Q42 / 180 · AIIMS 2016 · Chemistry
Four diatomic species are listed below in different sequences. Which of these represents the correct order of their increasing bond order?
- A$\mathrm{C}_2^{2-} < \mathrm{He}_2^{+} < \mathrm{NO} < \mathrm{O}_2^{-}$
- Correct answerB$\mathrm{He}_2^{+} < \mathrm{O}_2^{-} < \mathrm{NO} < \mathrm{C}_2^{2-}$
- C$\mathrm{O}_2^{-} < \mathrm{NO} < \mathrm{C}_2^{2-} < \mathrm{He}_2^{+}$
- D$\mathrm{NO} < \mathrm{C}_2^{2-} < \mathrm{O}_2^{-} < \mathrm{He}_2^{+}$
Solution
According to molecular orbital theory, the energy level of the given molecules are
$\begin{aligned} & \mathrm{C}_2^{2-} \rightarrow \sigma 1 s^2 \sigma^{\star} 1 s^2 \sigma 2 s^2 \sigma^{\star} 2 s^2 \pi 2 p_x^2 \pi 2 p_y^2 \sigma 2 p_z^2 \\ & \text { B.O. }=1 / 2[10-4]=3 \\ & \mathrm{He}_2^{+} \rightarrow \sigma 1 s^2 \sigma^{\star} 1 s^1 \\ & \text { B.O. }=1 / 2[2-1]=1 / 2=0.5 \\ & \mathrm{NO} \rightarrow \sigma 1 s^2 \sigma^{\star} 1 s^2 \sigma 2 s^2 \sigma^{\star} 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2 \\ & \quad \pi^{\star} 2 p_x^1\end{aligned}$
$\begin{aligned} & \text { B.O. }=1 / 2[10-5]=2.5 \\ & \mathrm{O}_2^{-} \rightarrow \sigma 1 s^2 \sigma^{\star} 1 s^2 \sigma 2 s^2 \sigma^{\star} 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2 \pi^{\star} 2 p_x^2 \\ & \text { B.O. }=1 / 2[10-7]=1.5 \\ & \pi^{\star} 2 p_y^1 \\ & \end{aligned}$
So, the correct order of their increasing bond order is $\mathrm{He}_2^{+} < \mathrm{O}_2^{-} < \mathrm{NO} < \mathrm{C}_2^{2-}$
Hence, the correct answer is option: 'B'.