1. RULES OF OXIDATION NUMBER ASSIGNMENT
STEPS IN ASSIGNNING OXIDATION NUMBERS
1) IF THE PERIODIC TABLE GIVES ONLY ONE OXIDATION STATE, USE THAT STATE. EXAMPLE Zn+2 .
2) APPLY THE RULES OF THIS POWERPOINT TO SOLVE FOR OXIDATION NUMBERS OF ELEMENTS WITH MULTIPLE OXIDATION STATES. THIS IS VALUABLE WITH TRANSITION METALS.
3) IF A MULTIPLE OXIDATION STATE ELEMENT HAS NO RULE, ASSIGN IT A VARIABLE AND SOLVE USING THE ALGEBRA RULE(SLIDE 3).

2. IF AN ELEMENT HAS MULTIPLE OXIDATION STATES, AS IN CARBON, YOU WILL HAVE TO USE THE ALGEBRA RULE TO SOLVE FOR THE STATE IN THE PARTICULAR COMPOND YOU ARE INVESTIGATING.

3. SLIDE-3 ALGEBRA RULE
THE SUM OF THE OXIDATION NUMBER (STATE ) OF EACH ATOM IN A COMPOUND EQUALS THE CHARGE OF THAT COMPOUND. KMnO GIVEN: K IS +1(GROUP 1 RULE ), O IS -2 (OXYGEN RULE) Mn HAS MULTIPLE STATES(PERIODIC TABLE) +1 + X + 4(-2) = 0
ALGEBRA RULE EXAMPLES.
Use group one to find other elements.
note: caution with oxygen in binary compounds
Li2O2
2(+1) + 2(X) = 0
X=O (oxygen) = -1
Use group one as a known for
muli oxidation stare elements
Na2SO4
2(+1) + (X) + 4(-2)= 0
X = S = +6 each S

4. NaOH 1 + -2 + 1 = 0 Each O is 2- rule OXYGEN RULE
Each Na is 1+, GROUP 1 RULE
Each H is 1+, rule
HYDROGEN RULE
NaOH
= 0

5. X = OXIDATION STATE OF Cr = 6+
EACH O is 2-
OXYGEN RULE
THE CHARGE OF THIS POLYATOMIC ION IS -2, ALL THE SUM OF THE OXIDATION STATES MUST EQUAL -2.
Cr HAS MULTIPLE OXIDATION STATES AND HAS NO RULE, SOLVE FOR IT AS THE VARIABLE X
Cr2O7 2-
2X + 7(-2) = -2
X = OXIDATION STATE OF Cr = 6+

6. GROUP ONE RULE EXAMPLES.
SLIDE-6 GROUP 1 RULE
GROUP ONE ELEMENTS ARE ALWAYS 1+ IN COMPOUNDS OR AS FREE (AQ) IONS. Ex in NaCl EACH Na has an oxidation state of in K2SO4 EACH K has an oxidation state of A Li+1 (aq) free ion has an oxidation number of 1+.
GROUP ONE RULE EXAMPLES.
Use group one as a known for
muli oxidation stare elements.
Na2SO4
2(+1) + (X) + 4(-2)= 0
X = S = +6 each S
Use group one to find other elements.
Li2O2
2(+1) + 2(X) = 0
X=O (oxygen) = -1 each

7. GROUP ONE RULE EXAMPLES.
SLIDE-7 GROUP 2 RULE
GROUP TWO ELEMENTS HAVE AN OXIDATION STATE OF 2+ IN COMPOUNDS AND AS FREE (AQ) IONS.   Ex in1) BaCl2 EACH Ba has an oxidation state of JJIBGO 2) In Ca3(PO4)2 EACH Ca has an oxidation state of A free Mg+2 ION (AQ) has an oxidation state of 2+.  
GROUP ONE RULE EXAMPLES.
Use group TWO as a known for
muli oxidation stare elements.
Ca3(PO4)2
3(+2) + 2(x) + 8(-2)=0
X = P = +5 each P
Use group TWO to find other elements.
BaO2
1(+2) + 2(X) = 0
X=O (oxygen) = -1 each

8. SLIDE-8 ELEMENTAL STATE RULE
ALL UNCOMBINED NEUTRAL (ELEMENTAL STATE) ATOMS HAVE AN OXIDATION STATE OF 0. THE NONPOLAR COVALENT DIATOMOIC MOLECULES OF THE PERIODIC TABLE ALSO HAVE AN OX# OF 0. EX: Au0, Fe0, Na0, etc. EX: H2, O2, N2, Cl2, Br2, I2, F2 are all in the 0 oxidation state

9. SLIDE-9 IONIC CHARGE RULE
THE OXIDATION NUMBER OF IONS IS THE IONIC CHARGE. EX: Au+2 = 2+, Fe+3 = 3+, Na+ = 1+, etc. --- USE THE CHARGE WRITTEN ON THE ION…NOT OTHER STATES ON P TABLE. EX: NO3 - HAS AN OXIDATION NUMBER OF 1- FOR THE (EACH) POLY ATOMIC ION GROUP.

