Q12.53P - Chemistry: Molecular Nature Of Matter And Change

Question

For pairs of molecules in the gas phase, average H-bond dissociation energies are 17 kJ/mol for ${\mathbf{NH}}_3$, 22 kJ/mol for ${\mathbf{H}}_2\mathbf{O}$ , and 29 kJ/mol for $\mathbf{HF}$ . Explain this increase in H-bond strength.

 

Step-by-Step Solution

Answer

These corresponding increasing values of H-bond strength for ${\mathbf{NH}}_3$  , ${\mathbf{H}}_2\mathbf{O}$ , and  $\mathbf{HF}$, are due to the increasing electronegativity strength of the atom which is as follows: $\mathbf{N}\mathbf{>}\mathbf{O}\mathbf{>}\mathbf{F}$ .

 

Step-1: H-bond dissociation energy

The H-bond dissociation energy is the energy required to break a bond which is due to the strong H-bonding interactions. Stronger the bond, more will be the energy required to break the bond and thus higher will be the bond dissociation energy.

Step-2: Effect of electronegativity

Considering the hydrides , ${\mathbf{NH}}_3$ , ${\mathbf{H}}_2\mathbf{O}$  and $\mathbf{HF}$  molecules in gas phase, N has least electronegativity, O is comparatively higher in strength and F is the most electronegative element. With the increasing electronegativity from N to F, the ionic interactions or specifically H- bonding interactions would be more for  $\mathbf{HF}$, than  ${\mathbf{H}}_2\mathbf{O}$ and  ${\mathbf{NH}}_3$ has least such interactions.

 

Step-3: Effect on H-bond dissociation energies

As the H- bonding interactions are lowest for ${\mathbf{NH}}_3$  and highest for $\mathbf{HF}$ , the energy needed to break this bond is thus more for $\mathbf{HF}$ , than  $\mathbf{HF}$,${\mathbf{H}}_2\mathbf{O}$ , and least for  ${\mathbf{NH}}_3$.