When determining the formula of manganese (IV) oxide, since it’s made up of $\ce{Mn^4+}$ and $\ce{O^2-}$ ions, why is the formula $\ce{MnO2}$ and not $\ce{Mn2O4}$?
Any help would be greatly appreciated
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$\begingroup$ Both are right, but MnO2 is simpler. $\endgroup$– Ivan NeretinJun 17, 2018 at 18:57
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$\begingroup$ Good question, be careful what do you mean is it solid MnO2 or manganese(VI) oxide in the gas form. I have no idea if manganese (VI) oxide can form a gas but I know that chromium(VI) oxide in the gas state is in fact Cr3O9 rather than CrO3 $\endgroup$– Nuclear ChemistJun 17, 2018 at 20:37
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2 Answers
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For ionic compounds there is really no such thing as a "molecule" of the compound. In the solid state the ions exist in a 3D crystalline structure and there is no unique way to associate a particular $\ce{Mn^{4+}}$ ion with a particular pair of $\ce{O^{2-}}$ ions. Therefore the simplest integer ratio is used as the "molecular formula."
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$\begingroup$ Quite important misconception here is assumption of presence of discrete ions in MnO2. It's not a good approximation as it's primarily covalent. $\endgroup$– MithoronJun 18, 2018 at 17:14
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When writing an ionic formula, the simplest ratio is the 'ionic formula.'
Any sample of manganese (IV) oxide will have many, many, many manganese and oxygen ions. The 'criss cross' method of translating the ion charge down to the opposite subscript gives you a starting point. For an ionic substance, you then reduce this ratio to the lowest whole number ratio.
To get a proper formula, the ionic charges have to neutralize, so $\ce{1 Mn^4+}$ ion neutralizes $\ce{2 O^2-}$ ions.
This does not work for molecular formulae, which have a structure based on atoms connected with covalent bonds.
answered Jun 17, 2018 at 19:07