Question

The B-B bond order in $B_2$ is ______.

Answer 1

Correct Answer: 1

To determine the bond order of $B_2$, we must analyze its molecular orbital (MO) configuration. Boron ($B$) has an atomic number of 5, thus $B_2$ comprises 10 electrons in total (5 per atom). We follow the standard electron filling order in molecular orbitals for diatomic molecules from the $2^{nd}$ period, particularly those with less than 14 electrons. The sequence is: σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², π(2p_x,y)², σ(2p)⁰. Here, the $π(2p)$ orbitals are filled before the $σ(2p)$ due to being lower in energy. For $B_2$, we fill up to π(2p_x,y) orbitals with 2 electrons:

• σ(1s)²
• σ*(1s)²
• σ(2s)²
• σ*(2s)²
• π(2p_x)¹, π(2p_y)¹

Next, we calculate the bond order using the formula: $\text{Bond Order} = \frac{1}{2}(\text{number of bonding electrons} - \text{number of antibonding electrons})$. In this case:

Bonding electrons: (σ(1s)², σ(2s)², π(2p_x)¹, π(2p_y)¹) = 6

Antibonding electrons: (σ*(1s)², σ*(2s)²) = 4

Thus, bond order = $\frac{1}{2}(6 - 4) = \frac{1}{2}(2) = 1$.

The bond order of $B_2$ is 1