Noble gas configuration and valence shell electrons [duplicate]

Question

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

Answer 1

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 m_l) 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? :)