## Octet Exception Question [ENDORSED]

Paula Dowdell 1F
Posts: 72
Joined: Tue Nov 15, 2016 3:00 am

### Octet Exception Question

What does size mean when referring to octet exceptions?

Joshua Hughes 1L
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### Re: Octet Exception Question

Yeah, I was a bit confused about that as well from one of the review sessions earlier today. I think by size she meant just as you go up in the shells (ie n=3 or 4) and the atom is bigger there are more orbitals for the electrons to fill, and specifically, the exceptions for the octet rule come into play with bigger atoms that have unfilled d-orbitals. So for example for carbon, n=2 and l could be 1 or zero(p or s orbital). Compare that to say phosphorous when n=3, l could be 2, 1, or 0 (which corresponds to the d-orbital, p-orbital, and s-orbital). So phosphorous after filling up its 3p orbital and forming an octet has the potential to then fill up the d orbital as well and go beyond an octet. I'm not really sure why the exception starts at Phosphorous, it may be possible for Si or other atoms to do that but maybe it just becomes more unstable because the nuclear charge isn't enough to keep extra electrons beyond the octet and would be unstable to the point that forming anything beyond an octet would be rare. IDK I'm kind of speculating on that last bit but someone else probably has a better answer.

Joshua Hughes 1L
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Joined: Sat Jul 22, 2017 3:01 am
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### Re: Octet Exception Question

I have an additional question that kind of follows up my previous post. If phosphorous or some other exception is filling up beyond an octet why would it not fill up the 4s orbital before going into the 3d orbital? I mean I know that 3d comes before 4s but at least when going by the periodic table after 3p fills up usually 4s would fill up before 3d. I may post this question as a separate subject if people don't see it here.

Nora Sharp 1C
Posts: 53
Joined: Sat Jul 22, 2017 3:00 am

### Re: Octet Exception Question  [ENDORSED]

I looked online for an answer and this is what I got:

"However, some of the third-period elements (Si, P, S, and Cl) have been observed to bond to more than four other atoms, and thus need to involve more than the four pairs of electrons available in an s2p6 octet. This is possible because for n=3, the d sublevel exists, and it has five d orbitals. Although the energy of empty 3d-orbitals is ordinarily higher than that of the 4s orbital, that difference is small and the additional d orbitals can accommodate more electrons. Therefore, the d orbitals participate in bonding with other atoms and an expanded octet is produced." (1)

From another website(2):
"Consider the electronic structure of neutral iron and iron (III).

The 4s electrons are lost first followed by one of the 3d electrons. This last bit about the formation of the ions is clearly unsatisfactory.

1. We say that the 4s orbitals have a lower energy than the 3d, and so the 4s orbitals are filled first.
2. We know that the 4s electrons are lost first during ionization. The electrons lost first will come from the highest energy level, furthest from the influence of the nucleus. So the 4s orbital must have a higher energy than the 3d orbitals.

Those statements are directly opposed to each other and cannot both be right.

The solution:
Potassium and calcium: The pattern is still working here. The 4s orbital has a lower energy than the 3d, and so fills next. That entirely fits with the chemistry of potassium and calcium.
The d-block elements: For reasons which are too complicated to go into at this level, once you get to scandium, the energy of the 3d orbitals becomes slightly less than that of the 4s, and that remains true across the rest of the transition series."

Therefore it would seem that the 4s subshell is not always lower energy than 3d, or there is significant overlap between the energy ranges of 4s and 3d. It is probably because of this ambiguity that website 1 claims that the difference in energy between 4s and 3d is small, and therefore electrons "prefer" to occupy the 3d orbital when phosphorus or sulfur bonds to multiple atoms. My guess is that using the 3d orbital in this specific situation is lower energy for reasons that have not yet been taught to us at this stage of chemistry: perhaps its favored for the 3d orbitals to have multiple bonded electrons than have the 4s and 3d orbitals form bonds with other electrons at the same time.

(1): https://courses.lumenlearning.com/bound ... ctet-rule/

(2): https://chem.libretexts.org/Core/Physic ... s_Orbitals