JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Chemical Bonding and Molecular Structure

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✓ Correct
(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

Answer: (A)

Solution

For the first statement,

BF X_{4} X^{-}

is tetrahedral,

SiF X_{4}

is tetrahedral and

XeF X_{4}

is square planar, and in each the symmetry makes every E\textendash F bond equal; only

SF X_{4}

, which is see-saw shaped, has unequal axial and equatorial S\textendash F bonds, so just one species has unequal bond lengths, not two, and the first statement is false. For the second statement, the bond orders are

O X_{2} X^{+}

2.5,

O X_{2} X^{-}

1.5,

O X_{2} X^{2 -}

1 and

F X_{2}

1, so the highest belongs to

O X_{2} X^{+}

, not

O X_{2} X^{-}

, and the second statement is false as well. Both statements are false.

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Chemical Bonding and Molecular Structure

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