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

Statement I: Let's analyze the geometry of the given species.
$B{F}_4^{-}$: $s{p}^3$ hybridization, tetrahedral geometry. All B-F bond lengths are equal.
$Si{F}_4$: $s{p}^3$ hybridization, tetrahedral geometry. All Si-F bond lengths are equal.
$Xe{F}_4$: $s{p}^3{d}^2$ hybridization, square planar geometry. All Xe-F bond lengths are equal.
$S{F}_4$: $s{p}^{3}d$ hybridization, see-saw geometry. It has axial and equatorial bonds. Axial bonds are longer than equatorial bonds due to greater repulsion. Thus, only $S{F}_4$ has unequal bond lengths. The number of species is one, not two. Statement I is false.
Statement II: Bond order calculation using Molecular Orbital Theory:
Total electrons in $O_2^{-}$ = 17, Bond Order = 1.5
Total electrons in $O_2^{2-}$ = 18, Bond Order = 1.0
Total electrons in $F_2$ = 18, Bond Order = 1.0
Total electrons in $O_2^{+}$ = 15, Bond Order = 2.5
Among these, $O_2^{+}$ has the highest bond order (2.5), not $O_2^{-}$. Statement II is false.
Both Statement I and Statement II are false.