Question:

In which pair or pairs is the stronger bond found in the first species? I: $O_2^{2-},\, O_2$
II: $N_2,\, N_2^+$
III: $NO^+,\, NO$

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Bond order trends:
Higher bond order $\Rightarrow$ shorter and stronger bond
Removal of antibonding electrons increases bond order
Addition of antibonding electrons decreases bond order

Updated On: Jan 9, 2026

I only
II only
I and III only
II and III only

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The Correct Option is D

Solution and Explanation

Step 1: Bond strength is directly proportional to bond order. Higher bond order $\Rightarrow$ stronger bond.
Step 2: Pair I: $O_2^{2-}$ and $O_2$ \[ \text{Bond order of } O_2 = 2 \] \[ \text{Bond order of } O_2^{2-} = 1 \] Hence, bond is stronger in $O_2$, not in the first species. \[ \Rightarrow \text{Pair I is incorrect} \]
Step 3: Pair II: $N_2$ and $N_2^+$ \[ \text{Bond order of } N_2 = 3 \] \[ \text{Bond order of } N_2^+ = 2.5 \] Hence, bond is stronger in $N_2$ (first species). \[ \Rightarrow \text{Pair II is correct} \]
Step 4: Pair III: $NO^+$ and $NO$ \[ \text{Bond order of } NO = 2.5 \] \[ \text{Bond order of } NO^+ = 3 \] Hence, bond is stronger in $NO^+$ (first species). \[ \Rightarrow \text{Pair III is correct} \]
Step 5: Therefore, the stronger bond is found in the first species for pairs II and III only.

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