Chemical Bonding and Molecular Structure — JEE Advanced · 2025 · 18 May 2025 · Paper 1 Question with Solution

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.✓ Correct
(C)Bond energy of $O_{2}^{+}$ is smaller than the bond energy of $O_{2}$ .✓ Correct
(D)Bond length of $Li_{2}$ is larger than the bond length of $B_{2}$

Answer: (B), (C)

Solution

(A) Correct:

Ne_{2}

has electronic configuration with equal bonding and antibonding electrons, giving bond order =

\frac{10 - 10}{2} = 0

.

(B) Incorrect: For

F_{2}

, the HOMO is

π^{*} 2 p

(antibonding pi orbital), not a

σ

-type orbital.

(C) Incorrect:

O_{2}^{+}

has bond order 2.5 (one electron removed from antibonding

π^{*}

), while

O_{2}

has bond order 2.0. Higher bond order means higher bond energy, so

O_{2}^{+}

has greater bond energy than

O_{2}

.

(D) Correct: Li is larger than B atom. As atomic size increases, bond length increases. Therefore,

Li_{2}

has larger bond length than

B_{2}

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · One or more correct options