Chemistry · Ch 6 — Redox Reactions
Rules to Assign Oxidation Number
Rules to Assign Oxidation Number
Eight rules fix the oxidation number of any atom in any compound or ion. Rule 1: an atom of an element in its free, uncombined state has oxidation number zero -- every atom in H2, Cl2, O3, S8, P4, or a lump of the metal Ca, is at zero. Rule 2: in a monoatomic ion, the oxidation number simply equals the ion's charge; this is why alkali metals (Group 1) are always +1 in every compound (NaCl, KCl, ...), alkaline-earth metals (Group 2) are always +2 (CaCO3, MgCl2, ...), and aluminium is always taken as +3. Rule 3: oxygen is normally -2 in its compounds, with three named exceptions -- in a peroxide (or peroxide ion, e.g. H-O-O-H) oxygen is -1, in a superoxide (e.g. KO2) each oxygen atom averages -1/2, and in OF2, the one compound where oxygen is bonded to the more electronegative fluorine, oxygen is +2. Rule 4: hydrogen's oxidation number is +1 when it is bonded to a non-metal (as in H-O-H, water), but -1 when it is bonded to a metal, i.e. in a metal hydride (as in Li-H or H-Ca-H). Rule 5: fluorine is always -1 in every one of its compounds, since it is the most electronegative element and can never be assigned a positive oxidation number; the other halogens (Cl, Br, I) are usually -1 in their halide compounds, but flip to +1 specifically when bonded to oxygen (as in Cl-O-Cl or H-O-Cl). Rule 6: the oxidation numbers of every atom in a neutral molecule must add up to exactly zero. Rule 7: the oxidation numbers of every atom in a polyatomic ion must add up to the ion's own net charge. Rule 8: when the same element appears more than once in one molecule or ion, the oxidation number quoted for it is the AVERAGE across all of that element's atoms present -- individual atoms of the same element can, in reality, sit at different oxidation states even th …
Worked out. Worked example deducing sulfur's oxidation number in two species. For SO2, a neutral molecule, the sum of oxidation numbers is zero: (ON of S) + 2 x (ON of O) = 0, and since O is -2 here, ON of S = 0 - 2(-2) = +4. For SO4²⁻, an ion, the sum of oxidation numbers equals the ion's charge, -2: (ON of S) + 4 x (ON of O) = -2, so ON of S = -2 - 4(-2) = -2 + 8 = +6. This shows the same element, sulfur, taking two different oxidation numbers (+4 and +6) depending on how many …