According to molecular orbital theory, which of the following is true with respect to `C_(2)^(2+)` and `C_(2)^(2-)` ?

Options

A. Both have same number of `sigma` bonds
B. `C_(2)^(2+)` has more number of `pi` bonds and `C_(2)^(2-)` has less number of `pi` bonds
C. `C_(2)^(2+)` is paramagnetic and `C_(2)^(2-)`is diamagnetic
D. Both have same number of `pi`bonds

Explanation

To solve the question regarding the molecular orbital theory of \( C_2^{2+} \) and \( C_2^{2-} \), we will follow these steps:

### Step 1: Determine the total number of electrons in each species.
- For \( C_2^{2+} \):
- Each carbon atom has 6 electrons, so \( C_2 \) has \( 2 \times 6 = 12 \) electrons.
- The \( 2+ \) charge means we remove 2 electrons, giving us \( 12 - 2 = 10 \) electrons.

- For \( C_2^{2-} \):
- Again, \( C_2 \) has 12 electrons.
- The \( 2- \) charge means we add 2 electrons, giving us \( 12 + 2 = 14 \) electrons.

### Step 2: Fill the molecular orbitals for \( C_2^{2+} \) and \( C_2^{2-} \).
- For \( C_2^{2+} \) (10 electrons):
- The molecular orbital filling order is:
- \( \sigma_{1s}^2 \)
- \( \sigma_{1s}^* \) (antibonding) \( 0 \)
- \( \sigma_{2s}^2 \)
- \( \sigma_{2s}^* \) (antibonding) \( 0 \)
- \( \pi_{2px}^2 \)
- \( \pi_{2py}^2 \)
- \( \sigma_{2pz}^0 \)
- The configuration is: \( \sigma_{1s}^2 \sigma_{1s}^* 0 \sigma_{2s}^2 \sigma_{2s}^* 0 \pi_{2px}^2 \pi_{2py}^2 \sigma_{2pz}^0 \)

- For \( C_2^{2-} \) (14 electrons):
- The molecular orbital filling order is:
- \( \sigma_{1s}^2 \)
- \( \sigma_{1s}^* \) \( 0 \)
- \( \sigma_{2s}^2 \)
- \( \sigma_{2s}^* \) \( 0 \)
- \( \pi_{2px}^2 \)
- \( \pi_{2py}^2 \)
- \( \sigma_{2pz}^2 \)
- The configuration is: \( \sigma_{1s}^2 \sigma_{1s}^* 0 \sigma_{2s}^2 \sigma_{2s}^* 0 \pi_{2px}^2 \pi_{2py}^2 \sigma_{2pz}^2 \)

### Step 3: Determine the magnetic properties.
- For \( C_2^{2+} \):
- The filling shows that there are unpaired electrons in the \( \pi_{2px} \) and \( \pi_{2py} \) orbitals.
- Therefore, \( C_2^{2+} \) is **paramagnetic**.

- For \( C_2^{2-} \):
- The filling shows that all electrons are paired.
- Therefore, \( C_2^{2-} \) is **diamagnetic**.

### Step 4: Conclusion
- The correct statement regarding \( C_2^{2+} \) and \( C_2^{2-} \) is that \( C_2^{2+} \) is paramagnetic and \( C_2^{2-} \) is diamagnetic.

### Final Answer
- **\( C_2^{2+} \) is paramagnetic and \( C_2^{2-} \) is diamagnetic.**

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