Question

The correct statement(s) about $N{H}_3$ and $P{H}_3$ is/are:

• A. ${\mu }_D$ of $P{H}_3<{\mu }_D$ of $N{H}_3$
• B. $P{H}_3$ is stronger Lewis base than $N{H}_3$
• C. $\angle H-N-H>\angle H-P-H$
• D. both have $s{p}^3$ hybridization

Answer 1

Answer: (A), (C)

The correct options are
A ${\mu }_D$ of $P{H}_3<{\mu }_D$ of $N{H}_3$
C $\angle H-N-H>\angle H-P-H$

Following are true for $N{H}_3$.
Hybridization of $N$ is $s{p}^3$. Bond angle $H\hat{N}H$ is less than ${109}^{\circ }$. Lone pair exists in hybrid orbitals.
Following are true for $P{H}_3$,
$P$ follows hybridization according to Drago's rule.
$\angle HPH$ is close to ${90}^{\circ }$. Dipole moment of pure orbital is zero. Also, $N{H}_3$ is stronger Lewis base than $P{H}_3$.
Thus, the statements A and C are correct and the statements B and D are correct.