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

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 ${{90}^{o}}$.