`{:(,"Column I",,"Column II "), ((A),B_(2),(p),"Paramagnetic"),((B),N_(2),(q),"Undergoes oxidation "),((C),O_(2)^(-),(r),"Undergoes reduction"),((D),O_(2) ,(s),"Bond order"ge 2 ),(,,(t),"Mixing of s and p-orbitals"):}`

Explanation

To solve the matching question between Column I and Column II, we will analyze each species in Column I and determine its corresponding characteristics from Column II.

### Step-by-Step Solution:

1. **Identify B₂:**
- B₂ has 10 electrons. According to Molecular Orbital Theory (MOT), the electron configuration is:
- σ1s² σ*1s² σ2s² σ*2s² σ2p² (which includes 2 in π2p_x and 1 in π2p_y).
- Since it has unpaired electrons (1 in π2p_x and 1 in π2p_y), B₂ is **paramagnetic**.
- Therefore, **A matches with P (Paramagnetic)**.

2. **Identify N₂:**
- N₂ has a bond order of 3 (since it has 10 electrons, the configuration is σ1s² σ*1s² σ2s² σ*2s² σ2p²).
- The bond order is greater than or equal to 2, and it also exhibits **sp orbital mixing** due to the proximity of 2s and 2p orbitals.
- Therefore, **B matches with S (Bond order ≥ 2)** and **T (Mixing of s and p-orbitals)**.

3. **Identify O₂⁻:**
- O₂ has 12 electrons, and the configuration is σ1s² σ*1s² σ2s² σ*2s² σ2p² (with 2 unpaired electrons).
- O₂ is **paramagnetic** and can undergo **oxidation** (O₂⁻ can lose an electron to become O₂).
- Therefore, **C matches with P (Paramagnetic)** and **Q (Undergoes oxidation)**.

4. **Identify O₂:**
- O₂ is also **paramagnetic** (with 2 unpaired electrons).
- It can undergo oxidation and has a bond order of 2 (due to the presence of a double bond).
- Therefore, **D matches with P (Paramagnetic)**, **Q (Undergoes oxidation)**, and **S (Bond order ≥ 2)**.

### Final Matches:
- A → P (Paramagnetic)
- B → S (Bond order ≥ 2) and T (Mixing of s and p-orbitals)
- C → P (Paramagnetic) and Q (Undergoes oxidation)
- D → P (Paramagnetic), Q (Undergoes oxidation), and S (Bond order ≥ 2)