N_2 forms NCl_3,whereas P can form both PCl_3 | vedclass

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Question

(A) . $N$ belongs to the second period and has only $2s$ and $2p$ orbitals available for bonding. It lacks $d$-orbitals,so it cannot expand its octet

Vedclass · Accepted Answer

$P$ has low-lying $3d$ orbitals,which can be used for bonding,but $N$ does not have low-lying $3d$ orbitals.

Vedclass · Answer

(A) . $N$ belongs to the second period and has only $2s$ and $2p$ orbitals available for bonding. It lacks $d$-orbitals,so it cannot expand its octet and thus cannot form $NCl_5$.$P$ belongs to the third period and has vacant $3d$ orbitals available. This allows $P$ to undergo $sp^3d$ hybridization and expand its octet,enabling it to form both $PCl_3$ and $PCl_5$.

$N_2$ forms $NCl_3$,whereas $P$ can form both $PCl_3$ and $PCl_5$. Why?

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