Nitrogen forms \(N_2\), phosphorus forms \(P_2\) but phosphorus is readily converted to \(P_4\) because:

\(p\pi -p\pi \) bonding in phosphorus is weak

The correct answer is option 2. \(p\pi -p\pi \) bonding in phosphorus is weak.

Let us dive into why phosphorus forms \(P_4\) instead of \(P_2\) by comparing it with nitrogen and understanding the bonding characteristics.

Atomic and Molecular Properties of Nitrogen (\(N_2\)):

Size and Orbital Overlap:

Nitrogen is a small atom with a relatively small atomic radius. Because of this, the p orbitals of nitrogen can overlap effectively with each other. The overlap of p orbitals leads to strong \(p\pi - p\pi\) bonding, which allows nitrogen atoms to form a stable triple bond in \(N_2\).

Structure of \(N_2\):

The nitrogen molecule (\(N_2\)) is held together by a strong triple bond, consisting of:

One sigma (\(\sigma\)) bond from head-on overlap of sp orbitals. Two pi (\(\pi\)) bonds from the sideways overlap of p orbitals. The bond dissociation energy of \(N_2\) is very high, making it extremely stable and less reactive under standard conditions.

Reason for Stability:

The strong \(p\pi - p\pi\) bonding in nitrogen is due to the effective overlap of p orbitals, leading to a high bond strength and a short bond length. Because of this strong triple bond, \(N_2\) is stable as a diatomic molecule and does not form larger clusters like \(N_4\) under normal conditions.

Atomic and Molecular Properties of Phosphorus (\(P_2\) and \(P_4\)):

Size and Orbital Overlap:

Phosphorus atoms are significantly larger than nitrogen atoms. The larger atomic size means that the p orbitals are farther apart, leading to less effective orbital overlap. The \(p\pi - p\pi\) overlap in phosphorus is much weaker than in nitrogen. Consequently, the pi bonds in a hypothetical \(P_2\) molecule would be weak and unstable.

Hypothetical \(P_2\):

If phosphorus were to form a diatomic molecule \(P_2\), it would need to have a double or triple bond similar to nitrogen. However, due to the weak \(p\pi - p\pi\) interaction, the multiple bonds would not be as strong or stable. The resulting \(P_2\) molecule would be highly reactive and not stable under normal conditions.

Formation of \(P_4\):

Instead of forming \(P_2\) with weak multiple bonds, phosphorus prefers to form \(P_4\), a tetrahedral molecule where each phosphorus atom forms single sigma bonds with three other phosphorus atoms. The \(P_4\) molecule is much more stable because it avoids the need for weak multiple bonds and instead relies on stronger single bonds. In \(P_4\), each phosphorus atom is sp\(^3\) hybridized, and the bonding is entirely sigma bonds, which are stronger and more effective at the larger bond angles required by the tetrahedral geometry.

Reason for \(P_4\) Stability:

The single bonds in \(P_4\) are stable because sigma bonds are generally stronger than pi bonds when orbital overlap is less effective.  The \(P_4\) structure is more stable than a hypothetical \(P_2\) structure because it maximizes bond strength and minimizes the strain that would result from weak multiple bonding.

Conclusion:

Phosphorus forms \(P_4\) instead of \(P_2\) primarily because the \(p\pi - p\pi\) bonding in phosphorus is weak due to the larger atomic size and less effective orbital overlap. This weak \(p\pi - p\pi\) bonding makes the formation of multiple bonds (such as in \(P_2\)) unfavorable. Instead, phosphorus forms \(P_4\) with stronger single bonds, resulting in a stable tetrahedral structure.

So, the correct explanation is: option 2: \(p\pi - p\pi\) bonding in phosphorus is weak.