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Nitrogen forms $N_2$, but phosphorus forms $P_4$ instead of $P_2$; the reason is : -
A
Triple bond present between phosphorus atoms
B
$p\pi - p\pi$ bonding is weak
C
$p\pi - p\pi$ bonding is strong
D
Multiple bonds form easily
Detailed Solution
Nitrogen is small, so its 2p orbitals overlap well sideways and form strong $p\pi - p\pi$ bonds. It exists as $N\equiv N$.
Phosphorus atoms are larger, so the sideways overlap of their 3p orbitals is poor and $p\pi - p\pi$ bonds are weak.
So instead of forming a triple bond ($P\equiv P$), each phosphorus forms three single P–P bonds, giving the tetrahedral $P_4$ molecule.
So the reason is that $p\pi - p\pi$ bonding is weak in phosphorus.
Phosphorus atoms are larger, so the sideways overlap of their 3p orbitals is poor and $p\pi - p\pi$ bonds are weak.
So instead of forming a triple bond ($P\equiv P$), each phosphorus forms three single P–P bonds, giving the tetrahedral $P_4$ molecule.
So the reason is that $p\pi - p\pi$ bonding is weak in phosphorus.