The hybridisation of $S$ in $S F_{6}$ molecule is

s p^{3}

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s p^{3} d^{2}

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s p^{3} d

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None

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Solution

The correct option is B $s p^{3} d^{2}$
Sulphur forms $6$ bond pairs. It thus needs $6$ hybrid orbitals and therefore, hybridization is $s p^{3} d^{2}$ this molecule has octahedral structure.

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Similar questions

What is the hybridization of P in $P C l_{5} (s)$ ?

X e F_{2}, X e F_{3}^{+}, X e F_{4}, X e F_{6}

In solid state of $P C l_{5}$ , the hybridsation of 'P'is

$M o l e c u l e H y b r i d i s a t i o n S h a p e$

$A . P O_{4}^{- 3} 1. s p^{2} a) D i s t o r t e d o c t a h e d r a l$

$B . X e F_{6} 2. s p^{3} b) P y r a m i d$

$C . S O_{3}^{- 2} 3. s p^{3} d c) T e t r a h e d r a l$

$D . N O_{3}^{- 1} 4. s p^{3} d^{3} d) T r i a n g u l a r p l a n a r$

The correct match is

A l H_{3}

A l H_{4}^{-}

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