Out of `O_(2)` and `O_(2)^(-)` ion which has smaller bond order ?

Answer

To determine which species has a smaller bond order between \( O_2 \) and \( O_2^- \), we will follow these steps:

### Step 1: Determine the electronic configuration of \( O_2 \)
Oxygen has 8 electrons, so \( O_2 \) has a total of \( 16 \) electrons. The electronic configuration for \( O_2 \) can be written as:
- \( \sigma 1s^2 \)
- \( \sigma^* 1s^2 \)
- \( \sigma 2s^2 \)
- \( \sigma^* 2s^2 \)
- \( \sigma 2p_z^2 \)
- \( \pi 2p_x^2 = \pi 2p_y^2 \)
- \( \pi^* 2p_x^0 = \pi^* 2p_y^0 \)

This gives us the complete configuration:
\[ \sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 \pi 2p_y^2 \]

### Step 2: Count the bonding and anti-bonding electrons for \( O_2 \)
- **Bonding Electrons**:
- \( 2 \) (from \( \sigma 1s \))
- \( 2 \) (from \( \sigma 2s \))
- \( 2 \) (from \( \sigma 2p_z \))
- \( 4 \) (from \( \pi 2p_x \) and \( \pi 2p_y \))
- Total = \( 10 \) bonding electrons.

- **Anti-bonding Electrons**:
- \( 2 \) (from \( \sigma^* 1s \))
- \( 2 \) (from \( \sigma^* 2s \))
- Total = \( 4 \) anti-bonding electrons.

### Step 3: Calculate the bond order for \( O_2 \)
The bond order is calculated using the formula:
\[ \text{Bond Order} = \frac{1}{2} \times (\text{Bonding Electrons} - \text{Anti-bonding Electrons}) \]
Substituting the values:
\[ \text{Bond Order} = \frac{1}{2} \times (10 - 4) = \frac{1}{2} \times 6 = 3 \]

### Step 4: Determine the electronic configuration of \( O_2^- \)
The \( O_2^- \) ion has one additional electron, giving it a total of \( 17 \) electrons. The electronic configuration for \( O_2^- \) can be written as:
- \( \sigma 1s^2 \)
- \( \sigma^* 1s^2 \)
- \( \sigma 2s^2 \)
- \( \sigma^* 2s^2 \)
- \( \sigma 2p_z^2 \)
- \( \pi 2p_x^2 = \pi 2p_y^2 \)
- \( \pi^* 2p_x^1 = \pi^* 2p_y^0 \)

### Step 5: Count the bonding and anti-bonding electrons for \( O_2^- \)
- **Bonding Electrons**:
- Same as \( O_2 \) = \( 10 \) bonding electrons.

- **Anti-bonding Electrons**:
- \( 2 \) (from \( \sigma^* 1s \))
- \( 2 \) (from \( \sigma^* 2s \))
- \( 1 \) (from \( \pi^* 2p_x \))
- Total = \( 5 \) anti-bonding electrons.

### Step 6: Calculate the bond order for \( O_2^- \)
Using the bond order formula again:
\[ \text{Bond Order} = \frac{1}{2} \times (10 - 5) = \frac{1}{2} \times 5 = 2.5 \]

### Conclusion
Now we can compare the bond orders:
- \( O_2 \): Bond Order = \( 2 \)
- \( O_2^- \): Bond Order = \( 1.5 \)

Thus, the species with the smaller bond order is \( O_2^- \).

### Final Answer
The \( O_2^- \) ion has a smaller bond order than \( O_2 \).

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