Which one of the following is not correct with respect to bond length of the species?

Options

A. `C_2 gt C_2^-`
B. `B_2^+ gtB_2`
C. `Li_2^+ gt Li`
D. `O_2 gt O^-`

Explanation

To determine which statement regarding bond lengths is incorrect, we need to analyze the bond orders of the given species and their corresponding bond lengths. The bond order is inversely related to bond length; a higher bond order corresponds to a shorter bond length, while a lower bond order corresponds to a longer bond length.

### Step-by-Step Solution:

1. **Understanding Bond Order and Bond Length**:
- Bond order is calculated using the formula:
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of anti-bonding electrons})}{2} \]
- Higher bond order means stronger bonds and shorter bond lengths.

2. **Analyzing C2 and C2⁻**:
- **C2**:
- Electronic configuration: \(1s^2 2s^2 2p^2\)
- Bonding electrons: 4 (2 in \(\sigma\) and 2 in \(\pi\))
- Anti-bonding electrons: 0
- Bond order: \(\frac{4 - 0}{2} = 2\)
- **C2⁻**:
- One additional electron goes into the \(\sigma^*\) orbital.
- Bonding electrons: 5
- Anti-bonding electrons: 1
- Bond order: \(\frac{5 - 1}{2} = 2\)
- **Conclusion**: C2 has a higher bond order than C2⁻, so C2 has a shorter bond length.

3. **Analyzing B2 and B2⁺**:
- **B2**:
- Electronic configuration: \(1s^2 2s^2 2p^1\)
- Bonding electrons: 2 (both in \(\pi\))
- Anti-bonding electrons: 0
- Bond order: \(\frac{2 - 0}{2} = 1\)
- **B2⁺**:
- One electron is removed from the \(\pi\) orbital.
- Bonding electrons: 1
- Anti-bonding electrons: 0
- Bond order: \(\frac{1 - 0}{2} = 0.5\)
- **Conclusion**: B2 has a higher bond order than B2⁺, so B2 has a shorter bond length.

4. **Analyzing Li2 and Li2⁺**:
- **Li2**:
- Electronic configuration: \(1s^2 2s^1\)
- Bonding electrons: 2
- Anti-bonding electrons: 0
- Bond order: \(\frac{2 - 0}{2} = 1\)
- **Li2⁺**:
- One electron is removed from the \(\sigma\) orbital.
- Bonding electrons: 1
- Anti-bonding electrons: 0
- Bond order: \(\frac{1 - 0}{2} = 0.5\)
- **Conclusion**: Li2 has a higher bond order than Li2⁺, so Li2 has a shorter bond length.

5. **Analyzing O2 and O2⁻**:
- **O2**:
- Electronic configuration: \(1s^2 2s^2 2p^4\)
- Bonding electrons: 6
- Anti-bonding electrons: 2
- Bond order: \(\frac{6 - 2}{2} = 2\)
- **O2⁻**:
- One additional electron goes into the \(\pi^*\) orbital.
- Bonding electrons: 6
- Anti-bonding electrons: 3
- Bond order: \(\frac{6 - 3}{2} = 1.5\)
- **Conclusion**: O2 has a higher bond order than O2⁻, so O2 has a shorter bond length.

6. **Final Comparison**:
- The bond lengths should be compared as follows:
- C2 < C2⁻
- B2 < B2⁺
- Li2 < Li2⁺
- O2 < O2⁻
- The incorrect statement is that O2 has a greater bond length than O2⁻, which contradicts our findings.

### Answer:
The incorrect statement is that O2 has a greater bond length than O2⁻. The correct option is **option number 4**.