JEE Advanced 2025 — Paper 1 — CHEM-II solution

JEE Advanced2025Paper 1CHEM-II

Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is(are)

A
Bond order of Ne $_{2}$ is zero.
B
The highest occupied molecular orbital (HOMO) of F $_{2}$ is $σ$ -type.
C
Bond energy of O $_{2}^{+}$ is smaller than the bond energy of O $_{2}$ .
D
Bond length of Li $_{2}$ is larger than the bond length of B $_{2}$ .

Solution

(A) The Ne $_{2}$ configuration fills both bonding and antibonding MOs through $σ^{*} 2 p_{z}$ : bond order $= \frac{6 - 6}{2} = 0$ . Statement is correct, so it is not the answer.

(B) For F $_{2}$ , the HOMO is the degenerate pair $π^{*} 2 p_{x} = π^{*} 2 p_{y}$ , which is $π$ -type, not $σ$ -type. Statement is incorrect.

(C) O $_{2}^{+}$ has one fewer antibonding electron than O $_{2}$ . Bond order: O $_{2}^{+}$ $= 2.5$ , O $_{2}$ $= 2$ . Higher bond order implies stronger and shorter bond, so O $_{2}^{+}$ has greater bond energy, not smaller. Statement is incorrect.

(D) Atomic radius decreases left to right across the period, so Li > B in size, leading to a longer bond in Li $_{2}$ than in B $_{2}$ . Statement is correct.

The incorrect statements are (B) and (C).

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