this theory are as follows:

* A covalent bond is formed by the overlapping of two half filled valence atomicorbitals of two different atoms.

* The electrons in the overlapping orbitals get paired and confined between thenuclei of two atoms.

* The electron density between two bonded atoms increases due to overlapping.This confers stability to the molecule.

* Greater the extent of overlapping, stronger is the bond formed.

* The direction of the covalent bond is along the region of overlapping of the atomicorbitals i.e., covalent bond is directional.

* There are two types of covalent bonds based on the pattern of overlapping asfollows:

The covalent bond formed due to overlapping of atomic orbital along the

inter nucleus axis is called -bond. It is a stronger bond and cylindrically symmetrical.

Depending on the types of orbitals overlapping, the -bond is divided into followingtypes:

s-s bond:

p-p bond:

s-p bond:

The covalent bond formed by sidewise overlapping of atomic orbitals iscalled - bond. In this bond, the electron density is present above and below the inter

nuclear axis. It is relatively a weaker bond since the electrons are not strongly attractedby the nuclei of bonding atoms.
http://www.adichemistry.com/general/chemicalbond/vbt/valence-bond-theory-hybridization.html#HYBRIDIZATIONhttp://www.adichemistry.com/general/chemicalbond/vbt/valence-bond-theory-hybridization.html#HYBRIDIZATION


14/68


Note: The 's' orbitals can only form -bonds, whereas the p, d & f orbitals can formboth and-bonds.

: H2 molecule:

* The electronic configuration of hydrogen atom in the ground state is 1s1.

* In the formation of hydrogen molecule, two half filled 1s orbitals of hydrogenatoms overlap along the inter-nuclear axis and thus by forming a s-sbond.

:Cl2 molecule:

* The electronic configuration of Cl atom in the ground state is[Ne]3s2 3px2 3py2 3pz1.

* The two half filled 3pz atomic orbitals of two chlorine atoms overlap along the

inter-nuclear axis and thus by forming a p-p bond.

: HCl molecule:

* In the ground state, the electronic configuration of hydrogen atom is 1s1.

* And the ground state electronic configuration of Cl atom is [Ne]3s2 3px2 3py2 3pz1.

* The half filled 1s orbital of hydrogen overlap with the half filled 3pz atomic orbitalof chlorine atom along the inter-nuclear axis to form a s-p bond.

: O2 molecule:


15/68


* The electronic configuration of O in the ground state is [He] 2s2 2px2 2py1 2pz1.

* The half filled 2py orbitals of two oxygen atoms overlap along the inter-nuclearaxis and form p-p bond.

* The remaining half filled 2pz orbitals overlap laterally to form a p-p bond.

* Thus a double bond (one p-p and one p-p) is formed between two oxygen atoms.

: N2 molecule:

* The ground state electronic configuration of N is [He] 2s2 2px1 2py1 2pz1.

* A p-p bond is formed between two nitrogen atoms due to overlapping of half filled2px atomic orbitals along the inter-nuclear axis.

* The remaining half filled 2py and 2pz orbitals form two p-p bonds due to lateraloverlapping. Thus a triple bond (one and two) is formed between two nitrogen atoms.

However the old version of valence bond theory is limited to diatomic moleculesonly. It could not explain the structures and bond angles of molecules with more thanthree atoms.

E.g. It could not explain the structures and bond angles of H2O, NH3 etc.,

However, in order to explain the structures and bond angles of molecules, LinusPauling modified the valence bond theory using hybridization concept.


16/68


The intermixing of two or more pure atomic orbitals of an atom with almost same

energy to give same number of identical and degenerate new type of orbitals is knownas hybridization.

The new orbitals formed are also known as hybrid orbitals.

The intermixing or hybridization of atomic orbitals is a mathematical concept based

on quantum mechanics. During this process, the wave functions, of atomic orbitals of

same atom are combined to give new wave functions corresponding to hybrid orbitals.

* The atomic orbitals of same atom with almost same energy can only participate inthe hybridization.

* The full filled or half filled or even empty orbitals can undergo hybridizationprovided they have almost equal energy.

No! The hybridization is the mixing of orbitals of same atom only. The combinationof orbitals belonging to different atoms is called bonding.

* The new orbitals that are formed due to intermixing of atomic orbitals are alsoknown as hybrid orbitals, which have mixed characteristics of atomic orbitals.

* The shapes of hybrid orbitals are identical. Usually they have one big lobeassociated with a small lobe on the other side.

* The hybrid orbitals are degenerate i.e., they are associated with same energy.

How many hybrid orbitals are formed?

* The number of hybrid orbitals formed is equal to the number of pure atomicorbitals undergoing hybridization.

E.g. If three atomic orbitals intermix with each other, the number of hybrid orbitalsformed will be equal to 3.


17/68


* The hybrid orbitals are filled with those electrons which were present in the pureatomic orbitals forming them.

* The filling up of electrons in them follows Pauli's exclusion principle and Hund'srule.

What is the use of hybrid orbitals?

* The hybrid orbitals participate in the bond formation with other atoms.

Why atomic orbitals in a given atom undergo hybridization?

* The hybrid orbitals are oriented in space so as to minimize repulsions between

them. This explains why the atomic orbitals undergo hybridization before bond

formation.

The reason for hybridization is to minimize the repulsions between the bonds thatare going to be formed by the atoms by using hybrid orbitals.

Remember that the hybridization is the process that occurs before bond formation.

* The bond angles in the molecule are equal to or almost equal to the angles between

the hybrid orbitals forming the bonds. The shape of the molecule is determined by the

type of hybridization, number of bonds formed by them and the number of lone pairs.

During hybridization, the atomic orbitals with different characteristics are mixedwith each other. Hence there is no meaning of hybridization between same type oforbitals i.e., mixing of two 's' orbitals or two 'p' orbitals is not called hybridization.

However orbital of 's' type can can mix with the orbitals of 'p' type or of 'd' type.

Based on the type and number of orbitals, the hybridization can be subdivided intofollowing types.

* Intermixing of one 's' and one 'p' orbitals of almost equal energy to give twoidentical and degenerate hybrid orbitals is called 'sp' hybridization.

* These sp-hybrid orbitals are arranged linearly at by making 180o of angle.

* They possess 50% 's' and 50% 'p' character.


18/68


: Beryllium Chloride (BeCl2):

* The electronic configuration of 'Be' in ground state is 1s2 2s2. Since there are no

unpaired electrons, it undergoes excitation by promoting one of its 2s electron intoempty 2p orbital.

Thus in the excited state, the electronic configuration of Be is 1s2 2s1 2p1.

If the beryllium atom forms bonds using these pure orbitals, the molecule might be

angular. However the observed shape of BeCl2 is linear. To account for this, following sphybridization was proposed.

* In the excited state, the beryllium atom undergoes 'sp' hybridization by mixing a 2s

and one 2p orbitals. Thus two half filled 'sp' hybrid orbitals are formed, which arearranged linearly.

* These half filled sp-orbitals form two bonds with two 'Cl' atoms.

* Thus BeCl2 is linear in shape with the bond angle of 180o.

: Acetylene (C2H2):

* The ground state electronic configuration of 'C' is 1s2 2s2 2px12py1. There are onlytwo unpaired electrons in the ground state. However, the valency of carbon is four i.e., it

forms 4 bonds. In order to form four bonds, there must be four unpaired electrons.

Hence carbon promotes one of its 2s electron into the empty 2pz orbital in the excitedstate.

