Q42 / 180 · COMEDK 2024 (Afternoon Shift) · Chemistry

The bond dissociation enthalpy of the species in their correct order is:

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

Solution

Bond order $= \dfrac{1}{2}(N_b - N_a)$. Higher bond order $\to$ higher bond dissociation enthalpy.

$O_2^+$ (15e): $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*1}$ $\to$ BO $= \dfrac{1}{2}(10-5) = 2.5$

$O_2$ (16e): $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*1} \pi_{2p_y}^{*1}$ $\to$ BO $= \dfrac{1}{2}(10-6) = 2.0$

$O_2^-$ (17e): $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*2} \pi_{2p_y}^{*1}$ $\to$ BO $= \dfrac{1}{2}(10-7) = 1.5$

$O_2^{2-}$ (18e): $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*2} \pi_{2p_y}^{*2}$ $\to$ BO $= \dfrac{1}{2}(10-8) = 1.0$

Order: $O_2^+ > O_2 > O_2^- > O_2^{2-}$

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

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