pBlock ElementsII 2 Question 12
12. The shape of $\mathrm{XeO}{2} \mathrm{~F}{2}$ molecule is
(2012)
(a) trigonal bipyramidal
(b) square planar
(c) tetrahedral
(d) see-saw
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Solution:
- In $\mathrm{XeO}{2} \mathrm{~F}{2}$, the bonding arrangement around the central atom $\mathrm{Xe}$ is
$4 \sigma$ bonds $+1.01 p=5$
Hybridisation of $\mathrm{Xe}=s p^{3} d$
$s p^{3} d$-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.
