Question

Why does `"Zn"^(2+)` have `18` electrons in its outer shell? Does it have to follow the octet rule?

Answer 1

It doesn't.

Explanation:

Generally, the octet rule spies to elements in groups 13 through 18, which react with jyst `s` and `p` electrons. And it does not apply, at least not in straightforward fashion, for the heavier elements even in these groups.

Answer 2

The valence shell of main-group elements is typically composed of the outermost (highest energy) `"s"` and `"p"` sublevels, which hold a total of `8` valence electrons. An octet refers to having `8` electrons in the valence shell, but transition metals are another story.

Explanation:

The valence shell of a main-group element is typically made up of the outermost (highest energy) `"s"` and `"p"` sublevels.

The ground state electron configuration of a neutral `"Zn"` atom is:

`"1s"^2"2s"^2"2p"^6"3s"^2"3p"^6"4s"^2"3d"^10`

The electron configuration of a `"Zn"^(2+)"` cation is:

`"1s"^2"2s"^2"2p"^6"3s"^2"3p"^6"3d"^10`

The removal of the `"4s"^2"` electrons leaves `18` electrons in the 3rd energy level, which should not be surprising. The `"3d"` orbitals allow for violation of the octet "rule", as in `"PF"_5`, `"SF"_6`, etc.