Q. The shape of $XeO_2F_2$ molecule is
IIT JEEIIT JEE 2012Chemical Bonding and Molecular Structure
Solution:
In $XeO_2F_2$, the bonding arrangement around the central
atom Xe is
$4 \sigma $ bonds + 1.01/7=5
Hybridisation of Xe = $sp^3d$
$sp^3$d-hybridisation corresponds to trigonal bipyramidal
geometry.
Also, in trigonal bipyramidal geometry, lone pairs remain
present on equatorial positions in order to give less
electronic repulsion.
NOTE According to Bent's rule, the more electronegative atoms must
be present on axial position. Hence, F are kept on axial
positions.
