$C F_{4}, S F_{4}$ and $X e F_{4}$ contain the following electronic structure on their central atoms. Which one is correct option?

1, 2 and 3 lone pairs of electrons respectively

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0, 1 and 2 lone pairs of electrons respectively

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1, 1 and 1 lone pairs of electrons respectively

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No lone pairs of electrons on any molecule

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Solution

The correct option is B 0, 1 and 2 lone pairs of electrons respectively
Formula to determine the lone pair is:-
$\frac{v a l a n c e e l e c t r o n o f c e n t r a l a t o m - n u m b e r o f s i g m a b o n d s}{2}$
So from this formula we can calculate that $C F_{4}$ , $S F_{4}$ , $X e F_{4}$ has $0, 1, 2$ lone pairs respectively.
For example: In $X e F_{4}$ , $X e$ has 8 valance electrons and 4 \sigma bonds.
So $l o n e p a i r s = \frac{8 - 4}{2}$
$∴ l o n e p a i r s = 2$
So the correct option is [B].

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Similar questions

The number of lone pairs of electrons on SF₄, CF₄ and XeF₄are:

The molecular shapes of $S F_{4}, C F_{4} a n d X e F_{4}$ are

S F_{4}

C F_{4}

X e F_{4}

S F_{4}

C F_{4}

X e F_{4}

{B F_{4}}^{-}

S F_{4}

C F_{4}

X e F_{4}

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