Electron Shell Capacities

  • #1
FeDeX_LaTeX
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Main Question or Discussion Point

Hello;

I couldn't find the answer to this anywhere, so I thought that I might ask here instead. What are the electron shell capacities for any given element? I think that any elements that are arranged in GROUPS have electron capacities that follow the pattern; 2,8,8,18,18,32,32... is this correct? And, as for elements that are not arranged in groups, i.e. actinides, lanthanides, and transition metals, will it follow the pattern of (3n2 - 2)? i.e. 2,8,18,32,50,72,98,128... ?

For example, the formula of gold bromide would be AuBr, because the charge of a gold ion is Au+ (because according to my pattern it has 1 electron in its outer shell) and because the charge of a bromine ion is Br-. So is this why the formula should be AuBr (as an example)?

EDIT: I looked up gold bromide on Wikipedia, which lists gold bromide as having the formula AuBr3. Why?
 

Answers and Replies

  • #2
FeDeX_LaTeX
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Or is it because a triple covalent bond is formed for AuBr3?
 
  • #3
sophiecentaur
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"For example, the formula of gold bromide would be AuBr, because the charge of a gold ion is Au+ (because according to my pattern it has 1 electron in its outer shell)"

That would imply, on a superficial level, that Gold is an alkali metal? Or, you might realise that things just ain't as simple as they appear, once you get more than two shells filled. Shells are only a simple, pictorial, model which you can't expect to apply throughout Chemistry. It's more about 'energy states' than orbits.
 
  • #4
FeDeX_LaTeX
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Okay... so can you explain why the formula for gold bromide would be AuBr3?
 
  • #5
sophiecentaur
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No, I can't; it's too complicated for me. But can you predict why it should be AuBr, now it's been pointed out that tings are not as simple as all that?
 
  • #6
SpectraCat
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Okay... so can you explain why the formula for gold bromide would be AuBr3?
Gold has two oxidation states [apart from the neutal atom, Au(0)], Au(I), with a +1 charge, and Au(III) with a +3 charge. The formula you found is for the Au(III) oxidation state.

The fact that you got "1 electron in the outer shell" for gold is almost certainly fortuitous, but can you please post the details of that calculation? It wasn't obvious from the context of your post.

Gold is nominally a 6s25d9 metal in its normal valence state. However, because of the anomalous stability of the filled d-subshell (i.e. the d10 configuration), 6s15d10 is actually the ground state configuration for the valence electrons. Since your model certainly doesn't allow for this detailed picture, it seems likely that your prediction was just blind luck.
 
  • #7
FeDeX_LaTeX
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Thank you for the replies. I made this topic when I was not aware of sub-orbitals or bonding working differently for transition metals. I have a few questions...

1) What is an "oxidation state"? How do you know that there are two oxidation states?
2) What is a valence state?
3) You said that it is nominally 6s25d9, but how did you know that it should be 6s15d10?
4) Could you explain what you mean by "ground state configuration for the valence electrons"?

Thanks.
 
  • #8
alxm
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1) See http://en.wikipedia.org/wiki/Oxidation_state" [Broken]
When he says it has two oxidation states, that really means that there are two oxidation states which are fairly stable, so you don't see other ones in stable compounds in practice. Naturally there's nothing physically stopping you from removing 2 electrons from a gold atom rather than 1 or 3.

2) By "valence state" he means how the electrons are distributed among the orbitals (or sub-shells) of the outermost (valence) shell. 4) The different ways of distributing the electrons are 'configurations' and the one with the lowest energy is the 'ground state configuration'.

3) These different sub-shells are close in energy (especially for transition metals) and so it can be difficult to predict. In most cases, the "s"-orbitals fill before the "d" ones do. But as an exception to this rule, filled 'd' sub-shells are extra stable, so it would rather be 6s15d10 than 6s25d9.

Although this rule and its exception-to-the-rule still doesn't explain why platinum has 6s15d9 rather than 6s25d8. (Which says something about how non-rigorous these "rules" are) But if you go left another step in the periodic table to Iridium, it does have 6s25d7 just as the rules would suggest. (Filled s-shell and partially empty d-shell)
The explanation for these things is way beyond the level at which you first learn them, so I'd say to just go with whatever your textbook tells you and just remember that you're only getting a general idea at this stage.
Learning chemistry is a bit like peeling an onion. You keep peeling back layers and having your previous ideas replaced by a more sophisticated picture. (and like an onion, it can cause tears on the way, :smile:)
 
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