Which of the following statements are correct?
I. Bond order of NO is 2.5
II. Bond order of `NO^(+)` is 3.0
III. Bond order of `O_(2)` is 1.5
IV. Bond order of CO is 3.0

Options

A. I, II, III
B. I, II, IV
C. I, III, IV
D. II, III, IV

Explanation

To solve the question regarding the bond orders of the given molecules, we will analyze each statement one by one.

### Step 1: Analyze the bond order of NO
1. **Determine the total number of electrons in NO**: Nitric oxide (NO) has 15 electrons (7 from nitrogen and 8 from oxygen).
2. **Construct the molecular orbital configuration**:
- σ(1s)²
- σ*(1s)²
- σ(2s)²
- σ*(2s)²
- σ(2p_z)²
- π(2p_x)²
- π(2p_y)²
- π*(2p_x)¹
3. **Count bonding and antibonding electrons**:
- Bonding electrons = 2 (σ(1s)) + 2 (σ(2s)) + 2 (σ(2p_z)) + 2 (π(2p_x)) + 2 (π(2p_y)) = 8
- Antibonding electrons = 2 (σ*(1s)) + 2 (σ*(2s)) + 1 (π*(2p_x)) = 5
4. **Calculate bond order**:
\[ \text{Bond Order} = \frac{\text{Bonding Electrons} - \text{Antibonding Electrons}}{2} = \frac{8 - 5}{2} = \frac{3}{2} = 2.5 \]
- **Conclusion**: Statement I is correct.

### Step 2: Analyze the bond order of NO⁺
1. **Determine the total number of electrons in NO⁺**: NO⁺ has 14 electrons (1 electron removed from NO).
2. **Construct the molecular orbital configuration**:
- The configuration remains the same as NO but without the π*(2p_x) electron.
3. **Count bonding and antibonding electrons**:
- Bonding electrons = 8 (same as above)
- Antibonding electrons = 4 (σ*(1s) + σ*(2s) + 0 (π*(2p_x))) = 4
4. **Calculate bond order**:
\[ \text{Bond Order} = \frac{8 - 4}{2} = \frac{4}{2} = 2 \]
- **Conclusion**: Statement II is incorrect (the bond order is 2, not 3).

### Step 3: Analyze the bond order of O₂
1. **Determine the total number of electrons in O₂**: O₂ has 16 electrons (8 from each oxygen).
2. **Construct the molecular orbital configuration**:
- σ(1s)²
- σ*(1s)²
- σ(2s)²
- σ*(2s)²
- σ(2p_z)²
- π(2p_x)²
- π(2p_y)²
- π*(2p_x)¹
- π*(2p_y)¹
3. **Count bonding and antibonding electrons**:
- Bonding electrons = 10 (2 from σ(1s), 2 from σ(2s), 2 from σ(2p_z), 4 from π(2p_x) and π(2p_y))
- Antibonding electrons = 2 (σ*(1s) + σ*(2s))
4. **Calculate bond order**:
\[ \text{Bond Order} = \frac{10 - 6}{2} = \frac{4}{2} = 2 \]
- **Conclusion**: Statement III is incorrect (the bond order is 2, not 1.5).

### Step 4: Analyze the bond order of CO
1. **Determine the total number of electrons in CO**: Carbon monoxide (CO) has 14 electrons (6 from carbon and 8 from oxygen).
2. **Construct the molecular orbital configuration**:
- σ(1s)²
- σ*(1s)²
- σ(2s)²
- σ*(2s)²
- σ(2p_z)²
- π(2p_x)²
- π(2p_y)²
3. **Count bonding and antibonding electrons**:
- Bonding electrons = 10
- Antibonding electrons = 4
4. **Calculate bond order**:
\[ \text{Bond Order} = \frac{10 - 4}{2} = \frac{6}{2} = 3 \]
- **Conclusion**: Statement IV is correct.

### Final Conclusion
- **Correct Statements**: I and IV
- **Incorrect Statements**: II and III