For pairs of molecules in the gas phase, average H-bond dissociation energies are 17 kJ/mol for NH3, 22 kJ/mol forH2O , and 29 kJ/mol forHF . Explain this increase in H-bond strength.

Short Answer

Expert verified

These corresponding increasing values of H-bond strength forNH3, H2O, and HF, are due to the increasing electronegativity strength of the atom which is as follows: N>O>F.

Step by step solution

01

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.

02

Step-2: Effect of electronegativity

Considering the hydrides ,NH3, H2O and 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 HF, than H2O and NH3 has least such interactions.

03

Step-3: Effect on H-bond dissociation energies

As the H- bonding interactions are lowest for NH3 and highest for HF, the energy needed to break this bond is thus more for HF, than HF,H2O, and least for NH3.

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