Volume 2

Oxidation State • Exercises
2.9.1. Oxidation State. Redox reactions are paramount in all areas of chemistry and biology. A large number of vital processes in living cells involve redox reactions, including those that govern the respiratory systems. Without redox chemical transformations, there would have been no electronics, cell phones, computers, steel, aluminum, plastics, and countless other products that we use on a daily basis.

The most basic information needed to use a chemical reaction is its stoichiometry. We recall from Volume 1 that stoichiometry refers to the ratio of reagents and products in a chemical reaction and is described by the coefficients of a balanced chemical equation for that reaction. As an example, the stoichiometry of the reaction 2 H2 + O2 = 2 H2O is 2 to 1 (reagents) to 2 (product), meaning that two molecules of hydrogen react with one molecule of oxygen to produce two molecules of water. By the way, the word "stoichiometry" originates from two Ancient Greek words, στοιχεῖον (stoicheion meaning "element") and μέτρον (metron meaning "measure").

We are talking about reaction stoichiometry because balancing redox reactions can be much trickier than most non-redox reactions, such as neutralization or ion exchange. This warning might surprise you since the balancing of the previously considered redox reactions of sodium with chlorine (Figure 2-100) and calcium with oxygen (Figure 2-101) was as easy as one-two-three. Now look at the coefficients in the redox equation presented in Figure 2-103.
Figure 2-103. A prodigious redox reaction.

The chemical equation shown in Figure 2-103 is where the complexity of the balancing of redox equations is taken to an extreme. In fact, this reaction has never been observed and was made up as a brain teaser. Nevertheless, balancing many actual redox reactions can be challenging.

To balance a redox equation, we need to know how each element involved changes its oxidation state during the reaction. The oxidation state, also known as the oxidation number of an atom, is the hypothetical net electric charge (positive, negative or zero) that an atom in a molecule would acquire if all of the covalent bonds to that atom were heterolytically cleaved by fully shifting the shared electron pairs to the more electronegative atoms. Figure 2-104 shows how the oxidation states are determined for the H and Cl atoms of HCl.
Figure 2-104. Stepwise determination of the oxidation states of H and Cl in HCl.

First we write the formula and a Lewis dot diagram for the compound, HCl. Next, we identify the more electronegative atom of the two held together by the covalent bond, which is Cl in this case. Finally, we cleave the bond heterolytically by moving the shared electron pair in its entirety from the less electronegative atom (H) to the more electronegative atom (Cl). As a result, the H now bears a single positive charge and the Cl a single negative charge. Thus, the oxidation states of H and Cl in HCl are +1 and -1, respectively.

Let us now determine the oxidation states of the N, O, and H atoms in the molecules of ammonia (NH3) and water (H2O), following the same procedure (Figure 2-105). Both O and N are more electronegative than H. The conceptual heterolytic cleavage of the N-H and O-H bonds gives the oxidation numbers of -3, -2, and +1 for the N, O, and H atoms, respectively.
Figure 2-105. Stepwise determination of the oxidation states of H, N, and O in NH3 and H2O.

When determining the oxidation state of an atom using this method, one should watch out for the presence of double and triple bonds in the molecule. Each double bond is two shared electron pairs, not one, and each triple bond is three shared electron pairs. To illustrate this point, let us calculate the oxidation states of the atoms in CO2 and HCN (Figure 2-106). All four shared electron pairs in CO2 are polarized toward the more electronegative oxygen atoms. Conceptually breaking both double C=O bonds heterolytically gives rise to two O2- ions and one C4+ ion, which means that the oxidation state of the O atoms is -2 and that of the C atom is +4. Similarly, we shift all three electron pairs of the C≡N triple bond in HCN entirely to the N atom because nitrogen is more electronegative than carbon. As hydrogen is less electronegative than carbon, we move the electron pair shared by the C and H to the carbon. The resultant oxidation states of the H, C, and N atoms in HCN are +1, +2, and -3, respectively.
Figure 2-106. Determination of the oxidation states of atoms in CO2 and HCN molecules featuring multiple bonds.

