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The hybridisation of $\mathrm{Xe}$ in $\mathrm{XeO}_3$ is
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$s p^3$
$\begin{aligned} & H=\frac{1}{2}(V+M-C+A)=\frac{1}{2}(8+0-0+0)=4\left(s p^3\right) \\ & {[\because V=\text { number of valence electrons of } \mathrm{Xe}=8} \\ & M=\text { number of monovalent atoms }=0 \\ & C=\text { number of cationic charge }=0 \\ & A=\text { number of anionic charge }=0]\end{aligned}$
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