JEE Advanced 2025 Paper 1 Q05 Chemistry Chemical Bonding Molecular Orbital Theory Medium

JEE Advanced 2025 Paper 1 · Q05 · Molecular Orbital Theory

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

  1. A. Bond order of $Ne_2$ is zero.
  2. B. The HOMO of $F_2$ is $\sigma$-type.
  3. C. Bond energy of $O_2^+$ is smaller than that of $O_2$.
  4. D. Bond length of $Li_2$ is larger than that of $B_2$.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:B, C

Solution

Question asks for INCORRECT options. (A) $Ne_2$: bonding electrons = anti-bonding electrons ⟹ bond order = 0. CORRECT statement (so NOT marked). (B) $F_2$ (14 valence $e^-$): MO order gives HOMO = $\pi^*(2p)$, which is $\pi$-type, not $\sigma$. INCORRECT statement (marked). (C) $O_2$ has bond order 2 (16 valence $e^-$); $O_2^+$ has bond order 2.5 (lose one anti-bonding $e^-$). Higher bond order ⟹ stronger and shorter bond ⟹ bond energy of $O_2^+ > O_2$. So saying $O_2^+$ has SMALLER bond energy is INCORRECT (marked). (D) $Li_2$: bond order 1 ($2s$ $\sigma$); $B_2$: bond order 1 (with two unpaired electrons in $2p$ $\pi$). Although both are bond order 1, $Li_2$ has weaker, longer bond ($Li-Li$ ~267 pm > $B-B$ ~159 pm). Statement is CORRECT (so NOT marked). Answer (incorrect statements): B, C.

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