Which of the following possess two lone pair of electrons on the central atom and square planar in shape?

(I) $S F_{4}$ (II) $X e O_{4}$ (III) $X e F_{4}$ (IV) $I C l_{4}^{-}$

I, II

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II, IV

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III, IV

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Solution

The correct option is C
III, IV

Firstly, calculate the value of electron pairs and geometry.

$S F_{4} \Rightarrow \frac{(V + L + A - C)}{2} = \frac{(6 + 4)}{2} = 5 \Rightarrow$ [Trigonal Bipyramidal]

$X e O_{4} \Rightarrow \frac{8}{2} = 4 \Rightarrow [A V S E P R] \Rightarrow$ Tetrahedral

$X e F_{4} \Rightarrow \frac{(8 + 4)}{2} = 6 \Rightarrow$ Octahedral

$I C l_{4}^{-} \Rightarrow \frac{(7 + 4 - 1)}{2} = 6 \Rightarrow$ Octahedral

$\Rightarrow S F_{4} \Rightarrow$ ep = 5 $\Rightarrow$ 1 lone pair

$\Rightarrow X e O_{4} \Rightarrow$ ep = 4 $\Rightarrow$ 0 lone pair

$\Rightarrow X e F_{4} \Rightarrow$ ep = 6 $\Rightarrow$ 2 lone pairs

$\Rightarrow I C l_{4}^{-} \Rightarrow$ ep = 6 $\Rightarrow$ 2 lone pairs

And structure of $X e F_{4}$ and $I C l_{4}^{-}$

Therefore, both $X e F_{4}$ and $I C l_{4}^{-}$ possess two lone pairs and has square planar shape.

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