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$\mathrm{XeF}_4$ is square planar while $\mathrm{XeF}_6$ has a distorted octahedral structure. What is the correct explanation for this observation?
Options:
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Verified Answer
The correct answer is:
$\mathrm{XeF}_4$ has two lone pairs of electrons on $\mathrm{Xe}^2 \mathrm{XeF}_6$ has one lone pair of electrons on Xe
$\mathrm{XeF}_4$ is square planar and $s p^3 d^2$-hybridised.
$4 \sigma+2 l p=6$-hybrid orbital $s p^3 d^2$-hybridisation
$\mathrm{XeF}_6$ has distorted octahedral structure and
$$
s p^3 d^3 \text {-hybridised. }
$$
$6 \sigma+1 l p=7$-hybrid orbital $\left(s p^3 d^3\right)$ (distorted octahedral) Thus, option (4) is correct.
$4 \sigma+2 l p=6$-hybrid orbital $s p^3 d^2$-hybridisation
$\mathrm{XeF}_6$ has distorted octahedral structure and
$$
s p^3 d^3 \text {-hybridised. }
$$
$6 \sigma+1 l p=7$-hybrid orbital $\left(s p^3 d^3\right)$ (distorted octahedral) Thus, option (4) is correct.
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