Hybridization of the central atom in `PF_3` is

Options

A. `sp`
B. `dsp^2`
C. `sp^2`
D. `sp^3`

Answer

D

Explanation

The ground-state electron configuration of central P atom is `[Ne]3s^2 3p^3`. Therefore, the P atom has five valence electrons. Three of the valence electrons are used to hold the three univalent F atoms and two nonbonding valence electrons form one lone pair.
`F-underset(F)underset(|)overset(..)P-F`
There are three bonded atoms and one lone pair on the central P atom. Thus, the steric number of `P` in `PF_3` is `3+1=4`. We conclude that P must be `sp^3`-hybridized. One of the `sp^3` hybrid orbitals is used to accomodate the lone pair on P while the other three `sp^3` hybrid orbitals from covalent `P-F` bonds with the `2p` orbitals of `F`. We predict the shape of the molecule to be pyramidal and the FPF bond angle should be somewhat less than `109.5^@`.