Match the following columns :

Column-I	Column-II
(a) ${\mathrm{XeF}}_5^{+}$	(p) Two lone pairs on central atom
(b) ${\mathrm{ICl}}_4^{-}$	(q) Planar
(c) ${\mathrm{TeCl}}_4$	(r) Non-planar
(d) ${\mathrm{I}}_3^{+}$	(s) sp3d2 (Hybridization of central atom)

Since the ${\mathrm{ICl}}_4^{-}$ 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, ${\mathrm{TeCl}}_4$ 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.