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Shape and hybridization of ${IF}_{5}$ respectively are

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Trigonal bipyramidal ${sp}^{3} d$

See saw, ${sp}^{3} d$

Square pyramidal, ${sp}^{3} d^{2}$

Pentagonal pyramidal, ${sp}^{3} d^{3}$

answer is C.

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Detailed Solution

Count the no. of bond pairs and lone-pair around the central atom.

${IF}_{5} :$ Summation of group number $= 42$

Bond pairs $= \frac{42}{8} = 5 (Residue 2)$

Lone pair $= \frac{2}{2} = 1$

5 bond pairs, 1 lone pair means the geometry is square pyramidal.

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