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.

Short 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-by-step solution

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 ${\mathbf{NH}}_3$.