Q34 / 120 · NEET 2012 (Mains) · Chemistry
Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them.
- A$\mathrm{NO} < \mathrm{O}_2^{-} < \mathrm{C}_2^{2-} < \mathrm{He}_2^{+}$
- B$\mathrm{O}_2^{-} < \mathrm{NO} < \mathrm{C}_2^{2-} < \mathrm{He}_2^{+}$
- C$\mathrm{C}_2^{2-} < \mathrm{He}_2^{+} < \mathrm{O}_2^{-} < \mathrm{NO}$
- Correct answerD$\mathrm{He}_2^{+} < \mathrm{O}_2^{-} < \mathrm{NO} < \mathrm{C}_2^{2-}$
Solution
Bond order $=\frac{N_b-N_a}{2}$
In NO, total electrons $=7+8=15$
$\therefore$ Configuration of $\mathrm{NO}$
$$
\begin{aligned}
& =K K, \sigma\left(2 s^2\right), \stackrel{*}{\sigma}\left(2 s^2\right), \sigma\left(2 p_z^2\right), \pi\left(2 p_x^2\right) \\
& \approx \pi\left(2 p_y^2\right) \stackrel{*}{\pi}\left(2 p_x^1\right) \\
&
\end{aligned}
$$
$\therefore$ Bond order $=\frac{8-3}{2}=\frac{5}{2}=2.5$
In $\mathrm{O}_2^{-}$, total electrons $=16+1=17$
$$
\begin{aligned}
& \therefore \text { Configuration of } \mathrm{O}_2^{-} \\
& =K K, \sigma\left(2 s^2\right), \stackrel{*}{\sigma}\left(2 s^2\right), \sigma\left(2 p_z^2\right), \pi\left(2 p_x^2\right) \\
& \approx \pi\left(2 p_y^2\right), \pi\left(2 p_x^2\right) \approx \pi\left(2 p_y^1\right) \\
& \therefore \text { Bond order }=\frac{8-5}{2}=\frac{3}{2}=1.5 \\
& \text { In } \mathrm{C}_2^{2-} \text {, total electrons }=12+2=14 \\
& \therefore \text { Configuration of } \mathrm{C}_2^{2-} \\
& =K K, \sigma\left(2 s^2\right), \stackrel{*}{\sigma}\left(2 s^2\right), \sigma\left(2 p_z^2\right), \approx \pi\left(2 p_x^2\right), \pi\left(2 p_y^2\right) \\
& \therefore \quad \text { Bond order }=\frac{8-2}{2} \\
& =\frac{6}{2}=3 \\
& \text { In } \mathrm{He}_2^{+} \text {, total electrons }=4-1=3 \\
& \therefore \text { Configuration of } \mathrm{He}_2^{+}=\sigma\left(1 s^2\right), \stackrel{*}{\sigma}\left(1 s^1\right) \\
& \therefore \quad \text { Bond order }=\frac{2-1}{2} \\
& =\frac{1}{2}=0.5 \\
&
\end{aligned}
$$
Hence, correct order of bond order is
$$
\mathrm{He}_2^{+} < \mathrm{O}_2^{-} < \mathrm{NO} < \mathrm{C}_2^{2-}
$$
Hence, the correct answer is option: 'D'.