Which of the following is the bond order of the diatomic molecule with 10 valence electrons, such as O2?

• A. 0
• B. 1
• C. 3
• D. 2

Answer: (D)

Verified step by step guidance

1. Identify the total number of valence electrons in the diatomic molecule. For O\_2, there are 10 valence electrons in the molecular orbitals.
2. Write the molecular orbital (MO) electron configuration for O\_2, considering the order of orbitals for molecules with atomic number greater than 7: \(\sigma_{2s}^2 \; \sigma_{2s}^*{}^2 \; \sigma_{2p_z}^2 \; \pi_{2p_x}^2 \; \pi_{2p_y}^2 \; \pi_{2p_x}^*{}^1 \; \pi_{2p_y}^*{}^1\).
3. Count the number of electrons in bonding orbitals and antibonding orbitals separately. Bonding orbitals are \(\sigma_{2s}\), \(\sigma_{2p_z}\), and \(\pi_{2p_x}\), \(\pi_{2p_y}\); antibonding orbitals are \(\sigma_{2s}^*\) and \(\pi_{2p_x}^*\), \(\pi_{2p_y}^*\).
4. Calculate the bond order using the formula: \(\text{Bond order} = \frac{(\text{number of bonding electrons}) - (\text{number of antibonding electrons})}{2}\).
5. Substitute the counted electrons into the formula to find the bond order, which will help determine the stability and bond strength of the molecule.