Question

The correct statements(s) about $N{H}_3$ and $P{H}_3$ is/are:
This question has multiple correct options
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. ∠H-N-H > ∠H-P-H
D. both have $s{p}^3$ hybridization.

Answer 1

Hint: The size of the P atom is greater than the N atom, down the group basicity of the atom decreases when the size of the atom increases and the electron density decreases. Though $P{H}_3$ is a larger molecule with greater dispersion forces than ammonia, $N{H}_3$ has very polar N-H bonds which lead to strong hydrogen bonding. This is the prevailing intermolecular force that results in a stronger attraction between molecules of $N{H}_3$ than between molecules of $P{H}_3$.

Complete step by step answer:
A. ${\mu }_D$ of $P{H}_3$ < ${\mu }_D$ of $N{H}_3$
Nitrogen (N) is more electronegative than phosphorus (P). As the electronegativity of the element increases, the dipole moment also increases.
Therefore, the dipole moment of $N{H}_3$ is more than $P{H}_3$.
Hence, this statement is correct about $N{H}_3$ and $P{H}_3$.
B. $P{H}_3$ is stronger Lewis base than $N{H}_3$
Basically, $N{H}_3$ is a base due to the presence of a lone pair of electrons on the nitrogen atom which is available for donation. The size of P atom is larger than N atom, hence the lone pair of electrons on P atom are less readily available for donation as compared to N atom, making $P{H}_3$ less basic than $N{H}_3$.
Therefore, $P{H}_3$ is a weaker Lewis base than $N{H}_3$.
Hence, this statement is incorrect about $N{H}_3$ and $P{H}_3$.
 C. ∠H-N-H > ∠H-P-H
Bond angle between H-N-H in $N{H}_3$ is about ${109}^o$ whereas bond angle between H-P-H in $P{H}_3$ is close to ${90}^o$.
Therefore, ∠H-N-H is greater than ∠H-P-H.
Hence, this statement is correct about $N{H}_3$ and $P{H}_3$ .
D. Both have $s{p}^3$ hybridization
Hybridization of N in $N{H}_3$ is $s{p}^3$ whereas hybridization of P in $P{H}_3$ is according to Drago's rule. In $P{H}_3$ hybridization does not take place. The pure p orbitals take part in bonding.
Therefore, hybridization of $N{H}_3$ and $P{H}_3$ is different.
Hence, this statement is incorrect about $N{H}_3$ and $P{H}_3$.
So, the correct answer is “Option A and C”.

Note: Always remember, in $P{H}_3$ hybridization does not take place. The pure p orbitals take part in bonding.
- Always remember, as the electronegativity of the element increases, the dipole moment also increases.
- Always remember, the bond angle between H-N-H in $N{H}_3$ is about ${109}^o$ whereas the bond angle between H-P-H in $P{H}_3$ is close to ${90}^o$.