Question

According to Molecular Orbital Theory, the bond order of O$_2$ is:

• A. 1
• B. 1.5
• C. 2
• D. 3

Hint:

A quick trick for finding bond orders of second-period diatomic species (total electrons $10$ to $18$):
$14$ electrons (like $\text{N}_2$) $\implies$ Bond Order = $3.0$
Each addition or subtraction of $1$ electron decreases the bond order by $0.5$.
For $\text{O}_2$ ($16$ electrons): $3.0 - (2 \times 0.5) = 2.0$.
For $\text{O}_2^+$ ($15$ electrons): $3.0 - 0.5 = 2.5$.

Answer 1

The Correct Option is C

Step 1: MO count for O2.
Bonding electrons=10, antibonding=6.
Step 2: Bond order.
$(10-6)/2$
\[ \boxed{2} \]