Thus in the excited state, the electronic configuration of carbon is

1s2 2s1 2px12py12pz1.


19/68


* Each carbon atom undergoes 'sp' hybridization by using a 2s and one 2p orbitals inthe excited state to give two half filled 'sp' orbitals, which are arranged linearly.

* The two carbon atoms form a sp-sp bond with each other by using sp-orbitals.

However there are also two unhybridized p orbitals i.e., 2py and 2pz on each carbon

atom which are perpendicular to the sp hybrid orbitals. These orbitals form two p-p bonds between the two carbon atoms.

Thus a triple bond (including one sp-sp bond & two p-p bonds ) is formed betweencarbon atoms.

* Each carbon also forms a sp-s bond with the hydrogen atom.

* Thus acetylene molecule is linear with 180o of bond angle.

* Intermixing of one 's' and two 'p' orbitals of almost equal energy to give threeidentical and degenerate hybrid orbitals is known as sp2 hybridization.

* The three sp2 hybrid orbitals are oriented in trigonal planar symmetry at angles of120o to each other.

* The sp2 hybrid orbitals have 33.3% 's' character and 66.6% 'p' character.


20/68


: Boron trichloride (BCl3):

* The electronic configuration of 'B' in ground state is 1s2 2s2 2p1 with only oneunpaired electron. Since the formation of three bonds with chlorine atoms require three

unpaired electrons, there is promotion of one of 2s electron into the 2p sublevel byabsorbing energy.

Thus Boron atom gets electronic configuration: 1s2 2s2 2px12py1.

However to account for the trigonal planar shape of this BCl3 molecule,sp2 hybridization before bond formation was put forwarded.

* In the excited state, Boron undergoes sp2 hybridization by using a 2s and two 2p

orbitals to give three half filled sp2 hybrid orbitals which are oriented in trigonal planarsymmetry.

* Boron forms three sp-p bonds with three chlorine atoms by

using its half filled sp2 hybrid orbitals. Each chlorine atom uses it'shalf filled p-orbital for the -bond formation.

* Thus the shape of BCl3 is trigonal planar with bond anglesequal to 120o.

: Ethylene (C2H4):

* During the formation of ethylene molecule, each carbon atom undergoes

sp2 hybridization in its excited state by mixing 2s and two 2p orbitals to give three halffilled sp2 hybrid orbitals oriented in trigonal planar symmetry.


21/68


There is also one half filled unhybridized 2pz orbital on each carbon perpedicular tothe plane of sp2 hybrid orbitals.

* The carbon atoms form a sp2-sp2 bond with each other by using sp2 hybrid

orbitals.

A p-p bond is also formed between them due to lateral overlapping of unhybridized2pz orbitals.

Thus there is a double bond (sp2-sp2 & p-p) between two carbon atoms.

* Each carbon atom also forms two sp2-s bonds with two hydrogen atoms.

* Thus ethylene molecule is planar with HCH & HCC bond angles equal to 120o.

* All the atoms are present in one plane.

* In sp3 hybridization, one 's' and three 'p' orbitals of almost equal energy intermixto give four identical and degenerate hybrid orbitals.

* These four sp3 hybrid orbitals are oriented in tetrahedral symmetry with 109o28'angle with each other.

* The sp3hybrid orbitals have 25% s character and 75% 'p' character.

: Methane (CH4):


22/68


* During the formation of methane molecule, the carbon atom undergoes

sp3hybridization in the excited state by mixing one 2s and three 2p orbitals to furnish

four half filled sp3 hybrid orbitals, which are oriented in tetrahedral symmetry in spacearound the carbon atom.

* Each of these sp3hybrid orbitals forms a sp3-s bond with one hydrogen atom. Thuscarbon forms four sp3-s bonds with four hydrogen atoms.

* Methane molecule is tetrahedral in shape with 109o28' bond angle.

: Ethane (C2H6):

* Just like in methane molecule, each carbon atom undergoes sp3 hybridization in

the excited state to give four sp3 hybrid orbitals in tetrahedral geometry.

* The two carbon atoms form a sp3-sp3bond with each other due to overlapping ofsp3 hybrid orbitals along the inter-nuclear axis.

Each carbon atom also forms three sp3-s bonds with hydrogen atoms.

* Thus there is tetrahedral symmetry around each carbon with HCH & HCC bondangles equal to 109o28'.

: Ammonia (NH3)

* The ground state electronic configuration of nitrogen atom is: 1s2 2s2 2px12py12pz1.

Since there are three unpaired electrons in the 2p sublevel, the nitrogen atom can form

three bonds with three hydrogen atoms. This will give ammonia molecule with 90o ofbond angles. However, the bond angles are reported to be 107o48'.

* Therefore, it was proposed that, the Nitrogen atom undergoes sp3 hybridization of

a 2s and three 2p orbitals to give four sp3 orbitals, which are arranged in tetrahedral


23/68


symmetry. It is clear that this arrangement will give more stability to the molecule dueto minimization of repulsions.

Among them three are half filled and one is full filled.

* Nitrogen atom forms 3 sp3-s bonds with three hydrogen atoms by using three half

filled sp3 hybrid orbitals. There is also a lone pair on nitrogen atom belonging to the fullfilled sp3 hybrid orbital. It occupied more space than the bond pairs.

* However, the HNH bond angle is not equal to normal tetrahedral angle: 109o28'.

The reported bond angle is 107o48'. The observed decrease in the bond angle is due to

the repulsion caused by lone pair over the bond pairs.That is why, ammonia molecule is trigonal pyramidal in shape with a lone pair on

nitrogen atom.

: Water molecule (H2O)

* The electronic configuration of oxygen is 1s2 2s2 2px22py12pz1. There are two

unpaired electrons in oxygen atom, which may form bonds with hydrogen atoms.However the the bond angles in the resulting molecule should be equal to 90o.

The experimental bond angles reported were equal to 104o28'. To account this,sp3 hybridization before the bond formation was proposed.

* During the formation of water molecule, the oxygen atom undergoes

sp3

hybridization by mixing a 2s and three 2p orbitals to furnish four sp3

hybrid orbitalsoriented in tetrahedral geometry.

Among them, two are half filled and the remaining two are completely filled.

* Now the oxygen atom forms two sp3-s bonds with hydrogen atoms by using half

filled hybrid orbitals.


24/68


* The reported bond angle is 104o28' instead of regular tetrahedral angle: 109o28'.It is again due to repulsions caused by two lone pairs on the bond pairs.

Thus water molecule gets angular shape (V shape).

* In sp3d hybridization, one 's', three 'p' and one 'd' orbitals of almost equal energy

intermix to give five identical and degenerate hybrid orbitals, which are arranged intrigonal bipyramidal symmetry.

Among them, three are arranged in trigonal plane and the remaining two orbitalsare present above and below the trigonal plane at right angles.

* The sp3d hybrid orbitals have 20% 's', 60% 'p' and 20% 'd' characters.

: Phosphorus pentachloride(PCl5)

* The ground state electronic configuration of phosphorus atom is:1s2 2s22p6 3s23px13py13pz1.

* The formation of PCl5 molecule requires 5 unpaired electrons. Hence the

phosphorus atom undergoes excitation to promote one electron from 3s orbital to oneof empty 3d orbital.

* Thus the electronic configuration of 'P' in the excited state is1s2 2s22p6 3s23px13py13pz1 3d1.

