Compare the relative stabilities of `O_(2)^(2-)and N_(2)^(2-)` and comment on their magnetic (paramagnetic) behavior.

Answer

To compare the relative stabilities of \( O_2^{2-} \) and \( N_2^{2-} \) and comment on their magnetic behavior, we will follow these steps:

### Step 1: Determine the number of electrons in each ion.
- **For \( O_2^{2-} \)**: Oxygen has 8 electrons, so \( O_2 \) has \( 8 + 8 = 16 \) electrons. The \( 2- \) charge means it has 2 additional electrons, giving a total of \( 16 + 2 = 18 \) electrons.
- **For \( N_2^{2-} \)**: Nitrogen has 7 electrons, so \( N_2 \) has \( 7 + 7 = 14 \) electrons. The \( 2- \) charge means it has 2 additional electrons, giving a total of \( 14 + 2 = 16 \) electrons.

### Step 2: Write the molecular orbital (MO) configuration for each ion.
- **For \( O_2^{2-} \)**:
- The MO configuration is:
\[ 1s^2 \sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 \pi 2p_y^2 \pi^* 2p_x^2 \pi^* 2p_y^2 \]

- **For \( N_2^{2-} \)**:
- The MO configuration is:
\[ 1s^2 \sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 \pi 2p_y^2 \pi^* 2p_x^1 \pi^* 2p_y^1 \]

### Step 3: Calculate the bond order for each ion.
- **For \( O_2^{2-} \)**:
- Bonding electrons: 10 (from \( \sigma \) and \( \pi \) orbitals)
- Antibonding electrons: 8
- Bond order = \( \frac{(10 - 8)}{2} = 1 \)

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

### Step 4: Determine the magnetic behavior of each ion.
- **For \( O_2^{2-} \)**:
- The last orbitals (the \( \pi^* \) orbitals) are fully filled, meaning all electrons are paired. Therefore, \( O_2^{2-} \) is **diamagnetic**.

- **For \( N_2^{2-} \)**:
- The last orbitals (the \( \pi^* \) orbitals) have unpaired electrons. Therefore, \( N_2^{2-} \) is **paramagnetic**.

### Step 5: Compare the relative stabilities.
- The stability of a molecule is directly proportional to its bond order. Since \( N_2^{2-} \) has a bond order of 2 and \( O_2^{2-} \) has a bond order of 1, we conclude that:
- \( N_2^{2-} \) is more stable than \( O_2^{2-} \).

### Final Conclusion:
- **Stability**: \( N_2^{2-} \) is more stable than \( O_2^{2-} \).
- **Magnetic Behavior**: \( O_2^{2-} \) is diamagnetic, while \( N_2^{2-} \) is paramagnetic.

---