The molecule which has highest bond order is

Options

A. `N_(2) `
B. `Li _(2)`
C. `H_(2)`
D. `O_(2)`

Answer

To determine which molecule has the highest bond order among N2, Li2, H2, and O2, we can follow these steps:

### Step 1: Understand Bond Order
Bond order is calculated using the formula:
\[ \text{Bond Order} = \frac{1}{2} (n_B - n_A) \]
where \( n_B \) is the number of electrons in bonding molecular orbitals and \( n_A \) is the number of electrons in anti-bonding molecular orbitals.

### Step 2: Determine the Molecular Orbital (MO) Configuration
For molecules with up to 14 electrons, the MO configuration is:
1. \( \sigma 1s^2 \)
2. \( \sigma^* 1s^2 \)
3. \( \sigma 2s^2 \)
4. \( \sigma^* 2s^2 \)
5. \( \pi 2p_x^2 = \pi 2p_y^2 \)
6. \( \sigma 2p_z^2 \)
7. \( \pi^* 2p_x \) and \( \pi^* 2p_y \)
8. \( \sigma^* 2p_z \)

### Step 3: Calculate Bond Order for Each Molecule

#### For Li2:
- Total electrons = 6 (2 from each Li)
- MO configuration: \( \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2 \)
- \( n_B = 4 \) (2 from \( \sigma 1s \) and 2 from \( \sigma 2s \))
- \( n_A = 2 \) (from \( \sigma^* 1s \))
- Bond Order = \( \frac{1}{2} (4 - 2) = 1 \)

#### For H2:
- Total electrons = 2 (1 from each H)
- MO configuration: \( \sigma 1s^2 \)
- \( n_B = 2 \) (from \( \sigma 1s \))
- \( n_A = 0 \) (no anti-bonding electrons)
- Bond Order = \( \frac{1}{2} (2 - 0) = 1 \)

#### For O2:
- Total electrons = 16 (8 from each O)
- MO configuration: \( \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2, \sigma 2p_z^2, \pi^* 2p_x^1, \pi^* 2p_y^1 \)
- \( n_B = 10 \) (2 from \( \sigma 1s \), 2 from \( \sigma 2s \), 4 from \( \pi 2p \), and 2 from \( \sigma 2p \))
- \( n_A = 6 \) (2 from \( \sigma^* 1s \), 2 from \( \sigma^* 2s \), and 2 from \( \pi^* 2p \))
- Bond Order = \( \frac{1}{2} (10 - 6) = 2 \)

#### For N2:
- Total electrons = 14 (7 from each N)
- MO configuration: \( \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2, \sigma 2p_z^2 \)
- \( n_B = 10 \) (2 from \( \sigma 1s \), 2 from \( \sigma 2s \), 4 from \( \pi 2p \), and 2 from \( \sigma 2p \))
- \( n_A = 4 \) (2 from \( \sigma^* 1s \) and 2 from \( \sigma^* 2s \))
- Bond Order = \( \frac{1}{2} (10 - 4) = 3 \)

### Step 4: Compare Bond Orders
- Li2: Bond Order = 1
- H2: Bond Order = 1
- O2: Bond Order = 2
- N2: Bond Order = 3

### Conclusion
The molecule with the highest bond order is **N2** with a bond order of 3.

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