What is the oxidation state of . . .

Problems 1 - 15

Exampes and Problems with no answers

**Problem #1:** What is the oxidation state of carbon in calcium oxalate?

The oxidation state of carbon in the oxalate ion is +3, based on two things: 1. the oxidation state of oxygen is usually -2 2. the sum of the oxidation states for a compound is zero, and for an ion, the sum is the charge on the ion.

**Problem #2:** C in CO_{3}^{2}¯

The O is −2 and three of them makes −6. Since −2 is left over, the C must be +4

**Problem #3:** Cr in CrO_{4}^{2}¯

The O is −2 and four of them makes −8. Since −2 must be left over, the Cr must be +6

**Problem #4:** Cr in Cr_{2}O_{7}^{2}¯

The O is −2 and seven of them makes −14. Since −2 is left over, the two Cr must be +12.What follows is an important point. Each Cr is +6. Each one. We do not speak of Cr

_{2}being +12. Each Cr atom is considered individually.

**Problem #5:** Fe in Fe_{2}O_{3}

The O is −2 and three of them makes −6. Each Fe must then be +3

**Problem #6:** Pb in PbOH^{+}

The O is −2 and the H is +1. In order to have a +2 for the formula, the Pb must be a +2.

**Problem #7:** V in VO_{2}^{+}

Two O gives a total of −4. To make the formula have a +1 charge, the V must be +5

**Problem #8:** V in VO^{2+}

The V is +4. This comes from the one O being −2 and the fact that a +2 must be present on the ion.

**Problem #9:** Mn in MnO_{4}¯

Four O totals to −8. The Mn is +7, leaving −1 left over.

**Problem #10:** Mn in MnO_{4}^{2}¯

Four O totals to −8. The Mn is +6, leaving −2 left over.

Hydrides are compounds of a metal and hydrogen. What are the oxidation states of the elements present in each formula given in 11-13?

**Problem #11:** NaH

We know from other compounds like NaCl that Na is a +1. The H is, therefore, a −1.

**Problem #12:** CaH_{2}

Ca = +2. Each H is −1.

**Problem #13:** SH_{6}

As of July 2017, this hydride has been predicted but not yet been discovered. The S is a +6 and each of the six hydrides is a −1.

Peroxides have the polyatomic ion O_{2}^{2}¯ present. Two problems:

**Problem #14:** H_{2}O_{2}

Each hydrogen is a +1, the peroxide is −2. That means that each oxygen in the peroxide is a −1. However, be careful. If you make reference to the −1 value, you must make it clear that you are referring to the oxygens in a peroxide. If you simply refer to an O without any context, the assumption is that you are referring to an oxide, the oxidation state of which is −2.

**Problem #15:** CaO_{2}

Calcium is +2, the peroxide is −2, when discussed as the group. If the question were to ask the oxidation state of each oxygen in a peroxide, the answer would be −1.Comment: PbO

_{2}is an example of possible confusion with identifying peroxides. Is the Pb = +4? In which case, the O_{2}portion refers to two oxides, each of which is −2. Or, is the Pb = +2? If that is the case, then the O_{2}portion is a peroxide and has a total charge of −2, each O being −1 in the peroxide.How do you tell which one is meant? You would do so by the context in which the formula is presented. if context is lacking, ask your teacher for help. If it's an online course and live help is hard (or impossible) to get, pick the Pb = +4 answer. If it's wrong, give the other one.

**Bonus Problem:** What is the oxidation state of Re in potassium nonahydridorhenate, which has the formula K_{2}ReH_{9}

We know that potassium always takes on a +1 oxidation state. Eliminate the potassium to get this:ReH_{9}^{2}¯That looks very much like a hydride, when hydrogen takes on a −1 oxidation state.

The fact that there are nine of them means Re must have an oxidation state of +7.

Another compound with Re = +7 is NaReO

_{4}.