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(C) The chemical formula for $Xenon$ $(II)$ fluoride is $XeF_2$. To find the hybridisation of $Xe$ in $XeF_2$,we calculate the steric number: $Steric

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$C\ell F_3, SF_4$

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(C) The chemical formula for $Xenon$ $(II)$ fluoride is $XeF_2$. To find the hybridisation of $Xe$ in $XeF_2$,we calculate the steric number: $Steric \ number = \frac{1}{2} 	imes (V + M - C + A) = \frac{1}{2} 	imes (8 + 2 - 0 + 0) = 5$. $A$ steric number of $5$ corresponds to $sp^3d$ hybridisation. Now,let's check the hybridisation of the given options: $A) XeO_3 (sp^3), SF_4 (sp^3d)$ $B) BrF_5 (sp^3d^2), PF_5 (sp^3d)$ $C) C\ell F_3 (sp^3d), SF_4 (sp^3d)$ $D) PCl_3 (sp^3), NH_3 (sp^3)$ Both molecules in option $C$ have $sp^3d$ hybridisation,which matches $XeF_2$.

Identify the pair of molecules which have same hybridisation as the hybridisation in $Xenon$ $(II)$ fluoride.

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