# MO diagram of BeH₂

I was learning about the MO diagrams of triatomic molecules and was left slightly confused about $$\ce{BeH2}$$.

A picture of the MO is shown below: When looking at the $$\mathrm{1s}$$ combinations, I saw that each of the hydrogen $$\mathrm{1s}$$ combinations only had $$1$$ electron, which confused me.

I thought each hydrogen atom in the $$\ce{H-H}$$ pair contributed one electron, so I expected two electrons on each line for the $$\ce{H-H}$$ LGO (ligand group pair).

Why is there only $$1$$ electron on each line?

It looks like that particular MO was created with the two possibilities of hydrogen combinations: Either the two 1s electrons are same phase or different phase. Combining those options, alongside the AOs for $$\ce{Be}$$ produces the MO for $$\ce{BeH2}$$

This picture would have been clearer if they had opted to use the MO of dihydrogen, (bonding AND antibonding) which should have ended up at the same result. But it has been way to long since I linearly combined anything so I'll leave it as an exercise to reader.

edit: further comment - I am also not 100% sure about the reason to put one electron in each - but from a purely statistical point of view: 1 in each is the average situation.

It is an odd rendering. I'm not sure what the actual mechanism is to create $$\ce{BeH2}$$ but the simplistic reaction would be

$$\ce{Be + H2 <=> BeH2}$$

The MO for $$\ce{H2}$$, which is shown in the figure below is taken from Wikipedia.

The right side of the diagram you showed neither represents a hydrogen molecule, nor two independent (and hence equivalent) hydrogen atoms.

• so the correct diagram is what you represented above? is that correct? – David Smith Apr 17 at 18:47
• The image in my answer if for the molecular bonding of the hydrogen molecule. Obviously for the two hydrogen atoms to be on opposite sides of the boron atom the hydrogen molecule must break apart. That is the sticky point on the image in your question. – MaxW Apr 17 at 19:46