Noble gas configuration and valence shell electrons [duplicate]

I know that the maximum number of electrons within a shell is equal to 2n^2. I would think that the noble gases would reflect this but that isn’t always the case

n = 1

2(1^2) = 2, this is the atomic number of He ✅

n = 2

2(2^2) = 8, plus the 2 e from n = 1, this is 10, which is the atomic number for Ne ✅

n = 3 2*(3^2) = 18, this is the atomic number for Ar, but I realized that the 3d orbital is empty. The 3rd energy level should have a total of 18 e and the total amount of electrons until the 3rd energy level should include 10 e from the previous 2. I understand that this is because of the aufbau principle, but it got me confused about how to consider noble gases. I thought they were supposed to have the maximum number of valence electrons, but then I thought what about group 12, Lutetium, and Lawrencium? Just based on their electron configuration, every orbital they have is completely occupied. I know these are very different from noble gases. I’m just wondering why, especially because these elements can be very reactive (right?).

Edit: I forgot to mention that after n = 3, nothing matches up at all.

n = 4

2(4^2) = 32, For Kr, Z = 36

n = 5

2(5^2) = 50, For Xe, Z = 54

n = 6

2(6^2) = 72, For Rn, Z = 86

And lastly, Idk if Oganesson is even considered a noble gas but it’s in the same group so I’ll put it here.

n = 7

2(7^2) = 98, For Og, Z = 118

• Noble gases were never about filling everything that is available. They were about filling everything until you need to increase the n. May 20 '21 at 16:47
• By increasing n, do you mean in the order electrons occupy orbitals?
– Ibby
May 20 '21 at 16:55
• Like you keep adding electrons just before the electrons can occupy the next energy level
– Ibby
May 20 '21 at 17:00
• @lbby Yes, because that's when we say a period in the periods table is changed. May 20 '21 at 17:01
• Like a wise person once said: don't forget the Aufbau principle... you may want to look at Ben Norris' or Brian's answers in this answer (other answers may go into too much advanced detail). There are many linked posts on the site you may want to browse as well. May 21 '21 at 6:34

The number of orbitals per shell (n) is n2. At most there are two electrons per orbital, each with opposite spin. However, due to nuclear screening, the subshells (n and l), and as a result the orbitals (n, l, and ml) are not of the same energy.

When you get to more complicated situations, such as the third row transition metals, and you see exceptions such as chromium and copper (and their respective group members) having ns1 (n-1)d5 and ns1 (n-1)d10 configurations, the added stability from half-filled and fully filled sub shells outweighs pairing the ns electron as it is more distant from the nucleus and less stabilized by screening.

With lutetium, the configuration is [Xe] 4f14 5d1 6s2; for lawrencium, the configuration is [Rn] 5f14 7s2 7p1.

In terms of reactivity, valence electrons further from the nucleus are easier to remove, so reactivity can vary. With regard to these rare earth metals you mentioned, I am not as familiar and have limited knowledge on them. But, in terms of filling orbitals, you are correct to follow Aufbau. (f will come before d and be (n-1) lower than the respective d, and d will be (n-1) lower than the respective s/p.

Does this help? :)

• Thanks for your answer. This helps with some other questions I have but there’s just one part I’m stuck on. The other person here told me that d and f group electrons are not part of the valence shell. I have verified this as true but Idk why this is. If it’s not too much trouble, could you explain please?
– Ibby
May 20 '21 at 22:28
• Also I forgot to mention that I thought Luterium and Lawerencium were the inner transition metals with f14s2 electron configuration but I was off by one in each case. I was trying to use them as examples to illustrate completely filled orbitals so thanks for pointing that out.
– Ibby
May 20 '21 at 22:33
• The valence (outer most) shell of lutetium is n = 6 and that of lawrencium is n = 7. Both the d and f orbitals for the respective elements have principal quantum numbers smaller than this. See above discussion. May 21 '21 at 0:14
• Thanks for elaborating. That makes so much sense. Idk why I didn’t consider that
– Ibby
May 21 '21 at 0:21