The molecular electronic configuration of `B_2` is

Options

A. `KK(sigma2s)^2(sigma^ast2s)^2(pi2p_x^1=pi2p_y^1)`
B. `KK(sigma2s)^2(sigma^ast2s)^2(pi2p_x)^2`
C. `KK(sigma2s)^2(sigma^ast2s)^2(sigma2p)^2`
D. `KK(sigma2s)^2(sigma^ast2s)^2(sigma2p)^1(pi2p)^1`

Answer

To determine the molecular electronic configuration of the diatomic molecule \( B_2 \) (Boron), we will follow these steps:

### Step 1: Identify the Atomic Number of Boron
Boron has an atomic number of 5. This means that a single boron atom has 5 electrons.

### Step 2: Determine the Total Number of Electrons in \( B_2 \)
Since \( B_2 \) consists of two boron atoms, the total number of electrons in \( B_2 \) is:
\[ 2 \times 5 = 10 \text{ electrons} \]

### Step 3: Write the Molecular Orbital Energy Level Diagram
According to Molecular Orbital Theory, the order of filling the molecular orbitals for diatomic molecules like \( B_2 \) is as follows:
1. \( \sigma_{1s} \)
2. \( \sigma^*_{1s} \)
3. \( \sigma_{2s} \)
4. \( \sigma^*_{2s} \)
5. \( \sigma_{2p_z} \)
6. \( \pi_{2p_x} \) and \( \pi_{2p_y} \) (degenerate orbitals)
7. \( \pi^*_{2p_x} \) and \( \pi^*_{2p_y} \)

### Step 4: Fill the Molecular Orbitals with Electrons
Now we fill the molecular orbitals with the 10 electrons:
- \( \sigma_{1s}^2 \) (2 electrons)
- \( \sigma^*_{1s}^2 \) (2 electrons)
- \( \sigma_{2s}^2 \) (2 electrons)
- \( \sigma^*_{2s}^2 \) (2 electrons)
- \( \sigma_{2p_z}^2 \) (2 electrons)

At this point, we have filled 10 electrons:
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \]

### Step 5: Write the Final Electronic Configuration
The molecular electronic configuration of \( B_2 \) can be summarized as:
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \]

### Step 6: Determine the Bond Order
To find the bond order, we use the formula:
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons}) - (\text{Number of antibonding electrons})}{2} \]
From the configuration:
- Bonding electrons: \( 2 + 2 + 2 = 6 \) (from \( \sigma_{1s}, \sigma_{2s}, \sigma_{2p_z} \))
- Antibonding electrons: \( 2 + 2 = 4 \) (from \( \sigma^*_{1s}, \sigma^*_{2s} \))

Calculating the bond order:
\[ \text{Bond Order} = \frac{6 - 4}{2} = 1 \]

### Conclusion
The molecular electronic configuration of \( B_2 \) is:
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \]
And the bond order is 1.