* In the excited state, intermixing of a 3s, three 3p and one 3d orbitals to give five

half filled sp3d hybrid orbitals, which are arranged in trigonal bipyramidal symmetry.

i.e., Three orbitals are arranged in trigonal planar symmetry, whereas the remainingtwo are arranged perpendicularly above and below this plane.


25/68


* By using these half filled sp3d orbitals, phosphorous forms five sp3d-p bonds with

chlorine atoms. Each chlorine atom makes use of half filled 3pzorbital for the bondformation.

* The shape of PCl5 molecule is trigonal bipyramidal with 120o and 90o ofCl - P - Clbond angles.

* Intermixing of one 's', three 'p' and two 'd' orbitals of almost same energy by givingsix identical and degenerate hybrid orbitals is called sp3d2hybridization.

* These six sp3d2 orbitals are arranged in octahedral symmetry by making 90o anglesto each other. This arrangement can be visualized as four orbitals arranged in a square

plane and the remaining two are oriented above and below this plane perpendicularly.

: Sulfur hexa flouride (SF6):

* The electronic configuration of 'S' in ground state is 1s2 2s22p6 3s23px23py13pz1.

* In SF6 molecule, there are six bonds formed by sulfur atom. Hence there must be 6unpaired electrons. However there are only 2 unpaired electrons in the ground state ofsulfur. Hence it promotes two electrons into two of the 3d orbitals (one from 3s and one

from 3px).

* Thus the electronic configuration of 'S' in its 2nd excited state is1s2 2s22p6 3s13px13py13pz13d2.


26/68


* In the second excited state, sulfur under goes sp3d2 hybridization by mixing a 3s,three 3p and two 3d orbitals. Thus formed six half filled sp3d2hybrid orbitals arearranged in octahedral symmetry.

Sulfur atom forms six sp3d2-p bonds with 6 fluorine atoms by using these

sp3d2 orbitals. Each fluorine atom uses is half-filled 2pz orbitals for the bond formation.SF6 is octahedral in shape with bond angles equal to 90o.

* In sp3d3 hybridization, one 's', three 'p' and three 'd' orbitals of almost same energy

intermix to give seven sp3d3 hybrid orbitals, which are oriented in pentagonalbipyramidal symmetry.

* Five among the sp3d3 orbitals are arranged in a pentagonal plane by making 72o of

angles. The remaining are arranged perpendicularly above and below this pentagonalplane.

: Iodine heptafluoride (IF7):


27/68


* The electronic configuration of Iodine atom in the ground state is: [Kr]4d105s25p5.

Since the formation of IF7 requires 7 unpaired electrons, the iodine atom promotesthree of its electrons (one from 5s orbital and two from 5p sublevel) into empty 5dorbitals. This state is referred to as third excited state.

* The electronic configuration of Iodine in the third excited state can be written as:[Kr]4d105s15p35d3.

In the third excited state, iodine atom undergoes sp3d3 hybridization to give 7 halffilled sp3d3 hybrid orbitals in pentagonal bipyramidal symmetry. These will form 7sp3d3-p bonds with fluorine atoms.

Thus the shape of IF7 is pentagonal bipyramidal. The F-I-F bond angles in thepentagonal plane are equal to 72o, whereas two fluorine are present perpendicularly tothe pentagonal plane above and below.


28/68


Introduction to Mole Concept

When carbon (C) reacts with oxygen (O), carbon dioxide is produced.

The chemical equation for the reaction is:

In this reaction, one atom of carbon combines with one molecule or two atoms of

oxygen to form one molecule of carbon dioxide. We can also say that in this chemicalreaction, 12 u of carbon atoms combine with 32 u of oxygen molecules to give 44 u of

carbon dioxide. Hence, the quantity of substances can be represented in terms of the

number of molecules or its mass. However, a chemical equation only indicates the

number of atoms or molecules taking part in the chemical reaction. Therefore, it is

easier to represent the quantity of a substance by the number of atoms or molecules

rather than its mass. In order to do the same, a new term was introduced.

The word is derived from the Latin word moles meaning heap or pile. It

was first introduced by Wilhelm Ostwald in 1896, but was accepted universally

in 1967 as a way of indicating the number of atoms or molecules in a sample.

The molecules of an element are composed of identical atoms. For example, an oxygen

molecule (O2) consists of two oxygen atoms, and a nitrogen molecule (N2) consists of

two nitrogen atoms. N2 and O2 are called . Thus, the atomicity of

nitrogen and oxygen is two When three atoms of oxygen combine, a molecule of ozone

(O3) is formed. Here, the atomicity of oxygen is three.

The atomicity of some common elements is given .

HeliumMonoatomic


29/68


Neon

Argon

Oxygen

Hydrogen

Nitrogen

Chlorine

Fluorine

Diatomic

Phosphorus Tetra-atomic

Sulphur Polyatomic(8 atoms per molecule)

This means that one mole

atom of any substance contains 6.022 1023 atoms. Similarly, one mole molecule of any

substance contains 6.022 1023 molecules and one mole ion of any substance contains

6.022 1023 ions. Hence, the mass of a particular substance is fixed.

The number 6.022 1023 is an experimentally obtained value and is known

as or (represented by ). It is named

after the Italian scientist, Amedeo Avogadro.

Thus, 1 mole of oxygen atoms (O) = 6.022 1023 oxygen atoms

1 mole of oxygen molecules (O2) = 6.022 1023 oxygen molecules

1 mole atoms of an element has a mass equal to the gram atomic mass of that element.

Hence, the mass of 1 mole molecules is equal to its molecular mass in grams. The atomic

mass of an element is the mass of its atom in atomic mass units (u). By taking the same

numerical value of atomic mass and changing the units from u to g, the mass of 1 mole

atoms of that element is obtained. The mass of one mole of atoms is known as the

, , or . For example, the atomic mass of

nitrogen is 14 u and the gram atomic mass of nitrogen is 14 g.


30/68


Similarly, the molecular mass of a substance gives the mass of a molecule of that

substance in atomic mass units (u). Therefore, as discussed earlier, by taking the same

numerical value of molecular mass and by changing its units from u to g, the mass of

one mole molecules of that substance can be obtained.

Therefore, the mass of one mole molecules of any substance is equal to the

of that substance.

For example, 12 u of carbon contains only 1 atom of carbon; and 12 g carbon contains 1

mole atoms of carbon i.e., 6.022 1023 number of carbon atoms. Similarly, the

molecular mass of oxygen (O2) is 32 u. Thus, its gram molecular mass is 32 g. Thus, 32 g

of O2 contains 1 mole molecules of O2 i.e., 6.022 1023 number of molecules of oxygen.

Initially, hydrogen atom was chosen as a standard unit and masses of other atoms were

compared with it. However, this created a problem as hydrogen had 3 isotopes.

Relative mass is obtained by relating mass of an atom of an element or a molecule of a

compound to the mass of lightest atom.

One twelfth mass of an isotope of a carbon atom is called atomic mass unit. Atomic

mass unit is only number and it has no units. On the basis of above unit, the atomic mass

of carbon atom is 12 amu.

It is the ratio of mass of one atom of an element to the mass of an atom of hydrogen

taken as unity.

It is the ratio of mass of one atom of an element to 1/12th mass of an atom of carbon.


31/68


It is the ratio of mass of one molecule of a substance to the mass of an atom of hydrogen

taken as unity.

It is the ratio of mass of one atom of an element to 1/12th mass of an atom of carbon.

The volume occupied by 1 gram molecule of a dry gas at S.T.P is called gram molecularvolume. The experimental value of 1 gram molecular volume of a gas is 22.4 litres at

S.T.P.

