How do you find the oxidation numbers of the elements in #MnO4^(2-)#?

1 Answer
Feb 20, 2017

Answer:

Well, it has a #VI+# oxidation state if it is actually #MnO_4^(2-)#.

Explanation:

#"Manganate ion is"# #MnO_4^(2-)#, #Mn(VI+)#.

#"Permanganate ion is"# #MnO_4^(-)#, #Mn(VII+)#.

In both instances, the sum of the oxidation numbers of the individual elements must equal the charge on the ion. The oxidation number of oxygen is usually #-II# in its compounds, and it is here in both these examples.

Thus for #MnO_4^(2-)#, #4xx(-2)+Mn_"oxidation number"=-2#.

And thus for #"permanganate"#, #Mn_"oxidation number"+4xx(-2)=-1#

I leave it up to you to solve for #Mn_"oxidation number"# in each instance.

Both are known, and both are accessible oxidation states.