[luiseduardo]

Hello,

I'm from Brazil, my english is not very good.

My question is: The distribution of orbital d (of a element) in figure is representing a ground or excited state?



My point: First, i thought that it's excited, but then I thought about Hund Rule and Pauli Exclusion and I guess that this distribution is respecting the two laws, so it must be ground state. Now, I'm confused. So, anyone could help me ?

Thanks

[Corribus]

Technically there's not enough information to tell based on what you've drawn. But if we simplify it: presuming all lower orbitals are filled, then we'd call this a ground state configuration. In the absence of other contributing factors - external magnetic field, crystal field, etc. - the orbital energy only depends on the quantum numbers n and l, not ml. Meaning, all the d-orbitals have the same energy. Therefore, it doesn't matter which one of them is drawn to have the extra paired electron.

(In truth, we'd consider each possibility or microstate to be part of the overall specified state. But the spin and orbital momenta interact/couple, so there are actually multiple possible states for the electronic configuration you've drawn, some of which will have lower energy than others. These are embodied in term symbols, but I'm guessing based on your question that you haven't learned these nuances yet, so calling it a ground state configuration would probably be sufficiently robust for your needs.)