The relationship between the mole, Avogadros number, and mass is summarised as

follows:

It helps in determination of atomicity of gases, which occur as elements.The number of atoms present in one molecule of an element is called its atomicity.

: Elements having one atom in their molecules such as helium and neon are

called monatomic while elements having two atoms in their molecules such as hydrogenand oxygen are called diatomic.


32/68


Elements having more than three atoms in their molecules such as sulphur molecule

and carbon molecule are called polyatomic.

Explaining Gay lussacs law Determination of relation between gram molecular mass and gram molecular volume

The mass of one mole molecules of any substance is equal to the of

that substance. The volume occupied by 1 gram molecule of a dry gas at S.T.P is called

gram molecular volume. The experimental value of 1 gram molecular volume of a gas is

22.4 litres at S.T.P.

On applying Avogadros law, we obtain

Gram molecular volume of gas = Gram molecular mass density of gas at S.T.P

Let us solve a numerical based on this concept.

Calculation of gram molecular volume of hydrogen

Mass of hydrogen = 2.016g

Density of hydrogen = 0.09g/

We know that,

Gram molecular volume of gas = Gram molecular mass density of gas at S.T.P

Therefore,

Gram molecular volume of hydrogen = 2.016 g /0.09 g/

= 22.4 at S.T.P

Relation between relative molecular mass {RMM} and vapour density{VD}Relative molecular mass is the ratio of mass of one molecule of a substance to the mass

of an atom of hydrogen taken as unity. Vapour density is the ratio of certain mass of a

gas or vapour to the mass of same volume.

On applying Avogadros law, we obtain

RMM = 2 VD



33/68


Calculation of relative molecular mass of phosphorous

Vapour density of phosphorous = 23

We know,

RMM = 2 VD

Therefore, RMM of phosphorous = 2 23

= 46 amu

Determination of molecular formula of a gasMolecular formula gives the actual number of atoms of the different elements present inone molecule of a compound.

:

Hence, the mole concept can provide us with the following information.

If one mole of a substance (atoms, molecules, or ions) is present, then the number ofelementary particles present in that substance is equal to 6.022 1023.

The mass of one mole of a substance (i.e., atoms, molecules, or ions) is equal to its molarmass.

While carrying out reactions, scientists required the number of atoms and molecules.This requirement was fulfilled by the use of the mole concept as:

1 mole = 6.022 1023 = Relative mass in grams

One mole of gas occupies a volume of 22.4 litres at STP (standard temperature and

pressure conditions i.e., 273 K and 1 atm). The volume occupied by 1 mole of gas at STP

is called the molar volume of gas.


34/68


The mole conceptcan be used in various calculations. The use of the mole concept can be

understood with the help of the following examples.

Interconversion among number of moles, mass and number of molecules


35/68


It refers to the percentage weight of an element present in the formula weight of a

chemical composition.


: Calculation of percentage composition of sulphur in copper sulphate

Gram molecular weight of copper sulphate = 1(Cu) + 1(S) + 4(O)

= 1(64) + 1(32) +4(16)

=160 g

Gram weight of element sulphur in copper sulphate = 1(S) = 132

= 32 g

We know,

Therefore,

Percentage composition of copper in copper sulphate = 32/ 160 100

= 20%

Molecular formula is the chemical formula, which represents the actual number of

atoms of each element present in one molecule of a compound.


36/68


: Sulphuric acid has molecular formula . It contains two hydrogen atoms,

1 sulphur atom, and 4 oxygen atoms.

Empirical formula is the chemical formula, which shows the simplest whole number

ratio between the atoms of various elements in the compound.

: Sulphuric acid has molecular formula . Therefore,

Thus, the empirical formula is 2:1:4.

Let us study this using an example.

To determine the empirical formula of iron oxide from the following percentages:

Fe = 72.41%, O = 27.59%

Atomic weight of Fe = 56

Atomic weight of O = 16

Percentage weight of Fe = 72.41%

Percentage weight of O = 27.59%

After obtaining this information, divide percentage weights of each element with its

atomic weight. Write the answer up till 2 decimal places. This ratio gives the number of

moles of each element.

Relative number of moles of iron = 72.41 / 56 = 1.29

Relative number of moles of oxygen = 27.59 / 16 = 1.72

Next, divide each ratio of relative number of moles obtained by the smallest ratio. This

gives simplest ratio of atoms present in a compound. If the ratios are not whole


37/68


numbers, then multiply them by the smallest integer so that the ratios are whole

numbers.

Here, we will divide each ratio by 1.29.

Simple ratio of iron = 1.29/1.29 = 1

Simple ratio of oxygen = 1.72/ 1.29 = 1.33

To make it a whole number, we will multiply them by the smallest integer so that the

ratios are whole numbers i.e., 3.

Therefore,

Simple ratio of iron = 13 = 3

Simple ratio of oxygen = 1.33 3 = 4

Thus, empirical formula of compound =

Determination of molecular formula of a compound from its percentage composition

Let us study this using an example.

To determine the molecular formula from the following percentage compositions such

that vapour density of compound is 30:

C = 40%, O = 53.33%, and H = 6.67%

Atomic weight of C = 12

Atomic weight of O= 16

Atomic weight of hydrogen = 1

Calculate relative number of moles and simple ratio of atoms as explained above.

Relative number of moles of carbon = 40 / 12 = 3.33

Relative number of moles of oxygen = 53.33/ 16 = 3.33

Relative number of moles of hydrogen = 6.67/1 = 6.67

Next, divide each ratio of relative number of moles obtained by the smallest ratio. This

gives the simplest ratio of atoms present in the compound. If the ratios are not whole


38/68


numbers, then multiply them by the smallest integer so that the ratios are whole

numbers.

Here, we will divide each ratio by 3.33.

Simple ratio of carbon = 3.33/3.33 = 1

Simple ratio of oxygen = 3.33/3.33 = 1

Simple ratio of hydrogen = 6.67 / 3.33 =3

Thus, empirical formula of compound =

Now, empirical formula weight of =1 (C) + 2(H) +1(O)

=1 12+ 2 1 +1 16 = 30

Molecular weight = 2 Vapour density

= 2 30 = 60

N = Molecular weight/ Empirical formula weight

= 60/30 = 2

Therefore, molecular formula = 2 Empirical formula

=

An example of a balanced chemical equation is given below.

From the above balanced chemical equation, the following information is obtained:

One mole of C3H8(g) reacts with five moles of O2(g) to give three moles of CO2(g) andfour moles of H2O(l).

One molecule of C3H8(g) reacts with five molecules of O2(g) to give three moleculesof CO2(g) and four molecules of H2O(l).

44 g of C3H8(g) reacts with (5 32 = 160) g of O2(g) to give (3 44 = 132) g ofCO2 and (4 18 = 72) g of H2O.


39/68


22.4 L of C3H8(g) reacts with (5 22.4) L of O2(g) to give (3 22.4) L of O2 and (4 22.4) L of H2O.

Nitric acid (HNO3) is commercially manufactured by reacting nitrogen

dioxide (NO2) with water (H2O). The balanced chemical equation is represented asfollows:

Calculate the mass of NO2 required for producing 5 moles of HNO3.

According to the given balanced chemical equation, 3 moles of NO2 will

produce 2 moles of HNO3.

Therefore, 2 moles of HNO3 require 3 moles of NO2.

