Question

Nitrogen forms $N_2$ , but phosphorus is converted into $P_4$ from $P$, the reason is

• A. Triple bond is present between phosphorus atom
• B. $p\pi -p\pi$ bonding is strong in nitrogen
• C. $p\pi -p\pi$ bonding is weak in nitrogen
• D. Multiple bond is formed easily

Answer 1

Answer: (B)

Nitrogen form $N_2$ (i.e. $N\equiv N$ ) but phosphorus form $P_4$, because in $P_2$,
$p\pi -p\pi$ bonding is present which is a weaker bonding.
Nitrogen is smaller in size hence it forms multiple bond with other nitrogen atom $(N\equiv N)$
Nitrogen has unique ability to form $p\pi -p\pi$ multiple bonds with itself and with other elements having small size and high electron negativity.
However heavier elements of this group do not form $p\pi -p\pi$ bonds as their atomic orbitals are so large and diffuse that they cannot have effective overlapping.
Thus nitrogen exist as diatomic molecule with a triple bonds.
The single $P$ - $P$ bond is stronger and has higher catenation tendency