According to Molecular Orbital Theory,

Options

A. `C_(2)^(2-)` is expected to be diamagnetic
B. `O_(2)^(2+)` is expected to have a longer bond length than `O_(2)`
C. `N_(2)^(+) and N_(2)^(-)` have the same bond order
D. `He_(2)^(+)` has the same energy as two isolated He atoms

Explanation

To solve the question regarding the molecular orbital theory and the properties of the species mentioned, we will follow a systematic approach.

### Step-by-Step Solution:

1. **Identify the Species and Their Electrons**:
- For \( C_2^{2-} \): It has 14 electrons.
- For \( O_2^{2+} \): It has 14 electrons.
- For \( N_2^{-} \): It has 15 electrons.
- For \( N_2^{+} \): It has 13 electrons.
- For \( He_2^{+} \): It has 3 electrons.

2. **Molecular Orbital Filling**:
- **For \( C_2^{2-} \)**:
- Fill the molecular orbitals in the order:
- \( \sigma 1s^2 \)
- \( \sigma 1s^* \)
- \( \sigma 2s^2 \)
- \( \sigma 2s^* \)
- \( \sigma 2p_z^2 \)
- \( \pi 2p_x^2 = \pi 2p_y^2 \)
- \( \pi 2p_x^* = \pi 2p_y^* \)
- \( \sigma 2p_z^* \)
- Total filling: \( 2 + 2 + 2 + 2 + 2 + 2 = 14 \) electrons.
- All electrons are paired, hence \( C_2^{2-} \) is **diamagnetic**.

3. **Bond Order Calculation for \( C_2^{2-} \)**:
- Bond order = \( \frac{(Number \, of \, bonding \, electrons - Number \, of \, anti-bonding \, electrons)}{2} \)
- Bonding electrons = 10, Anti-bonding electrons = 4.
- Bond order = \( \frac{10 - 4}{2} = 3 \).

4. **For \( O_2^{2+} \)**:
- Fill the molecular orbitals similarly:
- Total electrons = 14.
- Bonding electrons = 10, Anti-bonding electrons = 4.
- Bond order = \( \frac{10 - 4}{2} = 3 \).

5. **Bond Length Comparison**:
- Bond length is inversely proportional to bond order. Since \( O_2^{2+} \) has a higher bond order than \( O_2 \), it has a shorter bond length, making the statement "O2 2+ is expected to have a longer bond length than O2" **incorrect**.

6. **For \( N_2^{-} \)** and \( N_2^{+} \)**:
- \( N_2^{-} \): 15 electrons.
- Bonding electrons = 10, Anti-bonding electrons = 5.
- Bond order = \( \frac{10 - 5}{2} = 2.5 \).
- \( N_2^{+} \): 13 electrons.
- Bonding electrons = 9, Anti-bonding electrons = 4.
- Bond order = \( \frac{9 - 4}{2} = 2.5 \).
- Both have the same bond order, hence the statement is **correct**.

7. **For \( He_2^{+} \)**:
- Total electrons = 3.
- Bonding electrons = 2, Anti-bonding electrons = 1.
- Bond order = \( \frac{2 - 1}{2} = 0.5 \).
- Isolated He atoms have a bond order of 0, thus \( He_2^{+} \) is more stable than isolated He atoms, making the statement **incorrect**.

### Conclusion:
- The correct statements based on the analysis are:
- \( C_2^{2-} \) is diamagnetic.
- \( N_2^{-} \) and \( N_2^{+} \) have the same bond order.