Question

The hybridization of P exhibited in $PF_5$ is $sp_xd_y$. The value of y is _____ .

Answer 1

PF5 has 5 fluorine atoms bonded to phosphorus, and there are no lone pairs on phosphorus. The electron configuration of phosphorus is 1s² 2s² 2p⁶ 3s² 3p³, and when it forms bonds in PF5, one of its 3s electrons and three of its 3p electrons are used. The hybridization is determined by the number of orbitals involved in bonding, which is four in this case.
The hybridization can be represented as sp³d. The value of y in sp³dy is 1.

Therefore, the value of y is 1.