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

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Answer 1 · 2018-06-24T02:08:34

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

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

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

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