Hence, 5 moles of HNO3 require moles of NO2

= 7.5 moles of NO2

Molar mass of NO2 = (14 + 2 16) g mol1

= 46 g mol1

Thus, required mass of NO2 = (7.5 46) g mol1

= 345 g mol1

Limiting reagent or limiting reactant: Reactant which gets completely consumed when a reaction goes to completion

So called because its concentration limits the amount of the product formedLead nitrate reacts with sodium iodide to give lead iodide and sodium nitrate

in the following manner:

What amount of sodium nitrate is obtained when 30 g of lead nitrate reacts with 30g of

sodium iodide?

Molar mass of


40/68


= 331 g mol1

Molar mass of NaI = (23 + 127) = 150 g mol1

According to the given equation, 1 mole of Pb(NO3)2 reacts with 2 moles of NaI, i.e.,

331 g of Pb(NO3)2 reacts with 300 g of NaI to give PbI2 and NaNO3

Thus, Pb(NO3)2 is the limiting reagent.

Therefore, 30 g of Pb (NO3)2 mole

According to the equation, 0.09 mole of Pb(NO3)2 will give (2 0.09) mole of NaNO3 =

0.18 mole of NaNO3.

Ways for expressing the concentration of a solution

Mass per cent

4.4 g of oxalic acid is dissolved in 200 mL of a solution. What is the mass per

cent of oxalic acid in the solution? (Density of the solution = 1.1 g mL1)

Density of the solution = 1.1 g mL1

So the mass of the solution = (200 mL) (1.1 g mL1)

= 220 g

Mass of oxalic acid = 4.4 g

Therefore, mass per cent of oxalic acid in the solution


41/68


If a substance A dissolves in a substance B, then mole fraction of

A

Mole fraction of B

nA Number of moles of A

nB Number of moles of B

A solution is prepared by dissolving 45 g of a substance (molar mass = 25 g

mol1) in 235 g of a substance (molar mass = 18 g mol1). Calculate the mole fractions

of and .

Moles of X, nX =

= 1.8 mol

Moles of Y, nY =

= 13.06 mol

Therefore, mole fraction of X, nX

And, mole fraction of Y, nY= 1 nX

= 1 0.121


42/68


= 0.879

Molarity:

Number of moles of a solute in 1 L of a solution

Molarity (M) =

Molarity equation:

M1V1 = M2V2

M1 = Molarity of a solution when its volume is V1

M2 = Molarity of the same solution when its volume is V2

10g of HCl is dissolved in enough water to form 500 mL of the solution.

Calculate the molarity of the solution.

Molar mass of HCl = 36.5 g mol1

So the moles of HCl = mol

= 0.274 mol

Volume of the solution = 500 mL = 0.5 L

Therefore, molarity =

= 0.548 M

Commercially available concentrated HCl contains 38% HCl by mass. What

volume of concentrated HCl is required to make 2.5 L of 0.2 M HCl? (Density of the

solution = 1.19 g mL1)

38% HCl by mass means that 38g of HCl is present in 100 g of the solution.


43/68


Moles of HCl =

Volume of the solution

= 84.03 mL

= 0.08403L

Therefore, molarity of the solution =

= 12.38 M

According to molarity equation,

M1V1 = M2V2

Here,

M1 = 12.38 M

M2 = 0.2 M

V2 = 2.5 L

Now, M1V1 = M2V2

Hence, required volume of HCl = 0.0404 L


44/68


Number of moles of solute present in 1 kg of solvent

Molality (m) =

: What is the molality of a solution of glucose in water, which is labelled as

15% (w/w)?

15% (w/w) solution means that 15 g of glucose is present in 100 g of the

solution, i.e., (100 15) g = 85 g of water = 0.085 kg of water

Moles of glucose =

= 0.083 mol

Therefore, molality of the solution

= 0.976 m


45/68


The molecular mass of nitrogen is 28 u.

Mass = Molar mass Number of moles

Mass = 28 g 0.5 = 14 g

1 mole 6.022 1023

1 mole of nitrogen contains 6.022 1023 molecules.

Therefore, 3 moles of nitrogen contains = 6.022 1023 3 molecules =

18 1023 molecules

Molecular weight of lead nitrate

= 1 (Pb) + 2(N) + 6(O)

= 1 207 + 2 14 +6 16

= 331 amu

Thus, the molecular weight of lead nitrate is 331 amu.


46/68


Molecular weight of oxygen ( ) = 2 16 = 32 amu

Thus, gram molecular weight of oxygen ( ) = 32 g

Now, 32 g of oxygen will occupy at S.T.P = 22.4

Therefore, 6 g will occupy at S.T.P = 22.4 6 / 32

= 4.2

: What number of moles contains 3.011 1023 molecules of glucose?

1 mole of glucose is equivalent to 6.022 1023 molecules of glucose.

Hence, 3.011 1023 molecules of glucose will be present in

mol = 0.5 mol (of glucose)

Thus, 0.5 mole of glucose contains 3.011 1023 molecules of glucose.

What is the mass of a mole of fluorine molecule?

1 mole of fluorine molecule contains 6.022 1023 molecules and weighs 38 g.

Therefore, mass of a fluorine molecule = g

= 6.31 1023 g

: One million silver atoms weigh 1.79 x 1016

g. Calculate the gram atomic massof silver.

Number of silver atoms = 1 million = 1 x 106

Mass of one million Ag atoms = 1.79 x 1016 g

= 107.8 g

Gram atomic mass of silver is equal to the mass of 6.023 x 1023 atoms of silver. So, the

gram atomic mass of silver is 107.8 g.


47/68


: A sample of gaseous substance weighing 0.5 g occupies a volume of 1.12 litre

under NTP conditions. Calculate the molar mass of the substance.

1 mole of any gaseous substance at NTP occupies 22.4 L.1.12 L of gaseous substance = 0.5 g

The molar mass of the substance therefore is 10 g/mol.

Determine the mass in grams of 3.50 moles of Copper.

We know that moles are associated with mass in grams, by the relationship

Moles of a compound/element =

.

Thus, Moles of copper = 3.50 moles.

Molar mass of Cu = 63.55 grams/mol.

So, this can be written as: 63.55 g Cu / 1 mole of Cu

Multiplying the number of moles with this factor

= 3.50 moles of Cu x 63.55 g Cu/1 mole of Cu

222 grams of Copper

Determine the number of moles represented by 237 grams of Strontium.

Molar mass of Strontium = 38 grams/mol.

So, Molar mass of Strontium = 38 grams of Sr/ 1 mole of Sr.

Multiplying the number of grams by this factor:

237 g of Sr x 1 mole of Sr/38 grams of Sr

Moles of Strontium = 6.24 moles.

Find the mass in grams of 8.6 moles of Bromine atoms.

Molar mass of Bromine = 80 grams/mol.

So, Molar mass of Bromine = 80 grams of Br / 1 mole of Br.

Multiplying the number of moles by this factor,

= 8.6 moles of Br x 80 grams of Br/1 mole of Br

= 688 grams of Br.
http://chemistry.tutorvista.com/inorganic-chemistry/molar-mass.htmlhttp://chemistry.tutorvista.com/inorganic-chemistry/molar-mass.html


48/68



49/68


The Gas LawsOne of the most amazing things about gases is that, despite wide differences in chemical

properties, all the gases more or less obey the gas laws. The gas laws deal with how

gases behave with respect to pressure, volume, temperature, and amount.

