# 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? – Philipp Sep 9 '14 at 2:38

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.
• @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 Sep 9 '14 at 13:04