# 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.

$\left[R n\right] 7 {s}^{2} 5 {f}^{14} 6 {d}^{10} 7 {p}^{3}$ ($\text{Mc}$)
$\left[X e\right] 6 {s}^{2} 4 {f}^{14} 5 {d}^{10} 6 {p}^{3}$ ($\text{Bi}$) 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 $n \ge 4$ for ANY $f$ orbital, not only $4$.

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

• $n = 7$ for the $n s$ orbitals.
• $n - 2 = 5$ for the $\left(n - 2\right) f$ orbitals.
• There are $2 l + 1 = 2 \left(3\right) + 1 = \boldsymbol{7}$ of these $5 f$ orbitals.
• Each of these have $\boldsymbol{2}$ electrons max.

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

Therefore, since there are fourteen $4 f$ core electrons as well, we have that there are $\boldsymbol{28}$ total electrons in the $f$ orbitals.