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Question: The correct statements(s) about \(N{{H}_{3}}\) and \(P{{H}_{3}}\) is/are: This question has multip...

The correct statements(s) about NH3N{{H}_{3}} and PH3P{{H}_{3}} is/are:
This question has multiple correct options
A. μD{{\mu }_{D}} of PH3P{{H}_{3}} < μD{{\mu }_{D}} of NH3N{{H}_{3}}
B. PH3P{{H}_{3}} is stronger Lewis base than NH3N{{H}_{3}}
C. ∠H-N-H > ∠H-P-H
D. both have sp3s{{p}^{3}} hybridization.

Explanation

Solution

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 PH3P{{H}_{3}} is a larger molecule with greater dispersion forces than ammonia, NH3N{{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 NH3N{{H}_{3}} than between molecules of PH3P{{H}_{3}}.

Complete step by step answer:
A. μD{{\mu }_{D}} of PH3P{{H}_{3}} < μD{{\mu }_{D}} of NH3N{{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 NH3N{{H}_{3}} is more than PH3P{{H}_{3}}.
Hence, this statement is correct about NH3N{{H}_{3}} and PH3P{{H}_{3}}.
B. PH3P{{H}_{3}} is stronger Lewis base than NH3N{{H}_{3}}
Basically, NH3N{{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 PH3P{{H}_{3}} less basic than NH3N{{H}_{3}}.
Therefore, PH3P{{H}_{3}} is a weaker Lewis base than NH3N{{H}_{3}}.
Hence, this statement is incorrect about NH3N{{H}_{3}} and PH3P{{H}_{3}}.
C. ∠H-N-H > ∠H-P-H
Bond angle between H-N-H in NH3N{{H}_{3}} is about 109o{{109}^{o}} whereas bond angle between H-P-H in PH3P{{H}_{3}} is close to 90o{{90}^{o}}.
Therefore, ∠H-N-H is greater than ∠H-P-H.
Hence, this statement is correct about NH3N{{H}_{3}} and PH3P{{H}_{3}} .
D. Both have sp3s{{p}^{3}} hybridization
Hybridization of N in NH3N{{H}_{3}} is sp3s{{p}^{3}} whereas hybridization of P in PH3P{{H}_{3}} is according to Drago's rule. In PH3P{{H}_{3}} hybridization does not take place. The pure p orbitals take part in bonding.
Therefore, hybridization of NH3N{{H}_{3}} and PH3P{{H}_{3}} is different.
Hence, this statement is incorrect about NH3N{{H}_{3}} and PH3P{{H}_{3}}.
So, the correct answer is “Option A and C”.

Note: Always remember, in PH3P{{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 NH3N{{H}_{3}} is about 109o{{109}^{o}} whereas the bond angle between H-P-H in PH3P{{H}_{3}} is close to 90o{{90}^{o}}.