Properties of Diatomic Molecules Using Molecular Orbital Theory

To determine the correct matching between Column I and Column II, we analyze each species using Molecular Orbital Theory (MOT):

1. General Principles of MO Theory

• $s\text{-}p$ Mixing:
  For homonuclear diatomic molecules of elements with atomic number $Z \le 7$ (such as $\mathrm{B}$ and $\mathrm{N}$), the energy gap between the $2s$ and $2p$ atomic orbitals is relatively small. This leads to significant mixing between the $\sigma_{2s}$ and $\sigma_{2p_z}$ orbitals, resulting in the energy order:
  $\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < (\pi_{2p_x} = \pi_{2p_y}) < \sigma_{2p_z} < (\pi^*_{2p_x} = \pi^*_{2p_y}) < \sigma^*_{2p_z}$
  For $Z \ge 8$ (such as oxygen), the energy separation between $2s$ and $2p$ orbitals is large, so there is negligible $s\text{-}p$ mixing, and the orbital sequence is:
  $\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < (\pi_{2p_x} = \pi_{2p_y}) < (\pi^*_{2p_x} = \pi^*_{2p_y}) < \sigma^*_{2p_z}$

• Redox Activity: All given species can lose electrons (undergo oxidation) and gain electrons (undergo reduction) under suitable conditions:

  • Oxidation produces cations or oxides/higher oxidation states.
  • Reduction produces anions or hydrides/lower oxidation states.
    Therefore, all species satisfy both (q) and (r).

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2. Analysis of Each Species

(A) $\mathrm{B_2}$

• Total electrons: $10$
• Electronic configuration:
  \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 (\pi_{2p_x}^1 = \pi_{2p_y}^1)
• Paramagnetic nature: There are $2$ unpaired electrons in degenerate $\pi_{2p}$ orbitals $\implies$ Paramagnetic (p).
• Redox property: Can undergo oxidation (e.g., forming $\mathrm{B_2^+}$) $\implies$ (q) and reduction (e.g., forming $\mathrm{B_2^-}$) $\implies$ (r).
• Bond order:
  $\text{Bond Order} = \frac{N_b - N_a}{2} = \frac{6 - 4}{2} = 1 < 2 \implies \text{does not satisfy (s)}$
• $s\text{-}p$ mixing: Boron has $Z = 5 \le 7$, so $s\text{-}p$ mixing is present $\implies$ (t).

Hence, $\text{A} \to \text{p, q, r, t}$.

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(B) $\mathrm{N_2}$

• Total electrons: $14$
• Electronic configuration:
  \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 (\pi_{2p_x}^2 = \pi_{2p_y}^2) \sigma_{2p_z}^2
• Paramagnetic nature: All electrons are paired $\implies$ Diamagnetic (not p).
• Redox property: Can undergo oxidation (e.g., forming $\mathrm{N_2^+}$ or $\mathrm{NO}$) $\implies$ (q) and reduction (e.g., forming $\mathrm{NH_3}$ in Haber's process or $\mathrm{N_2^-}$) $\implies$ (r).
• Bond order:
  $\text{Bond Order} = \frac{N_b - N_a}{2} = \frac{10 - 4}{2} = 3 \ge 2 \implies \textbf{(s)}$
• $s\text{-}p$ mixing: Nitrogen has $Z = 7 \le 7$, so $s\text{-}p$ mixing is present $\implies$ (t).

Hence, $\text{B} \to \text{q, r, s, t}$.

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(C) $\mathrm{O_2^-}$

• Total electrons: $17$
• Electronic configuration:
  \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)
• Paramagnetic nature: Contains $1$ unpaired electron in a $\pi^*_{2p}$ orbital $\implies$ Paramagnetic (p).
• Redox property: Can undergo oxidation ($\mathrm{O_2^- \to O_2 + e^-}$) $\implies$ (q) and reduction ($\mathrm{O_2^- + e^- \to O_2^{2-}}$) $\implies$ (r).
• Bond order:
  $\text{Bond Order} = \frac{N_b - N_a}{2} = \frac{10 - 7}{2} = 1.5 < 2 \implies \text{does not satisfy (s)}$
• $s\text{-}p$ mixing: For oxygen ($Z = 8 > 7$), $s\text{-}p$ mixing is negligible $\implies$ does not satisfy (t).

Hence, $\text{C} \to \text{p, q, r}$.

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(D) $\mathrm{O_2}$

• Total electrons: $16$
• Electronic configuration:
  \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)
• Paramagnetic nature: Contains $2$ unpaired electrons in degenerate $\pi^*_{2p}$ orbitals $\implies$ Paramagnetic (p).
• Redox property: Can undergo oxidation (e.g., reaction with $\mathrm{PtF_6}$ to form $\mathrm{O_2^+[PtF_6]^-}$) $\implies$ (q) and reduction (acts as an oxidant to form $\mathrm{O_2^-}$, $\mathrm{O_2^{2-}}$, or $\mathrm{O^{2-}}$) $\implies$ (r).
• Bond order:
  $\text{Bond Order} = \frac{N_b - N_a}{2} = \frac{10 - 6}{2} = 2 \ge 2 \implies \textbf{(s)}$
• $s\text{-}p$ mixing: Negligible for oxygen ($Z = 8$) $\implies$ does not satisfy (t).

Hence, $\text{D} \to \text{p, q, r, s}$.

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Final Matching:

• $\text{A} \to \text{p, q, r, t}$
• $\text{B} \to \text{q, r, s, t}$
• $\text{C} \to \text{p, q, r}$
• $\text{D} \to \text{p, q, r, s}$