The hybridisation of Xe in XeO_3 is | vedclass

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(A) To determine the hybridisation $(H)$ of the central atom $Xe$ in $XeO_3$,we use the formula: $H = \frac{1}{2}(V + M - C + A)$ Where: $V$ = number

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$s p^3$

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(A) To determine the hybridisation $(H)$ of the central atom $Xe$ in $XeO_3$,we use the formula: $H = \frac{1}{2}(V + M - C + A)$ Where: $V$ = number of valence electrons of the central atom $(Xe)$ = $8$ $M$ = number of monovalent atoms attached = $0$ (Oxygen is divalent) $C$ = number of cationic charge = $0$ $A$ = number of anionic charge = $0$ Substituting the values: $H = \frac{1}{2}(8 + 0 - 0 + 0) = 4$ $A$ value of $4$ corresponds to $s p^3$ hybridisation.

The hybridisation of $Xe$ in $XeO_3$ is

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Balance the following equation: $XeF_{6} + H_{2}O \rightarrow XeO_{2}F_{2} + HF$

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