Among the following molecules/ions `C_(2)^(2-),N_(2)^(2-),O_(2)^(2-),O_(2)`
Which of one diamagnetic and has the shortest bond length?

Options

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

Answer

To solve the problem, we will analyze the given molecules/ions: \( C_2^{2-}, N_2^{2-}, O_2^{2-}, O_2 \) to determine which one is diamagnetic and has the shortest bond length.

### Step 1: Determine the number of electrons in each species
- **For \( C_2^{2-} \)**: Each carbon atom has 6 electrons. For 2 carbon atoms, we have \( 2 \times 6 = 12 \) electrons. Adding 2 more electrons due to the \( 2- \) charge gives us a total of \( 12 + 2 = 14 \) electrons.

- **For \( N_2^{2-} \)**: Each nitrogen atom has 7 electrons. For 2 nitrogen atoms, we have \( 2 \times 7 = 14 \) electrons. Adding 2 more electrons due to the \( 2- \) charge gives us \( 14 + 2 = 16 \) electrons.

- **For \( O_2^{2-} \)**: Each oxygen atom has 8 electrons. For 2 oxygen atoms, we have \( 2 \times 8 = 16 \) electrons. Adding 2 more electrons due to the \( 2- \) charge gives us \( 16 + 2 = 18 \) electrons.

- **For \( O_2 \)**: Each oxygen atom has 8 electrons. For 2 oxygen atoms, we have \( 2 \times 8 = 16 \) electrons.

### Step 2: Write the electronic configurations using Molecular Orbital Theory (MOT)
- **For \( C_2^{2-} \)** (14 electrons):
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \]
(No unpaired electrons, thus diamagnetic)

- **For \( N_2^{2-} \)** (16 electrons):
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi^*_{2p_x}^1 \pi^*_{2p_y}^1 \]
(Unpaired electrons, thus paramagnetic)

- **For \( O_2^{2-} \)** (18 electrons):
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi^*_{2p_x}^2 \pi^*_{2p_y}^2 \]
(No unpaired electrons, thus diamagnetic)

- **For \( O_2 \)** (16 electrons):
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi^*_{2p_x}^1 \pi^*_{2p_y}^1 \]
(Unpaired electrons, thus paramagnetic)

### Step 3: Determine bond order for each species
Bond order can be calculated using the formula:
\[ \text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2} \]

- **For \( C_2^{2-} \)**:
- Bonding electrons = 10
- Antibonding electrons = 4
\[ \text{Bond Order} = \frac{10 - 4}{2} = 3 \]

- **For \( N_2^{2-} \)**:
- Bonding electrons = 10
- Antibonding electrons = 6
\[ \text{Bond Order} = \frac{10 - 6}{2} = 2 \]

- **For \( O_2^{2-} \)**:
- Bonding electrons = 10
- Antibonding electrons = 8
\[ \text{Bond Order} = \frac{10 - 8}{2} = 1 \]

- **For \( O_2 \)**:
- Bonding electrons = 10
- Antibonding electrons = 6
\[ \text{Bond Order} = \frac{10 - 6}{2} = 2 \]

### Step 4: Conclusion
- The only diamagnetic species among the given options is \( C_2^{2-} \) and \( O_2^{2-} \).
- The bond order of \( C_2^{2-} \) is 3, which is the highest, indicating it has the shortest bond length.

Thus, the answer is:
**\( C_2^{2-} \) is diamagnetic and has the shortest bond length.**