Solved Problems of Chemical Bonding

Prob 1. Arrange the bonds in order of increasing ionic character in the molecules:

LiF, K₂O, N₂, SO₂ and ClF₃

Solution: N₂ < ClF₃ < SO₂ < K₂O < LiF

Prob 2. Arrange the following in order of increasing ionic character:

C-H, F-H, Br-H, Na-I, K-F and Li-Cl

Solution: C-H < Br-H < F-H < Na-I < Li-Cl < K-F

Prob 3. 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

Prob 4. Arrange the following in increasing order of stability O₂, O₂⁺, O₂^-, O₂^2-

Solution: O₂^2- < O₂^- < O₂ < O₂⁺

Calculate first the bond order which is as follows

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

Prob 5. Arrange the following:

(i) N₂, O₂, F₂, Cl₂ in increasing order of bond dissociation energy.

(ii) Increasing strength of hydrogen bonding (X – H – X):

O, S, F, Cl, N

Solution: (i) F₂ < Cl₂ < O₂ < N₂ (ii) Cl < S < N < O < F

Prob 6. Explain the following o - hydroxy benzaldehyde is liquid at room temperature while p - hydroxy benzaldehyde is high melting solid.

Solution: There is intramolecular H bonding in o - hydroxy benzaldehyde while intermolecular hydrogen bonding in p-hydroxy benzaldehyde

Prob 7. Explain why ClF₂^- is linear but ClF₂⁺ is a bent molecular ion?

Solution: Chlorine atom lies in sp³d hybrid state. Three lone pairs are oriented along the corners of triangular plane

Chlorine atom lies in sp³ hybrid state. Two lone pairs are oriented along two corners of tetrahedral

Prob 8. AlF₃ is ionic while AlCl₃ is covalent.

Prob 9. Write down the resonance structure of nitrous oxide

Solution:
 

Prob 10. Explain why BeH₂ molecule has zero dipole moment although the Be-H bonds are    polar.

Solution: BeH₂ is a linear molecule (H-Be-H) with bond angle equal to 180^o. Although the Be-H bonds are polar on account of electronegativity difference between Be and H atoms, the bond polarities cancel with each other. The molecule has resultant dipole moment of zero.

Prob 11. Why the bond angle of H – C – H in methane (CH₄) is 109° 28’ while H – N – H bond angle in NH₃ is 107° though both carbon and nitrogen are sp³ hybridized

“If the electronegativity of the peripheral atoms is more, then the bond angle will be less”.
For example if we consider NH₃ and NF₃, F – N – F bond angle will be lower than H – N – H bond angle. This is because in NF₃ the bond pair is displaced more towards F and in NH₃ it is displaced more towards N. So accordingly the b.p. – b.p. interaction is less in NF₃ and more in NH₃.

Prob 13. The bond angle of H₂O is 104° while that that of F₂O is 102°.

Solution:Both H₂O and F₂O have a lone pair of electrons. But fluorine being highly electronegative, the bond pair of electrons are drawn more towards F in F₂O, whereas in H₂O it is drawn towards O. So in F₂O the bond pairs being displaced away from the central atom, has very little tendency to open up the angle. But in H₂O this opening up is more as the bond pair electrons are closer to each other. So bond Ð of F₂O is less than H₂O.