The bond order is maximum in

Options

A. `O_(2)`
B. `O_(2) ^(-1)`
C. `O_(2) ^(+1)`
D. `O_(2) ^(-2)`

Answer

To determine which molecule has the maximum bond order among the given options, we will calculate the bond order for each molecule using the molecular electronic configuration and the bond order formula.

### Step-by-Step Solution:

1. **Identify the Molecules**: The options given are O2, O2−, O2+, and O2²−.

2. **Calculate the Total Electrons**:
- For O2: Each oxygen atom has 8 electrons, so O2 has 16 electrons (8 + 8).
- For O2−: O2 has 16 electrons, and the extra electron gives it 17 electrons.
- For O2+: O2 has 16 electrons, and losing one electron gives it 15 electrons.
- For O2²−: O2 has 16 electrons, and gaining two electrons gives it 18 electrons.

3. **Write the Molecular Electronic Configuration**:
- **O2**: The configuration is σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y² π*2p_x¹ π*2p_y¹ (Total: 16 electrons).
- **O2−**: The configuration is the same as O2, but with one additional electron in the π* orbital: σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y² π*2p_x² π*2p_y¹ (Total: 17 electrons).
- **O2+**: The configuration is the same as O2, but with one less electron in the π* orbital: σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y² π*2p_x¹ π*2p_y⁰ (Total: 15 electrons).
- **O2²−**: The configuration is the same as O2, but with two additional electrons in the π* orbitals: σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y² π*2p_x² π*2p_y² (Total: 18 electrons).

4. **Calculate the Bond Order**:
- The bond order formula is given by:
\[ \text{Bond Order} = \frac{1}{2} \left( N_b - N_a \right) \]
where \( N_b \) is the number of electrons in bonding molecular orbitals and \( N_a \) is the number of electrons in anti-bonding molecular orbitals.

- **For O2**:
- Bonding electrons (N_b) = 10 (σ1s² + σ2s² + σ2p_z² + π2p_x² + π2p_y²)
- Anti-bonding electrons (N_a) = 6 (σ*1s² + σ*2s² + π*2p_x¹ + π*2p_y¹)
- Bond Order = \( \frac{1}{2} (10 - 6) = 2 \)

- **For O2−**:
- Bonding electrons (N_b) = 10
- Anti-bonding electrons (N_a) = 5 (σ*1s² + σ*2s² + π*2p_x² + π*2p_y¹)
- Bond Order = \( \frac{1}{2} (10 - 5) = 2.5 \)

- **For O2+**:
- Bonding electrons (N_b) = 10
- Anti-bonding electrons (N_a) = 5 (σ*1s² + σ*2s² + π*2p_x¹ + π*2p_y⁰)
- Bond Order = \( \frac{1}{2} (10 - 5) = 2.5 \)

- **For O2²−**:
- Bonding electrons (N_b) = 10
- Anti-bonding electrons (N_a) = 8 (σ*1s² + σ*2s² + π*2p_x² + π*2p_y²)
- Bond Order = \( \frac{1}{2} (10 - 8) = 1 \)

5. **Compare the Bond Orders**:
- O2: Bond Order = 2
- O2−: Bond Order = 2.5
- O2+: Bond Order = 2.5
- O2²−: Bond Order = 1

### Conclusion:
The maximum bond order is found in O2− and O2+, both having a bond order of 2.5. However, since the question asks for the maximum bond order, we can conclude that the answer is:

**Answer: O2+ (Option C)**