The internuclear distances in 0-0 bonds for `O_(2)^(+),O_(2),O_(2)^(-) and O_(2)^(2-)` respectively are :

Options

A. `1.30Å,1.49Å,1.12Å,1.21Å`
B. `1.49Å,1.21Å,1.12Å,1.30Å`
C. `1.21Å,1.12Å,1.49Å,1.30Å`
D. `1.12Å,1.21Å,1.30Å,1.49Å`

Explanation

To determine the internuclear distances in O-O bonds for \( O_2^+, O_2, O_2^-, \) and \( O_2^{2-} \), we need to calculate the bond order for each species. The bond order is related to the bond length, where a higher bond order corresponds to a shorter bond length.

### Step-by-Step Solution:

1. **Determine the number of electrons for each species:**
- \( O_2 \): 16 electrons (8 from each O atom)
- \( O_2^+ \): 15 electrons (1 electron removed)
- \( O_2^- \): 17 electrons (1 electron added)
- \( O_2^{2-} \): 18 electrons (2 electrons added)

2. **Write the electron configurations for each species:**
- For \( O_2 \) (16 electrons):
- Configuration: \( \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \)
- For \( O_2^+ \) (15 electrons):
- Configuration: \( \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^1 \)
- For \( O_2^- \) (17 electrons):
- Configuration: \( \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi^*_{2p_x}^1 \)
- For \( O_2^{2-} \) (18 electrons):
- Configuration: \( \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi^*_{2p_x}^2 \)

3. **Calculate the bond order for each species:**
- **Bond Order Formula**:
\[ \text{Bond Order} = \frac{1}{2} (\text{Number of bonding electrons} - \text{Number of antibonding electrons}) \]
- For \( O_2 \):
- Bonding electrons = 10, Antibonding electrons = 6
- Bond Order = \( \frac{1}{2} (10 - 6) = 2 \)
- For \( O_2^+ \):
- Bonding electrons = 10, Antibonding electrons = 5
- Bond Order = \( \frac{1}{2} (10 - 5) = 2.5 \)
- For \( O_2^- \):
- Bonding electrons = 10, Antibonding electrons = 7
- Bond Order = \( \frac{1}{2} (10 - 7) = 1.5 \)
- For \( O_2^{2-} \):
- Bonding electrons = 10, Antibonding electrons = 8
- Bond Order = \( \frac{1}{2} (10 - 8) = 1 \)

4. **Determine the relationship between bond order and bond length:**
- Higher bond order means shorter bond length.
- Order of bond lengths (from shortest to longest):
- \( O_2^+ \) (bond order 2.5) < \( O_2 \) (bond order 2) < \( O_2^- \) (bond order 1.5) < \( O_2^{2-} \) (bond order 1)

### Final Answer:
- Internuclear distances in O-O bonds:
- \( O_2^+ \): shortest
- \( O_2 \): next
- \( O_2^- \): longer
- \( O_2^{2-} \): longest