Which out of two `O_(2)^(+)and O_(2)^(2+)` has higher bond order and why ?

Answer

To determine which of the two species, \( O_2^+ \) or \( O_2^{2+} \), has a higher bond order, we need to follow these steps:

### Step 1: Determine the number of electrons in each species.
- For \( O_2 \), the total number of electrons is 16 (8 from each oxygen atom).
- For \( O_2^+ \), one electron is removed, so it has 15 electrons.
- For \( O_2^{2+} \), two electrons are removed, so it has 14 electrons.

### Step 2: Write the molecular orbital (MO) configuration for each species.
- **For \( O_2^+ \)** (15 electrons):
- The MO configuration is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^1 \]
- **For \( O_2^{2+} \)** (14 electrons):
- The MO configuration is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \]

### Step 3: Count the number of bonding and antibonding electrons.
- **For \( O_2^+ \)**:
- Bonding electrons: \( 10 \) (from \( \sigma_{1s}^2, \sigma_{2s}^2, \sigma_{2p_z}^2, \pi_{2p_x}^2, \pi_{2p_y}^2 \))
- Antibonding electrons: \( 5 \) (from \( \sigma_{1s}^*, \sigma_{2s}^*, \pi_{2p_x}^1 \))

- **For \( O_2^{2+} \)**:
- Bonding electrons: \( 10 \) (same as above)
- Antibonding electrons: \( 4 \) (from \( \sigma_{1s}^*, \sigma_{2s}^* \))

### Step 4: Calculate the bond order for each species using the formula:
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} \]

- **For \( O_2^+ \)**:
\[ \text{Bond Order} = \frac{(10 - 5)}{2} = \frac{5}{2} = 2.5 \]

- **For \( O_2^{2+} \)**:
\[ \text{Bond Order} = \frac{(10 - 4)}{2} = \frac{6}{2} = 3 \]

### Step 5: Compare the bond orders.
- \( O_2^+ \) has a bond order of \( 2.5 \).
- \( O_2^{2+} \) has a bond order of \( 3 \).

### Conclusion:
- \( O_2^{2+} \) has a higher bond order than \( O_2^+ \) because it has fewer antibonding electrons (4 compared to 5 in \( O_2^+ \)). This results in a stronger bond.