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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, H2 , O2 , Cl2 , N2 ). 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) (b) 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. 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. CaF2 +2 + 2(-1) = 0 H2 SO4 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. NH4 + -3 + 4(+1) = +1 HPO3 2+1 + P + 3(-2) = -2 Find P = +3 for ox. no. of P