Question:

Which of the following molecules contains maximum number of electrons in antibonding molecular orbitals?

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Antibonding electrons increase as you move from $\text{N}_2 \to \text{O}_2 \to \text{F}_2$ in the second period.
Updated On: Apr 26, 2026
  • $\text{Li}_2$
  • $\text{N}_2$
  • $\text{O}_2$
  • $\text{F}_2$
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The Correct Option is D

Solution and Explanation

Step 1: Configuration Analysis
We count electrons in $\sigma^*$, $\pi^*$ orbitals. - $\text{Li}_2$ ($6e^-$): $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2$. Antibonding = 2. - $\text{N}_2$ ($14e^-$): $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2$. Antibonding = 4. - $\text{O}_2$ ($16e^-$): Same as $\text{N}_2$ but plus $(\pi^* 2p_x)^1 (\pi^* 2p_y)^1$. Antibonding = 6. - $\text{F}_2$ ($18e^-$): Same as $\text{O}_2$ but plus $(\pi^* 2p_x)^2 (\pi^* 2p_y)^2$. Antibonding = 8.
Step 2: Comparison
$\text{F}_2$ has the highest number of electrons in antibonding orbitals.
Final Answer: (D)
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