[djdato]

Hi. We are currently learning about the periodic table, to be more exact, the transition metals. I know that Manganese has a wide range of oxidation numbers. But why wouldn't we expect an oxidation number of 8+ ?

Thanks 

[AWK]

http://en.wikipedia.org/wiki/Valence_electron

[Quantum07]

An easy way to know the highest oxidation state for a transition metal is to look at it's GROUP NUMBER.

Manganese... Mn is in group VIIB (7B) so that will be its highest ox. numer possible meaning it will range from 1 up to 7. This is just a general rule for metals IIIB to VIIB (Sc to Mn). For Iron obviously it's ox. states are 2+ and 3+, Copper is always +1 or +2, and Zinc is always +2.

Manganese has electron configuration of

[Ar] 4s² 3d⁵. The [Ar] is the kernel which contains all the electrons up until the last 7. The valence for Mn is 7 meaning there are only 7 electrons that can be lost from Mn not 8. The first 2 electrons will be lost from the 4s shell and the following 5 would be lost from the 3d shell.
Hope this helped.

[rkaminski]

That's ok. what you said before me Quantum07 but the simplest answer is that the Mn on this oxidation state is simply so unstable that I won't exist:) It's because when you try to get rid of the electron from the so called "atomic core" it requires too much energy. Theoretically you can even calculate the electronic structure of Mn(8+) but in reality it don't exist, even when you try to stabilize it with some ligands. In my opinion thinking of "atomic cores" and "valence shells" it's better that the oxidation states. It someone is interested below it's an article which is introduction to such approach.

http://www.ch.pw.edu.pl/~gorski/PJChem-2005-79-1435.pdf