If you are to draw the energy level for carbon (with no bonding) the electrons pair before moving to another orbital. For example, 6C12 would have an electron distribution in the ground state of
1s2 2s2 2p2. I assume you are just starting chemistry; later on you will learn that one of the 2s2 electrons can be promoted to the empty 2p orbital to make hybrid orbitals (between s and p) so that in carbon compounds it is
1s2 2s1 2p3 and the sp3 hybridization is responsible for the tetrahedral nature of carbon compounds.
I was asked to fraw the energy level diagram for carbon. Carbon has 6 electrons so would the 2s and 2p orbitals both be half full?
1s = 2 electrons
2s= 1 electron
2p = 3 electrons
8 answers
My teacher was saying that its better for the shells to be half full rather than one being completely full and the other being not half full. He was referring to Hund's law. What do you think about this?
Generally, your instructor is correct. Hund's rule says that electrons will enter the unoccupied orbitals unpaired and pairing will not begin until all of the orbitals are half filled. That is the 2p orbitals can be 2px1, 2y1 and 2pz1 before they begin pairing up. The next three electrons would enter as 2px2 2py1 2pz2, then 2px2 2py2 2pz1 and finally the 6th one would enter the 2pz to make the final 2px2 2py2 2pz2. But the half filled rule, and the one you apparently used to fill the s half way and the p orbitals half way, ARE FOR THE SAME ORBITAL. So when you have the s with one electron in it, before a p can be filled, the next electron must pair with the s that's already there. Then the p orbitals can be filled. The s orbital is lower (generally) than p, therefore, the s must be filled before getting into the p. You will find later (for Cu and Cr) that the s,and d orbitals are so close together that it does happen that one will fill before the other. I hope this helps.
"But the half filled rule, and the one you apparently used to fill the s half way and the p orbitals half way, ARE FOR THE SAME ORBITAL"
So the 2S and 2P are both in the second orbital so therefore there is no electron promotion?
So the 2S and 2P are both in the second orbital so therefore there is no electron promotion?
First, remember that we are talking about the ground state configuration.
When I say the same orbital I mean s and p are different. It doesn't matter that they are both n = 2 to make 2s or 2p. It's the s orbital that can be half filled but the next electron must pair up with the s before the p orbital (a different orbital and not the same as the s) can begin.
The 2s orbital is lower in energy than the 2p orbital; therefore, the 2s may fill half way with the first electron, then will fill completely with the next electron before the third electron enters the 2p orbital. Promotion does occur, especially with carbon, so that we have sp, sp2, and sp3 orbital (not electron) configurations, but these are not for the ground state.
When I say the same orbital I mean s and p are different. It doesn't matter that they are both n = 2 to make 2s or 2p. It's the s orbital that can be half filled but the next electron must pair up with the s before the p orbital (a different orbital and not the same as the s) can begin.
The 2s orbital is lower in energy than the 2p orbital; therefore, the 2s may fill half way with the first electron, then will fill completely with the next electron before the third electron enters the 2p orbital. Promotion does occur, especially with carbon, so that we have sp, sp2, and sp3 orbital (not electron) configurations, but these are not for the ground state.
So we are following the Aufbau Principle?
Yes. In the ground state we are following the Aufbau Principle as well as Hund's rule. Here is a site that provides a useful mnemonic for remembering how the orbitals are filled (although I've see better ones drawn).
http://en.wikipedia.org/wiki/Aufbau_principle
http://en.wikipedia.org/wiki/Aufbau_principle
Thank you so much :)
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