Q. mo diagram for \( \mathrm{O_2} \).

To generate a MO (molecular orbital) diagram for oxygen gas \( \mathrm{O_2} \), we build the valence molecular orbitals using the ordering that comes from combining the atomic orbitals (AOs) of two oxygen atoms.

Step 1: Identify the total valence electrons

Each oxygen atom has electron configuration \( \mathrm{1s^2\,2s^2\,2p^4} \). The valence electrons are in the \(2s\) and \(2p\) orbitals.

So each oxygen has:

\( \mathrm{2s^2} \) and \( \mathrm{2p^4} \) which is \(6\) valence electrons per atom.

For \( \mathrm{O_2} \):

\[ \text{Total valence electrons} = 2 \times 6 = 12 \]

Step 2: List the relevant molecular orbitals (valence MO diagram)

The valence MOs formed from oxygen \(2s\) and \(2p\) orbitals (for \( \mathrm{O_2} \)) are commonly ordered as follows (standard ordering for oxygen and similar second-row diatomics):

\[ \sigma_{2s} \quad \sigma^{*}_{2s} \quad \sigma_{2p} \quad \pi_{2p} \quad \pi^{*}_{2p} \quad \sigma^{*}_{2p} \]

Equivalently, in the usual energy-level sketch from low to high energy:

\[ \sigma_{2s} < \sigma^{*}_{2s} < \sigma_{2p} < \pi_{2p} < \pi^{*}_{2p} < \sigma^{*}_{2p} \]

Step 3: Place the 12 valence electrons into the MOs

Rule reminders (very important):

1. Fill lowest-energy orbitals first.

2. Use the Pauli exclusion principle (max 2 electrons per orbital, opposite spins).

3. For degenerate orbitals (like the two \( \pi \) orbitals), place electrons to maximize total spins (Hund’s rule) before pairing.

Now fill from the bottom:

(1) \( \sigma_{2s} \)

This orbital holds 2 electrons.

After filling \( \sigma_{2s} \):

Electrons used: 2, remaining: \(12 – 2 = 10\).

(2) \( \sigma^{*}_{2s} \)

Holds 2 electrons.

Electrons used: 2 (now total 4), remaining: \(10 – 2 = 8\).

(3) \( \sigma_{2p} \)

Holds 2 electrons.

Electrons used: 2 (now total 6), remaining: \(8 – 2 = 6\).

(4) \( \pi_{2p} \)

There are two degenerate \( \pi_{2p} \) orbitals (a pair of orbitals). Together they can hold 4 electrons.

We have 6 electrons remaining to place into the higher \(2p\)-derived set. First fill \( \pi_{2p} \) with 4 electrons total.

Electrons used: 4 (now total 10), remaining: \(6 – 4 = 2\).

(5) \( \pi^{*}_{2p} \)

The two degenerate antibonding \( \pi^{*}_{2p} \) orbitals together can also hold 4 electrons. We have 2 electrons left, so place them here following Hund’s rule.

So \( \pi^{*}_{2p} \) gets 2 electrons (one in each \( \pi^{*} \) orbital with parallel spins).

Electrons now placed: 2 (now total 12). None remain.

Step 4: Write the final electron configuration in the MO diagram

The valence MO electron configuration for \( \mathrm{O_2} \) is:

\[ \sigma_{2s}^2 \;\; \sigma^{*}_{2s}^2 \;\; \sigma_{2p}^2 \;\; \pi_{2p}^4 \;\; \pi^{*}_{2p}^2 \]

And \( \sigma^{*}_{2p} \) has 0 electrons for \( \mathrm{O_2} \) (it is above the filled levels for the 12 valence electrons).

Step 5: (Optional but commonly included) Bonding vs antibonding count

To understand bond order, count bonding and antibonding electrons:

Bonding MOs (net “bonding”): \( \sigma_{2s}, \sigma_{2p}, \pi_{2p} \)

Antibonding MOs: \( \sigma^{*}_{2s}, \pi^{*}_{2p} \)

From the configuration:

Bonding electrons:

\( \sigma_{2s}^2 \) contributes 2, \( \sigma_{2p}^2 \) contributes 2, \( \pi_{2p}^4 \) contributes 4

Total bonding: \(2 + 2 + 4 = 8\).

Antibonding electrons:

\( \sigma^{*}_{2s}^2 \) contributes 2, \( \pi^{*}_{2p}^2 \) contributes 2

Total antibonding: \(2 + 2 = 4\).

Bond order:

\[ \text{Bond order} = \frac{8 – 4}{2} = 2 \]

Final MO Diagram (filled levels description)

Low to high energy ordering with electron filling:

\[ \sigma_{2s}^2 \;<\; \sigma^{*}_{2s}^2 \;<\; \sigma_{2p}^2 \;<\; \pi_{2p}^4 \;<\; \pi^{*}_{2p}^2 \;<\; \sigma^{*}_{2p}^0 \]

This is the molecular orbital (MO) diagram electron arrangement for \( \mathrm{O_2} \).