Indicate the oxidation number of oxygen for each compound in the following reaction: 2h2o2(aq)→ 2h2o(l) o2(g) select the choice that gives the oxidation numbers of h2o2, h2o(l), and o2(g), respectively.

select the choice that gives the oxidation numbers of , , and , respectively. −1, −2, 0 −1, 0, −2 0, −2, −1 −2, −1, 0

2H2O2(aq)→ 2H2O(l) O2(g) : The oxidation number of oxygen for each compound is -1, -2, 0

Explanation:

In peroxides the oxidation state of oxygen is -1, since one oxygen bonds to the other oxygen and a hydrogen and the bound oxygen captures the electron of the remaining hydrogen. Through a scheme would be

H --- O --- O --- H

We remember that oxygen needs two electrons to get to have the configuration of the nearest noble gas (Lewis octet rule). In Peroxides, the oxygen is linked by covalent bonds. If we take it strictly, peroxide is a grouping of two oxygen, having the whole valence -2. which is why it is usually said that it is when oxygen has a valence -1

As we said the oxidation state is -2, the one that appears in the water molecule, since Hydrogen acts with valence +1 and it is 2 atoms that give up electrons to compensate for oxygen.

In the O2 it acts with valence 0 since we talk about gas in its elementary state. All diatomic molecules in their elemental state, generally gases or metals in solid state, act with a valence of 0.

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