RULES FOR ASSIGNING OXIDATION NUMBERS
http://www.csun.edu/~hcchm001/IntroChemHandouts.html
1. All elements in their free state (uncombined with other elements) have an oxidation number of zero (for
example, Na, Cu, Mg, H22, O , Cl22, N ).
2. H is +1, except in metal hydrides, where it is -1 (for example, NaH, CaH2)
Na : H
Electrons in the bond are assigned to H, the more electronegative atom. Na appears to have “lost” one electron, so its oxidation number is +1.* H appears to have “gained” one electron, so its oxidation number is -1.*
3. O is -2, except in peroxides, where it is -1, and in OF2, where it is +2. @@ @@ H : O : O : H @@ @@
Electrons in the H-O bonds are assigned to the O. Electrons in the O-O bond are divided equally between the two O atoms. H appears to have “lost” one electron, so its oxidation number is +1.* Each O appears to have “gained” one electron, so its oxidation number is -1.*
@@ @@ @@ : F : O : F : @@ @@ @@
Electrons in the F-O bonds are assigned to the F. Both atoms of F appears to have “gained” 1 electron each, so the oxidation number for each is -1.* The oxygen appears to have “lost” 2 electrons, so its oxidation number is +2.*
*When compared to the electrically neutral atom.
4. (a) The metallic element in an ionic compound has a positive oxidation number. For monoatomic cations, the oxidation number is equal to the charge on the ion. For example, Na+, Ca2+, Al3+, Fe3+, etc.
(b) The nonmetallic element in an ionic compound has a negative oxidation number. For monoatomic anions, the oxidation number is equal to the charge on the ion. For example, Cl-, S2-, N3-, etc.
5. In covalent compounds, the negative oxidation number is assigned to the most electronegative atom:
F > O > N > Cl > Br > I > S > C > H >>> metals
6. The algebraic sum of the oxidation numbers of elements in a compound is zero.
CaF22H SO4 +2 + 2(-1) = 0 2(+1) + S + 4(-2) = 0 Find S = +6 for ox. no. of S
7. The algebraic sum of the oxidation numbers of the elements covalently bound into a polyatomic ion is equal to the charge of the ion.
+2- NH43HPO -3 + 4(+1) = +1 +1 + P + 3(-2) = -2 Find P = +3 for ox. no. of P