Problem 93 - A Complete Resource Book in Chemistry for JEE Main

Question

Nitrogen forms \(\mathrm{N}_{2}\), but phosphorous is converted into \(\mathrm{P}_{4}\) from \(\mathrm{P}\), the reason is (a) triple bond is present between phosphorous atom (b) \(\mathrm{p} \pi-\mathrm{p} \pi\) bonding is strong (c) p \(\pi-\mathrm{p} \pi\) bonding is weak (d) multiple bond is formed easily

Step-by-Step Solution

Answer

(c) p {c-p} {p} {c-p} bonding is weak.

Step 1: Understanding Nitrogen and Phosphorous Molecules

First, we identify the differences in molecular structures of nitrogen, {N}_{2}, and phosphorus, {P}_{4}. In {N}_{2}, each nitrogen atom shares three electrons to form a triple bond. However, phosphorus tends not to form similar multi-atom structures with strong triple bonds.

Step 2: Analyzing Bond Types and Stability

Nitrogen forms a diatomic molecule {N}_{2} because of the strong triple bond involving {p} {c-p} {c-p} {c-p} (pi) bonding. This bond is significant due to the higher energy availability of electrons in the {p} orbital.

Step 3: Evaluating Phosphorous Bonding Tendencies

Phosphorus prefers to form {P}_{4} tetrahedral structures due to weak {p} {c-p} {p} {c-p} (pi) bonding. Since these bonds are weaker, phosphorus does not generally form strong double or triple bonds like nitrogen.

Step 4: Final Selection of the Correct Reason

Considering the bond strength, {c} (p {c-p} {p} {c-p} bonding is weak) is the most accurate explanation. Phosphorus forms {P}_{4} instead of a diatomic molecule because its abilities to form strong multiple bonds are limited, which forces it to form a less strong but stable {P}_{4}.

Key Concepts

Triple Bond in Nitrogen

Nitrogen, due to its small size and highly effective orbital overlap, forms a strong triple bond in its diatomic molecule, \( \mathrm{N}_2 \). Each nitrogen atom contributes three electrons, creating a stable configuration with a shared pair of electrons across a single sigma \((\sigma)\) bond and two pi \((\pi)\) bonds. The triple bond in nitrogen provides an incredible amount of stability and results in the molecule being quite inert unless significant energy is applied. Thus, \( \mathrm{N}_2 \) requires a lot of energy to break these bonds, making it highly unreactive under normal conditions. This property of nitrogen is central to its role in various chemical processes, as it doesn’t easily engage in reactions without an energy input.

P4 Tetrahedral Structure

Phosphorus, on the other hand, does not form similar diatomic molecules. Instead, phosphorus atoms prefer forming a \( \mathrm{P}_4 \) tetrahedral structure. This structure is a consequence of phosphorus’s larger atomic size and the relative weakness of its pi bonds. In the \( \mathrm{P}_4 \) arrangement, the atoms form a stable three-dimensional shape, with each phosphorus atom connecting to three others. This setup is less about the formation of multiple bonds and more about spatial arrangement, as the tetrahedral configuration maximizes bonding efficiency without requiring strong pi bonds. As a result, \( \mathrm{P}_4 \) becomes the more stable configuration for phosphorus at the atomic level.

Pi Bonding Strength

Pi \((\pi)\) bonding refers to the overlapping of electron clouds in p orbitals on both sides of the internuclear axis, which results in lateral (side-to-side) overlap. In the case of nitrogen, the strength of these pi bonds contributes to the overall stability of the \( \mathrm{N}_2 \) triple bond. This strength is due to the effective overlap made possible by nitrogen's small size, enhancing the likelihood of electron sharing between atoms.
In contrast, the larger size of phosphorus atoms leads to less effective overlap, making pi bonds weaker. Phosphorus's pi bonds are insufficient to stabilize a diatomic molecule through multiple bonding like nitrogen, therefore, it opts for forming \( \mathrm{P}_4 \) using the more stable single bonds. Thus, the relative strength and weakness of pi bonds in different elements underline why nitrogen and phosphorus have distinct structural preferences.