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MCQEasyJEE 2026Molecular Orbital Theory

JEE Chemistry 2026 Question with Solution

Pair of species among the following having same bond order as well as paramagnetic character will be:

  • A

    O2−\mathrm{O_2^-}, N2−\mathrm{N_2^-}

  • B

    O2+\mathrm{O_2^+}, N22−\mathrm{N_2^{2-}}

  • C

    O2−\mathrm{O_2^-}, N2+\mathrm{N_2^+}

  • D

    O2+\mathrm{O_2^+}, N2−\mathrm{N_2^-}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: We need the pair of species having the same bond order and also paramagnetic character.

Find: Which option contains two species with equal bond order and one unpaired electron each.

From the solution:

Step 1: Bond order calculation. Bond order of O2+\mathrm{O_2^+} = 2.52.5 and bond order of N2−\mathrm{N_2^-} = 2.52.5.

Step 2: Magnetic nature. Both species contain one unpaired electron, hence both are paramagnetic.

Step 3: Conclusion. Thus, O2+\mathrm{O_2^+} and N2−\mathrm{N_2^-} have same bond order and are paramagnetic.

Therefore, the correct option is D.

Common mistakes

  • Assuming only odd-electron species are paramagnetic without checking bond order. The question asks for both same bond order and paramagnetic character, so both conditions must be verified.

  • Using an incorrect molecular orbital ordering for nitrogen species. A wrong MO sequence gives an incorrect bond order, so use the proper MO filling before counting bonding and antibonding electrons.

  • Confusing O2+\mathrm{O_2^+} with O2−\mathrm{O_2^-}. Removing an electron and adding an electron affect bond order and magnetic nature differently, so track the electron count carefully.

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