The species having diamagnetic nature and bond order 1.0 is

Options

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

Answer

To solve the question of identifying the species that has a diamagnetic nature and a bond order of 1.0, we will follow these steps:

### Step 1: Understand Diamagnetic and Bond Order
- **Diamagnetic species** are those that have all their electrons paired. This means that there are no unpaired electrons present in the species.
- **Bond order** is calculated using the formula:
\[ \text{Bond Order} = \frac{(\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons})}{2} \]

### Step 2: Analyze the Options
We will analyze the provided options to find out which one meets the criteria of being diamagnetic and having a bond order of 1.0.

#### Option A: \(O_2^{2-}\)
1. **Electronic Configuration**: For \(O_2\), the electronic configuration is:
- \(O: 1s^2 2s^2 2p^4\) (for each oxygen)
- For \(O_2^{2-}\), we add 2 electrons: \(1s^2 2s^2 2p^6\)
2. **Filling the Molecular Orbitals**:
- The molecular orbital filling will be: \(\sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2\)
3. **Bonding and Antibonding Electrons**:
- Bonding electrons = 6 (from \(\sigma_{1s}, \sigma_{2s}, \sigma_{2p_z}, \pi_{2p_x}, \pi_{2p_y}\))
- Antibonding electrons = 0
4. **Calculate Bond Order**:
\[ \text{Bond Order} = \frac{(6 - 0)}{2} = 3 \]
5. **Check for Diamagnetism**: All electrons are paired, so it is diamagnetic.

#### Option B: \(O_2^+\)
1. **Electronic Configuration**: For \(O_2^+\), we remove one electron:
- Configuration: \(1s^2 2s^2 2p^3\)
2. **Filling the Molecular Orbitals**:
- The filling will be: \(\sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^1 \pi_{2p_y}^1\)
3. **Bonding and Antibonding Electrons**:
- Bonding electrons = 5
- Antibonding electrons = 1
4. **Calculate Bond Order**:
\[ \text{Bond Order} = \frac{(5 - 1)}{2} = 2 \]
5. **Check for Diamagnetism**: There are unpaired electrons, so it is paramagnetic.

#### Option C: \(O_2^{2+}\)
1. **Electronic Configuration**: For \(O_2^{2+}\), we remove two electrons:
- Configuration: \(1s^2 2s^2 2p^2\)
2. **Filling the Molecular Orbitals**:
- The filling will be: \(\sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2\)
3. **Bonding and Antibonding Electrons**:
- Bonding electrons = 6
- Antibonding electrons = 0
4. **Calculate Bond Order**:
\[ \text{Bond Order} = \frac{(6 - 0)}{2} = 3 \]
5. **Check for Diamagnetism**: All electrons are paired, so it is diamagnetic.

#### Option D: \(O_2\)
1. **Electronic Configuration**: For \(O_2\):
- Configuration: \(1s^2 2s^2 2p^4\)
2. **Filling the Molecular Orbitals**:
- The filling will be: \(\sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^1 \pi_{2p_y}^1\)
3. **Bonding and Antibonding Electrons**:
- Bonding electrons = 6
- Antibonding electrons = 2
4. **Calculate Bond Order**:
\[ \text{Bond Order} = \frac{(6 - 2)}{2} = 2 \]
5. **Check for Diamagnetism**: There are unpaired electrons, so it is paramagnetic.

### Conclusion
From the analysis:
- The only species that is diamagnetic and has a bond order of 1.0 is **Option A: \(O_2^{2-}\)**.