Explain why rubidium tends only to form a +1 ion

Rubidium is a Group 1 metal with the following electron configuration.  

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1  

Like all other Group 1 (or alkali) metals, it has 1 valence electron. Atoms tend to be most stable when they have 8 valence electrons (because doing so allows them to have a completely full s and p sublevel in their valence shell). Could rubidium gain 7 electrons to boost its valence electron count up to 8? Not really; rubidium's valence shell is far enough away from its nucleus that the atom simply doesn't have that kind of pull on other electrons. The other option is for rubidium to drop the one valence electron it has. Doing so would give it the following electron configuration:  

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6  

Simply by dropping an electron rubidium can have 8 valence electrons (because the 4th energy level becomes the valence shell once the 5s electron is gone. This makes rubidium much more stable, but it also causes the atom to become a positive ion (also called a cation). Rubidium still has 37 protons in its nucleus but now it only has 36 electrons. The end result is that a rubidium ion's charge is +1.  

All of the atoms in group 1 have one valence electron, and in every case it is more energetically feasible for those atoms to lose their single valence electron than it would be to gain seven more; therefore, all elements in group 1 behave in similar ways. The same is true for elements in any other group; all of the elements in a group have the same number of valence electrons and therefore similar chemical properties.  

I hope that helps. Good luck!

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