Gases are the only state of matter that can be compressed very tightly or expanded to fill

a very large space. Pressure is force per unit area, calculated by dividing the force by

the area on which the force acts. The earth's gravity acts on air molecules to create a

force, that of the air pushing on the earth. This is called atmospheric pressure.

The units of pressure that are used are pascal (Pa), standard atmosphere (atm), andtorr. 1 atm is the average pressure at sea level. It is normally used as a standard unit ofpressure. The SI unit though, is the pascal. 101,325 pascals equals 1 atm.

For laboratory work the atmosphere is very large. A more convient unit is the torr. 760torr equals 1 atm. A torr is the same unit as the mmHg (millimeter of mercury). It is thepressure that is needed to raise a tube of mercury 1 millimeter.

When seventeenth-century scientists began studying the physical properties of gases,they noticed some simple relationships between some of the measurable properties ofthe gas. Take pressure ( ) and volume ( ), for example. Scientists noted that for agiven amount of a gas (usually expressed in units of moles [ ]), if the temperature ( )of the gas was kept constant, pressure and volume were related: As one increases, theother decreases. As one decreases, the other increases. We say that pressure andvolume are .

There is more to it, however: pressure and volume of a given amount of gas at constanttemperature are related. If you take the pressure value and multiply it bythe volume value, the product is a constant for a given amount of gas at a constanttemperature:

= constant at constant and

If either volume or pressure changes while amount and temperature stay the same, then

the other property must change so that the product of the two properties still equals


50/68


that same constant. That is, if the original conditions are labeled 1 and 1 and the newconditions are labeled 2 and 2 , we have

1 1 = constant = 2 2

where the properties are assumed to be multiplied together. Leaving out the middlepart, we have simply

1 1 = 2 2 at constant and

This equation is an example of a gas law. A gas law is a simple mathematical formulathat allows you to model, or predict, the behavior of a gas. This particular gas law iscalled Boyle ' s law , after the English scientist Robert Boyle, who first announced it in1662. Figure shows two representations of how Boyle ' s law works.

A piston having a certain pressure and volume (left piston) will have half the volumewhen its pressure is twice as much (right piston). One can also plot versus for agiven amount of gas at a certain temperature; such a plot will look like the graph on theright.

Boyle ' s law is an example of a second type of mathematical problem we see inchemistry-one based on a mathematical formula. Tactics for working with mathematicalformulas are different from tactics for working with conversion factors. First, most ofthe questions you will have to answer using formulas are word-type questions, so the


51/68


first step is to identify what quantities are known and assign them to variables. Second,in most formulas, some mathematical rearrangements (i.e., algebra) must be performedto solve for an unknown variable. The rule is that to find the value of the unknownvariable, you must mathematically isolate the unknown variable

of one side of the equation. Finally, units must be consistent. For example, in

Boyle ' s law there are two pressure variables; they must have the same unit. There arealso two volume variables; they also must have the same unit. In most cases, it won ' tmatter the unit is, but the unit must be the on both sides of the equation.

Rita has two cylinders. One is empty and the other contains compressed

nitrogen at 25 atm. She wants to distribute the gas in the two cylinders. To do so, she

connects the two cylinders. If the volume of the cylinder containing the gas is 50 L and

that of the empty one is 80 L, then what will be the pressure inside the two cylinders?

According to Boyles law,

Given,p1 = 25 atm

V1 = 50 L

V2 = ( 50 + 80) L = 130 L

Now, 25 atm 50 L = p2 130 L

Hence, the pressure inside the cylinders is 9.62 atm.

A sample of gas has an initial pressure of 2.44 atm and an initial volume of

4.01 L. Its pressure changes to 1.93atm. What is the new volume if temperature and

amount are kept constant?

First, determine what quantities we are given. We are given an initial pressureand an initial volume, so let these values be P 1 and V 1:

P 1 = 2.44 atm and V 1 = 4.01 L

We are given another quantity, final pressure of 1.93 atm, but not a final volume. This

final volume is the variable we will solve for.


52/68


P 2 = 1.93 atm and V 2 = ? L

Substituting these values into Boyle ' s law, we get

(2.44 atm)(4.01 L) = (1.93 atm) V 2

To solve for the unknown variable, we isolate it by dividing both sides of the equationby 1.93 atm-both the number and the unit:

Note that, on the left side of the equation, the unit atm is in the numerator and thedenominator of the fraction. They cancel algebraically, just as a number would. On theright side, the unit atm and the number 1.93 are in the numerator and the denominator,so the entire quantity cancels:

What we have left is

Now we simply multiply and divide the numbers together and combine the answer withthe L unit, which is a unit of volume. Doing so, we get

V 2 = 5.07 L

Does this answer make sense? We know that pressure and volume are inversely related;as one decreases, the other increases. Pressure is decreasing (from 2.44 atm to 1.93atm), so volume should be increasing to compensate, and it is (from 4.01 L to 5.07 L). Sothe answer makes sense based on Boyle ' s law.

As mentioned, you can use any units for pressure or volume, but both pressures must beexpressed in the same units, and both volumes must be expressed in the same units.

A sample of gas has an initial pressure of 722 torr and an initial volume of

88.8 mL. Its volume changes to 0.663L. What is the new pressure?

We can still use Boyle ' s law to answer this, but now the two volumequantities have different units. It does not matter which unit we change, as long as weperform the conversion correctly. Let us change the 0.663 L to milliliters:

Now that both volume quantities have the same units, we can substitute into Boyle ' slaw:

The mL units cancel, and we multiply and divide the numbers to get

2 = 96.7 torr

The volume is increasing, and the pressure is decreasing, which is as expected for Boyle

' s law.


53/68


There are other measurable characteristics of a gas. One of them is temperature ( ).Perhaps one can vary the temperature of a gas sample and note what effect it has on the

other properties of the gas. Early scientists did just this, discovering that if the amountof a gas and its pressure are kept constant, then changing the temperature changes thevolume ( ). As temperature increases, volume increases; as temperature decreases,volume decreases. We say that these two characteristics are .

A mathematical relationship between and should be possible except for onethought: what temperature scale should we use? We know from

We can modify this equation as we modified Boyle ' s law: the initialconditions 1 and 1 have a certain value, and the value must be the same when theconditions of the gas are changed to some new conditions 2 and 2 , as long as

pressure and the amount of the gas remain constant. Thus, we have another gas law:

This gas law is commonly referred to as Charles ' s law , after the French scientistJacques Charles, who performed experiments on gases in the 1780s. The tactics forusing this mathematical formula are similar to those for Boyle ' s law. To determine anunknown quantity, use algebra to isolate the unknown variable by itself and in thenumerator; the units of similar variables must be the same. But we add one more tactic:all temperatures must be expressed in the absolute temperature scale (Kelvin). As areminder, we review the conversion between the absolute temperature scale and theCelsius temperature scale:

K = C +273

where K represents the temperature in kelvins, and C represents the temperature indegrees Celsius.

Figure shows two representations of how Charles ' s law works.


54/68


A piston having a certain volume and temperature (left piston) will have twice the

volume when its temperature is twice as much (right piston). One can also

plot versus for a given amount of gas at a certain pressure; such a plot will look like

the graph on the right.

A sample of gas has an initial volume of 34.8 mL and an initial temperature of

315 K. What is the new volume if the temperature is increased to 559 K? Assume

constant pressure and amount for the gas.

