Question

Among the species O$_2^+$, N$_2^-$, N$_2^{2-}$ and O$_2^-$ which have same bond order as well as paramagnetic in nature.

• A. O$_2^+$, N$_2^-$
• B. O$_2^-$, N$_2^-$
• C. O$_2^+$, O$_2$
• D. O$_2^-$, N$_2$

Hint:

Same bond order and paramagnetism require identical electron occupation in antibonding orbitals.

Answer 1

The Correct Option is B

Step 1: Concept of bond order

Bond order of a molecule is given by:

Bond Order = ( Number of bonding electrons − Number of antibonding electrons ) / 2

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Step 2: Bond order of individual species

(i) O₂⁺

Electronic configuration:
1σ² 1σ*² 2σ² 2σ*² π_2px² π_2py² π*_2px¹

Bond order = (10 − 5)/2 = 2.5

Unpaired electrons = 1 → Paramagnetic

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(ii) N₂⁻

Electronic configuration:
1σ² 1σ*² 2σ² 2σ*² π_2px² π_2py² π*_2px¹

Bond order = (10 − 7)/2 = 1.5

Unpaired electrons = 1 → Paramagnetic

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(iii) N₂²⁻

Electronic configuration:
1σ² 1σ*² 2σ² 2σ*² π_2px² π_2py² π*_2px¹ π*_2py¹

Bond order = (8 − 4)/2 = 2

No unpaired electrons → Diamagnetic

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(iv) O₂⁻

Electronic configuration:
1σ² 1σ*² 2σ² 2σ*² π_2px² π_2py² π*_2px¹ π*_2py¹

Bond order = (10 − 7)/2 = 1.5

Unpaired electrons = 1 → Paramagnetic

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Step 3: Final conclusion

The species having the same bond order and showing paramagnetic behavior are:

O₂⁻ and N₂⁻