Question

The hybrid states of phosphorous atoms in $PC{l}_5$ and $PB{r}_3$ in gaseous phase are $s{p}^{3}d$. But in solid $PC{l}_5$, phosphorous shows $s{p}^3{d}^2$ hybrid state. While $P$ in $PBr$ is in $s{p}^3$ hybrid state. This is because :

• A. $PC{l}_5$ in solid form exists as $[PC{l}_4{]}^{+}C{l}^{-}$
• B. $PB{r}_5$ in solid form exists as $[PB{r}_4{]}^{+}B{r}^{-}$
• C. $PC{l}_5$ in solid form exists as $\left[PC{l}_4{]}^{+}\right[PC{l}_6{]}^{-}$
• D. $PB{r}_5$ in solid form exists as $\left[PC{l}_4{]}^{+}\right[PB{r}_6{]}^{-}$

Answer 1

Answer: (B), (C)

Both $PC{l}_5$ and $PB{r}_5$ have triagonal bipyramidal geometry this is not a regular structure and is not very stable.

Therefore $PC{l}_5$ splits up into two more stable octahedral and tetrahedral structures .

$PC{l}_5$ in solid form exists as $\left[PC{l}_4{]}^{+}\right[PC{l}_6{]}^{-}$. The hybridisation state of $P$ in $[PC{l}_4{]}^{+}$ and $[PC{l}_6{]}^{-}$ is

$s{p}^3$ and $s{p}^3{d}^2$ respectively.