Question

G What is the hybridization of the central atom in the chlorotetrafluoride cation

Answer 1

Answer : The hybridization of the central atom chlorine is, $sp^3d$

Explanation :

Formula used  :

$\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

Now we have to determine the hybridization of the given molecules.

The given molecule is chlorotetrafluoride cation, $ClF_4^+$

As, the chlorine is more electropositive than the fluorine. So, chlorine is a central atom and fluorine is neighboring atoms.

$\text{Number of electrons}=\frac{1}{2}\times [7+4-1]=5$

The number of electron pair are 5 that means the hybridization will be $sp^3d$ and the electronic geometry of the molecule will be trigonal bipyramidal.

But as there are 4 atoms around the central chlorine atom, the 5th position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be sea-saw.

Hence, the hybridization of the central atom chlorine is, $sp^3d$