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

2 Answers
Dec 18, 2017

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.

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.