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Part I

Textbook for Class XI

Page 133


molecular orbital, the electron waves cancel
each other due to destructive interference. As
a result, the electron density in a bonding
molecular orbital is located between the nuclei
of the bonded atoms because of which the
repulsion between the nuclei is very less while
in case of an antibonding molecular orbital,
most of the electron density is located away
from the space between the nuclei. Infact, there
is a nodal plane (on which the electron density
is zero) between the nuclei and hence the
repulsion between the nuclei is high. Electrons
placed in a bonding molecular orbital tend to
hold the nuclei together and stabilise a
molecule. Therefore, a bonding molecular
orbital always possesses lower energy than
either of the atomic orbitals that have combined
to form it. In contrast, the electrons placed in
the antibonding molecular orbital destabilise
the molecule. This is because the mutual
repulsion of the electrons in this orbital is more
than the attraction between the electrons and
the nuclei, which causes a net increase in

It may be noted that the energy of the
antibonding orbital is raised above the energy
of the parent atomic orbitals that have
combined and the energy of the bonding
orbital has been lowered than the parent
orbitals. The total energy of two molecular
orbitals, however, remains the same as that
of two original atomic orbitals.

4.7.2 Conditions for the Combination of
Atomic Orbitals

The linear combination of atomic orbitals to
form molecular orbitals takes place only if the
following conditions are satisfied:

1.The combining atomic orbitals must
have the same or nearly the same energy.
This means that 1s orbital can combine with
another 1s orbital but not with 2s orbital
because the energy of 2s orbital is appreciably
higher than that of 1s orbital. This is not true
if the atoms are very different.

2.The combining atomic orbitals must
have the same symmetry about the
molecular axis. By convention z-axis is

taken as the molecular axis. It is important
to note that atomic orbitals having same
or nearly the same energy will not combine
if they do not have the same symmetry.
For example, 2p

orbital of one atom can

combine with 2p
orbital of the other atom

but not with the 2p
or 2p

orbitals because

of their different symmetries.

3.The combining atomic orbitals must
overlap to the maximum extent. Greater
the extent of overlap, the greater will be the
electron-density between the nuclei of a
molecular orbital.

4.7.3 Types of Molecular Orbitals

Molecular orbitals of diatomic molecules are
designated as σ (sigma), π (pi), δ (delta), etc.

In this nomenclature, the sigma (σσσσσ)
molecular orbitals are symmetrical around
the bond-axis while pi (πππππ) molecular orbitals
are not symmetrical. For example, the linear
combination of 1s orbitals centered on two
nuclei produces two molecular orbitals which
are symmetrical around the bond-axis. Such
molecular orbitals are of the σ type and are
designated as σ1s and σ*1s [Fig. 4.20(a),page
124]. If internuclear axis is taken to be in
the z-direction, it can be seen that a linear
combination of 2p

- orbitals of two atoms

also produces two sigma molecular orbitals
designated as σσσσσ2pz and σσσσσ*2pz. [Fig. 4.20(b)]

Molecular orbitals obtained from 2px and
2py orbitals are not symmetrical around the
bond axis because of the presence of positive
lobes above and negative lobes below the
molecular plane. Such molecular orbitals, are
labelled as π and π* [Fig. 4.20(c)]. A π bonding
MO has larger electron density above and
below the inter -nuclear axis. The π*
antibonding MO has a node between the nuclei.

4.7.4 Energy Level Diagram for Molecular

We have seen that 1s atomic orbitals on two
atoms form two molecular orbitals designated
as σ1s and σ*1s. In the same manner, the 2s
and 2p atomic orbitals (eight atomic orbitals
on two atoms) give rise to the following eight
molecular orbitals:

Page 266


surroundings when additional solvent is
added to the solution. The enthalpy of dilution
of a solution is dependent on the original
concentration of the solution and the amount
of solvent added.

