Question

The bond order of a homodiatomic molecule is 3. If the number of bonding electrons in it is 10, the number of antibonding electrons will be

• A. 4
• B. 5
• C. 6
• D. 3

Hint:

For nitrogen ($\text{N}_2$), the total number of electrons is 14.
Its molecular orbital configuration has $N_b = 10$ and $N_a = 4$, giving a bond order of 3.
This matches the typical triple-bond description of nitrogen gas.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question is based on Molecular Orbital Theory (MOT).
We need to find the number of antibonding electrons in a homonuclear diatomic molecule with a known bond order and number of bonding electrons.

Step 2: Key Formula or Approach:
The bond order (B.O.) of a diatomic molecule is calculated using 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.

Step 3: Detailed Explanation:

• We are given:
Bond Order = 3
Number of bonding electrons ($N_b$) = 10

• Let us substitute these values into the bond order formula:
\[ 3 = \frac{10 - N_a}{2} \]

• Multiplying both sides by 2:
\[ 6 = 10 - N_a \]

• Solving for $N_a$:
\[ N_a = 10 - 6 = 4 \]

• Thus, the number of antibonding electrons is 4.


Step 4: Final Answer:
The number of antibonding electrons in the homodiatomic molecule is 4.