6.6 – NOTES Oxidation Numbers

B. Naming ions and ionic compounds 1. “ate” vs. “ite”; “per” vs. “hypo” - ate has one more O than ite - hypo has one less O than ite while per has one more O than ate - predict phosphite if phosphate is known

per ___ ate persulfate: SO5-2 perbromate: BrO4- ___ ate sulfate: SO4-2 bromate: BrO3- ___ ite sulfite: SO3-2 bromite: BrO2- hypo__ ite hyposulfite: SO2-2 hypobromite: BrO- *decrease by one oxygen as go down

2. Oxidation Number: The charge (real or hypothetical) assigned to an atom. - Charge an atom would have IF e- were transferred completely;

Oxidation number rules: Pure elements uncombined state (diatomics, P4, S8, He, etc.) = 0 Monatomic ions - ions composed of only 1 atom; equals the charge Specific nonmetals - Oxygen for most compounds is -2; in H2O2 and other peroxides (O2-2) = -1 Hydrogen +1; unless bonded to metals in binary compounds (LiH, NaH, CaH2) = -1 Fluorine -1 in ALL compounds; other halogens will be (-) as halides, when combined w/ oxygen could have (+) oxidation #s Sum of oxidation numbers in a compound = 0 Sum of oxidation numbers in a polyatomic ion = charge found on ion

Examples: Determine the oxidation number of the underlined atom: Na2SO3, MnO4-, Cr2O7-2, Ca(NO2)2, HCl, HClO, Na2S2O3, H2SO3, H2SO4

The unknown is x, multiply each atom by its charge and the number of the atom present, then set equal to zero or charge. Na2SO3 = Na has a charge of +1, S is unknown and O has a charge of -2. 2(+1) + x + 3(-2) = 0 so x = 4 MnO4- = x + 4(-2) = -1 so x = 7 Cr2O7-2 = 2x + 7(-2) = -2 so x = 6 Ca(NO2)2 = 1(+2) + 2x + 4(-2) = 0 so x = 3 HCl x = - 1 HClO x = 1 Na2S2O3 x = 2 H2SO3 x = 4 H2SO4 x = 6

Example #2: Determine the oxidation number of the transition metal: (Hint: when given a polyatomic ion, use the number and charge of the ion rather than individual atoms.) Cu(NO2)2, Sn(SO4)2 Cu(NO2)2 = x + 2(-1) = 0 so x = 2 Sn(SO4)2 = x + 2(-2) = 0 so x = 4