JEE Advanced 2025 · Paper 1 · Chemistry Q5

ChemistryMultiple correct+4 / -2Question 37 of 48

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

Bond order of

\ce{Ne2}

is zero.

The highest occupied molecular orbital (HOMO) of

\ce{F2}

\sigma

-type.

Bond energy of

\ce{O2+}

is smaller than the bond energy of

\ce{O2}

Bond length of

\ce{Li2}

is larger than the bond length of

\ce{B2}

Answer and worked solution

Official answer: B, C

(A) Correct. $\ce{Ne2}$ has equal bonding and antibonding electrons, so its bond order is 0.
(B) Incorrect. For $\ce{F2}$ the HOMO is $\pi^*2p$ , not a σ orbital.
(C) Incorrect. $\ce{O2+}$ has bond order 2.5, higher than 2 for $\ce{O2}$ , so its bond energy is larger.
(D) Correct. $\ce{Li2}$ (about 267 pm) is longer than $\ce{B2}$ (about 159 pm).

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

JEE Advanced 2025 · Paper 1, Chemistry question 5