Question

If hybridisation of "P" in transition state during hydrolysis of $PCl_5$ is $sp^xd^y$ then find out value of $\frac{x}{y}$.

Answer 1

1.5

Answer 2

The hydrolysis of $PCl_5$ proceeds through a nucleophilic attack by water on the phosphorus atom. The phosphorus atom in $PCl_5$ is $sp^3d$ hybridized and has a trigonal bipyramidal geometry. The attack of water, a nucleophile, occurs at the electrophilic phosphorus center. This attack leads to the formation of a transition state where the phosphorus atom is simultaneously bonded to the five chlorine atoms and the oxygen atom of the attacking water molecule.

In this transition state, the phosphorus atom is surrounded by 6 atoms/groups (5 Cl atoms and 1 O atom from water). The coordination number of phosphorus in this transition state is 6. According to VSEPR theory and the concept of hybridization, a central atom surrounded by 6 electron domains (representing sigma bonds or lone pairs) typically adopts an octahedral geometry, and the hybridization associated with octahedral geometry is $sp^3d^2$.

The formation of the P-O bond is in progress, and a P-Cl bond is in the process of breaking. Even though it's a transition state, the arrangement around the phosphorus atom involves significant interaction with 6 surrounding atoms. This expanded coordination sphere suggests the involvement of phosphorus's vacant 3d orbitals in bonding. The hybridization required to form 6 sigma interactions in an approximately octahedral arrangement is $sp^3d^2$.

The question states the hybridization of P in the transition state is $sp^xd^y$. Comparing $sp^3d^2$ with $sp^xd^y$: The power of p is $x=3$. The power of d is $y=2$.

We need to find the value of $\frac{x}{y}$. $\frac{x}{y} = \frac{3}{2} = 1.5$.