Question

In case of nitrogen, $NC{l}_3$ is possible but not $NC{l}_5$ while in case of phosphorus, $PC{l}_3$ as well as $PC{l}_5$ are possible. It is due to: ​

• A. Availability of vacant $d$-orbitals in​ $P$ but not in $N$​
• B. Lower electronegativity of $P$ than that of $N$​
• C. Lower tendency of $H$-bond formation​ in $P$ than that in $N$​
• D. Occurrence of $P$ in solid state but $N$ in gaseous state at room temperature​

Answer 1

The correct option is A Availability of vacant $d$-orbitals in​ $P$ but not in $N$​

Due to the absence of $d$-orbital in nitrogen, it doesn't form $NC{l}_5.$ But phosphorus can expand its octet due to the presence of $d$-orbitals and form $PC{l}_5.$
Hence, option (a) is the correct answer.