## LiCl is hydrated, but NaCl is always anhydrous. Why ?

Q: (a)   LiCl is hydrated, but NaCl is always anhydrous. Why ?

(b)   Why is it that although Li+ is far smaller than the other metals ions, it moves through a solution less rapidly than the others ?

Solution: (a)   Li+ is smaller in size and hence it is more hydrated than Na+ which has a larger size.

(b)   Smaller the size of cation, larger the charge, then greater the hydration. The dense charge of Li+ attracts several layers of water molecules around it. They increase the effective size of the ion, thus slowing it down.

## Why Al(OH)3 is amphoteric in nature ?

Q: Why Al(OH)3 is amphoteric in nature ?

Solution:Since Al – O and O – H bond strength are almost equal, there exist almost equal probability of their breaking making them amphoteric in nature.

## The C=C bond dissociation energy is 146 kcal/mole whereas C – C bond dissociation energy is 83 kcal/mole then whey paraffins are less reactive than olefins .

Q: The C=C bond dissociation energy is 146 kcal/mole whereas C – C bond dissociation energy is 83 kcal/mole then whey paraffins are less reactive than olefins .

Solution: In most olefin reactions only 1 bond is broken & it is the π bond that is broken. This bond has dissociation energy of 146 – 83 = 63 kcal, which is less than the C – C  σ bond energy of 83 kcal/mole. This is why olefins are more reactive than parafins.

## Interpret the non- linear shape of H2S molecule and non – planar shape of PCl3 using valence shell electron pair repulsion theory (VSEPR)

Q: Interpret the non- linear shape of H2S molecule and non – planar shape of PCl3 using valence shell electron pair repulsion theory (VSEPR)

Solution:   In H2S, two bonded pair & two lone pairs are present i.e. sulphur is in sp3 hybridized state. The angle is less than 109°28I as contraction occurs due to presence of lone pairs. Thus H2S have V shaped structure. In PCl3 three bond pairs & one lone pair are present i.e. phosphorus is also in sp3 hybridized state but it has pyramidal structure.

## A diatomic molecule has dipole moment of 1.2 D. If the bond distance is 1.0 A° . What fraction of an electronic charge ‘e’ exists on each atom?

Q: A diatomic molecule has dipole moment of 1.2 D. If the bond distance is 1.0 A°. What fraction of an electronic charge ‘e’ exists on each atom ?

Sol: $\large Partial \; Charge = \frac{Dipole \;moment}{Bond \;distance}$

$\large = \frac{1.2 \times 10^{-18} esu \;cm}{1.0 \times 10^{-8}cm}$

= 1.2 × 10-10 esu

The fraction of an electronic Charge is

$\large = \frac{1.2 \times 10^{-10}}{4.8 \times 10^{-10}}$

= 0.25

= 25% of ‘e’

## Arrange the following in increasing order of stability O2, O2^+, O2^-, O2^2-

Q: Arrange the following in increasing order of stability O2, O2+, O2, O22-

Solution: O22- < O2 < O2 < O2+

Calculate first the bond order which is as follows

O2 – 2, O2+ – 2.5, O2 – 1.5, O22- -1 & then arrange according to increasing bond order.

## Predict the shapes of the following ions

Q: Predict the shapes of the following ions

(a)   BeF3

(b) BF4

(c) IF4

(d)   IBr2

Solution:

(a)   Triangular

(b)   Tetrahedral

(c)   Square planar

(d)   Linear