The species with fractional bond order is :

Options

A. `O_(2)^(+)`
B. `O_(2)^(2+)`
C. CO
D. `He_(2)`

Explanation

To determine the species with fractional bond order, we need to analyze the electron configuration of various molecules and their bond orders. Here’s a step-by-step solution:

### Step 1: Understand Bond Order
Bond order is defined as the difference between the number of bonding electrons and the number of antibonding electrons, divided by two. The formula is:
\[ \text{Bond Order} = \frac{(N_b - N_a)}{2} \]
where \( N_b \) is the number of bonding electrons and \( N_a \) is the number of antibonding electrons.

### Step 2: Identify Odd Electron Species
A fractional bond order typically occurs in species with an odd number of total electrons. Examples of odd electron counts include 11, 13, 15, etc.

### Step 3: Analyze O2+ (Oxygen Dication)
1. **Total Electrons**: O2+ has 15 electrons (O2 has 16 electrons, and removing one gives us 15).
2. **Molecular Orbital Configuration**:
- The configuration for O2+ is:
- \( \sigma_{1s}^2 \)
- \( \sigma^*_{1s}^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma^*_{2s}^2 \)
- \( \sigma_{2p_z}^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \pi^*_{2p_x}^1 \) (1 unpaired electron)
3. **Count Bonding and Antibonding Electrons**:
- Bonding electrons: 10 (from \( \sigma_{1s}^2, \sigma_{2s}^2, \sigma_{2p_z}^2, \pi_{2p_x}^2, \pi_{2p_y}^2 \))
- Antibonding electrons: 5 (from \( \sigma^*_{1s}^2, \sigma^*_{2s}^2, \pi^*_{2p_x}^1 \))
4. **Calculate Bond Order**:
\[ \text{Bond Order} = \frac{(10 - 5)}{2} = \frac{5}{2} = 2.5 \]

### Step 4: Compare with Other Species
- **CO (Carbon Monoxide)**: 14 electrons, bond order = 3 (integer).
- **O2 (Dioxygen)**: 16 electrons, bond order = 2 (integer).
- **He2 (Helium Dimer)**: 4 electrons, bond order = 0 (does not exist).

### Conclusion
The only species with a fractional bond order is **O2+**, which has a bond order of 2.5.

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