Question

Identify the incorrect statement related to $\text{PC}{{\text{l}}_{5}}$ from the following:
A. Three equatorial $\text{P}-\text{Cl}$ bonds make an angle of ${{120}^{0}}$ with each other.
B. Two axial $\text{P}-\text{Cl}$ bonds make an angle of ${{180}^{\text{0}}}$ with each other.
C. Axial $\text{P}-\text{Cl}$ bonds are longer than equatorial $\text{P}-\text{Cl}$ bonds.
D. $\text{PC}{{\text{l}}_{5}}$ molecules is non-reactive

Answer 1

According to VSEPR theory, gaseous and molten $\text{PC}{{\text{l}}_{5}}$ is a neutral molecule with trigonal bi-pyramidal as its geometry. This trigonal bi-pyramidal structure persists in nonpolar solvents like $\text{CC}{{\text{l}}_{4}}$. Look at the structure of $\text{PC}{{\text{l}}_{5}}$ and the location of the atoms in the molecule to find out the answer.

Complete step by step answer:
$\text{PC}{{\text{l}}_{5}}$ is a pentavalent compound of phosphorus connected to five chloride ions. In this three chloride ions are in one plane and two left chloride ions are in the perpendicular plane, one above and one being placed below. The structure of $\text{PC}{{\text{l}}_{5}}$ is

The features of this structure are:
- The three chloride ions in one plane forms a triangle forming central angle of ${{120}^{\text{o}}}$ with the central atom phosphorus means the $\text{Cl}-\text{P}-\text{Cl}$ bond is ${{120}^{\text{o}}}$. These chlorine atoms are said to be placed in the equatorial region.

- The two chlorine atoms placed above the plane and another below the plane forms an angle of ${{90}^{\text{o}}}$ with the phosphorus atom. The chlorine atoms are said to be placed in the axial region. So, the angle between the chlorine atoms which are in the perpendicular plane is ${{180}^{\text{o}}}$ taking phosphorus as the centre of measure. So, $\angle \left( \text{Cl}-\text{P}-\text{Cl} \right)$ is ${{180}^{\text{o}}}$.

- The axial bonds are longer than equatorial bonds because the bond pair-bond pair repulsions occur between them due to the presence of lone pairs and angle difference is just ${{90}^{\text{o}}}$. These repulsions make the axial bond pairs to extend their bond length to suffer less repulsion. This makes $\text{PC}{{\text{l}}_{5}}$ highly reactive which breaks into $\text{PC}{{\text{l}}_{5}}\to \text{PC}{{\text{l}}_{3}}+\text{C}{{\text{l}}_{2}}$. These axial chlorine atoms are removed as chlorine gas. The repulsion between equatorial bond pairs is negligible as the bond angle is long enough $\left( {{120}^{\text{o}}} \right)$.
The correct answer of this question is option ‘d’ that $\text{PC}{{\text{l}}_{5}}$ molecules is non-reactive. So, the correct answer is “Option D”.

Note: In the solid state, $\text{PC}{{\text{l}}_{5}}$ exists in ionic form like ${{\left[ \text{PC}{{\text{l}}_{4}} \right]}^{+}}{{\left[ \text{PC}{{\text{l}}_{6}} \right]}^{-}}$. The cation ${{\left[ \text{PC}{{\text{l}}_{4}} \right]}^{+}}$ has the tetrahedral geometry and anion ${{\left[ \text{PC}{{\text{l}}_{6}} \right]}^{-}}$ has octahedral geometry. It was thought that $\text{PC}{{\text{l}}_{5}}$ also forms a dimeric structure ${{\text{P}}_{2}}\text{C}{{\text{l}}_{10}}$ , but this is not true.