In this chapter we have considered two kinds of bonds, ionic and covalent. We have seen (Section 8.5) that when a covalent bond forms between two atoms with different electronegativities, the bonding electrons lie somewhat closer to the more electronegative atom. For example, in HCl, the bonding is polar covalent with the electrons in the HCl bond displaced toward the more electronegative Cl atom:
The H atom therefore carries a partial positive charge, and the Cl atom a partial negative one.
In keeping track of electrons, it is sometimes helpful to assign charges to atoms by assigning shared electrons to the more electronegative atom. In HCl this procedure gives Cl eight valence-shell electrons, one more than the neutral atom. We have, in effect, given a -1 charge to the chlorine. Hydrogen, stripped of its electron, is assigned a charge of +1.
Charges assigned in this fashion are called oxidation numbers or oxidation states. The oxidation number of an atom is the charge that results when the electrons in a covalent bond are assigned to the more electronegative atom; it is the charge an atom would possess if the bonding were ionic. In HCl the oxidation number of H is +1 and that of Cl is -1. (In writing oxidation numbers, we will write the sign before the number to distinguish them from actual electronic charges, which we write with the number first.)
Oxidation numbers do not correspond to real charges on the atoms, except in the special case of simple ionic substances. Nevertheless, they are useful in a variety of situations, such as in naming compounds, in balancing chemical equations for reactions in which changes in oxidation numbers occur, and in examining trends in chemical properties.
We determine oxidation numbers using the following set of rules.
1. The oxidation number of an element in its elemental form is zero. For an isolated atom--like an N atom, where there is no bonding and no net charge--the oxidation state must be 0. It is also zero for any elemental substance in which there is bonding. For example, in Na metal, N2, Cl2, and P4 the bonding electrons are shared equally between identical atoms, giving each atom an oxidation state of 0.
2. The oxidation number of a monoatomic ion is the same as its charge. For example, the oxidation number of sodium in Na+ is +1, and that of sulfur in S2- is -2.
3. In binary compounds (those with two different elements) the element with greater electronegativity is assigned a negative oxidation number equal to its charge in simple ionic compounds of the element. For example, consider, the oxidation state of Cl in PCl3. Cl is more electronegative than P. In its simple ionic compounds, chlorine appears as the chloride ion, Cl-. Thus, in PCl3, Cl is assigned an oxidation number of -1.
4. The sum of the oxidation numbers equals zero for an electrically neutral compound and equals the overall charge for an ionic species. For example, PCl3 is a neutral molecule. Thus, the sum of the oxidation numbers of the P and Cl atoms must equal zero. Because the oxidation number of each Cl in this compound is -1 (rule 3), the oxidation number of P must be +3. In like manner, the sum of the oxidation numbers of C and O in CO32- must equal -2. The oxidation number of each O in this ion is -2 because O is more electronegative than C and -2 is the charge on the oxide ion (rule 3). Therefore, the oxidation number on C must be +4 because +4 + 3(-2) = -2.
The periodic table provides us with many additional guidelines for assigning oxidation numbers. As shown in Figure 8.13, oxidation numbers exhibit periodic trends. Some observations are particularly helpful. The alkali metals (group 1A) exhibit only the oxidation state of +1 in their compounds. The alkaline earth metals (group 2A) are always found in compounds in the +2 oxidation state. The most commonly encountered element in group 3A, Al, is always found in the +3 oxidation state.
FIGURE 8.13 Common nonzero oxidation numbers for elements with atomic numbers 3 through 39. Notice that the maximum and minimum oxidation states (through which lines have been drawn for emphasis) are a periodic function of atomic number.
The most electronegative element, F, is always found in compounds in the -1 oxidation state. Oxygen in compounds is nearly always in the -2 oxidation state. The only common exception to this rule occurs in peroxides. In the peroxide ion, O22-, and in molecular peroxides, such as H2O2, oxygen has an oxidation number of -1. Hydrogen has an oxidation number of +1 when it is bonded to a more electronegative element (most nonmetals) and of -1 when bonded to a less electronegative element (most metals).
Determine the oxidation state of sulfur in each of the following: (a) H2S; (b) S8; (c) SCl2; (d) Na2SO3; (e) SO42-.
