JEE Main 2026 — Chemical Bonding and Molecular Structure Question with Solution

The steric number $H$ is calculated as $H=\frac{1}{2}(V+M-C+A)$, where $V$ is the number of valence electrons on the central atom, $M$ is the number of monovalent atoms, $C$ is the cationic charge, and $A$ is the anionic charge.

For $s{p}^{3}d$ hybridization, $H=5$.

BrF${}_5$: $H=\frac{1}{2}(7+5)=6$ ($s{p}^3{d}^2$)

XeF${}_5^{-}$: $H=\frac{1}{2}(8+5+1)=7$ ($s{p}^3{d}^3$)

BF${}_4^{-}$: $H=\frac{1}{2}(3+4+1)=4$ ($s{p}^3$)

ICl${}_4^{-}$: $H=\frac{1}{2}(7+4+1)=6$ ($s{p}^3{d}^2$)

XeF${}_4$: $H=\frac{1}{2}(8+4)=6$ ($s{p}^3{d}^2$)

SF${}_4$: $H=\frac{1}{2}(6+4)=5$ ($s{p}^{3}d$)

NH${}_4^{+}$: $H=\frac{1}{2}(5+4-1)=4$ ($s{p}^3$)

ClF${}_3$: $H=\frac{1}{2}(7+3)=5$ ($s{p}^{3}d$)

XeF${}_2$: $H=\frac{1}{2}(8+2)=5$ ($s{p}^{3}d$)

ICl${}_2^{-}$: $H=\frac{1}{2}(7+2+1)=5$ ($s{p}^{3}d$)

The species with $s{p}^{3}d$ hybridization are SF${}_4$, ClF${}_3$, XeF${}_2$, and ICl${}_2^{-}$.

The number of such species is $4$.

Answer: $4$