JEE Main 2026 — Jan 28, Shift 1 — Chemistry solution

JEE Main2026Jan 28, Shift 1Chemistry

Given below are two statements:

Statement I: The number of species among $BF_{4}^{-}$ , $SiF_{4}$ , $XeF_{4}$ and $SF_{4}$ , that have unequal E-F bond lengths is two. Here, E is the central atom.

Statement II: Among $O_{2}^{-}$ , $O_{2}^{2 -}$ , $F_{2}$ and $O_{2}^{+}$ , $O_{2}^{-}$ has the highest bond order.

In the light of the above statements, choose the correct answer from the options given below

A
Both Statement I and Statement II are false
B
Both Statement I and Statement II are true
C
Statement I is true but Statement II is false
D
Statement I is false but Statement II is true

Solution

Statement I: Geometries and bond lengths:

BF $_{4}^{-}$ : tetrahedral, all four B-F equal.

SiF $_{4}$ : tetrahedral, all four Si-F equal.

XeF $_{4}$ : square planar, all four Xe-F equal.

SF $_{4}$ : see-saw (trigonal bipyramidal with one lone pair); the two axial S-F bonds differ in length from the two equatorial S-F bonds. So only one species has unequal E-F bond lengths, not two. False.

Statement II: Bond orders from MO theory:

O $_{2}^{+}$ : 2.5, O $_{2}^{-}$ : 1.5, O $_{2}^{2 -}$ : 1, F $_{2}$ : 1.

The highest bond order belongs to O $_{2}^{+}$ , not O $_{2}^{-}$ . False.

$∴$ Both statements are false. Option (1).

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