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 $$P$$ but not in $$N$$
B.
lower electronegativity of $$P\,{\text{than}}\,N$$
C.
lower tendency of $$H$$-bond formation in $$P\,{\text{than}}\,N$$
D.
occurrence of $$P$$ in solid while $$N$$ in gaseous state at
room temperature.
Answer :
availability of vacant $$d$$ orbitals in $$P$$ but not in $$N$$
Solution :
$${}_7N = 1{s^2}2{s^2}2{p^3};\,\,\,\,\,{}_{15}P = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^3}$$
NOTE: In phosphorous the $$3d$$- orbitals are available. Hence phosphorous can from pentahalides also but nitrogen can not form pentahalide due to absence of $$d\, - $$orbitals