Match the following columns :
Column-I Column-II
(a) ${XeF}_{5}^{+}$ (p) Two lone pairs on central atom
(b) ${ICl}_{4}^{-}$ (q) Planar
(c) ${TeCl}_{4}$ (r) Non-planar
(d) $I_{3}^{+}$ (s) sp³d² (Hybridization of central atom)

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

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

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

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

answer is B.

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Detailed Solution

Since the ${ICl}_{4}^{-}$ ion includes four bond pairs in addition to lone pairs, it takes the shape of a square planar molecular structure. It has octahedral geometry.
Since not all of its atoms reside in the same plane, ${TeCl}_{4}$ is not a planar substance.
The bond pairs occupy the equatorial plane and the axial plane. Therefore, we can say that the plane in which the bond pairs are present is planar.

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