Given the neutral oxide, $M n_{3} O_{4}$ $Mn_3O_4$ ...what are the probable oxidation states of the manganese centres given that the metal expresses ONLY two oxidation states in this species?

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Given the neutral oxide, $M n_{3} O_{4}$ $Mn_3O_4$ ...what are the probable oxidation states of the manganese centres given that the metal expresses ONLY two oxidation states in this species?

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Answer 1 · 2018-01-05T09:02:53

The sum of the oxidation numbers equals the charge on the ion...

Explanation:

Of course, here, the parent species is NEUTRAL, and therefore the sum of the oxidation numbers EQUALS ZERO.

The oxidation number of oxygen, as oxide, is typically $- I I$ $-II$ ...and so $3 \times M n_{oxidation number} - 8 = 0$ $3xxMn_"oxidation number"-8=0$ ... $M n_{average oxidation number} = + \frac{8}{3}$ $Mn_"average oxidation number"=+8/3$ ...and so we gots $2 \times M n^{+ I I I} + M n^{+ I I}$ $2xxMn^(+III)+Mn^(+II)$ ...

The solid state structure does correspond to a formulation of $M n O \cdot M n_{2} O_{3}$ $MnO*Mn_2O_3$ ...the which does reflect the oxidation states we assigned. Capisce?

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Given the neutral oxide, $M n_{3} O_{4}$ $Mn_3O_4$ ...what are the probable oxidation states of the manganese centres given that the metal expresses ONLY two oxidation states in this species?

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Explanation:

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Science

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Humanities

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Given the neutral oxide, Mn3O4Mn_3O_4...what are the probable oxidation states of the manganese centres given that the metal expresses ONLY two oxidation states in this species?

1 Answer

Explanation:

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Impact of this question

Given the neutral oxide, $M n_{3} O_{4}$ $Mn_3O_4$ ...what are the probable oxidation states of the manganese centres given that the metal expresses ONLY two oxidation states in this species?