In `Be_(2)` the bond order is

Options

A. one
B. zero
C. two
D. one-half

Explanation

To determine the bond order of the molecule \( \text{Be}_2 \), we can follow these steps:

### Step 1: Determine the total number of electrons
Beryllium (Be) has an atomic number of 4, which means each beryllium atom has 4 electrons. Since there are 2 beryllium atoms in \( \text{Be}_2 \), the total number of electrons is:
\[ \text{Total electrons} = 2 \times 4 = 8 \]

### Step 2: Write the molecular orbital configuration
The molecular orbital (MO) configuration for \( \text{Be}_2 \) can be filled based on the energy levels of the molecular orbitals. The order of filling for the first 8 electrons is as follows:
- \( \sigma_{1s}^2 \)
- \( \sigma^*_{1s}^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma^*_{2s}^2 \)

Thus, the complete molecular orbital configuration for \( \text{Be}_2 \) is:
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \]

### Step 3: Count the number of electrons in bonding and anti-bonding orbitals
- **Bonding molecular orbitals (BMO)**:
- \( \sigma_{1s} \): 2 electrons
- \( \sigma_{2s} \): 2 electrons
- Total in bonding MOs, \( n_B = 2 + 2 = 4 \)

- **Anti-bonding molecular orbitals (ABMO)**:
- \( \sigma^*_{1s} \): 2 electrons
- \( \sigma^*_{2s} \): 2 electrons
- Total in anti-bonding MOs, \( n_A = 2 + 2 = 4 \)

### Step 4: Calculate the bond order
The bond order can be calculated using the formula:
\[ \text{Bond Order} = \frac{1}{2} (n_B - n_A) \]
Substituting the values we found:
\[ \text{Bond Order} = \frac{1}{2} (4 - 4) = \frac{1}{2} \times 0 = 0 \]

### Conclusion
The bond order of \( \text{Be}_2 \) is 0, indicating that there is no net bond between the two beryllium atoms.