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As I know, $\ce{MnO2}$ is called manganese(IV) oxide.

Why can't we name it manganese peroxide since manganese(II) exists too?

In other words, how do we know that the oxidation number of manganese in $\ce{MnO2}$ is $4$ and not $2$?

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Peroxides contain two oxygens connected by a single bond. X-ray or neutron diffraction will show that the oxygens in $\ce{MnO2}$ are too far apart to be bonded, and therefore it is not a peroxide.

Peroxide is well known as a ligand to transition metals (though whether it is better described as peroxide or superoxide is sometimes not always clear, e.g. in some cobalt species). An example which you might come across in the lab is chromium(VI) oxide peroxide, $\ce{CrO5}$.

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The question’s logical premise is skewed. $\ce{CO2}$ is carbon(IV) oxide or carbon dioxide. But carbon(II) also exists and so does carbon(II) oxide (carbon monoxide, $\ce{CO}$). Just because there are twice as many oxygens in a compound does not mean that it can be a peroxide with a lower oxidation state of the central atom.

In fact, peroxides are decidedly rare. There is $\ce{H2O2}$, $\ce{Na2O2}$, $\ce{[Cr(O)(O2)2]}$ (the chromium butterfly), $\ce{mCPBA}$, $\ce{tBuOOX}$, $\ce{(PhCOO)2}$, artemisinin and I’m already at a loss. Compare that to the vast number of oxides out there and you get what I’m getting at. This is because the $\ce{O-O}$ bond in peroxides is very energetic, and they readily oxidise or reduce other compounds for oxygen to achieve its more stable oxidation states $\mathrm{-II}$ or $\pm 0$ (the former typically more common and stable than the latter).

Thus, peroxides can only form if electrons are there to reduce atmospheric oxygen, but not enough to reduce it to an oxide (or water or an alcohol group etc.). And especially in the case of metals with many available oxidation states, it is very unlikely that a peroxide should form for a given one — especially if the metal is not fully oxidised (remember that the highest common oxidation state of manganese is $\mathrm{+VII}$). If a peroxide accidentally did form, it will typically be a good oxidising or reducing agent. But $\ce{MnO2}$ is neither, it is the thermodynamic hole of ionic manganese compounds. Glow a manganese-containing salt in the flame of a bunsen burner, and given enough time you will arrive at $\ce{MnO2}$. These conditions of formation show us that it is very, very, very, very unlikely for $\ce{MnO2}$ to be a peroxide. And it also hardly reacts to any other compounds, further showing that it is most likely not one.

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The reason is oxygen bonding in the molecules. Actually peroxide means oxygen - oxygen should be in single bond but in MnO2 oxygens are double bonded with mn there is no bond between oxygens. For example in case of H2O2 it has oxygen - oxygen single bond so it is named as hydrogen peroxide but in MnO2, both the oxygens are bonded with Mn there is no bond between oxygens.

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    $\begingroup$ $\ce{MnO2}$ is an ionic lattice with several different polymorphs. Describing the oxygens as being double bonded to the manganese is not really true. $\endgroup$ – bon Jul 2 '16 at 17:36
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Peroxides are compounds in which the oxidation state of Oxygen atoms is -1. Since you know that MnO2 is Mn(IV) oxide, which means Oxygen atoms are in -2 oxidation state. Hence it is not called a peroxide.

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    $\begingroup$ Circular logic... $\endgroup$ – orthocresol Jul 2 '16 at 16:10
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    $\begingroup$ Circular reasoning works because circular reasoning works and because circular reasoning works! (CC @ortho ;)) $\endgroup$ – Jan Jul 2 '16 at 18:09
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    $\begingroup$ Don't answer when you don't know clearly about it but please don't apply circular logic just for the sake of answering something. $\endgroup$ – user5764 Jul 3 '16 at 3:07
  • $\begingroup$ not so circular, as we have methods to determine oxidation of Mn by spectroscopy $\endgroup$ – MolbOrg Jul 3 '16 at 3:45

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