# Homework Help: Electrons and their quantum numbers for the oxygen atom

1. Nov 28, 2011

### fluidistic

1. The problem statement, all variables and given/known data
Write down the electonic configuration for the O atom. Write down the quantum numbers $(n,l,m_l,m_s)$ for each one of the electrons.

2. Relevant equations
Pauli exclusion principle. Least energy principle for filling the shells.

3. The attempt at a solution
$1s^22s^22p^4$.
I have trouble for the electrons on the 2p shell.
The quantum numbers of the electrons are:
(1,0,0,1/2)
(1,0,0,-1/2)
(2,0,0,1/2)
(2,0,0,-1/2)
Here I am not sure:
(2,1,0,1/2)
(2,1,0,-1/2)
And 2 more electrons but I really don't know which quantum numbers to choose. They all have the same energy (degenerated). An external magnetic field would solve some degeneracy (the $m_l$ one, not the $m_s$ one).
So...
$(2,1,\pm 1 , \pm 1/2)$ and $(2,1,0 , \pm 1/2)$ are all possible for the 4 electrons.
However if I apply an external magnetic field, I can discard $m_l=1$, right?

2. Nov 28, 2011

### ehild

The energy levels for a given l are filled with one electron on each m first. The oxygen atom is known to have two unpaired outer electrons with equal spins.

ehild

Last edited: Nov 29, 2011
3. Nov 29, 2011

### fluidistic

Thank you ehild.
So it means that I have the choice? For example here are my arbitrary choice for the quantum numbers for the 4 electrons on the 2p shell:
1----(2,1,0,-1/2)
2----(2,1,-1,-/1/2)
3----(2,1,1,-1/2)
4----(2,1,0,1/2).
The 2 unpaired electrons are 2 and 3.
However if I apply an external magnetic field the answer would change since I have to apply Aufbau principle, i.e. "fill lowest energy levels first". And in reality I think there's always an external magnetic field (as tiny as it may be) and so it seems that the paired 2p electrons must have their $m_l$ quantum number equal to -1.
So I'm tempted to choose:
1----(2,1,0,-1/2)
2----(2,1,-1,-/1/2)
3----(2,1,-1,1/2)
4----(2,1,0,1/2).
But I have a problem here. Aubfau principle seems to be in contradiction with Hund's rule. One electron should have the $m_l=1$ quantum number according to Hund's rule, while if I do this, I'd give it more energy than if it had $m_l=0$, so I go in counter of Aubfau principle.
So I'm unsure how to answer the problem.

4. Nov 30, 2011

### ehild

The second configuration is not correct for the free oxygen atom. Do not forget that there is magnetic momentum assigned to spin, too. Two electrons on the same level (with the same m) and only the spins opposite repel each other more than attract because of their opposite spins, and a weak magnetic field does not change it. The effect of the magnetic field appears in the spins of the electrons occupying the m=1 and m=-1 levels.

I know the electron configuration of oxygen from Chemistry. It has two
electrons with paired spins - with m=0, and two unpaired electrons
which take part in covalent bonds -paired with a lone electron of an other atom.
See this: http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Electronic_Configurations#Hund's_Rule

ehild

5. Nov 30, 2011

### fluidistic

Ok thank you.
I've read the link you provided, but I don't know where it's written that the 2 paired electrons of the 2p shell have their quantum numbers $m_l=0$.
Also in the link we can see that all lone electrons have their spin always up. Is it a convention or it must be this way, always? (didn't find the part that explain this). If it must always be like this, this would also invalidate my first choice in my last post, namely:
1----(2,1,0,-1/2)
2----(2,1,-1,-/1/2)
3----(2,1,1,-1/2)
4----(2,1,0,1/2).
Because the unpaired electrons 2 and 3 have their spin down.

6. Nov 30, 2011

### ehild

I do not know. To find the energy belonging to a configuration is not that simple. In weak magnetic fields the spin is coupled to the the orbital momentum, and their resultant determines the magnetic potential energy. I do not know enough about this field.

ehild

7. Nov 30, 2011

### fluidistic

Ok thank you for all the information.