Q18 / 180 · COMEDK 2023 (Memory Based) · Chemistry

Which of these represents the correct order of decreasing bond order?

  • Correct answer
    A
    $\mathrm{C}_2^{2-}>\mathrm{O}_2^{2+}>\mathrm{O}_2^{-}>\mathrm{He}_2^{2+}$
  • B
    $\mathrm{O}_2^{-}>\mathrm{O}_2^{+}>\mathrm{He}_2^{+}>\mathrm{C}_2^{2-}$
  • C
    $\mathrm{He}_2^{+}>\mathrm{O}_2^{-}>\mathrm{C}_2^{2-}>\mathrm{O}_2^{+}$
  • D
    $\mathrm{He}_2^{+}>\mathrm{O}_2^{2-}>\mathrm{O}_2^{+}>\mathrm{O}_2^{-}$

Solution

$\begin{aligned} & \mathrm{He}_2^{+}: \sigma 1 s^2, \sigma 1 s^1 \\ & \mathrm{BO}=\frac{1}{2}\left(N_b-N_a\right)=\frac{1}{2}(2-1)=\frac{1}{2}=0.5 \\ & \mathrm{C}_2^{2-}: \sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2,\left(\pi 2 p_x^2 \approx \pi 2 p_y^2\right), \sigma 2 p_z^2 \\ & \mathrm{BO}=\frac{1}{2}(10-4)=\frac{6}{2}=3.0 \\ & \mathrm{O}_2^{+}=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2 \sigma 2 p_z^2\left(\pi 2 p_x^2 \approx \pi 2 p_y^2\right), \\ & \mathrm{BO}=\frac{1}{2}(10-5)=\frac{5}{2}=2.5 \\ & \mathrm{O}_2^{-}=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2\left(\pi 2 p_x^2 \approx \pi 2 p_y^2\right), \\ & \mathrm{BO}=\frac{1}{2}(10-7)=\frac{3}{2}=1.5 \quad\left(\pi^* 2 p_x^2 \approx \pi^* 2 p_y^1\right)\end{aligned}$ Thus, the correct order of decreasing bond order is $\mathrm{C}_2^{2-}>\mathrm{O}_2^{+}>\mathrm{O}_2^{-}>\mathrm{He}_2^{+}$

Hence, the correct answer is option: 'A'.

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