ASSIGNING OXIDATION NUMBERS

14.2. ASSIGNING OXIDATION NUMBERS

In Sec. 5.5, electron dot diagrams were introduced. The electrons shared between atoms were counted as “belonging” to both atoms. We thus counted more valence electrons than we actually had. For oxidation numbers, however, we can count each electron only once. If electrons are shared, we arbitrarily assign “control” of them to the more electronegative atom. For atoms of the same element, each atom is assigned one-half of the shared electrons. The oxidation number is then defined as the number of valence electrons in the free atom minus the number “controlled” by the atom in the compound. If we actually transfer the electrons from one atom to another, the oxidation number equals the resulting charge. If we share the electrons, the oxidation number does not equal the charge; there may be no charge. In this case, control is not meant literally, but is just a term to describe the

counting procedure. For example, the electron dot diagram of CO 2 may be written as O C O

Since O is to the right of C in the second period of the periodic table, O is more electronegative, and we assign control of all eight shared electrons to the two O atoms. (It does not really have complete control of the electrons; if it did, the compound would be ionic.) Thus, the oxidation number of each atom is calculated as follows:

OXIDATION AND REDUCTION

[ CHAP. 14

C Each O atom

Number of valence electrons in free atom

− Number of valence electrons “controlled”

Oxidation number

Like the charge on an ion, each atom is assigned an oxidation number. Do not say that oxygen in CO 2 has an oxidation number of −4 because the two oxygen atoms together control four more electrons in the compound than they would in the free atoms. Each oxygen atom has an oxidation number of −2.

Although the assignment of control of electrons is somewhat arbitrary, the total number of electrons is accurately counted, which leads to a main principle of oxidation numbers:

The total of the oxidation numbers of all the atoms (not just all the elements) is equal to the net charge on the molecule or ion.

For example, the total of the three oxidation numbers in CO 2 is 4 + 2(−2) = 0, and the charge on CO 2 is 0. One principal source of student errors is due to confusion between the charge and oxidation number. Do not confuse them. In naming compounds or ions, use Roman numerals to represent positive oxidation numbers. (The Romans did not have negative numbers.) In this book, charges have the numeral first, followed by the sign; oxidation numbers have the sign first, followed by the numeral. In formulas, Arabic numeral superscripts represent charges. While working out answers, you might want to write oxidation numbers encircled and under the symbol for the element. Individual covalently bonded atoms do not have easily calculated charges, but they do have oxidation numbers. For example, the oxidation number of each element and the charge on the ion for

SO 3 2− and for Cl − are shown below.

Charges SO 3 2 − Cl −

Oxidation numbers: +4 −2

It is too time consuming to calculate oxidation numbers by drawing electron dot diagrams each time. We can speed up the process by learning the following simple rules:

1. The sum of all the oxidation numbers in a species is equal to the charge on the species.

2. The oxidation number of uncombined elements is equal to 0.

3. The oxidation number of every monatomic ion is equal to its charge.

4. In its compounds, the oxidation number of every alkali metal and alkaline earth metal is equal to its group number.

5. The oxidation number of hydrogen in compounds is +1 except when the hydrogen is combined with active metals; then it is −1.

6. The oxidation number of oxygen in its compounds is −2, with some much less important exceptions. The oxidation number of oxygen in peroxides is −1, in superoxides it is − 1 2 , and in OF 2 and O 2 F 2 it is positive. The peroxides and superoxides generally occur only with other elements in their maximum oxidation states. You will be able to recognize peroxides or superoxides by the presence of pairs of oxygen atoms and by the fact that if the compounds were normal oxides, the other element present would have too high an oxidation number (Sec. 14.3).

Na 2 O 2 sodium peroxide

oxidation number of sodium = 1

BaO 2 barium peroxide

oxidation number of barium = 2

SnO 2 tin(IV) oxide

oxidation number of tin = 4 (permitted)

KO 2 potassium superoxide

oxidation number of potassium = 1

H 2 O 2 hydrogen peroxide

oxidation number of hydrogen = 1

CHAP. 14]

OXIDATION AND REDUCTION

7. The oxidation number of every halogen atom in its compounds is −1 except for a chlorine, bromine, or iodine atom combined with oxygen or a halogen atom higher in the periodic table. For example, the chlorine atoms in each of the following compounds have oxidation numbers of −1:

SnCl 2 SiCl 4 KCl

PCl 3 HCl

The chlorine atom in each of the following has an oxidation number different from −1:

F has an oxidation number of −1.) With these rules, we can quickly and easily calculate the oxidation numbers of an element most of the time

Cl

2 O 3 ClO 2 ClF 3 (

from the formulas of its compounds.

EXAMPLE 14.1. What are the oxidation numbers of fluorine and iodine in IF 7 ?

Ans. F has an oxidation state of −1 (rule 7). I has an oxidation number of +7 (rule 1). (I is not −1, because it is combined with a halogen higher in the periodic table.)

EXAMPLE 14.2. Calculate the oxidation number of S in (a) SO 2 , (b) SO 3 2− , and (c) SO 3 .

Ans. Let x = oxidation number of S in each case: (a)

x + 3( −2) = −2 x = +4 Oxidation

Oxidation Three Oxidation Total number of atoms number of

number of atoms number of charge sulfur

charge on

oxygen on ion

(c) x + 3( −2) = 0

x = +6

EXAMPLE 14.3. Calculate the oxidation number of (a) Cr in Cr 2 O 7 2− and (b) S in S 2 O 3 2− .

Ans. (a)

number of

oxygen

number of

(b) 2x + 3(−2) = −2

x = +2

Oxidation numbers are most often, but not always, integers. If there are nonintegral oxidation numbers, there must be multiple atoms so that the number of electrons is an integer.

EXAMPLE 14.4. Calculate the oxidation number of N in NaN 3 , sodium azide.

Ans. Na has an oxidation number of +1. Therefore, the total charge on the three nitrogen atoms is 1−, and the average charge, which is equal to the oxidation number, is 1 3 − . (Three nitrogen atoms, times − 1 3 each, have a total of oxidation numbers equal to an integer, −1.)