Chapter 7-2: Oxidation Numbers Coach Kelsoe Chemistry Pages 232-235

Section 7-2 Objectives List the rules for assigning oxidation numbers. Give the oxidation number for each element in the formula of a chemical compound. Name binary molecular compounds using oxidation numbers and the Stock system.

Oxidation Numbers The charges on the ions composing an ionic compound reflect the electron distribution of the compound. In order to indicate the general distribution of electrons among the bonded atoms in a molecular compound or a polyatomic ion, oxidation numbers, or oxidation states are assigned to the atoms composing the compound or ion.

Oxidation Numbers Oxidation numbers do not have an exact physical meaning. However they do come in handy when naming compounds, writing formulas, and balancing chemical equations. As a general rule in assigning oxidation numbers, shared electrons are assumed to belong to the more-electronegative atom in each bond. Here are the rules for assigning oxidation numbers.

Rules for Assigning Oxidation Numbers The atoms in a pure element have an oxidation number of zero. For example, the oxygens in O2 and the sulfur in S8. The more-electronegative element in a binary molecular compound is assigned the number equal to the negative charge it would have as an anion. The less-electronegative atom is assigned the number equal to the positive charge it would have as a cation.

Rules for Assigning Oxidation Numbers Fluorine has an oxidation number of -1 in all of its compounds because it is the most electronegative element. Oxygen has an oxidation number of -2 in almost all compounds. Exceptions include when it is in peroxides, such as H2O2, in which its oxidation number is -1, and when it is in compounds with halogens, such as OF2, in which its oxidation number is +2.

Rules for Assigning Oxidation Numbers Hydrogen has an oxidation number of +1 in all compounds containing elements that are more-electronegative than it; it has an oxidation number of -1 in compounds with metals. The algebraic sum of the oxidation numbers of all atoms in a neutral compound is equal to zero.

X Y Here’s How It Works… O = +2 H = +1 O = -2 H = -1 Most electronegative Least electronegative O = +2 H = +1 O = -2 H = -1

Rules for Assigning Oxidation Numbers The algebraic sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge of the ion. Although rules 1-7 apply to covalently bonded atoms, oxidation numbers can also be assigned to atoms in ionic compounds. For example, the ions Na+, Ca2+, and Cl- have oxidation numbers of +1, +2, and -1, respectively.

Oxidation Numbers Let’s examine hydrogen fluoride and water. In HF, the bond is polar with a partial negative charge on the fluorine and a partial positive charge on the hydrogen. If HF were an ionic compound in which an electron was fully transferred to the fluorine ion, H would have a 1+ charge and F would have a 1- charge. Thus the oxidation numbers of H and F in hydrogen fluoride are +1 and -1 respectively.

Oxidation Numbers Let’s examine hydrogen fluoride and water. In a water molecule, the oxygen atom is more electronegative than the hydrogen atoms. If H2O were an ionic compound, the oxygen would have a charge of 2- and the hydrogen ions would each have a charge of 1+. The oxidation numbers of H and O in water are therefore +1 and -2 respectively. Because the sum of the oxidation numbers of the atoms in a compound must satisfy rule 6 or 7 of the guidelines, it is often possible to assign oxidation numbers when they are not known.

Sample Problem 1 Assign oxidation numbers to each atom in the following compounds or ions: UF6 H2SO4 ClO3-

Solutions to Sample Problem 1 UF6 F has an O.N. of -1 and there are 6, so to make the sum be zero, U must have an oxidation number of +6. H2SO4 H has an O.N. of +1 giving the total H’s a +2, and O has an O.N. of -2 giving the total O’s a -8. So S must have a O.N. of +6. ClO3- O has an O.N. of -2, there are 3 giving the total O’s a -6. To keep the -1 charge, Cl must have a +5 charge.

Sample Problem 2 Assign oxidation numbers to each atom in the following compounds or ions: CF4 N2O5 HClO3

Solutions to Sample Problem 2 CF4 F has an O.N. of -1 and there are 4, so to make the sum be zero, U must have an oxidation number of +4. N2O5 O has an O.N. of -2 and there are 5, so to make the sum be zero, N must have an oxidation number of +5. HClO3 H has an O.N. of +1 giving the total H’s a +1, and O has an O.N. of -2 giving the total O’s a -6. So Cl must have a O.N. of +5.

Using Oxidation Numbers for Formulas and Names Many nonmetals can have more than one oxidation number (See page 903). These numbers can sometimes be used in the same manner as ionic charges to determine formulas. Oxidation numbers alone, however, cannot be used to predict the existence of a compound. For example, sulfur (+4, +6) and oxygen can form either SO2 or SO3. This is not enough!

Using Oxidation Numbers for Formulas and Names We used Roman numerals to denote ionic charges in the Stock system of naming ionic compounds. The Stock system is actually based on oxidation numbers, and it can be used as an alternative to the prefix system for naming binary molecular compounds. For example, SO2 is called sulfur dioxide in the prefix system and sulfur(IV) oxide in the Stock system.

Using Oxidation Numbers for Formulas and Names The international body that governs nomenclature has endorsed the Stock system, which is more practical for complicated compounds. Prefix-based names and Stock-system names are still used interchangeably for many simple compounds, however.

Examples of Names in Both Systems Prefix system Stock system PCl3 Phosphorus trichloride Phosphorus(III) chloride PCl5 Phosphorus pentachloride Phosphorus(V) chloride N2O Dinitrogen monoxide Nitrogen(I) oxide NO Nitrogen monoxide Nitrogen(II) oxide PbO2 Lead dioxide Lead(IV) oxide Mo2O3 Dimolybdenum trioxide Molybdenum(III) oxide