`O_(2)^(2-)` will have .

Options

A. bond order equal to `H_(2)` and diamagnetic
B. bond order equal to `H_(2)` but diamagnetic
C. bond order equal to `N_(2)` and diamagnetic
D. bond order higher than `O_(2)`

Explanation

To determine the bond order of the \( O_2^{2-} \) ion, we will follow these steps:

### Step 1: Write the Electronic Configuration
The electronic configuration of the \( O_2 \) molecule is:
\[ \sigma_{1s}^2 \, \sigma_{1s}^*^2 \, \sigma_{2s}^2 \, \sigma_{2s}^*^2 \, \sigma_{2p_x}^2 \, \pi_{2p_y}^2 \, \pi_{2p_z}^2 \, \pi_{2p_y}^*^0 \, \pi_{2p_z}^*^0 \]
For \( O_2^{2-} \), we add two additional electrons to the \( O_2 \) configuration, filling the antibonding orbitals:
\[ \sigma_{1s}^2 \, \sigma_{1s}^*^2 \, \sigma_{2s}^2 \, \sigma_{2s}^*^2 \, \sigma_{2p_x}^2 \, \pi_{2p_y}^2 \, \pi_{2p_z}^2 \, \pi_{2p_y}^*^2 \, \pi_{2p_z}^*^2 \]

### Step 2: Count Bonding and Antibonding Electrons
- **Bonding Electrons (NB)**:
- \( \sigma_{1s}^2 \) = 2
- \( \sigma_{2s}^2 \) = 2
- \( \sigma_{2p_x}^2 \) = 2
- \( \pi_{2p_y}^2 \) = 2
- \( \pi_{2p_z}^2 \) = 2
- Total = \( 2 + 2 + 2 + 2 + 2 = 10 \)

- **Antibonding Electrons (NA)**:
- \( \sigma_{1s}^*^2 \) = 2
- \( \sigma_{2s}^*^2 \) = 2
- \( \pi_{2p_y}^*^2 \) = 2
- \( \pi_{2p_z}^*^2 \) = 2
- Total = \( 2 + 2 + 2 + 2 = 8 \)

### Step 3: Calculate Bond Order
The bond order is calculated using the formula:
\[ \text{Bond Order} = \frac{1}{2} (NB - NA) \]
Substituting the values:
\[ \text{Bond Order} = \frac{1}{2} (10 - 8) = \frac{1}{2} \times 2 = 1 \]

### Step 4: Determine Magnetic Properties
Since all molecular orbitals are completely filled, \( O_2^{2-} \) is diamagnetic.

### Final Answer
The bond order of \( O_2^{2-} \) is 1, and it is diamagnetic.

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