Question

The highest occupied molecular orbital for \(Ne_2\) is

• A. \(\sigma^{*}_{2p}\)
• B. \(\pi_{2p}\)
• C. \(\sigma_{2p}\)
• D. \(\pi^{*}_{2p}\)

Hint:

For molecules after nitrogen (O, F, Ne): Orbital order is \[ \sigma_{2p} \lt \pi_{2p} \lt \pi^{*}_{2p} \lt \sigma^{*}_{2p} \] For \(Ne_2\), the highest occupied orbital is always \[ \sigma^{*}_{2p} \] Memorize MO ordering carefully for chemistry exams.

Answer 1

The Correct Option is A

Step 1: Understand what HOMO means.
The highest occupied molecular orbital is the orbital of greatest energy that still contains electrons.
Step 2: Count the electrons in \(Ne_2\).
Each Ne atom has 10 electrons, so \(Ne_2\) has 20 electrons.
Step 3: Recall the filling order.
For these heavier diatomics the order is \(\sigma_{1s}, \sigma^{*}_{1s}, \sigma_{2s}, \sigma^{*}_{2s}, \sigma_{2p}, \pi_{2p}, \pi^{*}_{2p}, \sigma^{*}_{2p}\).
Step 4: Fill in the 20 electrons.
\((\sigma_{1s})^2 (\sigma^{*}_{1s})^2 (\sigma_{2s})^2 (\sigma^{*}_{2s})^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi^{*}_{2p})^4 (\sigma^{*}_{2p})^2\).
Step 5: Identify the last filled orbital.
The final electrons go into \(\sigma^{*}_{2p}\), which is the highest-energy occupied orbital.
Step 6: State the HOMO.
Therefore the HOMO of \(Ne_2\) is \(\sigma^{*}_{2p}\).
\[ \boxed{\sigma^{*}_{2p}} \]