Which of the following has sp^3 hybridisation | vedclass

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(A) To determine the hybridization,we calculate the steric number using the formula: $	ext{Steric Number} = \frac{1}{2} (V + M - C + A)$,where $V$ is

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$XeO_3$

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(A) To determine the hybridization,we calculate the steric number using the formula: $	ext{Steric Number} = \frac{1}{2} (V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$1$. For $XeO_3$: $Xe$ has $8$ valence electrons. Oxygen is divalent,so $M = 0$. $	ext{Steric Number} = \frac{1}{2} (8 + 0) = 4$. $A$ steric number of $4$ corresponds to $sp^3$ hybridization.$2$. For $BCl_3$: $B$ has $3$ valence electrons. $Cl$ is monovalent,so $M = 3$. $	ext{Steric Number} = \frac{1}{2} (3 + 3) = 3$. This corresponds to $sp^2$ hybridization.$3$. For $XeF_4$: $Xe$ has $8$ valence electrons. $F$ is monovalent,so $M = 4$. $	ext{Steric Number} = \frac{1}{2} (8 + 4) = 6$. This corresponds to $sp^3d^2$ hybridization.$4$. For $BBr_3$: $B$ has $3$ valence electrons. $Br$ is monovalent,so $M = 3$. $	ext{Steric Number} = \frac{1}{2} (3 + 3) = 3$. This corresponds to $sp^2$ hybridization.Therefore,the correct option is $A$.

Which of the following has $sp^3$ hybridisation?

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