Question

Ammonia is more basic than phosphine because

• A. In phosphine, the H-P-H bond angle is ${90}^{\circ }$ while the H-N-H bond angle is ${107}^{\circ }$
  
• B. P is greater in size than N
  
• C. N has greater electronegativity than P
  
• D. Of all the above factors

Answer 1

The correct option is A

In phosphine, the H-P-H bond angle is ${90}^{\circ }$ while the H-N-H bond angle is ${107}^{\circ }$

Bond angle of ${90}^{\circ }$ in: $P{H}_3$ indicates that in $P{H}_3$ formation of P-H bonds involves pure orbitals of P. Thus its three p orbitals form bonds with three H atoms and the electron pair of s orbitals remains unused. So here the nonbonding electrons belongs to s orbital, while in: $N{H}_3,$ the nonbonding electron pair belongs to $s{p}^3$ orbital. Now since lesser the s character of an orbital, farther it is away from the nucleus and hence s electrons are readily available for promotion. Hence, $N{H}_3$ will be more basic than $P{H}_3$.