### Redox RulesWhat is the oxidation state of . . .Problems 11 - 30

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: CaH2

Ca = +2. Each H is −1.

Problem #13: SH6

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 O22¯ present. Two problems:

Problem #14: H2O2

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: CaO2

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: PbO2 is an example of possible confusion with identifying peroxides. Is the Pb = +4? In which case, the O2 portion refers to two oxides, each of which is −2. Or, is the Pb = +2? If that is the case, then the O2 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.

Problem #16: HIO3, the iodine has an oxidation state of:

The three oxygens total to −6 and the hydrogen adds a +1, leaving −5 to be offset by the iodine. The iodine has an oxidation state of +5.

Problem #17: S2O32¯, thiosulfate:

Three oxygens, each of which is −2, total to −6. Since −2 is left over, a total of −4 must be offset by the two sulfurs. Each sulfur has an oxidation state of +2.

Problem #18: What is the oxidation state of all free elements? Another way to ask this is to say that the elements are in their uncombined state.

Zero.

By the way, molecular elements (the seven diatomics plus O3, P4, and S8) are considered to be uncombined. Elements are only considered combined when they are bonded to at least one other (different) element.

Problem #19: Determine the oxidation state of P in P2O5.

The five oxygens total to −10. Each P is +5.

Problem #20: Determine the oxidation state of S in Al2S3.

Aluminum takes on a +3 oxidation state in all common aluminum-containing compounds. Examples include AlCl3 and Al2O3.

Sulfur can take on several different oxidation states. When it is alone and in the nonmetal position of a binary compound, it takes on a −2 oxidation state. An example of this is Na2S.

In this compound, there are two Al, for a total of +6. The three S must, therefore, total to −6. That means each S is a −2.

Problem #21: Determine the oxidation state of Te in Sc2Te3.

The compound H2Te is known to exist, which means that Te has an oxidation state of −2 when acting as a nonmetal.

The presence of three Te means a total negative oxidation state of −6.

To offset this (and make the formula have a charge of zero), each Sc must be a +3.

Problem #22: Determine the oxidation state of Si in SiF62¯.

The F is a −1. We know this from the existence of HF. Six of the yields a total of −6. In order to have −2 left over, the Si must be a +4

Problem #23: All monatomic ions have oxidation states equal to the ______.

(a) difference of protons and electrons
(b) number of protons
(c) number of electrons
(d) difference of protons and neutrons

The answer is (a). Examples include Na+, Ca2+ and F¯.

Problem #24: On rare occasions, oxidation states do not have to be whole numbers. Determine the oxidation state for S in S4O62¯.

There are six O, which contribute a total of −12 in oxidation state. Since −2 is left over, that leaves −10 to be offset by the four sulfurs.

Each sulfur is, therefore, +104. Reduced would be +52 (also seen as +2.5).

In truth, the four sulfurs would have a total of +10 oxidation state and each oxidation state on the four atoms would be a whole value. For example, if three S were each +2 and one S was a +4, this would total up to +10. Chemical substances do not have fractional oxidation states.

Problem #25: What are the oxidation states for each Mn in Mn3O4? For each Fe in Fe3O4?

Four oxygens gives a total of −8 in oxidation state.

The three Mn atoms must each have an integer oxidation state. This is satisfied if two Mn are each a +3 and the third Mn is a +2.

For iron (II, III) oxide, the answer is the same: two iron atoms are each a +3 and one Fe atom is a +2.

Problem #26: In which compound does hydrogen have an oxidation state of −1?

(a) NH3
(b) KH
(c) HCl
(d) H2O

Solution:

1) The correct answer is (b).

2) Some discussion:

You might get led into a wrong conclusion by ammonia, since the hydrogen comes last in the formula. However, coming last in the formula is a historical accident and must not be taken to indicate that hydrogen is a −1 in ammonia. Remember that hydrides are composed of a metal and hydrogen. Nitrogen is a non-metal. CH4 is an example of another compound that is not a hydride, even though hydrogen is written last.

KH is the hydride, a chemical compound of a metal and hydrogen. NaH and CaH2 are other examples of hydrides.

Problem #27: In which compound does oxygen have an oxidation state of −1?

(a) O2
(b) H2O2
(c) H2O
(d) OF2
(e) KO2

Solution:

1) The correct answer is (b).

2) Some discussion:

Peroxide is O22¯, so each oxygen is considered to have a −1 oxidation state.

A possible answer is superoxide, O2¯. Note that in superoxide, each oxygen is considered to have an oxidation state of −12 as opposed to one oxygen having a −1 and one oxygen having an oxidation state of zero.

Superoxide is a good example of oxidation states simply being bookkeepping. The two oxygens in superoxide do not literally each have a charge of −12, yet we are forced to use −12 if we ask the oxidation state on each oxygen.

More correctly, the two oxygen entity that is superoxide has a −1 charge that is distributed over the entire unit of two oxygens and is not the sole property of one or the other oxygen atoms. see below for a bit more discussion.

By the way, the charge on the oxygen in OF2 is +2. Fluorine is stronger than oxygen in attracting electrons to itself and so forces oxygen into playing the positive role in oxidation states. The next halogen down (chlorine) is weaker than oxygen in attracting electrons, so its formula is Cl2O. Oxygen has an oxidation state of −1 and chlorine is pushed into having an oxidation state of +1.

Problem #28: Superoxide is not commonly mentioned in introductory chemistry classes. It is the polyatomic ion O2¯. KO2 is the most well-known of the superoxides. What are the oxidation states on each oxygen in the superoxide anion?

Solution:

It appears that, within the superoxide, each oxygen must have an oxidation state of −12.

The ChemTeam thinks, but is not 100% sure, that the electronic structure of the superoxide anion is such that one O has an oxidation state of zero and the other O has an oxidation state of −1. The fact that there are, then, two resonance structures of superoxide leads to an average of −12 for each oxygen atom.

Problem #29: Some chemical compounds have one (or more) atoms in them where the oxidation state is zero. An example of this is glucose, C6H12O6.

Solution:

The six oxygens (each at an oxidation state of −2) total up to −12.

The twelve hydrogens (each at an oxidation state of +1) total up to +12.

+12 and −12 add to zero.

Therefore, the six carbons are zero for their total oxidation state contribution.

The reality is a bit more complicated.

Problem #30: What is the oxidation state of P in HPO42¯?

Solution:

H contributes +1.

Four O contribute a total of −8.

−2 must be left over, so P must be +5

Bonus Problem: What is the oxidation state and coordination number of rhodium in the coordination compound K[RhCl(OH)(C2O4)2] 6H2O?

Solution:

The coordination number (the number of coordinate covalent bonds between the ligands and the Rh ion) is 6. There are two bonds to each oxalate ion and one to OH¯ and one to the Cl¯ ions. (This isn't part of the redox rules material, so I won't delve into it further.)

The complex ion (everything inside the square brackets) has a charge of −1. We know this from the fact that it bonds with one potassium, which always has an oxidation state of +1.

Since the sum of oxidation states must equal the charge on the ion:

Rh + Cl¯ + OH¯ + 2C2O42¯ = −1

Rh + (−1) + (−1) + (−4) = −1

Rh = +5