Question

Bond angle in $P{H_3}$ is:
(A) much less than $N{H_3}$
(B) much less than $P{H_3}$
(C) slightly more than $N{H_3}$
(D) much more than $P{H_3}$

Answer 1

In this question, we will use the concept of hybridisation. Hybridisation is a method of combining two or more atomic orbitals to form an entirely new orbital different from its components. The bond angle decreases due to the presence of lone pairs of electrons and increases along with electronegativity.

Complete step by step solution:
There are certain molecules in which there is no hybridisation occurring. This is known as Drago's rule. According to Drago’s rule, for hybridisation the molecule must follow these conditions,
The central atom should have at least one lone pair.
The electronegativity of a terminal atom should be less than carbon.
The central atom should belong to the third or higher period.
Instead of hybridisation, these atoms involve pure p orbitals in bond formation.
In $N{H_3}$ , nitrogen has a lone pair and it forms three bonds with a hydrogen atom. Hence the hybridisation of ammonium molecules is $s{p^3}$ . Similarly, phosphorus has one lone pair and forms three bonds with hydrogen atoms in $P{H_3}$ yet it does not show hybridisation as it obeys all the conditions of Drago’s rule. And hence the bond angle of phosphine is not the same as that of ammonia. The bond angle observed in ammonia is ${107^ \circ }$ and the bond angle of phosphine is ${93.5^ \circ }$ .
Therefore, the bond angle of $P{H_3}$ is much less than $N{H_3}$ .
Hence, the correct answer is option A.

Note:
Phosphine is a colourless, flammable and toxic gas having a rotten fish-like smell. Ammonia is a colourless gas with a pungent smell. The prominent use of ammonia is in the preparation of fertilizers.