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Q.

Hybridization of central atom in $S F_{4}$ molecule and the shape of molecule are

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a

$S P^{3}$ ; tetrahedral

b

$S P^{3} d^{2}$ ; square pyramidal

c

$S P^{3} d$ ; see saw

d

$S P^{3} d$ ; square planar

answer is A.

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Detailed Solution

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$H = \frac{1}{2} [V + M] = \frac{1}{2} [6 + 4] = 5 - s p^{3} d .$ One L.p and 4 B.p. Shape is see saw (R) distorted tetrahedral

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