Among the following species, how many are both square planar in shape and have a zero dipole moment? XeF4, SF4, ICl4^-, [Ni(CN)4]^2-, [NiCl4]^2-…

Examine each species using VSEPR or ligand-field arguments. $\mathrm{XeF_4}$: four bond pairs and two lone pairs, octahedral electron geometry with the lone pairs trans, so square planar and non-polar. $\mathrm{SF_4}$: one lone pair, see-saw shape, polar. $\mathrm{ICl_4^-}$: iodine has four bond pairs and two lone pairs, square planar, non-polar. $[\mathrm{Ni(CN)_4}]^{2-}$: $d^8$ with strong-field $\mathrm{CN^-}$, $dsp^2$ square planar, symmetric, non-polar. $[\mathrm{NiCl_4}]^{2-}$: weak-field chloride, $sp^3$ tetrahedral, so not square planar. $\mathrm{BrF_5}$: square pyramidal, polar. $\mathrm{ClF_3}$: T-shaped, polar. $[\mathrm{PtCl_4}]^{2-}$: $5d^8$ platinum(II), always square planar and non-polar. Four species qualify: $\mathrm{XeF_4}$, $\mathrm{ICl_4^-}$, $[\mathrm{Ni(CN)_4}]^{2-}$ and $[\mathrm{PtCl_4}]^{2-}$.