# How to find the highest and lowest oxidation state of transitional metal

My teacher taught me the formulas 'nS' and 'nS+(n-1)d' to find the highest and lowest oxidation state of transition elements. He gave the example of $\ce{Mn}$ and showed that the numbers were +2 and +7. But I did not understand it. So can please you give me a good explanation.

• Welcome to chemistry.se! Please have a look at the help center and take the tour. Could you explain, what the variables in your formula stand for, please. – Martin - マーチン May 9 '15 at 11:41
• @Martin Although OP should confirm, the "n" stands for the principal quantum number as "s" and "d" refer to the s and d orbitals. – Binary Geek May 9 '15 at 12:21
• Electron configuration [Ar] 3d5 4s2 - could you be more specific about your lack of understanding? ;) – Mithoron May 9 '15 at 12:51
• Well … what about $\ce{Mn2(CO)10}$ and $\ce{[Mn(CO)5]-}$? – Jan Jun 9 '15 at 21:53