Match the species in Column I with the bond order in Column II
`{:(" Column I Column II " ),((i) NO " (a)" 1.5),((ii) CO" (b)" 2.0),((iii)O_(2)^(-)" "(c) 2.5),((iv) O_(2)" (d)" 3.0):}`

Explanation

To solve the problem of matching the species in Column I with their bond order in Column II, we will calculate the bond order for each species using the known electron configurations and the bond order formula.

### Step-by-Step Solution:

1. **Identify the species and their total number of electrons:**
- (i) NO: Nitrogen (N) has 7 electrons, and Oxygen (O) has 8 electrons. Total = 7 + 8 = 15 electrons.
- (ii) CO: Carbon (C) has 6 electrons, and Oxygen (O) has 8 electrons. Total = 6 + 8 = 14 electrons.
- (iii) O₂⁻: Each Oxygen (O) has 8 electrons, and with an additional electron due to the negative charge, Total = 8 + 8 + 1 = 17 electrons.
- (iv) O₂: Each Oxygen (O) has 8 electrons. Total = 8 + 8 = 16 electrons.

2. **Calculate the bond order for each species:**
- **Bond Order Formula:** Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2.
- For the given trick:
- 14 electrons correspond to a bond order of 3.
- 15 electrons correspond to a bond order of 2.5.
- 16 electrons correspond to a bond order of 2.
- 17 electrons correspond to a bond order of 1.5.

3. **Match the bond orders:**
- (i) NO (15 electrons) → Bond order = 2.5 → Match with (c).
- (ii) CO (14 electrons) → Bond order = 3 → Match with (d).
- (iii) O₂⁻ (17 electrons) → Bond order = 1.5 → Match with (a).
- (iv) O₂ (16 electrons) → Bond order = 2 → Match with (b).

### Final Matches:
- (i) NO → (c) 2.5
- (ii) CO → (d) 3.0
- (iii) O₂⁻ → (a) 1.5
- (iv) O₂ → (b) 2.0

### Summary of Matches:
- (i) NO → (c) 2.5
- (ii) CO → (d) 3.0
- (iii) O₂⁻ → (a) 1.5
- (iv) O₂ → (b) 2.0