# What is the electron configuration for Fr?

Aug 9, 2016

$\left[R n\right] 7 {s}^{1}$

#### Explanation:

From the Periodic Table, $Z$, the atomic number of $F r = 87$

And thus,

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{10} 4 {s}^{2} 4 {p}^{6} 4 {d}^{10} 5 {s}^{2} 5 {p}^{6} 4 {f}^{14} 5 {d}^{10} 6 {s}^{2} 6 {p}^{6} 7 {s}^{1}$

Well, I have got 87 electrons there at least. Of course, all the action, all the chemistry takes place at $7 {s}^{1}$, i.e. $F r$ is univalent and would react as an alkali metal. It is extremely rare element and gram quantities would be unknown.