How many electrons are in orbitals with #l = 3# within moscovium?

1 Answer
May 5, 2017

Ooh, moscovium.

Well, #l = 3# corresponds to the #f# orbitals, as you know. In principle, the valence electron configuration of moscovium is expected to be similar to that of bismuth.

#[Rn] 7s^(2) 5f^(14) 6d^(10) 7p^(3)# (#"Mc"#)
#[Xe] 6s^(2) 4f^(14) 5d^(10) 6p^(3)# (#"Bi"#)

http://www.ptable.com/

But, for moscovium, we thus have that in general, the valence orbitals have #n ne 4# (we're not at the #g# orbitals yet!). In fact, we have that #n >= 4# for ANY #f# orbital, not only #4#.

We are actually at the 7th row in the periodic table, so:

  • #n = 7# for the #ns# orbitals.
  • #n - 2 = 5# for the #(n-2)f# orbitals.
  • There are #2l+1 = 2(3) + 1 = bb(7)# of these #5f# orbitals.
  • Each of these have #bb(2)# electrons max.

You do therefore have a total of fourteen #5f# valence electrons. However, we need to also consider the core #4f# electrons. The core orbitals are presumed all occupied.

Therefore, since there are fourteen #4f# core electrons as well, we have that there are #bb(28)# total electrons in the #f# orbitals.