1. Oxidation numbers

2. Oxidation numbers are used to describe the distribution of electrons among bonded atoms. Oxidation numbers are used to describe the distribution of electrons among bonded atoms. Covalent bonds involve sharing of electrons, but oxidation numbers show what the distribution would be if the electrons were completely transferred. Covalent bonds involve sharing of electrons, but oxidation numbers show what the distribution would be if the electrons were completely transferred.

3. Guidelines for Assigning Oxidation Numbers ( see p. 180 for a complete list) The oxidation of any free (uncombined) element is zero. The oxidation of any free (uncombined) element is zero. The oxidation number of a monatomic ion is equal to the charge of the ion. The oxidation number of a monatomic ion is equal to the charge of the ion. e.g. The oxidation number of K + is +1.e.g. The oxidation number of K + is +1. The oxidation number of each hydrogen atom is +1, unless it is combined with a metal, then it has a state of -1. The oxidation number of each hydrogen atom is +1, unless it is combined with a metal, then it has a state of -1. The oxidation number of fluorine is always -1. The oxidation number of fluorine is always -1. The oxidation number of each oxygen atom in most of its compounds is -2. The oxidation number of each oxygen atom in most of its compounds is -2.

4. Guidelines for Assigning Oxidation Numbers ( see p. 180 for a complete list) The algebraic sum of the oxidation numbers for all the atoms in a compound is zero. The algebraic sum of the oxidation numbers for all the atoms in a compound is zero. The algebraic sum of the oxidation numbers for all the atoms in a polyatomic ion is equal to the charge on that ion. The algebraic sum of the oxidation numbers for all the atoms in a polyatomic ion is equal to the charge on that ion.

5. Example: Determine the oxidation numbers for each atom in KMnO 4 This compound is made up of a K + cation and an MnO 4 - anion. This compound is made up of a K + cation and an MnO 4 - anion. The K + is a monatomic ion with a charge of +1, so its oxidation number is +1. The K + is a monatomic ion with a charge of +1, so its oxidation number is +1. Assume that each O atom has an oxidation number of -2. Assume that each O atom has an oxidation number of -2. The MnO 4 - has a total charge of -1. There are 4 O atoms, each with an oxidation number of -2 The MnO 4 - has a total charge of -1. There are 4 O atoms, each with an oxidation number of -2 The oxidation number of the Mn may be found by the equation: The oxidation number of the Mn may be found by the equation: Mn + 4(-2) = -1Mn + 4(-2) = -1 Therefore, the oxidation number of Mn in this compound is +7. Therefore, the oxidation number of Mn in this compound is +7.

6. Example: Determine the oxidation numbers for each atom in Co(NO 2 ) 2 The anion is the nitrite ion, NO 2 -. The anion is the nitrite ion, NO 2 -. Assume that each O atom in the nitrite ion has an oxidation number of -2. Assume that each O atom in the nitrite ion has an oxidation number of -2. The NO 2 - has a total charge of -1. There are 2 O atoms, each with an oxidation number of -2 The NO 2 - has a total charge of -1. There are 2 O atoms, each with an oxidation number of -2 The oxidation number of the N may be found by the equation: The oxidation number of the N may be found by the equation: N + 2(-2) = -1N + 2(-2) = -1 Therefore, the oxidation number of N in this compound is +3. Therefore, the oxidation number of N in this compound is +3. Since there are two nitrate ions, each with a charge of -1, the charge on the Co must be +2. As an ion, its oxidation number is equal to its charge. Since there are two nitrate ions, each with a charge of -1, the charge on the Co must be +2. As an ion, its oxidation number is equal to its charge.