Question

The sum of all $s{p}^3,s{p}^{2}andsp$ units present in solid $PC{l}_5$, solid $PB{r}_5$, and solid $N_2{O}_5$ are :

Answer 1

Solid $PC{l}_5\to [PC{l}_4{]}^{+}+[PC{l}_6{]}^{-}$
Hybridisation of $[PC{l}_4{]}^{+}=s{p}^3$
Hybridisation of $[PC{l}_6{]}^{-}=s{p}^3{d}^2$

Solid $PB{r}_5\to [PB{r}_4{]}^{+}+[Br{]}^{-}$
Hybridisation of $[PB{r}_4{]}^{+}=s{p}^3$

Solid $N_2{O}_5\to [N{O}_2^{+}{]}^{+}+[N{O}_3^{+}{]}^{-}$
Hybridisation of $\left[N{O}_2^{+}\right]=sp$
Hybridisation of $[N{O}_3^{+}{]}^{-}=s{p}^2$
Hence, the sum of all $s{p}^3,s{p}^{2}andsp$ units present in solid $PC{l}_5$, solid $PB{r}_5$, and solid $N_2{O}_5$ is $4$.