SOLUTION (a) This is a binary (two-element) compound. Because S is more electronegative than H, S must have a negative oxidation number equal to its ionic charge (rule 3). Thus, S has an oxidation number of -2. This is consistent with the oxidation number of +1 expected for H when it is bonded to a nonmetal.
(b) Because this is an elemental form of sulfur, the oxidation state of S is 0 (rule 1).
(c) Chlorine is more electronegative than sulfur. Therefore, the Cl in this binary compound must have a negative oxidation number equal to its ionic charge (rule 3), which is -1. The sum of the oxidation numbers must equal zero (rule 4). Letting x equal the oxidation number of S, we have x + 2(-1) = 0. Consequently, the oxidation state of S, x, must be +2.
(d) Sodium, an alkali metal, is always found in compounds in the +1 oxidation state. Oxygen has a common oxidation state of -2. If we let x equal the oxidation number of S, we have 2(+1) + x + 3(-2) = 0. Therefore, the oxidation number of S in this compound is +4.
(e) The oxidation state of O is -2. The sum of the oxidation numbers equals -2, the net charge of the SO42- ion. Letting x equal the oxidation number of S, we have x + 4(-2) = -2. From this relation we conclude that the oxidation state of S is +6.
These examples illustrate the range of oxidation states exhibited by sulfur, from -2 to +6. In general, the most negative oxidation state of a nonmetal corresponds to the number of electrons that must be added to the atom to give it an octet. In this case, S, which belongs to the periodic group 6A, has six valence-shell electrons. Thus, two more are needed to give it an octet, as in the S2- ion. The most positive oxidation state, in this case, +6, corresponds to loss of all valence-shell electrons.
What is the oxidation state of the underlined element in each of the following: (a) 2O5; (b) Na ; (c) 2O72-; (d) Br4; (e) Ba2? Answers: (a) +5; (b) -1; (c) +6; (d) +4; (e) -1
In Section 8.6 we introduced the idea of formal charge as a way of assigning charge to atoms in a Lewis structure. It is natural to ask what are the differences between oxidation numbers and formal charges, and why do we use them both to keep track of electrons? In calculating formal charges, we assume that all atoms have the same electronegativity; that is, all bonds are treated as nonpolar covalent bonds. In calculating oxidation numbers, we assume that electrons transfer completely to the more electronegative atom. In practice, formal charges generally reflect the actual charges on atoms more accurately than do oxidation numbers. However, we will find oxidation numbers to be more useful in dealing with oxidation-reduction reactions. (For more information, see Section 4.6) We will discuss this relationship in greater detail in Chapter 20.
We saw in Section 2.7 that there are two general approaches to naming binary compounds (compounds composed of two elements): one used for ionic compounds and the other for molecular ones. In both approaches the name of the less electronegative element is given first, followed by the name of the more electronegative element, modified to have an -ide ending. Compounds that are ionic are given names based on their component ions, including the charge of the cation if it is variable. Those that are molecular are named using the prefixes listed in Table 2.6 to indicate the number of atoms of each kind in the substance:
|MgH2||magnesium hydride||H2S||dihydrogen sulfide|
|FeF2||iron(II) fluoride||OF2||oxygen difluoride|
|Mn2O3||manganese(III) oxide||Cl2O3||dichlorine trioxide|
The dividing line between the two approaches, however, is not always clear, and both approaches are often applied to the same substances. For example, TiO2, which is a commercially important white paint pigment, is sometimes referred to as titanium(IV) oxide but is more commonly called titanium dioxide.
One reason for the overlap in the two approaches to nomenclature is that compounds of metals in higher oxidation states tend to be molecular rather than ionic. For example, SnCl4 [tin tetrachloride or tin(IV) chloride] is a colorless liquid that freezes at -33°C and boils at 114°C; Mn2O7 [dimanganese heptoxide or manganese(VII) oxide] is a green liquid that freezes at 5.9°C. Recall that ionic compounds are solids at room temperature. When we see the formula of a compound containing a metal in a high oxidation state (above +3), we should not be surprised that it does not exhibit the general properties of ionic compounds. We will make wide use of oxidation numbers in later chapters to explain trends in chemical behavior.