6.6(c) Entropy and Second Law of

We know that for an isolated system the change
in energy remains constant. Therefore,
increase in entropy in such systems is the
natural direction of a spontaneous change.
This, in fact is the second law of
thermodynamics. Like first law of
thermodynamics, second law can also be
stated in several ways. The second law of
thermodynamics explains why spontaneous
exothermic reactions are so common. In
exothermic reactions heat released by the
reaction increases the disorder of the
surroundings and overall entropy change is
positive which makes the reaction

6.6(d) Absolute Entropy and Third Law of

Molecules of a substance may move in a
straight line in any direction, they may spin
like a top and the bonds in the molecules may
stretch and compress. These motions of the
molecule are called translational, rotational
and vibrational motion respectively. When
temperature of the system rises, these motions
become more vigorous and entropy increases.
On the other hand when temperature is
lowered, the entropy decreases. The entropy
of any pure crystalline substance
approaches zero as the temperature
approaches absolute zero. This is called
third law of thermodynamics. This is so
because there is perfect order in a crystal at
absolute zero. The statement is confined to pure
crystalline solids because theoretical
arguments and practical evidences have shown
that entropy of solutions and super cooled
liquids is not zero at 0 K. The importance of
the third law lies in the fact that it permits the
calculation of absolute values of entropy of

pure substance from thermal data alone. For
a pure substance, this can be done by


increments from 0 K to 298 K.

Standard entropies can be used to calculate
standard entropy changes by a Hess’s law type
of calculation.

UNIT VII: Equilibrium

7.12.1 Designing Buffer Solution

Knowledge of pK
, pK

and equilibrium

constant help us to prepare the buffer solution
of known pH. Let us see how we can do this.

Preparation of Acidic Buffer
To prepare a buffer of acidic pH we use weak
acid and its salt formed with strong base. We
develop the equation relating the pH, the
equilibrium constant, K

of weak acid and ratio

of concentration of weak acid and its conjugate
base. For the general case where the weak acid
HA ionises in water,

HA + H

O+ + A–

For which we can write the expression

+ –[H O ] [A ]


Rearranging the expression we have,


[H O ]

–[A ]

Taking logarithm on both the sides and
rearranging the terms we get -

–[A ]
p pH log



–[A ]

pH= p log


–[Conjugate base, A ]

pH= p log


Page 267


The expression (A-2) is known as
Henderson–Hasselbalch equation. The

–[A ]

is the ratio of concentration of

conjugate base (anion) of the acid and the acid
present in the mixture. Since acid is a weak
acid, it ionises to a very little extent and
concentration of [HA] is negligibly different from
concentration of acid taken to form buffer. Also,
most of the conjugate base, [A—], comes from
the ionisation of salt of the acid. Therefore, the
concentration of conjugate base will be
negligibly different from the concentration of
salt. Thus, equation (A-2) takes the form:

H= log

p pK

In the equation (A-1), if the concentration
of [A—] is equal to the concentration of [HA],
then pH = pK

because value of log 1 is zero.

Thus if we take molar concentration of acid and
salt (conjugate base) same, the pH of the buffer
solution will be equal to the pK

of the acid. So

for preparing the buffer solution of the required
pH we select that acid whose pK

is close to the

required pH. For acetic acid pK
value is 4.76,

therefore pH of the buffer solution formed by
acetic acid and sodium acetate taken in equal
molar concentration will be around 4.76.

A similar analysis of a buffer made with a
weak base and its conjugate acid leads to the

+[Conjugate acid,BH ]

pOH=p +log



pH of the buffer solution can be calculated
by using the equation pH + pOH =14.

We know that pH + pOH = pK


+ pK

= pK

On putting these values in

equation (A-3) it takes the form as follows:

w w

[Conjugate acid,BH ]
p - pH=p p log



[Conjugate acid,BH ]

pH = p log


If molar concentration of base and its

conjugate acid (cation) is same then pH of the
buffer solution will be same as pK

for the base.


value for ammonia is 9.25; therefore a
buffer of pH close to 9.25 can be obtained by
taking ammonia solution and ammonium
chloride solution of equal molar concentration.
For a buffer solution formed by ammonium
chloride and ammonium hydroxide, equation
(A-4) becomes:

[Conjugate acid,BH ]

pH=9.25 log

pH of the buffer solution is not affected by
dilution because ratio under the logarithmic
term remains unchanged.

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