MCQEasyJEE 2023Molecular Orbital Theory

JEE Chemistry 2023 Question with Solution

The bond order and magnetic property of acetylide ion are same as that of:

  • A

    NO+\mathrm{NO}^+.

  • B

    O2+\mathrm{O}_2^+.

  • C

    O2\mathrm{O}_2^-.

  • D

    N2+\mathrm{N}_2^+.

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: We compare the bond order and magnetic property of acetylide ion, C22\mathrm{C}_2^{2-}.

Find: Which given species has the same bond order and magnetic property.

For C22\mathrm{C}_2^{2-}, total electrons are 1414. Its molecular orbital configuration is:

σ(1s)2σ(1s)2σ(2s)2σ(2s)2σ(2pz)2π(2px)2π(2py)2\sigma(1s)^2\, \sigma^*(1s)^2\, \sigma(2s)^2\, \sigma^*(2s)^2\, \sigma(2p_z)^2\, \pi(2p_x)^2\, \pi(2p_y)^2

Now calculate bond order:

Bond Order=bonding electronsantibonding electrons2\text{Bond Order} = \frac{\text{bonding electrons} - \text{antibonding electrons}}{2}

Here, bonding electrons =10= 10 and antibonding electrons =4= 4, so

Bond Order=1042=3\text{Bond Order} = \frac{10 - 4}{2} = 3

All electrons are paired, so C22\mathrm{C}_2^{2-} is diamagnetic.

Among the given species, NO+\mathrm{NO}^+ also has 1414 electrons, bond order 33, and is diamagnetic.

Therefore, the correct option is A, i.e. NO+\mathrm{NO}^+.

The solution shows option C, but the worked explanation and final answer clearly conclude NO+\mathrm{NO}^+, so the worked solution is taken as authoritative.

Common mistakes

  • Counting the total electrons in C22\mathrm{C}_2^{2-} incorrectly. This leads to the wrong molecular orbital filling. Use the electron count consistent with the worked solution before comparing bond order and magnetism.

  • Using only bond order and ignoring magnetic property. Two species can have similar bond order but different numbers of unpaired electrons. Always check both criteria.

  • Confusing the option label with the actual species because the solution says option C. The detailed working and final statement identify NO+\mathrm{NO}^+, so rely on the concluded species from the solution.

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