First, we assign the given values to their variables. The initial volume is 1 ,so 1 = 34.8 mL, and the initial temperature is 1 , so 1 = 315 K. The temperature isincreased to 559 K, so the final temperature 2 = 559 K. We note that the temperaturesare already given in kelvins, so we do not need to convert the temperatures.Substituting into the expression for Charles ' s law yields

We solve for 2 by algebraically isolating the 2 variable on one side of the equation.We do this by multiplying both sides of the equation by 559 K (number and unit). Whenwe do this, the temperature unit cancels on the left side, while the entire 559 K cancelson the right side:

The expression simplifies to


55/68


By multiplying and dividing the numbers, we see that the only remaining unit is mL, soour final answer is

2 = 61.8 mL

Does this answer make sense? We know that as temperature increases, volumeincreases. Here, the temperature is increasing from 315 K to 559 K, so the volumeshould also increase, which it does.

It is more mathematically complicated if a final temperature must be calculated becausethe variable is in the denominator of Charles ' s law. There are several mathematicalways to work this, but perhaps the simplest way is to take the reciprocal of Charles ' slaw. That is, rather than write it as

Write the equation as

It is still an equality and a correct form of Charles ' s law, but now the temperaturevariable is in the numerator, and the algebra required to predict a final temperature issimpler.

A sample of a gas has an initial volume of 34.8 L and an initial temperature of

-67C. What must be the temperature of the gas for its volume to be 25.0 L?

Here, we are looking for a final temperature, so we will use the reciprocal form

of Charles ' s law. However, the initial temperature is given in degrees Celsius, notkelvins. We must convert the initial temperature to kelvins:

-67C +273 = 206 K

In using the gas law, we must use 1 = 206 K as the temperature. Substituting into thereciprocal form of Charles ' s law, we get

Bringing the 25.0 L quantity over to the other side of the equation, we get

The L units cancel, so our final answer is

2 = 148 K

This is also equal to -125C. As temperature decreases, volume decreases, which it doesin this example.

It is desired to increase the volume of 5 L of a gas by 40% without changing

the pressure. To what temperature should the gas be heated, if its initial temperature is

298 K?


56/68


Desired increase in the volume of gas = 40% of 5 L

= 2 L

Therefore, final volume of the gas = (5 + 2) L = 7 L

Applying Charles law,

Now, V1 = 5 L

T1 = 298 K

V2 = 7 L

Therefore,

You may notice in Boyle ' s law and Charles ' s law that we actually refer to four physicalproperties of a gas: pressure ( ), volume ( ), temperature ( ), and amount (inmoles; ). We do this because these are the only four independent physical propertiesof a gas. There are other physical properties, but they are all related to one (or more) ofthese four properties.

Boyle ' s law is written in terms of two of these properties, with the other two beingheld constant. Charles ' s law is written in terms of two different properties, with theother two being held constant. It may not be surprising to learn that there are other gaslaws that relate other pairs of properties-as long as the other two are held constant.Here we will mention a few.


57/68


relates pressure with absolute temperature. In terms of two sets ofdata, Gay-Lussac ' s law is

Note that it has a structure very similar to that of Charles ' s law, only with differentvariables-pressure instead of volume.

: An iron tank contains helium at a pressure of 3.0 atm at 300 K. The tank

can withstand a maximum pressure of 12.0 atm. The building in which the tank has

been placed catches fire. Predict whether the tank will blow up first or melt. (Given,

melting point of iron is 1808 K)

Solution: To calculate the pressure built up in the tank at the melting point of iron, we

have to apply Gay-Lussacs law.

According to Gay-Lusaacs Law,

Here, p1 = 3.0 atm

T1 = 300 K

T2 = 1808 K

Now,

= 18.08 atm

It is found that pressure of the gas in the tank is much more than 12 atm at the melting

point. Hence, the tank will blow up before reaching the melting point.


58/68


introduces the last variable for amount. The original statement ofAvogadro ' s law states that equal volumes of different gases at the same temperature

and pressure contain the same number of particles of gas. Because the number ofparticles is related to the number of moles (1 mol = 6.022 10 23 particles), Avogadro 's law essentially states that equal volumes of different gases at the same temperatureand pressure contain the same (moles, particles) of gas. Put mathematically intoa gas law, Avogadro ' s law is

(First announced in 1811, it was Avogadro ' s proposal that volume is related to thenumber of particles that eventually led to naming the number of things in a mole asAvogadro ' s number.) Avogadro ' s law is useful because for the first time we are seeingamount, in terms of the number of moles, as a variable in a gas law.

A 2.45 L volume of gas contains 4.5 10 21 gas particles. How many gas

particles are there in 3.87 L if the gas is at constant pressure and temperature?

We can set up Avogadro ' s law as follows:

We algebraically rearrange to solve for 2 :

The L units cancel, so we solve for 2 :

2 = 7.1 10 21 particles

The variable in Avogadro ' s law can also stand for the number of moles of gas inaddition to number of particles.

One thing we notice about all the gas laws is that, collectively, volume and pressure arealways in the numerator, and temperature is always in the denominator. This suggeststhat we can propose a gas law that combines pressure, volume, and temperature. Thisgas law is known as the combined gas law , and its mathematical form is

This allows us to follow changes in all three major properties of a gas. Again, the usualwarnings apply about how to solve for an unknown algebraically (isolate it on one sideof the equation in the numerator), units (they must be the same for the two similarvariables of each type), and units of temperature must be in kelvins.

A sample of gas at an initial volume of 8.33 L, an initial pressure of 1.82 atm,

and an initial temperature of 286 K simultaneously changes its temperature to 355 K

and its volume to 5.72 L. What is the final pressure of the gas?

We can use the combined gas law directly; all the units are consistent witheach other, and the temperatures are given in Kelvin. Substituting,


59/68


We rearrange this to isolate the 2 variable all by itself. When we do so, certain unitscancel:

Multiplying and dividing all the numbers, we get

2 = 3.29 atm

Ultimately, the pressure increased, which would have been difficult to predict becausetwo properties of the gas were changing.

As with other gas laws, if you need to determine the value of a variable in thedenominator of the combined gas law, you can either cross-multiply all the terms or justtake the reciprocal of the combined gas law. Remember, the variable you are solving formust be in the numerator and all by itself on one side of the equation.

It is clear from Boyles law and Charles law that the volume of a given mass of gas

depends upon the temperature and pressure of the gas. Thus, for comparing the masses

or densities of two or more gases having same volume, we need to standardise the

temperature and pressure at which the volume of the gases is measured.

The standard temperature and standard pressure for all gases is known as the

and . In short form, it is written as

The standard pressure-temperature is taken as 0C or 273 K.

The standard pressure is taken as 1 atm or 760 mm of mercury.

One mole of gas occupies a volume of 22.4 litres at STP (standard temperature and

pressure conditions i.e. 273 K and 1 atm).

Here is an application of this:

A gas X occupies a volume of 512 cm3. at S.T.P. What will be the volume occupied by X

at 300 K and at a pressure of 720 mm?

According to the gas equation,

Here, initial pressure of the gas, p1 = standard pressure

= 760 mm of mercury


60/68


Final pressure of the gas, Pp2 = 720 mm

Initial volume of the gas, V1 = 512 cm3

Final volume of the gas, V2 = ?

Initial temperature of the gas, T1 = standard temperature

= 273 K

Final temperature of the gas, T2 = 300 K

Now,

Hence, the volume of the gas occupied by X at 300 K and at a pressure of 720 mm will

be 593.89 cm3.


61/68


So far, the gas laws we have considered have all required that the gas change itsconditions; then we predict a resulting change in one of its properties. Are there any gas

laws that relate the physical properties of a gas at any given time?

