[matt1234]

Hello
I have a question about the configuration of gold
[xe] 4f14 5d10 6s1  this element has a total of 79 electrons.   are the lower levels of f d and s full aswell?  Because if they were it would be more then 79 electrons would it not?

Sorry im a new chem student taking gr 11 chem. 

 I

[renge ishyo]

The orbitals don't necessarily fill in numerical order. For instance, the 4f orbital doesn't fill until after the 6s orbital has filled. Have a look at the aufbau chart to get an idea of the filling order of the orbitals:

http://chemistry.about.com/od/electronicstructure/ss/aufbau_2.htm

After some time you'll be able to deduce the order of filling directly from the periodic table, but this takes a bit more work to see.

[matt1234]

Here is what the teacher wrote on the board and it confuses me. 

[renge ishyo]

All of that information is correct, but it won't help you answer your original question. Which part of the notes do you find confusing? Remember that each orbital holds a maximum of two electrons. There are three p orbitals, so in the 2p or the 3p rows you can fit 6 electrons. On the other hand, there are five d orbitals, so in the 3d orbital you can fit a total of 10 electrons. Remember two electrons per orbital. and just memorize how many orbitals an "s", "p", or "d" has and you can figure out how many electrons fit in there. Keep at it. This stuff is quite confusing at first!

[matt1234]

Thank you for your help,
Could someone confirm the maximum number of orbitals there are in  "s", "p" "d" and "f"? 
I was told this but not sure:

s = max of 1
p = max of 3
d = 5
and
f = 7


With 2 electrons per each orbital.

Is the above correct?

I dont think its correct because this page says other wise:
http://www.chemguide.co.uk/atoms/properties/atomorbs.html

Also looking at gold the ending says 6s1, so i no longer understand.

[matt1234]

I get it!!


c and P


Gold atoms contains six energy levels, as shown below:

1s2
2s2 2p6
3s2 3p6 3d10
4s2 4p6 4d10 4f14
5s2 5p6 5d10
6s1

The first number is the energy level. The lowercase letter is the sub-shell. And, the last number is the number of electrons in a sub-shell.

Gold has 2 electrons in level 1, 8 electrons in level 2, 18 in level 3, 32 in level 4, 18 in level 5 and 1 electron in level 6.

So, the the long electron configuration of gold is:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s1 4f14 5d10



It all makes sense, the reason the d's come before the s's is because of the aufbau chart, and the steps it fills in, less energy is required to fill an s level than a d orbital. cool!

[matt1234]

My next question,

Why does the electron proceed to the 6th level and not a f level instead?

[renge ishyo]

Glad to see you are making progress!

To try and analyze what is going on with the order of fillings, consider the "n+l" rule for orbital filling. Early on in the periodic table, orbital locations with higher "n+l" numbers are higher in energy and fill after the lower energy orbitals. Basically how this works is the distance from the nucleus is determined by the "n". For instance, for 2s the n is equal to 2. The 2s orbital is located further away from the nucleus than the 1s orbital as indicated by its higher n. The 2s orbital fills after the 1s orbital. Next, you have to consider the "l" part. I am going to deviate a bit from the orthodox way of explaining things by saying that "l" relates to the complexity of the orbitals. So, if you are comparing 2s to 2p, p has a higher "l" value than s. It is more complex, so even though both orbitals have n=2, the 2p orbitals are higher in energy due to their added complexity and fill after the 2s orbitals.

The values of l for the different orbitals are:

s = 0
p = 1
d = 2
f = 3

So you can use this to predict the order of filling (it is just another tool to come up with the chart I linked to earlier). Here are some examples to show how it works:

#1
----
2s: n+l = 2 + 0 = 2
2p: n+l = 2+ 1 = 3 (the n+l is higher for 2p so it fills after 2s)

#2
----
4s: n + l = 4 + 0 = 4
3d: n + l = 3 +2 = 5 (the n+l is higher for 3d so it fills after 4s even though 4s has the higher "n" value)

However, if the difference in "n" values is greater than 1 unit (as can be the case towards the bottom of the periodic table), you can have the lower n+l values fill first because the difference in n outweighs the n+l rule. This is the case for the example you were curious about:

#3
----
6s: n + l = 6+0
4f: n + l = 4 +3 = 7 (you would predict that the 4f orbital would be favored over 6s, but in actuality the difference in energy levels between 6 and 4 of two whole units overcomes previous trends so 6s actually fills before 4f. In fact, once you get down to this level on the periodic table a lot of the previous trends start breaking down because all the orbitals get very close to one another in energy).

[Borek]

consider the "n+l" rule for orbital filling

Just remember - effective as it is, it is only a rule of thumb.