Question

$\text{PC}{{\text{l}}_{\text{5}}}$ and $\text{PB}{{\text{r}}_{\text{5}}}$ exist in $\text{s}{{\text{p}}^{\text{3}}}\text{d}$ hybrid state in gaseous phase but in solid state, which of the following statement is true?

A. P in $\text{PC}{{\text{l}}_{\text{5}}}$exists in $\text{s}{{\text{p}}^{\text{3}}}$ - hybridization state, while P in $\text{PB}{{\text{r}}_{\text{5}}}$ exists in $\text{s}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$ and $\text{s}{{\text{p}}^{\text{3}}}$ - hybridization states
B. P in $\text{PC}{{\text{l}}_{\text{5}}}$and $\text{PB}{{\text{r}}_{\text{5}}}$ exists in $\text{s}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$ and $\text{s}{{\text{p}}^{\text{3}}}$- hybridization state
C. P in $\text{PC}{{\text{l}}_{\text{5}}}$ exists in $\text{s}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$ and $\text{s}{{\text{p}}^{\text{3}}}$- hybridization states, while P in $\text{PB}{{\text{r}}_{\text{5}}}$ exists in $\text{s}{{\text{p}}^{\text{3}}}$- hybridization state.
D. P in $\text{PC}{{\text{l}}_{\text{5}}}$ and $\text{PB}{{\text{r}}_{\text{5}}}$ exists in $\text{s}{{\text{p}}^{\text{3}}}$- hybridization state

Answer 1

Hint: Existence of species in what form depends on the given condition in which it is placed and whether it is stable in that condition or not. Now let us see what happens to hybridization of $\text{PC}{{\text{l}}_{\text{5}}}$ and $\text{PB}{{\text{r}}_{\text{5}}}$ in solid state.

Complete step by step solution:
$\text{PC}{{\text{l}}_{\text{5}}}$ exists in covalent form in a gaseous state. Formation of ions is difficult in gaseous state and charge cannot be stabilized in gaseous state so it exists in covalent form.
In solid state ions can be stabilized by lattice energy so, $\text{PC}{{\text{l}}_{\text{5}}}$ splits into $\text{PC}{{\text{l}}_{\text{5}}}\rightleftarrows \text{ }\\!\\![\\!\\!\text{ PC}{{\text{l}}_{\text{4}}}{{\text{ }\\!\\!]\\!\\!\text{ }}^{\text{+}}}\text{ + }\\!\\![\\!\\!\text{ PC}{{\text{l}}_{\text{6}}}{{\text{ }\\!\\!]\\!\\!\text{ }}^{\text{-}}}$. Shape of ${{\text{ }\\!\\![\\!\\!\text{ PC}{{\text{l}}_{\text{4}}}\text{ }\\!\\!]\\!\\!\text{ }}^{\text{+}}}$ is tetrahedral and its hybridization is $\text{s}{{\text{p}}^{\text{3}}}$, shape of ${{\text{ }\\!\\![\\!\\!\text{ PC}{{\text{l}}_{\text{6}}}\text{ }\\!\\!]\\!\\!\text{ }}^{\text{-}}}$ is octahedral and its hybridization is $\text{s}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$. Therefore, in solid state P shows hybridization as $\text{s}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}\text{ and s}{{\text{p}}^{\text{3}}}$.
In solid state $\text{PB}{{\text{r}}_{\text{5}}}$ splits into stable tetrahedral structure as $\text{PB}{{\text{r}}_{\text{5}}}\rightleftarrows {{\text{ }\\!\\![\\!\\!\text{ PB}{{\text{r}}_{\text{4}}}\text{ }\\!\\!]\\!\\!\text{ }}^{\text{+}}}\text{ + }\\!\\![\\!\\!\text{ Br}{{\text{ }\\!\\!]\\!\\!\text{ }}^{\text{-}}}$. Shape of ${{\text{ }\\!\\![\\!\\!\text{ PB}{{\text{r}}_{\text{4}}}\text{ }\\!\\!]\\!\\!\text{ }}^{\text{+}}}$ is tetrahedral and its hybridization is $\text{s}{{\text{p}}^{\text{3}}}$. Therefore, in solid state hybridization of P in $\text{PB}{{\text{r}}_{\text{5}}}$ is $\text{s}{{\text{p}}^{\text{3}}}$.
So, we can conclude that in solid state P in $\text{PC}{{\text{l}}_{\text{5}}}$ exists in $\text{s}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$ and $\text{s}{{\text{p}}^{\text{3}}}$- hybridization states, while P in $\text{PB}{{\text{r}}_{\text{5}}}$ exists in $\text{s}{{\text{p}}^{\text{3}}}$- hybridization state.

Note: The splitting of $\text{PB}{{\text{r}}_{\text{5}}}$ is different from $\text{PC}{{\text{l}}_{\text{5}}}$ because Br atoms are large and six atoms of Br cannot be easily accommodated around smaller P atom.