With the help of molecular orbital theory predict which of the following species are diamagnetic ?
`H_(2)^(+),O_(2),O_(2)^(2+)`

Answer

To determine which of the given species are diamagnetic using molecular orbital theory, we will analyze the electron configurations of each species: \(H_2^+\), \(O_2\), and \(O_2^{2+}\).

### Step 1: Analyze \(H_2^+\)
1. **Determine the number of electrons**: \(H_2^+\) has 1 electron (since it is a hydrogen molecule with one electron removed).
2. **Write the molecular orbital configuration**: The configuration for \(H_2^+\) is:
\[ \sigma_{1s}^1 \]
3. **Check for unpaired electrons**: There is 1 unpaired electron in \(H_2^+\).
4. **Conclusion**: Since there is an unpaired electron, \(H_2^+\) is **paramagnetic**.

### Step 2: Analyze \(O_2\)
1. **Determine the number of electrons**: \(O_2\) has 16 electrons (8 from each oxygen atom).
2. **Write the molecular orbital configuration**: The configuration for \(O_2\) is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^1 \pi_{2p_y}^1 \]
3. **Check for unpaired electrons**: The last two electrons in the \(\pi^*\) orbitals are unpaired.
4. **Conclusion**: Since there are unpaired electrons, \(O_2\) is **paramagnetic**.

### Step 3: Analyze \(O_2^{2+}\)
1. **Determine the number of electrons**: \(O_2^{2+}\) has 14 electrons (16 from \(O_2\) minus 2).
2. **Write the molecular orbital configuration**: The configuration for \(O_2^{2+}\) is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \]
3. **Check for unpaired electrons**: All electrons are paired in this configuration.
4. **Conclusion**: Since there are no unpaired electrons, \(O_2^{2+}\) is **diamagnetic**.

### Final Answer
- **Diamagnetic Species**: \(O_2^{2+}\)
- **Paramagnetic Species**: \(H_2^+\) and \(O_2\)