JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Chemical Bonding and Molecular Structure

Classifying the central atoms: $[{\mathrm{I}}_3{]}^{+}$ has two bonding pairs and two lone pairs, so it is $s{p}^3$ (bent); $[\mathrm{Si}{\mathrm{O}}_4{]}^{4-}$ is tetrahedral $s{p}^3$; $\mathrm{S}{\mathrm{O}}_2\mathrm{C}{\mathrm{l}}_2$ is tetrahedral $s{p}^3$; $\mathrm{Ni}(\mathrm{CO}{)}_4$ is tetrahedral $s{p}^3$; and $\mathrm{SOC}{\mathrm{l}}_2$ has three bonds with one lone pair, $s{p}^3$. The others are not $s{p}^3$: $\mathrm{Xe}{\mathrm{F}}_2$, $\mathrm{S}{\mathrm{F}}_4$, $\mathrm{Cl}{\mathrm{F}}_3$ and $\mathrm{Xe}{\mathrm{O}}_2{\mathrm{F}}_2$ are $s{p}^{3}d$; $\mathrm{Xe}{\mathrm{F}}_4$ is $s{p}^3{d}^2$; and $[\mathrm{PtC}{\mathrm{l}}_4{]}^{2-}$ is square planar ($ds{p}^2$). That gives five species with an $s{p}^3$ central atom.