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Chapter 3: Problem 55

In many of its compounds, oxygen has an oxidation state of \(-2 .\) However, there are exceptions. What is the oxidation state of oxygen in each of the following compounds? (a) \(\mathrm{OF}_{2} ;\) (b) \(\mathrm{O}_{2} \mathrm{F}_{2} ;\) (c) \(\mathrm{CsO}_{2}\); (d) \(\mathrm{BaO}_{2}\)

Short Answer

Expert verified
The oxidation states of oxygen in the given compounds are: (a) +2 in \(OF_2\), (b) +1 in \(O_2F_2\), (c) -0.5 in \(CsO_2\), and (d) -1 in \(BaO_2\).

Step by step solution

01

Oxidation state in \(OF_2\)

For \(OF_2\), the oxidation state of fluorine (F) is always -1. Since the overall charge of the neutral molecule must be zero, the sum of the oxidation number of oxygen and fluorine must equal zero. Therefore, the oxidation number of oxygen is +2.
02

Oxidation state in \(O_2F_2\)

For \(O_2F_2\), the oxidation state of two fluorine atoms is -2 since the oxidation state of each fluorine atom is -1. To balance the charge, two oxygen atoms should have an oxidation number of +2, hence each oxygen atom will have an oxidation number (+2/2) of +1.
03

Oxidation state in \(CsO_2\)

For \(CsO_2\), the oxidation state of cesium (Cs) is always +1. Then since we have two oxygen atoms and the overall charge of the molecule is zero, the sum of the oxidation states of two oxygen atoms should be -1. Hence, the oxidation state of each oxygen atom is -1/2, or -0.5.
04

Oxidation state in \(BaO_2\)

For \(BaO_2\), the oxidation state of barium (Ba) is always +2. Because it is a neutral compound, the sum of the oxidation states of the two oxygen atoms should equal -2. Hence, the oxidation state of each oxygen atom is -1.

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