What is the bond order of the diatomic molecule B_2?

• A. 3
• B. 0
• C. 1
• D. 2

Answer: (D)

Verified step by step guidance

1. Recall that bond order is calculated using the formula: \(\text{Bond order} = \frac{1}{2} (\text{number of bonding electrons} - \text{number of antibonding electrons})\).
2. Determine the total number of valence electrons in the B\(_2\) molecule. Each boron atom has 5 valence electrons, so B\(_2\) has \$5 \times 2 = 10$ valence electrons.
3. Write the molecular orbital (MO) electron configuration for B\(_2\). For molecules with atomic number less than 8, the order of MOs is: \(\sigma_{2s}\), \(\sigma^*_{2s}\), \(\pi_{2p_x} = \pi_{2p_y}\), \(\sigma_{2p_z}\), \(\pi^*_{2p_x} = \pi^*_{2p_y}\), \(\sigma^*_{2p_z}\).
4. Fill the MOs with the 10 valence electrons following the Aufbau principle and Hund's rule. Count how many electrons occupy bonding MOs and how many occupy antibonding MOs.
5. Apply the bond order formula using the counted electrons to find the bond order of B\(_2\).