We also need to know how to calculate the oxidation state of an atom in an ion. If we deal with a monoatomic cation or anion, the charge of the ion is the oxidation state of the atom. For example, the oxidation state of fluorine for the fluoride anion, F-, is -1, of sulfur for S2- is -2, of aluminum for Al3+ is +3, etc.

But, what do we do if the ion is polyatomic? Let us consider the metaphosphate anion, PO3-. This anion is produced on dissociation of metaphosphoric acid, HPO3. Let us recall how to draw structural formulas of oxygen-containing acids (Figure 2-73) and draw the structure of HPO3 and its anion (Figure 2-107).
Figure 2-107. Determination of the oxidation state of the P atom in PO3-.

The Lewis structure of HPO3 displays all of the shared electron pairs in the molecule. On dissociation of HPO3, the H on the O atom departs in the form of a proton, leaving the prviously shared electron pair to the oxygen. Consequently, the negative charge produced upon the dissociation is located on the oxygen of the PO3- (Figure 2-107). Moving all of the shared electron pairs to the more electronegative oxygen atoms gives rise to the oxidation states of -2 for the O atoms and +5 for the P atom.

The oxidation state of the central atom in an oxygen-containing acid, its anion(s), and salts is always the same. For example, in H2SO4, SO42-, HSO4-, Na2SO4, KHSO4, Al2(SO4)3, etc. the oxidation state of sulfur is +6. Similarly, the oxidation state of Mn in KMnO4 (potassium permanganate), HMnO4 (permanganic acid), and MnO4- (permanganate anion) is +7.

Next we consider NH4+, a polyatomic cation. On breaking heterolytically the first N-H bond, a molecule of ammonia, NH3, and H+ are produced (Figure 2-108). Next we determine the oxidation state of the N atom in ammonia, the way we did it before (Figure 2-105). Note that the lone electron pair on the N atom after the removal of the first proton (Figure 2-108, top) ) is omitted in the structure at the bottom of Figure 2-108. The oxidation state of nitrogen in both NH3 and NH4+ is the same, -3.
Figure 2-108. Determination of the oxidation state of the N atom in NH4+.

The sum of the oxidation numbers of all atoms for a neutral compound is zero, and for a polyatomic ion it equals the charge of the ion. Use this rule to double-check the numbers when calculating oxidation states of atoms in molecules and polyatomic anions and cations.

The oxidation state of an element in a simple substance is always zero. There are no less electronegative or more electronegative elements in simple substances that are always composed of atoms of the same type, such as Cl2, N2, and O2 (Figure 2-109). We therefore leave the equally shared electron pairs where they are, right in the middle between the two identical atoms. As a result, the charges on the atoms are zero and so are their oxidation states.
Figure 2-109. The oxidation state of an atom in a simple substance is always zero.

The same idea is applied to the oxidation state of an atom covalently bonded to another atom of the same element in a symmetric molecule, such as hydrogen peroxide, H2O2 (Figure 2-110). Conceptually shifting the electron pairs shared between the O and H atoms to the O atoms leads to two protons and one peroxide anion O22-, in which each oxygen bears a single negative charge. The electron pair for the O-O covalent nonpolar bond is not to be shifted to either atom due to their equivalence. The oxidation state of each oxygen atom in H2O2 is therefore -1.
Figure 2-110. Determination of the oxidation state of O atoms in H2O2.

The oxidation state of -1 for oxygen is rare. The most common oxidation state of oxygen is -2 because in most of its compounds, oxygen forms covalent bonds to atoms of lower electronegativity.

Many texts and chemical education websites such as this one provide a set of rules for determination of the oxidation state of an atom in a compound. Although everyone is free to memorize and use rules, I strongly believe in the benefit of understanding and logical thinking over straight memory work. Now that we are aware of the idea behind the oxidation state concept, let us quickly go over the rules and make some comments.

