# How do you draw the MO diagram for "O"_2 from scratch?

May 9, 2016

To draw a molecular orbital (MO) diagram, you need to consider which atomic orbitals (AOs) the molecule has.

ATOMIC ORBITAL DIAGRAM FOR OXYGEN ATOM

Oxygen atom is on period 2, so it has access to its $1 s$, $2 s$, and $2 p$ AOs. Their relative energies are $\setminus m a t h b f \left(2 p > 2 s\right)$ $\setminus m a t h b f \left(\text{>>}\right)$ $\setminus m a t h b f \left(1 s\right)$.

(The $1 s$ is much, much lower in energy than the $2 s$, and usually is not even on the MO diagram if done to-scale).

Then, note that oxygen atom has $8$ total electrons (including core), so you fill the AOs with $8$ electrons according to:

• Hund's rule (maximize the number of unpaired electrons).
• The Aufbau principle (fill orbitals from lowest to highest energy).
• The Pauli Exclusion Principle (paired electrons are opposite in "spin", or direction).

If you are unsure about any of these rules or principles, you should ask for further clarification, as I assume you already are familiar with these.

At this point, we have the AO diagram as follows for both oxygen atoms:

DETERMINING ATOMIC ORBITAL INTERACTIONS

Next, consider which orbitals can interact with each other. That means which ones can overlap to create an effective bond.

For a homonuclear diatomic molecule like ${\text{O}}_{2}$, this is simple; just choose the orbitals that are alike in look and energy. So, the $2 s$ interacts with the $2 s$, and the $2 p$ interacts with the $2 p$, etc.

These interactions generate what are called molecular orbitals, and they will conserve the number of orbitals. That means with $\setminus m a t h b f \left(10\right)$ AOs in, you get $\setminus m a t h b f \left(10\right)$ MOs out---$5$ per oxygen. So, we will be drawing $10$ MOs soon.

TYPES OF MOS TO DRAW

Next, recall that there exist bonding ($\sigma$, $\pi$, etc) and antibonding (${\sigma}^{\text{*}}$, ${\pi}^{\text{*}}$, etc) orbitals. The former is stabilized (lower in energy) relative to the AOs, and the latter is destabilized (higher in energy) relative to the AOs.

When we consider these interactions, we'll see the $n s$ (sigma, head-on) interactions:

and the slightly more complicated $n p$ (pi, sidelong, and sigma, head-on) interactions:

With ${\text{O}}_{2}$, the orbital ordering is normal (with ${\text{N}}_{2}$ it is weirder, and you should ask your teacher about an "orbital mixing effect" if you want to know why). "Normal" means it is just like what I'm showing you here.

So, when we combine what we see here into the full diagram, we first get:

And then when we fill the MOs with the $16$ total electrons contributed from both oxygens, according to the same three rules and principles defined for the AO diagram, we get:

Remember that that was for ${\text{O}}_{2}$!

• To modify the diagram for ${\text{O}}_{2}^{+}$, simply remove one of the highest-energy electrons.
• To modify the diagram for ${\text{O}}_{2}^{-}$, simply add one electron to the highest-energy MO.

If you have any questions still, and if this is still confusing, just ask.