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

Statement I: Checking geometry of each species:
$S{F}_4$: 4 bp + 1 lp → see-saw
$N{H}_4^{+}$: 4 bp → tetrahedral
$[NiC{l}_4{]}^{2-}$: $N{i}^{2+}$ (d⁸) with weak field $C{l}^{-}$ → $s{p}^3$ → tetrahedral
$Xe{F}_4$: 4 bp + 2 lp → square planar
$[PtC{l}_4{]}^{2-}$: $P{t}^{2+}$ (d⁸) → $ds{p}^2$ → square planar
$Se{F}_4$: 4 bp + 1 lp → see-saw
$[Ni(CN{)}_4{]}^{2-}$: $N{i}^{2+}$ with strong field $C{N}^{-}$ → $ds{p}^2$ → square planar
Only 2 species are tetrahedral. Statement I is false.
Statement II: $N{O}_2$ (odd electron), $Be{H}_2$ (4e⁻), $B{F}_3$ (6e⁻), $AlC{l}_3$ (6e⁻) all have incomplete octet. Statement II is true.