Question

Please,why is MnO2-4 i.e manganese(vi)oxide not MnO3?...i still dont get it..i mean if u calculate the oxidation number with the latter it'll still be +6

Answer 1

The formula for manganese(VI) oxide is `MnO_3`

Explanation:

I'm not sure if I understand the question completely, but I'll verify the oxidation number and formula for `MnO_2` and `MnO_3`

We know that the oxidation number for O is -2
Let's set the oxidation number for Mn as x

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Manganese(VI) oxide is indeed `MnO_3`

`Mn^"6+"` `O^"2-"` => `Mn_2O_6` => `MnO_3`

Oxidation number for Mn in `MnO_3`: x + 3(-2) = 0 => x = +6

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Manganese(IV) oxide, on the other hand is `MnO_2`

`Mn^"4+"` `O^"2-"` => `Mn_2O_4` => `MnO_2`

Oxidation number for Mn in `MnO_2`: x + 2(-2) = 0 => x = +4