# How much energy does it cost to have electron configurations that are not in accordance with Hund's rules?

What is a ballpark figure for the difference in energy for an atom that follows Hund's rule vs one that has two electrons with opposite spins? I'd be interested to know carbon and nitrogen. Is there a bigger difference when there are two electrons with one spin and one electron with the opposite spin vs two electrons with opposite spins?

• May I ask: What electron configurations were the ones you actually asked for? Commented Sep 9, 2014 at 2:38

Each energy level of a given electron configuration is described by a Russell-Saunders term symbol, assuming LS coupling. The ground state term symbol is predicted by Hund's rules. Tables of atomic energy levels identified by their term symbols have been compiled by NIST. So, the only thing you have to do, is to compare the energies of the terms describing the electron configurations you are looking for with their respective ground states.

In the following pictures the notation $m_{S} = + \frac{1}{2} \overset{\scriptsize{\text{def}}}{=} \, \uparrow$ and $m_{S} = - \frac{1}{2} \overset{\scriptsize{\text{def}}}{=} \, \downarrow$ will be used for the spin quantum number $m_{S}$.

The data for carbon can be found here. The ground state for carbon is ${}^{3}\mathrm{P}_{0}$ - it is assigned the energy $0.00 \, \mathrm{cm}^{-1}$. From your question it's not 100 % clear what exact electron configuration you want. I assume it is either ${}^{3}\mathrm{P}_{2}$ or ${}^{1}\mathrm{S}_{0}$ therefore I will give the values for both of them (for comparison: the thermal energy at room temperature is about $200 \, \mathrm{cm}^{-1}$).

The data for nitrogen can be found here. The ground state for nitrogen is ${}^{4}\mathrm{S}_{3/2}$ - it is assigned the energy $0.00 \, \mathrm{cm}^{-1}$. From your question it's not 100 % clear what exact electron configuration you want. I assume it is either ${}^{2}\mathrm{D}_{5/2}$ or ${}^{2}\mathrm{P}_{3/2}$ therefore I will give the values for both of them.

• How is a 3p0 not equivalent to 3p2? Is there a electron-spin coupling with the nucleus or something? Just off the cuff I would think that these would be identical if considering only this representation of the electron configuration. Commented Sep 8, 2014 at 20:14
• @LordStryker The energy difference is quite small compared to the other ones. Maybe it is a result of spin-orbit coupling. Commented Sep 8, 2014 at 20:23
• @LordStryker The $^3P_0$ and $^3P_2$ differs in J, i.e. the mutual orientation of L and S. This energy difference is coming from the spin-orbit coupling (which is per definition proportional with $LS$). Since spin-orbit coupling is small for light elements that is why the energy gap is so small.
– Greg
Commented Sep 9, 2014 at 13:04
• Question on carbon: First, in your diagrams, shouldn't singlet S zero be a singlet D two configuration? Second, in the link to your source for carbon transitions, why is there no transition for triplet S one in carbon - that is, for the valence carbon 2p subshell, an electron in ml=+1 and an electron in ml= -1? Thanks. Commented Dec 1, 2016 at 14:33