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According to the rules for naming ionic compounds, Fe2O5 would be referred to as Iron (V) oxide.

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11y ago
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amin abazarsa

Lvl 1
2y ago
Fe2O5 is iron(II,III) oxide if the ratio of iron(II)/iron(III) were 1:1.
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amin abazarsa

Lvl 1
2y ago
Fe2O5 has the iron(II)/iron(III) ratio of 1:1 which is incorrect. It has to be 1:2. So the only true formula for iron(II,III) oxide is Fe3O4.
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amin abazarsa

Lvl 1
2y ago
Also, Iron(V) oxide doesn't exist (at least in standard conditions).
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amin abazarsa

Lvl 2
2y ago

It's not a correct formula and I tell you why.

We know that iron(II) oxide has the formula FeO and iron(III) oxide has the formula Fe2O3. While iron(IV) exists it isn't a dominant cation; so iron(IV) oxide doesn't exist in standard conditions.

Iron(II,III) oxide has the formula Fe3O4. Some people say it is a combination of iron(II) oxide and iron(III) oxide because FeO + Fe2O3 = Fe3O4. But this assumption isn't correct. Because Fe3O4 is an independent molecule consisting of both iron(II) and iron(III) ions. So the true formula is Fe(Fe2O4), with the first Fe being iron(II) and the second Fe being iron(III). So the ratio of iron(II)/iron(III) in this compound is 1:2 and this ratio never changes.

Note: If we want to find the oxidation state of iron in Fe3O4 without considering the iron(II) and iron(III) we obtain 8/3 as our answer which is approximately 2.66 (so it is between 2 and 3).

If we assume the formula Fe2O5, we are assuming the said ratio to be 1:1 which is incorrect.

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Q: Is Fe2O5 a correct formula of compound?
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