1. "The oxidation number of a free element is always 0." (Here the author means a simple substance, one of those that are composed of just one element, such as O2, Cl2, P4, etc.) Yes, it is 0 (zero) because all shared electron pairs in a molecule of a simple substance are shared equally by identical atoms.

2. "The oxidation number of a monatomic ion equals the charge of the ion." A monoatomic ion does not have shared electron pairs. There are neither shared electrons nor other atoms to shift them to.

3. "The oxidation number of H is +1, but it is -1 in [sic] when combined with less electronegative elements." Hydrogen conventionally forms covalent bonds with nonmetals. As hydrogen is less electronegative than other nonmetals, a covalent bond between an H atom and an atom of another nonmetal is polarized away from the H, meaning the oxidation state of +1 for the H. Compounds where a hydrogen atom is bonded to a metal, called metal hydrides, are well-known, albeit less frequently encountered. Since most metals are less electronegative than hydrogen, the oxidation state of the H atom in a metal hydride is -1. The oxidation state of H in H2 is obviously zero.

4. "The oxidation number of O in compounds is usually -2, but it is -1 in peroxides." Not only do we know that, but we also understand why the oxidation state of oxygen is -1 in peroxides, such as H2O2 (Figure 2-110). In O2, the oxidation state of O is zero. In all other oxygen compounds, the oxidation state of oxygen is invariably -2, except for those featuring an O-F bond. Fluorine is more electronegative than oxygen, and is, in fact, the only element that is more electronegative than oxygen. For example, in oxygen difluoride (OF2), the oxidation state of the oxygen atom is +2.

5. "The oxidation number of a Group 1 element in a compound is +1. The oxidation number of a Group 2 element in a compound is +2." The alkali (Group 1) and alkaline-earth (Group 2) metals are the least electronegative elements that willingly donate their one (Group 1) or two (Group 2) valence electrons. Hence the indicated oxidation states.

6. "The oxidation number of a Group 17 element in a binary compound is -1." (A binary compound is a chemical compound composed of only two elements.) That is true but not always. All halogen atoms need one electron to attain the stable octet configuration of the outermost shell. This and the high electronegativity of the halogens explain the oxidation state of -1. However, there are binary halogen compounds where the oxidation state of a halogen is other than -1. For instance, the oxidation state of iodine in ICl is +1, of Cl in Cl2O7 is +7, of Br in BrF3 is +3, and of I in I2O5 is +5. These and many more examples, which I can provide, are yet another piece of evidence for the benefit of understanding rules rather than simply memorizing them.

7. "The sum of the oxidation numbers of all of the atoms in a neutral compound is 0." This is self-explanatory if you understand the oxidation state idea rather than memorize the rules. When, in our mind, we shift the shared electron pairs in a molecule (molecules are always neutral) to more electronegative atoms, electrons are neither added nor taken away.

8. "The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion." Same as item 7 above. No matter how we move electrons around within an ion, the overall charge remains the same.

Although oxidation states are hypothetical numbers, they are broadly used in chemistry for a variety of purposes, including the balancing of redox reaction equations. We will learn that in the next subsection.

2.9.2. Exercises.

1. Determine the oxidation states of all elements in the following molecules and ions: (a) SO2; (b) K+; (c) F2; (d) F-; (e) H2CO3; (f) HNO3; (g) Na2SO4; (h) KHSO4; (i) P4; (j) KMnO4; (k) MnO2; (l) CuI; (m) CO2; (n) CO; (o) CuSO4; (p) K2Cr2O7; (q) CrO3; (r) CrO42-; (s) KClO3; (t) Ca2+; (u) SO32-; (v) Fe2O3; (w) FeO; (x) Fe3O4; (y) Li2S; (z) HOCl (structure: H-O-Cl). (I would give you more but there are only 26 letters in the English alphabet.) Answer