(A) O₂⁻ , N₂⁺

(B) O₂⁺ , N₂²⁻

(C) O₂⁻ , N₂⁻

(D) O₂⁺ , N₂⁻

Correct Answer: O₂⁺ , N₂⁻

Explanation

First consider O₂⁺.

Total electrons in O₂ = 16 So, electrons in O₂⁺ = 15

Molecular orbital configuration of O₂⁺ leaves one unpaired electron.

Bond order of O₂⁺:

Bond order = (Bonding electrons − Antibonding electrons) / 2 = (10 − 5) / 2 = 2.5

Since O₂⁺ has an unpaired electron, it is paramagnetic.

Now consider N₂⁻.

Total electrons in N₂ = 14 So, electrons in N₂⁻ = 15

Molecular orbital configuration of N₂⁻ also contains one unpaired electron.

Bond order of N₂⁻:

Bond order = (Bonding electrons − Antibonding electrons) / 2 = (10 − 5) / 2 = 2.5

Since N₂⁻ has an unpaired electron, it is also paramagnetic.

Thus, O₂⁺ and N₂⁻ have the same bond order (2.5) and both are paramagnetic.

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