Question

What is the oxidation number of the metal in `"manganate ion"`, `MnO_4^(2-)`?

Answer 1

This is the manganate ion, where manganese has a `VI+` oxidation state.

Explanation:

The oxdiation number of manganese in the anion `MnO_4^(2-)` is clearly `VI+`, utilizing standard oxidation numbers of `-II` for oxygen. What is the oxidation state of the metal in permanganate ion, `MnO_4^-`?

Answer 2

`Mn^"+6"`

Explanation:

The first thing you need to realize is there are two bonds in this formula: the ionic bond between `NH_4^+` and the `MnO_4^"2-"` and the covalent bond in the `MnO_4^"2-"` ion. You can only do this by knowing at first look which one is the cation and which is the anion. As the cations are only a handful, it would be good if you can memorize them.

With this kind of substances, it would be helpful if you separate the ions first since it would break down your equation into a more simpler one.

Hence,
`(NH_4)_2MnO_4` `rarr` `2NH_4^+` + `MnO_4^"2-"`

Now that you separated the ions, you can safely ignore the ammonium ion for the time being, as it would not affect your calculations for the oxidation state of `Mn` atom.

Thus,

`MnO_4^"2-"` = -2 (oxidation state based on the charge)

`O` atoms almost always have a charge of -2 (with one possible exception in peroxide ions).

Thus we have the equation:

`color (red) x` + [(4) (-2)] = - 2 where `color (red) x` is the oxidation state on `Mn` atom

`color (red) x` + (-8) = - 2

`color (red) x` = - 2 + (+8)

`color (red) x` = +6

Therefore, the oxidation state of `Mn` atom is +6.

Still in doubt? Let's try solving `x`, this time, including the `NH_4^+` ion.

`(NH_4)_2MnO_4` = 0 since the substance is neutrally charge

[(+1) (2)] + `color (red) x` + [(4) (-2)] = 0

(+2) + `color (red) x` + (-8) = 0

`color (red) x` + (-6) = 0

`color (red) x` = +6

Same answer. :)