Question:

What is the bond order for diatomic carbon according to molecular orbital theory?

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To calculate the bond order using molecular orbital theory, use the formula: \[ \text{Bond order} = \frac{N_b - N_a}{2} \] where \( N_b \) is the number of bonding electrons and \( N_a \) is the number of antibonding electrons.
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the electronic configuration of \( C_2 \).
The electronic configuration of \( C_2 \) (diatomic carbon) can be written as: \[ (\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2px})^2(\pi_{2py})^2 \] Here, we have 8 bonding electrons and 4 antibonding electrons.
Step 2: Applying the bond order formula.
The bond order is given by the formula: \[ \text{Bond order} = \frac{N_b - N_a}{2} \] where \( N_b \) is the total number of bonding electrons and \( N_a \) is the total number of antibonding electrons. For \( C_2 \): \[ N_b = 8 \quad \text{and} \quad N_a = 4 \] Thus, the bond order is: \[ \text{Bond order} = \frac{8 - 4}{2} = 2 \] Step 3: Conclusion.
Therefore, the bond order for diatomic carbon (\( C_2 \)) is 2, which corresponds to option (B).
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