MCQ
Bond order of $1.5$ is shown by-
- A$O_{2}^+$
- ✓$O_{2}^-$
- C$O_2^{2-}$
- D$O_2$
✓
Answer
Correct option: B.
$O_{2}^-$
b
Bond order $=\frac{\text { bonding M.O. } {-} \text {antibonding M.O. }}{2}$
Bond order $=\frac{\text { bonding M.O. } {-} \text {antibonding M.O. }}{2}$
Bond order of $\mathrm{O}_{2}^{+}=\frac{10-5}{2}=2.5$
Bond order of $\mathrm{O}_{2}^{-}=\frac{10-7}{2}=1.5$
Bond order of $\mathrm{O}_{2}^{2-}=\frac{10-8}{2}=1.0$
Bond order of $\mathrm{O}_{2}=\frac{10-6}{2}=2$
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