To solve this matching problem, we will analyze each diatomic molecule in Column I using Molecular Orbital Theory (MOT) and determine its properties from Column II.

Properties Definitions:

• (P) Paramagnetic: The molecule has one or more unpaired electrons.
• (Q) Undergoes oxidation: The molecule can lose one or more electrons.
• (R) Undergoes reduction: The molecule can gain one or more electrons.
• (S) Bond order ≥ 2: The calculated bond order is 2 or greater.
• (T) Mixing of s and p orbitals: This occurs for diatomic molecules of elements up to Nitrogen (Li₂ to N₂), affecting the energy order of molecular orbitals.

Step-by-step Analysis of Each Molecule:

1. (A) B₂ (Boron molecule)

• Total electrons: $5 + 5 = 10$.
• MOT Configuration: For B₂, s-p mixing occurs. The configuration is: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} (\pi_{2p_x}^1 = \pi_{2p_y}^1)$.
• (P) Paramagnetic: Yes, it has two unpaired electrons in the $\pi_{2p}$ orbitals.
• (Q) Undergoes oxidation: Yes, it can lose an electron to form $B_2^+$.
• (R) Undergoes reduction: Yes, it can gain an electron to form $B_2^-$.
• (S) Bond order ≥ 2: Bond Order = $\frac{1}{2}(\text{Bonding e}^- - \text{Antibonding e}^-) = \frac{1}{2}(6 - 4) = 1$. This is not $\ge 2$.
• (T) Mixing of s and p orbitals: Yes, s-p mixing is significant for B₂.
• Conclusion for (A): Matches with (P), (Q), (R), (T).

2. (B) N₂ (Nitrogen molecule)

• Total electrons: $7 + 7 = 14$.
• MOT Configuration: s-p mixing occurs. The configuration is: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} (\pi_{2p_x}^2 = \pi_{2p_y}^2) \sigma_{2p_z}^2$.
• (P) Paramagnetic: No, all electrons are paired, so it is diamagnetic.
• (Q) & (R) Redox: While N₂ can be oxidized or reduced, it is extremely stable and unreactive due to its high bond dissociation energy. So, these are not its characteristic properties.
• (S) Bond order ≥ 2: Bond Order = $\frac{1}{2}(10 - 4) = 3$. This is $\ge 2$.
• (T) Mixing of s and p orbitals: Yes, s-p mixing occurs.
• Conclusion for (B): Its most defining properties from the list are (S) and (T).

3. (C) O₂⁻ (Superoxide ion)

• Total electrons: $8 + 8 + 1 = 17$.
• MOT Configuration: No significant s-p mixing for oxygen. The configuration is: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 (\pi_{2p_x}^2 = \pi_{2p_y}^2) (\pi_{2p_x}^{*2} \pi_{2p_y}^{*1})$.
• (P) Paramagnetic: Yes, it has one unpaired electron in a $\pi^*$ orbital.
• (Q) Undergoes oxidation: Yes, it can lose an electron to form the stable O₂ molecule ($O_2^- \to O_2 + e^-$). This is a common reaction.
• (R) Undergoes reduction: Yes, it can gain an electron to form the peroxide ion, $O_2^{2-}$.
• (S) Bond order ≥ 2: Bond Order = $\frac{1}{2}(10 - 7) = 1.5$. This is not $\ge 2$.
• (T) Mixing of s and p orbitals: No.
• Conclusion for (C): Matches with (P), (Q), and (R). Given the options, (P) and (Q) are listed.

4. (D) O₂ (Oxygen molecule)

• Total electrons: $8 + 8 = 16$.
• MOT Configuration: No significant s-p mixing. The configuration is: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 (\pi_{2p_x}^2 = \pi_{2p_y}^2) (\pi_{2p_x}^{*1} = \pi_{2p_y}^{*1})$.
• (P) Paramagnetic: Yes, it has two unpaired electrons in the $\pi^*$ orbitals.
• (Q) Undergoes oxidation: Yes, it can be oxidized to $O_2^+$, although it requires a very strong oxidizing agent.
• (R) Undergoes reduction: Yes, this is a very characteristic property. O₂ is a strong oxidizing agent, meaning it readily gets reduced.
• (S) Bond order ≥ 2: Bond Order = $\frac{1}{2}(10 - 6) = 2$. This is $\ge 2$.
• (T) Mixing of s and p orbitals: No.
• Conclusion for (D): Matches with (P), (Q), (R), (S).

Evaluating the Options:

Let's summarize our findings and compare them with the given options.

• (A) B₂: (P), (Q), (R), (T)
• (B) N₂: (S), (T) (most characteristic)
• (C) O₂⁻: (P), (Q), (R)
• (D) O₂: (P), (Q), (R), (S)

Now we check the options:

• A: (A)→(P),(Q),(R),(T); (B)→(S),(T); (C)→(P),(Q); (D)→(P),(Q),(S)
  • (A) is a perfect match.
  • (B) is a good match, focusing on the key properties.
  • (C) is a plausible match, as oxidation to stable O₂ is a key feature.
  • (D) matches (P) and (S) correctly. It includes (Q) which is possible, but incorrectly omits (R), a very important property of O₂. Despite this flaw, it's the best fit among other options.
• B: (A)→(P),(S),(R),(T) - Incorrect for (A). Bond order of B₂ is 1, so (S) is false.
• C: (B)→(P),(T) - Incorrect for (B). N₂ is diamagnetic, so (P) is false.
• D: (C)→(S),(Q) - Incorrect for (C). Bond order of O₂⁻ is 1.5, so (S) is false.

Options B, C, and D contain clear factual errors. Option A is the most consistent and correct choice, despite the ambiguity regarding the redox properties of O₂. Therefore, option A is the correct answer.