Question

How to calculate hybridisation of XeO2F2 ? What is its structure ?

Answer 1

look... you can solve it by VSEPR thoery ..

look from
Xe==8 elec.
2×O==12 elec.
2×F==14 elec.
------------------------
total elec.==34
Bond elec.==5×2==10 elec.

Answer 2

Answer: See saw

Explanation: Formula used

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

where, V = number of valence electrons present in central atom i.e. xenon = 8

N = number of monovalent (fluorine) atoms bonded to central atom=2

C = charge of cation = 0

A = charge of anion = 0

$XeO_2F_2:$

${\text{Number of electrons}} =\frac{1}{2}[8+2-0+0]=5$

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

But as there are four atoms around the central carbon, the fifth 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 see saw.