Question

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

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

Answer 1

Answer: (A)

${}_7\mathrm{N}=1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^3;{}_{15}\mathrm{P}=1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^3$

In phosphorous the 3d-orbitals are available. Hence phosphorus can form pentahalides also but nitrogen cannot form pentahalide due to absence of d-orbitals.