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According to the book for a binary compound, first we assign the element with greater electronegativity its oxidation number (oxygen always -2 except in peroxides). So, we have -2 x2 =-2. So, each oxygen would have an oxidation number of -2. The book says the answer is -1.

The only thing I can think of, is since barium is an alkaline metal, it has an oxidation number of +2, so each oxygen would be -1. But, it contradicts itself because in the rules. It says for a binary compound, "the element with greater electronegativity is assigned a negative oxidation number equal to its charge in simple ionic compounds of the element."

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    $\begingroup$ Did you look on the web for barium dioxide to find information about the bonding? $\endgroup$
    – LDC3
    Nov 9, 2014 at 17:02
  • $\begingroup$ In compounds, barium is universally encountered as Ba(II). I'm unaware of any compounds where barium has any other oxidation state. $\endgroup$
    – Abel Friedman
    Nov 9, 2014 at 20:25

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"oxygen always -2 except in peroxides"

This exception is in effect here, because $\ce{BaO2}$ is a peroxide consisting of $\ce{Ba^{2+}}$ and $\ce{O2^{2-}}$ ions. The oxidation number of Ba is +II, and the oxidation number of each of the oxygens in the peroxide anion is -I. This fits with the charge of the peroxide anion ($2 \times -1 = -2$), and as $\ce{BaO2}$ is a neutral compound, the sum of all oxidation numbers is 0. Moreover, there are even more exceptions to the rule of thumb cited above, for example, the superoxide radical anion $\ce{O2-}$ with a fractional oxidation numer of $-\frac{1}{2}$, or the dioxygenyl cation $\ce{O2+}$ with a formal oxidation number of $+\frac{1}{2}$ for each oxygen.

In contrast to peroxides, "simple ionic compounds" of oxygen are oxides, which contain the anion $\ce{O^{2-}}$. The charge of the oxide anion is $2-$, so the oxidation number of oxygen in those compounds is -II.

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  • $\begingroup$ I think we also have to account for the negative half oxidation number in case of super oxides in the general law at the head of your answer. $\endgroup$
    – stochastic13
    Nov 11, 2014 at 3:56
  • $\begingroup$ @SatwikPasani I have edited my answer to take this aspect into account. $\endgroup$
    – Jannis Andreska
    Nov 13, 2014 at 18:03
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Oxidation numbers are an extremely useful form of chemical accounting, but should not be considered as having any actual reality. For instance fractional oxidation numbers are perfectly possible, but you cannot have half an electron. Anoxidation number of +7 is assigned to chlorine in perchlorates, but the idea that you could possibly strip 7 electrons off a chlorine atom is unthinkable (add up the first 7 ionisation potentials). It is the second part of the quote that you give is misleading and in many cases incorrect. The point to be born in mind first is that the oxidation number of a given element in a compound is the charge it would have IF the compound was entirely ionic. The sign of the charge is based on relative electronegativity of the bonding atoms. Group I and II metals (except in very rare cases) are assigned charges of +1 and +2 respectively. Oxygen is most commonly -2, but in the case of peroxides it is assigned an oxidation number of -1 in order that the combined oxidation numbers come to zero in this case or to the total charge on the group of atoms if they happened to form an ion.

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