Question

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

Answer 1

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.