Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

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.

Answer 1

The Correct Option is B

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).

Answer 2

Step 1: Diatomic carbon (C₂)'s molecular orbital filling gives 8 electrons in bonding orbitals and 4 electrons in antibonding orbitals.

Step 2: Bond order = (bonding electrons − antibonding electrons) ÷ 2 = (8 − 4) ÷ 2 = 2.

Step 3: So the bond order of C₂ is 2 — option (B).

Answer 3

Valence electron count approach: Each carbon atom contributes 4 valence electrons, so C\(_2\) has 8 valence electrons to place into the 2s and 2p molecular orbitals, since the filled 1s orbitals cancel out and don't affect bonding. Filling \( \sigma_{2s}, \sigma_{2s}^{*}, \pi_{2px}, \pi_{2py} \) with these 8 electrons leaves the 2s pair cancelled by its antibonding partner and both pi orbitals fully bonding. That's 2 net bonding pairs, giving a bond order of 2, option (B).