Question

Explain why (a) \(\mathrm{NH}_{3}\) is more basic than \(\mathrm{PH}_{3},\) (b) \(\mathrm{NH}_{3}\) has a higher boiling point than \(\mathrm{PH}_{3},\) (c) \(\mathrm{PCl}_{5}\) exists but \(\mathrm{NCl}_{5}\) does not, and (d) \(\mathrm{N}_{2}\) is more inert than \(\mathrm{P}_{4}\).

Short answer

(a) NH_3 is more basic due to available lone pairs on nitrogen. (b) NH_3 has hydrogen bonds raising its boiling point. (c) PCl_5 exists, NCl_5 does not due to orbital availability for bonding. (d) N_2 is more inert due to its triple bond.

Step-by-step solution

Examining Basicity

The basicity of a compound is influenced by the availability of lone pairs on the central atom to accept protons. Ammonia ( NH_3 ) has a lone pair on nitrogen, which is more electronegative than phosphorus. This makes the lone pair more readily available for protonation, thus making NH_3 more basic than PH_3 . In contrast, the lone pairs in PH_3 are less available due to the larger size and lower electronegativity of phosphorus.

Evaluating Boiling Points

The boiling point of a substance is influenced by intermolecular forces. Ammonia ( NH_3 ) can form hydrogen bonds due to the presence of polar N-H bonds and lone pairs, resulting in stronger intermolecular interactions. In contrast, PH_3 forms weaker van der Waals forces because it cannot form hydrogen bonds. This makes NH_3 have a higher boiling point than PH_3 due to the presence of hydrogen bonding.

Analyzing Existence of PCl_5 and Absence of NCl_5

The existence of PCl_5 and the absence of NCl_5 are due to the availability of orbitals for bonding. Phosphorus can expand its valence shell to include d orbitals, allowing it to form PCl_5 with sp3d hybridization. In contrast, nitrogen has only s and p orbitals available and cannot expand its valence shell, making it unable to form NCl_5 with five bonds.

Comparing Inertness of N_2 and P_4

Inertness is related to bond strength and stability. Nitrogen molecules ( N_2 ) have a triple bond, one of the strongest bonds due to high bond energy and short bond length, contributing to its inertness. P_4 , on the other hand, consists of single bonds in a tetrahedral structure, which are weaker and more reactive. This makes N_2 more inert than P_4 due to the presence of a strong triple bond.

Key concepts

Basicity of Ammonia and Phosphine

Ammonia (\(\text{NH}_3\)) is much more basic than phosphine (\(\text{PH}_3\)) due to the nature of nitrogen and phosphorus atoms. Basicity refers to a compound's ability to donate a pair of electrons to accept a proton. The nitrogen atom in ammonia has a lone pair of electrons that are highly available for bonding. This is because nitrogen is more electronegative than phosphorus.
An electronegative atom holds its electrons more tightly, making them more available to coordinate with a proton (usually \(\text{H}^+\)).
In contrast, the lone pair on phosphorus in phosphine is less available due to phosphorus being larger and less electronegative than nitrogen. Consequently, this difference in electronegativity results in ammonia being a stronger base than phosphine.
Also, the compact size of nitrogen allows the lone pair to be more concentrated and thus more reactive compared to the more dispersed lone pair in the larger phosphorus atom.

Boiling Points and Intermolecular Forces

Ammonia has a significantly higher boiling point compared to phosphine primarily due to the strength and type of intermolecular forces present. In ammonia (\(\text{NH}_3\)), hydrogen bonding occurs due to the presence of nitrogen-to-hydrogen (N-H) polar bonds coupled with the lone pair on nitrogen.
Hydrogen bonds are much stronger than the van der Waals forces which are present in phosphine (\(\text{PH}_3\)).
Hydrogen bonds cause ammonia molecules to attract to each other more powerfully, requiring more thermal energy to break these interactions effectively. This results in a higher boiling point.
In contrast, phosphine cannot form hydrogen bonds. Instead, it relies on weaker dispersion forces because the differences in electronegativity in the P-H bonds are not sufficient to form hydrogen bonds.

Existence and Stability of Compounds

The existence of \(\text{PCl}_5\) and the non-existence of \(\text{NCl}_5\) can be explained by the electron configuration and orbital availability of phosphorus and nitrogen.
Phosphorus can expand its valence shell due to the presence of d orbitals, allowing it to accommodate more than eight electrons. This enables it to form \(\text{PCl}_5\) with sp3d hybridization.
On the other hand, nitrogen lacks d orbitals and thus cannot expand beyond its octet, limiting its compound formation to less than five bonds.
This restriction results in nitrogen being unable to form \(\text{NCl}_5\), as doing so would exceed the capacity of its valence shell. Therefore, the absence of available d orbitals in nitrogen makes \(\text{NCl}_5\) impractical.

Inertness of Nitrogen and Phosphorus

The inertness of substances like nitrogen (\(\text{N}_2\)) and phosphorus (\(\text{P}_4\)) can be understood through their molecular structures and bonding types.
Nitrogen is highly inert due to its triple bond, consisting of one sigma and two pi bonds, which is very short and strong. This triple bond results in high bond energy, making breaking these bonds energetically unfavorable and thus making nitrogen quite unreactive under most normal conditions.
In contrast, phosphorus exists as \(\text{P}_4\), which has a tetrahedral structure involving only single bonds. These bonds are longer and weaker compared to the triple bonds in nitrogen, leading to greater reactivity. Therefore, phosphorus readily participates in chemical reactions more than nitrogen.
This bond strength disparity highlights why \(\text{N}_2\) is much more chemically inert than \(\text{P}_4\).