Which of the following have fractional bond order and is(are) paramagnetic ?

Options

A. `C_(2)^(+)`
B. `O_(2)^(-)`
C. `NO`
D. `CO`

Explanation

To determine which of the given species have fractional bond order and are paramagnetic, we will analyze each option systematically using Molecular Orbital Theory (MOT).

### Step-by-Step Solution:

1. **Understanding Bond Order**:
- The bond order can be calculated using the formula:
\[ \text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2} \]
- A fractional bond order indicates that the bond is not a single or double bond but somewhere in between.

2. **Analyzing C₂⁺**:
- C₂⁺ has 11 electrons (C has 6 electrons, so C₂ has 12, and removing one gives us 11).
- Filling the molecular orbitals:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², σ(2p)², π(2p)², and π*(2p)¹.
- Bonding electrons = 6 (σ(1s) + σ(2s) + σ(2p)), Antibonding electrons = 4 (σ*(1s) + σ*(2s)).
- Bond Order = (6 - 4) / 2 = 1.
- **Conclusion**: C₂⁺ has a bond order of 1 (not fractional) and is not paramagnetic (no unpaired electrons).

3. **Analyzing O₂⁻**:
- O₂ has 16 electrons, and O₂⁻ has 17 electrons.
- Filling the molecular orbitals:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², σ(2p)², π(2p)⁴, and π*(2p)¹.
- Bonding electrons = 10, Antibonding electrons = 7.
- Bond Order = (10 - 7) / 2 = 1.5 (fractional).
- There is one unpaired electron in π*(2p), making it paramagnetic.
- **Conclusion**: O₂⁻ has a fractional bond order of 1.5 and is paramagnetic.

4. **Analyzing NO**:
- NO has 15 electrons (7 from N and 8 from O).
- Filling the molecular orbitals:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², σ(2p)², π(2p)⁴, and π*(2p)¹.
- Bonding electrons = 10, Antibonding electrons = 5.
- Bond Order = (10 - 5) / 2 = 2.5 (fractional).
- There is one unpaired electron in π*(2p), making it paramagnetic.
- **Conclusion**: NO has a fractional bond order of 2.5 and is paramagnetic.

5. **Analyzing CO**:
- CO has 14 electrons (6 from C and 8 from O).
- Filling the molecular orbitals:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², σ(2p)², π(2p)⁴.
- Bonding electrons = 10, Antibonding electrons = 4.
- Bond Order = (10 - 4) / 2 = 3 (not fractional).
- There are no unpaired electrons, making it diamagnetic.
- **Conclusion**: CO has a bond order of 3 (not fractional) and is not paramagnetic.

### Final Answer:
The species with fractional bond order and are paramagnetic are **O₂⁻ and NO**.