10. Fluorine is always 1- in its compounds
SLIDE-10 OXYGEN RULE
THE OXYGEN RULES: Each oxygen USUALLY has an oxidation state of 2- in compounds, 0 in O2. EX, in H2O oxygen is In H2SO4 EACH oxygen is 2- IMP -- EXCEPTION #1– when oxygen is contained in a PEROXIDE, it has an oxidation state of 1-. PEROXIDES - contain the peroxide ion (O2-2), EACH oxygen is peroxides are when 2 group one elements combine with the peroxide ion H2O2, Li2O2, K2O2 are peroxides, each O is 1- BaO2, CaO2, MgO2 are peroxides, each O is 1- IMP -- EXCEPTION #2 – when oxygen is combined with FLOURINE, its oxidation state is 1+ or 2+. Fluorine is electronegative enough to oxidize oxygen. EXAMPLE : in FO the oxygen is +1, in F2O each oxygen is 2+
USE CAUTION IN BINARY COMPOUNDS OF OXYGEN AND HYDROGEN -- CHECK FOR RULE EXCEPTIONS.
Fluorine is always 1- in its compounds

11. SLIDE-11 HYDROGEN RULE
THE HYDROGEN RULE: THE OXIDATION STATE OF HYDROGEN IS USUALLY 1+   EX in H2O each hydrogen is 1+   IMP-EXCEPTION – in GROUP 1 METAL HYDRIDES the hydrogen is 1-, hydrogen is electronegative enough to oxidize group one metals. EX: NaH, KH etc the hydrogen is 1-  
USE CAUTION IN BINARY COMPOUNDS OF OXYGEN AND HYDROGEN -- CHECK FOR RULE EXCEPTIONS.

12. SLIDE-12 HALOGEN RULE
THE HALOGEN RULE: THE OXIDATION STATE OF HALOGENS IS USUALLY 1- IN BINARY COMPOUNDS.   EX in MgF2 each F is 1- AS INDICATED BY THIS RULE   IMP-EXCEPTION – HALOGEN IN POLYATOMIC IONS WILL EXHIBIT MULTIPLE OXIDATION STATES…USE ALGEBRA RULE TO SOLVE FOR THE HALOGEN. SEE NEXT SLIDE…  

13. X = OXIDATION STATE OF Cl IS 6+
EACH O is 2-
OXYGEN RULE
THE CHARGE OF THIS POLYATOMIC ION IS -2, THE SUM OF THE OXIDATION STATES MUST EQUAL -2.
Cr HAS MULTIPLE OXIDATION STATES AND HAS NO RULE, SOLVE FOR IT AS THE VARIABLE X
ClO4 2-
X + 4(-2) = -2
X = OXIDATION STATE OF Cl IS 6+

14. THE SOLUBLE SALT RULE: 1) MANY TERNARY SALTS CONTAIN 2 ELEMENTS WITH MULTIPLE OXIDATION STATES AND THAT HAVE NO RULE DEFINNING THE OXIDATION NUMBER IN THAT PARTICULAR SALT.
2) AS YOU CANNOT HAVE TWO VARIABLES AT ONCE YOU CAN SOLVE ONE AT A TIME, THE METAL BY DISSOCIATION AND THE NONMETAL BY ALGEBRA.
FOR EXAMPLE THE SALT Ni(NO3)3 HAS Ni AND n, BOTH WITH MULTIPLE STATES. THE SHORT FORM FOR GETTING THE Ni IS:  
Ni (NO3)3
+3 (-1) * 3
CHARGE -1 CHARGE ON NITRATE ISFROM TABLE E

15. Ni(NO3)3 NITROGEN HAS MULTIPLE OXIDATION STATES, SOLVE FOR THIS ALSO.
NI HAS MULTIPLE OXIDATION STATES AND HAS NO RULE, SOLVE FOR IT AS THE VARIABLE X
OXYGEN IS 2- IN MOST COMPOUNDS …OXYGEN RULE.
Ni(NO3)3
THIS SALT CONTAINS 2 ELEMENTS WITH MULTIPLE OXIDATION STATES (Ni AND N). YOU HAVE TWO WAYS TO SOLVE.
USE THE DISSOCIATION REACTION AND BALANCE FOR CHARGE TO GET THE METAL (Ni) OXIDATION STATE. THEN SOLVE FOR NITROGEN.
OR SOLVE FOR NITROGEN IN THE NITRATE ION FIRST, THEN SOLVE FOR Ni. SEE SOLUTIONS IN NEXT TWO SLIDES…

16. X = OXIDATION STATE OF N = 5+
THIS SALT CONTAINS 2 ELEMENTS WITH MULTIPLE OXIDATION STATES (Ni AND N).
WRITE THE DISSOCIATION REACTION AND BALANCE FOR CHARGE TO GET THE METAL (Ni) OXIDATION STATE. THEN SOLVE FOR NITROGEN.
CHARGE FROM TABLE E
0 = X (-1)
X = Ni = +3
Ni(NO3)3  Ni + 3 NO3-
NITROGEN HAS MULTIPLE OXIDATION STATES
Ni HAS A OXIDATION STATE OF 3+ FROM THE DISSOCIATION ABOVE
OXYGEN IS 2- IN MOST COMPOUNDS …OXYGEN RULE.
Ni(NO3)3
(X) + 9(-2) = 0
X = OXIDATION STATE OF N = 5+

17. X = OXIDATION STATE OF N = 5+ X = OXIDATION STATE OF NI = 3+
THIS SALT CONTAINS 2 ELEMENTS WITH MULTIPLE OXIDATION STATES (Ni AND N).
OR SOLVE FOR NITROGEN IN THE NITRATE ION FIRST, THEN SOLVE FOR Ni. SEE SOLUTIONS IN NEXT TWO SLIDES…
CHARGE FROM TABLE E
NO3-
X + 3(-2) = -1
X = OXIDATION STATE OF N = 5+
NITROGEN HAS AN OXIDATION STATE OF 5+ FROM THE DISSOCIATION CALCULATION
OXYGEN IS 2- IN MOST COMPOUNDS …OXYGEN RULE.
Ni HAS MULTIPLE OXIDATION STATES, SOLVE FOR IT.
Ni(NO3)3
X + 3(+5) + 9(-2) = 0
X = OXIDATION STATE OF NI = 3+