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: Recall the bond order formula.
Bond order is half the difference between bonding and antibonding electrons: \[ \text{Bond Order} = \dfrac{N_b - N_a}{2} \]
Step 2: Plug in the known values.
We are told the bond order is 3 and $N_b = 10$, so $3 = \dfrac{10 - N_a}{2}$.
Step 3: Solve for $N_a$ directly.
Cross multiplying gives $N_b - N_a = 6$, and substituting $N_b=10$ gives $N_a = 10 - 6$.
\[ \boxed{N_a = 4} \]