Question

The formation of $P{H}_4^{+}$ is difficult as compared to $N{H}_4^{+}$ because:

• A. The lone pair of phosphorous resides in almost a pure $s-$ orbital
• B. The lone pair of phosphorous resides in almost a pure $p-$ orbital
• C. The lone pair of phosphorous resides at a $s{p}^3$ orbital
• D. The lone pair of phosphorous is optically inert

Answer 1

The correct option is A The lone pair of phosphorous resides in almost a pure $s-$ orbital

$\stackrel{..}{P}{H}_3+H^{+}\to P{H}_4^{+}$
According to Drago's rule, the lone pair on phosphorous resides in almost a pure $s-$ orbital. Due to the non-directional nature of $s$ orbital, its overlapping tendency is greatly reduced in comparison to a lone pair present in a hybrid orbital, which is directional as present in $\stackrel{..}{N}{H}_3$.