Four diatomic species are listed below the different sequences. Which of these presents the correct order of…
Four diatomic species are listed below the different sequences. Which of these presents the correct order of their increasing bond order?
- $\mathrm{He}_2{ }^{+} < \mathrm{O}_2^{-} < \mathrm{NO} < \mathrm{C}_2{ }^{2-}$
- $\mathrm{O}_2^{-} < \mathrm{NO} < \mathrm{C}_2{ }^{2-} < \mathrm{He}_2{ }^{+}$
- $\mathrm{NO} < \mathrm{C}_2^{2-} < \mathrm{O}_2^{-} < \mathrm{He}_2^{+}$
- $\mathrm{C}_2{ }^{2-} < \mathrm{He}_2{ }^{+} < \mathrm{NO} < \mathrm{O}_2{ }^{-}$
Solution
\(\begin{aligned}
& \text {Key Idea : Bond order }=\frac{N_b-N_a}{2} \\
& \mathrm{He}_2^{+}:(\sigma 1 s)^2\left(\sigma^* 1 s\right)^1 \\
& \text {Bond order }=\frac{2-1}{2}=\frac{1}{2} \\
& \mathrm{C}_2^{2-}: \\
& K K(\sigma 2 s)^2(\sigma * 2 s)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^2\left(\sigma 2 p_z\right)^2 \\
& \text {Bond order }=\frac{8-2}{2}=3 \\
& \mathrm{O}_2^{-}: K K(\sigma 2 s)^2(\sigma 2 s)^2\left(\sigma 2 p_z\right)^2\left(\pi 2 p_x\right)^2 \\
& \left(\pi 2 p_y\right)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^1 \\
& \text {Bond order }=\frac{8-5}{2}=1 \frac{1}{2} \\
& \text {NO : } K K(\sigma 2 s)^2(\sigma 2 s)^2\left(\sigma 2 p_z\right)^2\left(\pi 2 p_x\right)^2 \\
& \left.\left(\pi 2 p_y\right)^2(\pi) p_x\right)^1 \\
& \text {Bond order }=\frac{8-3}{2}=2 \frac{1}{2}
\end{aligned}\)
Hence, the order of increasing bond order is as:
\(\underset{\left(\frac{1}{2}\right)}{\mathrm{He}_2^{+}}<\underset{\left(1 \frac{1}{2}\right)}{\mathrm{O}_2^{-}}<\underset{\left(2 \frac{1}{2}\right)}{\mathrm{NO}}<\mathrm{C}_2^{2-}\)
Asked in: NEET 2008 (Mains)
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