On the basic of the bond order, predict which of the following species is the most stable.
`(i)O_(2)^(-)(ii)O_(2)(iii)Ne_(2)^(+)`

Answer

To determine which of the species \( O_2^-, O_2, \) and \( Ne_2^+ \) is the most stable based on bond order, we will follow these steps:

### Step 1: Determine the number of electrons in each species
- **For \( O_2^- \)**: Oxygen has 8 electrons, so \( O_2 \) has \( 8 + 8 = 16 \) electrons. The extra negative charge adds 1 more electron, giving us a total of 17 electrons.

- **For \( O_2 \)**: As calculated, \( O_2 \) has 16 electrons.

- **For \( Ne_2^+ \)**: Neon has 10 electrons, so \( Ne_2 \) has \( 10 + 10 = 20 \) electrons. The positive charge removes 1 electron, giving us a total of 19 electrons.

### Step 2: Write the electronic configurations
- **For \( O_2^- \)** (17 electrons):
\[ 1s^2 \sigma^* 1s^2 \sigma^2 2s^2 \sigma^* 2s^2 \sigma^2 2p_z^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi^*_{2p_x}^1 \]

- **For \( O_2 \)** (16 electrons):
\[ 1s^2 \sigma^* 1s^2 \sigma^2 2s^2 \sigma^* 2s^2 \sigma^2 2p_z^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi^*_{2p_x}^1 \pi^*_{2p_y}^0 \]

- **For \( Ne_2^+ \)** (19 electrons):
\[ 1s^2 \sigma^* 1s^2 \sigma^2 2s^2 \sigma^* 2s^2 \sigma^2 2p_z^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi^*_{2p_x}^2 \pi^*_{2p_y}^2 \sigma^*_{2p_z}^1 \]

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

- **For \( O_2^- \)**:
- Bonding electrons = 10 (from \( \sigma \) and \( \pi \) orbitals)
- Antibonding electrons = 7 (from \( \sigma^* \) and \( \pi^* \) orbitals)
- Bond Order = \( \frac{10 - 7}{2} = 1.5 \)

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

- **For \( Ne_2^+ \)**:
- Bonding electrons = 10
- Antibonding electrons = 9
- Bond Order = \( \frac{10 - 9}{2} = 0.5 \)

### Step 4: Compare the bond orders
- \( O_2^- \): Bond Order = 1.5
- \( O_2 \): Bond Order = 2
- \( Ne_2^+ \): Bond Order = 0.5

### Step 5: Determine stability based on bond order
The stability of a molecule is directly related to its bond order; higher bond order indicates greater stability.

Thus, the most stable species among the given options is:
\[ \text{The most stable species is } O_2 \text{ because it has the highest bond order of } 2. \]