The hybridization of Xe-atom in $X e O_{6}^{4 -}$ is same with :

Xe in

X e F_{6}

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Te in

T e O_{6}^{6 -}

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P in

P F_{6}^{-}

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Si in

S i F_{6}^{2 -}

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Solution

The correct options are
B P in $P F_{6}^{-}$
C Te in $T e O_{6}^{6 -}$
D Si in $S i F_{6}^{2 -}$
Xe have oxidation no +8 in $X e O_{6}^{4 -}$ .
here, Xe have hybridization $s p^{3} d^{2}$ .
And
Te have hybridization $s p^{3} d^{2}$ in $T e O_{6}^{6 -}$
P have hybridization $s p^{3} d^{2}$ in $P F_{6}^{-}$
Si have hybridization $s p^{3} d^{2}$ in $S i F_{6}^{2 -}$

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Similar questions

X e

X e F_{6}

X e

X e O_{6}^{4 -}

X e F_{6}

X e F_{6}

s p^{3} d^{3}

s p^{3} d

X e F_{2}, X e F_{4}, X e F_{6}, X e O F_{2}, X e O_{3}

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