How do you find the oxidation numbers of the elements in $M n O 4^{2 -}$ $MnO4^(2-)$ ?

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How do you find the oxidation numbers of the elements in $M n O 4^{2 -}$ $MnO4^(2-)$ ?

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Answer 1 · 2017-02-20T18:05:12

Well, it has a $V I +$ $VI+$ oxidation state if it is actually $M n O_{4}^{2 -}$ $MnO_4^(2-)$ .

Explanation:

$Manganate ion is$ $"Manganate ion is"$ $M n O_{4}^{2 -}$ $MnO_4^(2-)$ , $M n (V I +)$ $Mn(VI+)$ .

$Permanganate ion is$ $"Permanganate ion is"$ $M n O_{4}^{-}$ $MnO_4^(-)$ , $M n (V I I +)$ $Mn(VII+)$ .

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 $- I I$ $-II$ in its compounds, and it is here in both these examples.

Thus for $M n O_{4}^{2 -}$ $MnO_4^(2-)$ , $4 \times (- 2) + M n_{oxidation number} = - 2$ $4xx(-2)+Mn_"oxidation number"=-2$ .

And thus for $permanganate$ $"permanganate"$ , $M n_{oxidation number} + 4 \times (- 2) = - 1$ $Mn_"oxidation number"+4xx(-2)=-1$

I leave it up to you to solve for $M n_{oxidation number}$ $Mn_"oxidation number"$ in each instance.

Both are known, and both are accessible oxidation states.

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How do you find the oxidation numbers of the elements in $M n O 4^{2 -}$ $MnO4^(2-)$ ?

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