Consider a further extension of the combined gas law to include . By analogy toAvogadro ' s law, is positioned in the denominator of the fraction, opposite thevolume. So

Because pressure, volume, temperature, and amount are the only four independentphysical properties of a gas, the constant in the above equation is truly a constant;indeed, because we do not need to specify the identity of a gas to apply the gas laws, thisconstant is the same for all gases. We define this constant with the symbol , so theprevious equation is written as

which is usually rearranged as

=

This equation is called the ideal gas law . It relates the four independent properties of agas at any time. The constant is called the ideal gas law constant. Its value depends onthe units used to express pressure and volume.

0.08205

62.36

8.314

The ideal gas law is used like any other gas law, with attention paid to the unit and

making sure that temperature is expressed in Kelvin. However,. The ideal gas law implies that if

you know any three of the physical properties of a gas, you can calculate the fourthproperty.

A 4.22 mol sample of Ar has a pressure of 1.21 atm and a temperature of

34C. What is its volume?

The first step is to convert temperature to kelvins:

34 +273 = 307 K


62/68


Now we can substitute the conditions into the ideal gas law:

The unit is in the numerator of both sides, so it cancels. On the right side of theequation, the and units appear in the numerator and the denominator, so theycancel as well. The only unit remaining is , which is the unit of volume that we are

looking for. We isolate the volume variable by dividing both sides of the equation by1.21:

Then solving for volume, we get

= 87.9 L

: At a given temperature, 0.00332 g of Hg in the gas phase has a pressure of

0.00120 mmHg and a volume of 435 L. What is its temperature?

: We are not given the number of moles of Hg directly, but we are given a mass.We can use the molar mass of Hg to convert to the number of moles.

Pressure is given in units of millimeters of mercury. We can either convert this toatmospheres or use the value of the ideal gas constant that includes the mmHg unit. Wewill take the second option. Substituting into the ideal gas law,

The mmHg, L, and mol units cancel, leaving the K unit, the unit of temperature.Isolating all by itself on one side, we get

Then solving for K, we get

= 1,404 K

The ideal gas law can also be used in stoichiometry problems.

What volume of H 2 is produced at 299 K and 1.07 atm when 55.8 g of Zn

metal react with excess HCl?

Zn(s) +2HCl(aq) -ZnCl 2 (aq) +H 2 (g)

Here we have a stoichiometry problem where we need to find the number ofmoles of H 2 produced. Then we can use the ideal gas law, with the given temperatureand pressure, to determine the volume of gas produced. First, the number of moles ofH 2 is calculated:

Now that we know the number of moles of gas, we can use the ideal gas law todetermine the volume, given the other conditions:

All the units cancel except for L, for volume, which means


63/68


= 19.6 L

It should be obvious by now that some physical properties of gases depend strongly onthe conditions. What we need is a set of standard conditions so that properties of gases

can be properly compared to each other. Standard temperature and pressure (STP) isdefined as exactly 100 kPa of pressure (0.986 atm) and 273 K (0C). For simplicity, wewill use 1 atm as standard pressure. Defining STP allows us to compare more directlythe properties of gases that differ from each other.

One property shared among gases is a molar volume. The molar volume is the volume of1 mol of a gas. At STP, the molar volume of a gas can be easily determined by using theideal gas law:

All the units cancel except for L, the unit of volume. So

= 22.4 L

Note that we have not specified the identity of the gas; we have specified only that thepressure is 1 atm and the temperature is 273 K. This makes for a very usefulapproximation: ; that is, the molarvolume at STP is 22.4 L/mol ( Figure6.4, " Molar Volume " ). This molar volume makes auseful conversion factor in stoichiometry problems if the conditions are at STP. If theconditions are not at STP, a molar volume of 22.4 L/mol is not applicable. However, ifthe conditions are not at STP, the combined gas law can be used to calculate what thevolume of the gas would be if at STP; then the 22.4 L/mol molar volume can be used.

A mole of gas at STP occupies 22.4 L, the volume of a cube that is 28.2 cm on a side.


64/68


The ideal gas law can also be used to determine the densities of gases. Density, recall, isdefined as the mass of a substance divided by its volume:

Assume that you have exactly 1 mol of a gas. If you know the identity of the gas, you candetermine the molar mass of the substance. Using the ideal gas law, you can also

determine the volume of that mole of gas, using whatever the temperature and pressureconditions are. Then you can calculate the density of the gas by using

What is the density of N 2 at 25C and 0.955 atm?

First, we must convert the temperature into kelvins:

25 +273 = 298 K

If we assume exactly 1 mol of N 2 , then we know its mass: 28.0 g. Using the ideal gas law,we can calculate the volume:

All the units cancel except for L, the unit of volume. So

= 25.6 L

Knowing the molar mass and the molar volume, we can determine the density ofN 2 under these conditions:

: A vessel of 200 mL capacity contains a certain amount of gas at 27C and 0.9bar pressure. The gas is then transferred into another vessel of capacity 150 mL at 27C.

What would be the pressure of the gas in the vessel of capacity 150 mL?

According to combined gas law,

Here, initial pressure of the gas, p1 = 0.9 bar

Final pressure of the gas, p2 = ?

Initial volume of the gas, V1 = 200 mL

Final volume of the gas, V2 = 150 mL

Initial temperature of the gas, T1 = (27 + 273) K = 300 K

Final temperature of the gas, T2 = (27 + 273) K = 300 K


65/68


Now,

= 1.2 bar

Hence, the pressure of the gas in the vessel of capacity 150 mL would be 1.2 bar.

How many grams of nitrogen are present in an 8.21 L sample of a gas at 5 atm

and 23C?

It is given that,

V= 8.21 L

p= 5 atm

T= (23 + 273)K = 250 K

Here, R = 0.0821 L atm K1 mol1

From the ideal gas equation, we have

pV= nRT

=

= 2 mol

Molecular mass of nitrogen = 28 g

This means that 1 mole of nitrogen molecules contains 28 g of nitrogen.

Therefore, 2 moles of nitrogen molecules contain 2 28 g, i.e., 56 g of nitrogen.


66/68


Thus, at 5 atm and 23C, 56 g of nitrogen gas are present in an 8.21 L sample of the

gas.

Gases were among the first substances studied in terms of the modern scientificmethod, which was developed in the 1600s. It did not take long to recognize that gasesall shared certain physical behaviors, suggesting that all gases could be described byone all-encompassing theory. Today, that theory is thekinetic theory of gases . It isbased on the following statements:

1. Gases consist of tiny particles of matter that are in constant motion.2. Gas particles are constantly colliding with each other and the walls of a

container. These collisions are elastic; that is, there is no net loss of energy fromthe collisions.

3.

Gas particles are separated by large distances, with the size of a gas particle tinycompared to the distances that separate them.4. There are no interactive forces (i.e., attraction or repulsion) between the

particles of a gas.5. The average speed of gas particles is dependent on the temperature of the gas.

Figure shows a representation of how we mentally picture the gas phase.


67/68


The kinetic theory of gases describes this state of matter as composed of tiny particlesin constant motion with a lot of distance between the particles.

This model of gases explains some of the physical properties of gases. Because most of agas is empty space, a gas has a low density and can expand or contract under the

appropriate influence. The fact that gas particles are in constant motion means that twoor more gases will always mix, as the particles from the individual gases move andcollide with each other.

An ideal

16 class 10 semester 2 chemistry

Documents