Q.Nitrogen exists as diatomic molecule and phosphorus as P4. Why?

Step 1: Nitrogen's multiple bonding capability.

Because of its small atomic size, two nitrogen atoms can approach closely enough for their p-orbitals to overlap effectively sideways, forming strong $\pi$ bonds in addition to the $\sigma$ bond. This gives the very strong $N \equiv N$ triple bond (one $\sigma$ + two $\pi$), and nitrogen exists as the stable diatomic molecule $N_2$.

Step 2: Phosphorus's inability to form multiple bonds.

Phosphorus atoms are considerably larger. At the internuclear distance corresponding to a P–P single bond, the p-orbitals on adjacent P atoms are too far apart / too diffuse to achieve significant sideways ($p\pi$–$p\pi$) overlap. Consequently, phosphorus cannot form a stable $\pi$ bond, and hence cannot form a $P \equiv P$ triple bond analogous to $N_2$.

Step 3: Phosphorus catenates instead.

Since only a single ($\sigma$) P–P bond can form, each phosphorus atom instead forms three single bonds to three different neighbouring P atoms (using its three unpaired 3p electrons), building up the tetrahedral $P_4$ cage (each P at a vertex, bonded to the other three).

Step 4: Conclusion. …

State the size-dependence of pπ–pπ overlap, show N is small enough to form a strong triple bond (hence N2) while P is not, and explain P4's tetrahedral cage …

• Do not just restate the fact (N2 diatomic, P as P4) without explaining the underlying pπ–pπ overlap/size argument — this is the crux of the answer.
• P₄ is a tetrahedral (4-atom) cage held by six P–P single bonds, not a chain — describe the geometry correctly, not as a simple diatomic analogue. …