Question

Among $ NH{}_3,H{}_2O $ , and $ HF $ , which would you expect to have the highest magnitude of hydrogen bonding and why?

Answer 1

Hint: A small and highly electronegativity element forms a stronger hydrogen bond. The order of size of nitrogen, oxygen and fluorine is $ N < O < F $ .and so the order of strength of hydrogen bond is $ F > O > N $.

Complete step by step solution:
The extent of hydrogen bonding depends on the electronegativity and the number of hydrogen atoms which are available for bonding. Among N, F, and O, the increasing order of their electronegativity are $ N < O < F $
The expected order of hydrogen bonding is $ HF > H{}_2O > NH{}_3 $
However the actual order is:
 $ H{}_2O > HF > NH{}_3 $
Here in the $ HF $ molecule there is a shortage of $ HF $ , whereas there is the right number of hydrogen in water. As a result there is formation of straight chain bonding takes place. On the other hand there is a ring like structure for oxygen that is formed because of its high ability of hydrogen bonding. When it comes to $ NH{}_3 $ hydrogen bonding is limited as it has only one lone pair. Electronegativity of fluorine is the highest, therefore magnitudes of positive charge present on hydrogen and there is negative charge on fluorine is highest in $ HF $ and hence, electrostatic attraction of hydrogen bonding is the strongest in $ HF $ .

Note:
Hydrogen bond is an intermolecular attractive force where hydrogen atom is covalently bonded to a highly electronegative attracted to a lone pair of electrons on an atom. Hydrogen bonds are very strong in comparison to dipole interactions. The strength of a hydrogen bond is nearly $ 5\% $ of that og a covalent bond.