To determine which of the two species, \( O_2^{-} \) (superoxide ion) and \( O_2^{2-} \) (peroxide ion), has a higher bond order, we need to analyze their molecular orbital configurations and calculate their bond orders.
### Step-by-Step Solution:
1. **Identify the Total Number of Electrons**:
- For \( O_2^{-} \): Oxygen has 8 electrons, and since there are 2 oxygen atoms, the total is \( 8 + 8 = 16 \) electrons. However, since it has a -1 charge, we add one more electron, giving us a total of **17 electrons**.
- For \( O_2^{2-} \): Again, we start with 16 electrons from the two oxygen atoms. With a -2 charge, we add two more electrons, resulting in a total of **18 electrons**.
2. **Write the Molecular Orbital Configuration**:
- For \( O_2^{-} \):
- The molecular orbital configuration is:
\[
\sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \pi_{2p_y}^*^1
\]
- For \( O_2^{2-} \):
- The molecular orbital configuration is:
\[
\sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^2 \pi_{2p_y}^*^2
\]
3. **Count Bonding and Antibonding Electrons**:
- For \( O_2^{-} \):
- Bonding electrons: \( 10 \) (from \( \sigma_{1s}, \sigma_{2s}, \sigma_{2p_z}, \pi_{2p_x}, \pi_{2p_y} \))
- Antibonding electrons: \( 7 \) (from \( \sigma_{1s}^*, \sigma_{2s}^*, \pi_{2p_x}^*, \pi_{2p_y}^* \))
- For \( O_2^{2-} \):
- Bonding electrons: \( 10 \) (same as above)
- Antibonding electrons: \( 8 \) (one additional electron in the antibonding orbitals)
4. **Calculate the Bond Order**:
- The bond order formula is given by:
\[
\text{Bond Order} = \frac{(\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons})}{2}
\]
- For \( O_2^{-} \):
\[
\text{Bond Order} = \frac{(10 - 7)}{2} = \frac{3}{2} = 1.5
\]
- For \( O_2^{2-} \):
\[
\text{Bond Order} = \frac{(10 - 8)}{2} = \frac{2}{2} = 1
\]
5. **Conclusion**:
- Comparing the bond orders, \( O_2^{-} \) has a bond order of \( 1.5 \) while \( O_2^{2-} \) has a bond order of \( 1 \). Therefore, \( O_2^{-} \) has a higher bond order than \( O_2^{2-} \).
### Final Answer:
**\( O_2^{-} \) has a higher bond order than \( O_2^{2-} \) because it has fewer antibonding electrons, resulting in a bond order of 1.5 compared to 1 for \( O_2^{2-} \).**
