Chemistry - CHEMICAL BONDING Question with Solution | TestHub

${\mathrm{XeF}}_4$ is ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridised with two lone pairs of electrons, hence, it is square planar.

${\mathrm{SF}}_4$ is ${\mathrm{sp}}^3\mathrm{d}$ hybridised with one lone pairs of electrons, hence, it is See saw.

${\mathrm{SiF}}_4$ is ${\mathrm{sp}}^3$ hybridised with zero lone pairs of electrons, hence, it is Tetrahedral

${{\mathrm{BF}}_4}^{-}$ is ${\mathrm{sp}}^3$ hybridised with zero lone pairs of electrons, hence, it is Tetrahedral

$\left[\mathrm{Cu}\right({\mathrm{NH}}_3{)}_4]{ }^{2+}$is ${\mathrm{dsp}}^2$ hybridised and is Square planar

$\left[{\mathrm{FeCl}}_4\right]{ }^{2–}$ is ${\mathrm{sp}}^3$ hybridised and is Tetrahedral

$\left[{\mathrm{PtCl}}_4\right]{ }^{2–}$is ${\mathrm{dsp}}^2$ hybridised and is Square planar

${{\mathrm{BrF}}_4}^{-}$ is ${\mathrm{sp}}^3{\mathrm{d}}^2$ with  Square planar geometry.

So, $4$ square planer shape compounds are present.