Bond order of 2.5 is shown by

Options

A. `O_(2)^(2-)`
B. `O_(2)`
C. `O_(2)^(+)`
D. `O_(2)^(-)`

Answer

To find the species with a bond order of 2.5, we will analyze the molecular orbital (MO) configurations of different oxygen species. The bond order can be calculated using the formula:

\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} \]

### Step 1: Determine the number of electrons for the species
For \( O_2^+ \) (Oxygen molecule with a positive charge):
- Oxygen has an atomic number of 8, so \( O_2 \) has \( 8 + 8 = 16 \) electrons.
- Since it is \( O_2^+ \), it loses one electron, resulting in \( 15 \) electrons.

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

### Step 3: Count the bonding and antibonding electrons
- **Bonding Electrons**:
- From \( \sigma_{1s}^2 \): 2
- From \( \sigma_{2s}^2 \): 2
- From \( \sigma_{2p_z}^2 \): 2
- From \( \pi_{2p_x}^2 \): 2
- From \( \pi_{2p_y}^2 \): 2
- Total bonding electrons = \( 2 + 2 + 2 + 2 + 2 = 10 \)

- **Antibonding Electrons**:
- From \( \sigma_{1s}^*^2 \): 2
- From \( \sigma_{2s}^*^2 \): 2
- From \( \pi_{2p_x}^*^1 \): 1
- Total antibonding electrons = \( 2 + 2 + 1 = 5 \)

### Step 4: Calculate the bond order
Using the bond order formula:
\[ \text{Bond Order} = \frac{(10 - 5)}{2} = \frac{5}{2} = 2.5 \]

### Conclusion
Thus, the species that shows a bond order of 2.5 is \( O_2^+ \).

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