What is the electron configuration for #Fr#?

1 Answer
Aug 9, 2016

Answer:

#[Rn]7s^1#

Explanation:

From the Periodic Table, #Z#, the atomic number of #Fr=87#

And thus,

#1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)4s^(2)4p^(6)4d^(10)5s^(2)5p^(6)4f^(14)5d^(10)6s^(2)6p^(6)7s^1 #

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