Q27 / 90 · KVPY 2010 (SB/SX) · Chemistry
The bond order in $\mathrm{O}_{2}{ }^{2-}$ is-
- A2
- B3
- C$1.5$
- Correct answerD1
Solution
Bond order of $\mathrm{O}_{2}{ }^{2-}$
Total electron $=18$
Configuration $=\mathrm{KK} \sigma(2 \mathrm{~s})^{2} \sigma^{*}(2 \mathrm{~s})^{2} \sigma\left(2 \mathrm{p}_{z}\right)^{2} \pi\left(2 \mathrm{p}_{x}\right)^{2} \pi\left(2 p_{y}\right)^{2} \pi^{*}\left(2 \mathrm{p}_{\mathrm{x}}\right)^{2} \pi^{*}\left(2 \mathrm{p}_{y}\right)^{2}$
Bond order $=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{8-6}{2}=1.0$
Hence, the correct answer is option: 'D'.