Which of the following has fractional bond order ?

Options

A. `O_(2)^(2-)`
B. `H_(2)^(-)`
C. `F_(2)`
D. None of these

Answer

To determine which of the given species has a fractional bond order, we will calculate the bond order for each species using the formula:

\[ \text{Bond Order} = \frac{1}{2} (n_B - n_A) \]

where \( n_B \) is the number of bonding electrons and \( n_A \) is the number of anti-bonding electrons.

### Step 1: Analyze O2^2−
1. **Electronic Configuration**:
- For O2^2−, the molecular orbital configuration is:
- \(\sigma_{1s}^2\)
- \(\sigma^*_{1s}^2\)
- \(\sigma_{2s}^2\)
- \(\sigma^*_{2s}^2\)
- \(\sigma_{2p_z}^2\)
- \(\pi_{2p_x}^2\)
- \(\pi_{2p_y}^2\)
- \(\pi^*_{2p_x}^0\)
- \(\pi^*_{2p_y}^0\)

2. **Count Electrons**:
- Bonding electrons (\(n_B\)): 10 (2 from \(\sigma_{1s}\), 2 from \(\sigma_{2s}\), 2 from \(\sigma_{2p_z}\), 2 from \(\pi_{2p_x}\), 2 from \(\pi_{2p_y}\))
- Anti-bonding electrons (\(n_A\)): 4 (2 from \(\sigma^*_{1s}\), 2 from \(\sigma^*_{2s}\))

3. **Calculate Bond Order**:
\[ \text{Bond Order} = \frac{1}{2} (10 - 4) = \frac{1}{2} \times 6 = 3 \]

### Step 2: Analyze H2^−
1. **Electronic Configuration**:
- For H2^−, the molecular orbital configuration is:
- \(\sigma_{1s}^2\)
- \(\sigma^*_{1s}^1\)

2. **Count Electrons**:
- Bonding electrons (\(n_B\)): 2 (from \(\sigma_{1s}\))
- Anti-bonding electrons (\(n_A\)): 1 (from \(\sigma^*_{1s}\))

3. **Calculate Bond Order**:
\[ \text{Bond Order} = \frac{1}{2} (2 - 1) = \frac{1}{2} \times 1 = \frac{1}{2} \]

### Step 3: Analyze F2
1. **Electronic Configuration**:
- For F2, the molecular orbital configuration is:
- \(\sigma_{1s}^2\)
- \(\sigma^*_{1s}^2\)
- \(\sigma_{2s}^2\)
- \(\sigma^*_{2s}^2\)
- \(\sigma_{2p_z}^2\)
- \(\pi_{2p_x}^2\)
- \(\pi_{2p_y}^2\)
- \(\pi^*_{2p_x}^0\)
- \(\pi^*_{2p_y}^0\)

2. **Count Electrons**:
- Bonding electrons (\(n_B\)): 10 (same as O2^2−)
- Anti-bonding electrons (\(n_A\)): 4 (same as O2^2−)

3. **Calculate Bond Order**:
\[ \text{Bond Order} = \frac{1}{2} (10 - 4) = \frac{1}{2} \times 6 = 3 \]

### Conclusion:
The only species with a fractional bond order is **H2^−**, which has a bond order of \(\frac{1}{2}\).