What is the oxidation number of . . .

Problems 11 - 25

Hydrides are compounds of a metal and hydrogen. What are the oxidation numbers 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 examples:

**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 number 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 number 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.

**Problem #16:** HIO_{3}, the iodine has an oxidation number 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 number of +5.

**Problem #17:** S_{2}O_{3}^{2}¯, 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 number of +2.

**Problem #18:** What is the oxidation number 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 O

_{3}, P_{4}, and S_{8}) 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 number of P in P_{2}O_{5}.

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

**Problem #20:** Determine the oxidation number of S in Al_{2}S_{3}.

Aluminum takes on a +3 oxidation number in all common aluminum-containing compounds. Examples include AlCl_{3}and Al_{2}O_{3}.Sulfur can take on several different oxidation numbers. When it is alone and in the nonmetal position of a binary compound, it takes on a −2 oxidation number. An example of this is H

_{2}S.In this compound, Al is a +3 and S is a −2.

This link goes to a Yahoo Answers question about the oxidation number of S in Al

_{2}(SO_{3})_{3}.

**Problem #21:** Determine the oxidation number of Te in Sc_{2}Te_{3}.

The compound H_{2}Te is known to exist, which means that Te has an oxidation number of −2 when acting as a nonmetal.The presence of three Te means a total negative oxidation number 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 number of Si in SiF_{6}^{2}¯.

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 numbers equal to the ______.

(a) difference of protons and electrons

(b) number of protons

(c) number of electrons

(d) difference of protons and neutronsThe answer is (a). Examples include Na

^{+}, Ca^{2+}and F¯.

**Problem #24:** On rare occasions, oxidation numbers do not have to be whole numbers. Determine the oxidation number for S in S_{4}O_{6}^{2}¯.

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

^{10}⁄_{4}. Reduced would be +^{5}⁄_{2}(also seen as +2.5).In truth, the four sulfurs would have a total of +10 oxidation number and each oxidation number 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 numbers.

**Problem #25:** What are the oxidation numbers for each Mn in Mn_{3}O_{4}? For each Fe in Fe_{3}O_{4}?

Four oxygens gives a total of −8 in oxidation number.The three Mn atoms must each have an integer oxidation number. 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 number of −1?

(a) NH_{3}

(b) KH

(c) HCl

(d) H_{2}O

**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 ametaland hydrogen. Nitrogen is a non-metal. CH_{4}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 CaH

_{2}are other examples of hydrides.

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

(a) O_{2}

(b) H_{2}O_{2}

(c) H_{2}O

(d) OF_{2}

(e) KO_{2}

**Solution:**

1) The correct answer is (b).

2) Some discussion:

Peroxide is O_{2}^{2}¯, so each oxygen is considered to have a −1 oxidation number.A possible answer is superoxide, O

_{2}¯. Note that in superoxide, each oxygen is considered to have an oxidation number of −^{1}⁄_{2}as opposed to one oxygen having a −1 and one oxygen having an oxidation number of zero.Superoxide is a good example of oxidation numbers simply being bookkeepping. The two oxygens in superoxide do not literally each have a charge of −

^{1}⁄_{2}, yet we are forced to use −^{1}⁄_{2}if we ask the oxidation number oneachoxygen.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 OF

_{2}is +2. Fluorine is stronger than oxygen in attracting electrons to itself and so forces oxygen into playing the positive role in oxidation numbers. The next halogen down (chlorine) is weaker than oxygen in attracting electrons, so its formula is Cl_{2}O. Oxygen has an oxidation number of −1 and chlorine is pushed into having an oxidation number of +1.

**Problem #28:** Superoxide is not commonly mentioned in introductory chemistry classes. It is the polyatomic ion O_{2}¯. KO_{2} is the most well-known of the superoxides. What are the oxidation numbers on each oxygen in the superoxide anion?

**Solution:**

It appears that, within the superoxide, each oxygen must have an oxidation number of −^{1}⁄_{2}.The ChemTeam thinks, but is not 100% sure, that the electronic structure of the superoxide anion is such that one O has an oxidation number of zero and the other O has an oxidation number of −1. The fact that there are, then, two resonance structures of superoxide leads to an average of −

^{1}⁄_{2}for each oxygen atom.

**Problem #29:** Some chemical compounds have one (or more) atoms in them where the oxidation number is zero. An example of this is glucose, C_{6}H_{12}O_{6}.

**Solution:**

The six oxygens (each at an oxidation number of −2) total up to −12.The twelve hydrogens (each at an oxidation number of +1) total up to +12.

+12 and −12 add to zero.

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

**Problem #30:** This link takes you to a question on Yahoo Answers:

What is the oxidation number of Cl in HClO_{4}?