Pblock Elementsii Result Question 12
12. The shape of $\mathrm{XeO} _2 \mathrm{~F} _2$ molecule is square planar
(2012)
(a) trigonal bipyramidal
(b) square planar
(c) tetrahedral
(d) see-saw
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Answer:
Correct Answer: 12. ( a )
Solution:
$\text { In } \mathrm XeO _2 \mathrm F _2 \text {, the bonding arrangement around the central atom } \mathrm{Xe} \text { is }$
$4 \sigma$ bonds $+1.01 p=4$
Hybridisation of $\mathrm{Xe}=s p^3 d^2$
$s p^3 d$-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.
According to Bent’s rule, the more electronegative atoms should be present on axial position. Hence, $F$ are kept on axial plane positions
BETA
