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 beryllium molecule \( \text{Be}_2 \), we will follow these steps:

### Step 1: Understand the Molecular Orbital Configuration
Beryllium (Be) has an atomic number of 4, which means each beryllium atom has 4 electrons. In \( \text{Be}_2 \), there are a total of 8 electrons (4 from each Be atom).

### Step 2: Fill the Molecular Orbitals
The molecular orbital (MO) configuration for \( \text{Be}_2 \) is as follows:
- The order of filling the molecular orbitals for diatomic molecules like \( \text{Be}_2 \) is:
- \( \sigma_{1s} \)
- \( \sigma_{1s}^* \) (anti-bonding)
- \( \sigma_{2s} \)
- \( \sigma_{2s}^* \) (anti-bonding)
- \( \sigma_{2p_z} \)
- \( \pi_{2p_x} = \pi_{2p_y} \)
- \( \pi_{2p_x}^* = \pi_{2p_y}^* \) (anti-bonding)
- \( \sigma_{2p_z}^* \) (anti-bonding)

For \( \text{Be}_2 \), the 8 electrons will fill the molecular orbitals as follows:
- \( \sigma_{1s}^2 \) (2 electrons)
- \( \sigma_{1s}^* \) (0 electrons)
- \( \sigma_{2s}^2 \) (2 electrons)
- \( \sigma_{2s}^* \) (0 electrons)
- \( \sigma_{2p_z}^0 \) (0 electrons)
- \( \pi_{2p_x}^0 \) (0 electrons)
- \( \pi_{2p_y}^0 \) (0 electrons)

### Step 3: Count the Electrons in Bonding and Anti-bonding Orbitals
- **Bonding Molecular Orbitals (BMOs)**:
- \( \sigma_{1s} \): 2 electrons
- \( \sigma_{2s} \): 2 electrons
- Total bonding electrons = 2 + 2 = 4

- **Anti-bonding Molecular Orbitals (ABMOs)**:
- \( \sigma_{1s}^* \): 0 electrons
- \( \sigma_{2s}^* \): 0 electrons
- Total anti-bonding electrons = 0 + 0 = 0

### Step 4: Calculate the Bond Order
The bond order is calculated using the formula:
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of anti-bonding electrons})}{2} \]
Substituting the values we found:
\[ \text{Bond Order} = \frac{(4 - 0)}{2} = \frac{4}{2} = 2 \]

### Conclusion
The bond order of \( \text{Be}_2 \) is **0**.