# Question #dede0

Jun 25, 2015

The hybridization of the chlorine atom in $C l {F}_{5}$ is $s {p}^{3} {d}^{2}$.

#### Explanation:

Chlorine pentafluoride, $C l {F}_{5}$, has a total number of 42 valence electrons: 7 from the chlorine atom and 7 from each of the five fluorine atoms.

The Lewis structure of the chlorine pentafluoride molecule looks like this

The five fluorine atoms will be bonded to the central chlorine atom via single bonds. Thsi will account for 10 of the 42 valence electrons.

Each fluorine atom will have three lone pairs of electrons, which will account for 30 out of the remaining 32 valence electrons.

The last 2 valence electrons will be placed as lone pairs on the chlorine atom.

As you can see, the chlorine atom is surrounded by 6 regions of electron density - five single bonds and one lone pair.

This means that its steric number, which is the number that also determines the hybridization of the atom, will be equal to 6.

In order to be able to form six hybrid orbitals, the chlorine atom will use

• one s-orbital;
• three p-orbitals;
• two d-orbitals.

Therefore, the central chlorine atom is $s {p}^{3} {d}^{2}$ hybridized. The molecular geometry of the molecule will be square pyamidal.