1. 6.6 – NOTES Oxidation Numbers

2. 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

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

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

5. 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

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

7. 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

8. 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