Match the following columns :
Column-I Column-II
(a) ${IF}_{2}^{-}$ (p) sp³d
(b) ${CIF}_{3}$ (q) Polar
(c) ${XeO}_{3} F_{2}$ (r) One of the bond angles Is 180°
(d) ${SF}_{4}$ (s) Only one lone pair

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(a-p, r); (b-p, q); (c-p, r); (d-p, q, s)

(a-p, q); (b-p, r); (c-p, r); (d-p, q, s)

(a-p, q); (b-p, q); (c-p, r); (d-r, p, s)

(a-p, s); (b-p, q); (c-p, r); (d-p, q, r)

answer is B.

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Detailed Solution

Since iodine is a member of the halogen family, the hybridization of the I atom is sp³d. Its valence electrons number seven. Two monovalent fluorine atoms make up the IF₂ ion, which has a -1 charge. A polar molecule, it.
There is a net dipole and polarity as a result of the large disparity in electronegativity. A polar molecule is ClF₃.
Around the center atom, the geometry of Sf₄ will have an asymmetric distribution of electron regions (S). This molecule is polar as a result.

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