Question

If the bond order in \( C_2 \) is \( x \), then the bond order in \( B_2 \) and \( O_2 \), respectively are:

• A. \( \frac{1}{2} x, 2x \)
• B. \( x, x \)
• C. \( \frac{1}{2} x, x \)
• D. \( x, 2x \)

Hint:

The bond order can be calculated as: \[ \text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2} \]

Answer 1

The Correct Option is C

Bond order is related to the number of bonding and antibonding electrons in a molecule. For molecules like \( C_2 \), \( B_2 \), and \( O_2 \), the bond order can be calculated using the molecular orbital theory.
For \( C_2 \), the bond order is \( x \), which indicates that the total bonding electrons contribute to the bond strength. The molecular orbital theory states that in \( B_2 \) and \( O_2 \), bond orders are proportional to the bonding electron count relative to the number of electrons in the antibonding orbitals.
Thus, if the bond order in \( C_2 \) is \( x \), we find that the bond order in \( B_2 \) is \( \frac{1}{2} x \), and the bond order in \( O_2 \) is \( x \).

Answer 2

Step 1: Understand bond order in molecular orbital theory.
Bond order is calculated using the formula:
\[ \text{Bond order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} \]

Step 2: Bond order in \( C_2 \).
The electronic configuration of \( C_2 \) (12 electrons) in molecular orbitals is:
\[ (1\sigma)^2(1\sigma^*)^2(2\sigma)^2(2\sigma^*)^2(1\pi)^4 \] Bonding electrons = 8, Antibonding electrons = 4
\[ \text{Bond order of } C_2 = \frac{8 - 4}{2} = 2 \quad \Rightarrow x = 2 \]

Step 3: Bond order in \( B_2 \).
\( B_2 \) has 10 electrons:
Configuration: \((1\sigma)^2(1\sigma^*)^2(2\sigma)^2(2\sigma^*)^2(1\pi)^2\)
Bonding = 6, Antibonding = 4
\[ \text{Bond order of } B_2 = \frac{6 - 4}{2} = 1 \] \[ \text{So, Bond order of } B_2 = \frac{1}{2} \times x = \frac{1}{2} \times 2 = 1 \]

Step 4: Bond order in \( O_2 \).
\( O_2 \) has 16 electrons:
Configuration: \((1\sigma)^2(1\sigma^*)^2(2\sigma)^2(2\sigma^*)^2(1\pi)^4(1\pi^*)^2\)
Bonding = 10, Antibonding = 6
\[ \text{Bond order of } O_2 = \frac{10 - 6}{2} = 2 = x \]

Step 5: Conclusion.
Bond order in \( B_2 \) = \( \frac{1}{2} x \), and in \( O_2 \) = \( x \)
Hence, the correct answer is: \( \frac{1}{2} x, x \).