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. THE SUM OF THE OXIDATION NUMBER (STATE ) OF EACH ATOM IN A COMPOUND EQUALS THE CHARGE OF THAT COMPOUND. KMnO 4 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 as a known for muli oxidation stare elements Na 2 SO 4 2(+1) + (X) + 4(-2)= 0 X = S = +6 each S Use group one to find other elements. note: caution with oxygen in binary compounds Li 2 O 2 2(+1) + 2(X) = 0 X=O (oxygen) = -1 SLIDE-3 ALGEBRA RULE

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

5. 2X + 7(-2) = -2 X = OXIDATION STATE OF Cr = 6+ Cr HAS MULTIPLE OXIDATION STATES AND HAS NO RULE, SOLVE FOR IT AS THE VARIABLE X 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 2 O 7 2-

6. GROUP ONE ELEMENTS ARE ALWAYS 1+ IN COMPOUNDS OR AS FREE (AQ) IONS. Ex in NaCl EACH Na has an oxidation state of 1+. in K 2 SO 4 EACH K has an oxidation state of 1+. 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. Na 2 SO 4 2(+1) + (X) + 4(-2)= 0 X = S = +6 each S Use group one to find other elements. Li 2 O 2 2(+1) + 2(X) = 0 X=O (oxygen) = -1 each SLIDE-6 GROUP 1 RULE

7. GROUP TWO ELEMENTS HAVE AN OXIDATION STATE OF 2+ IN COMPOUNDS AND AS FREE (AQ) IONS. Ex in1) BaCl 2 EACH Ba has an oxidation state of 2+ JJIBGO 2) In Ca 3 (PO 4 ) 2 EACH Ca has an oxidation state of 2+. 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. Ca 3 (PO 4 ) 2 3(+2) + 2(x) + 8(-2)=0 X = P = +5 each P Use group TWO to find other elements. BaO 2 1(+2) + 2(X) = 0 X=O (oxygen) = -1 each SLIDE-7 GROUP 2 RULE

8. 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: Au 0, Fe 0, Na 0, etc. EX: H 2, O 2, N 2, Cl 2, Br 2, I 2, F 2 are all in the 0 oxidation state SLIDE-8 ELEMENTAL STATE RULE

9. 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: NO 3 - HAS AN OXIDATION NUMBER OF 1- FOR THE (EACH) POLY ATOMIC ION GROUP. SLIDE-9 IONIC CHARGE RULE

10. THE OXYGEN RULES: Each oxygen USUALLY has an oxidation state of 2- in compounds, 0 in O 2. EX, in H 2 O oxygen is 2- In H 2 SO 4 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 (O 2 -2 ), EACH oxygen is 1-. - peroxides are when 2 group one elements combine with the peroxide ion H 2 O 2, Li 2 O 2, K 2 O 2 are peroxides, each O is 1- BaO 2, CaO 2, MgO 2 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 F 2 O each oxygen is 2+ USE CAUTION IN BINARY COMPOUNDS OF OXYGEN AND HYDROGEN -- CHECK FOR RULE EXCEPTIONS. Fluorine is always 1- in its compounds SLIDE-10 OXYGEN RULE

11. THE HYDROGEN RULE : THE OXIDATION STATE OF HYDROGEN IS USUALLY 1+ EX in H 2 O 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. SLIDE-11 HYDROGEN RULE

12. THE HALOGEN RULE : THE OXIDATION STATE OF HALOGENS IS USUALLY 1- IN BINARY COMPOUNDS. EX in MgF 2 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… SLIDE-12 HALOGEN RULE

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

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(NO 3 ) 3 HAS Ni AND n, BOTH WITH MULTIPLE STATES. THE SHORT FORM FOR GETTING THE Ni IS: Ni (NO 3 ) 3 +3 (-1) * 3 +3 -3 CHARGE -1 CHARGE ON NITRATE ISFROM TABLE E

15. NI HAS MULTIPLE OXIDATION STATES AND HAS NO RULE, SOLVE FOR IT AS THE VARIABLE X NITROGEN HAS MULTIPLE OXIDATION STATES, SOLVE FOR THIS ALSO. OXYGEN IS 2- IN MOST COMPOUNDS …OXYGEN RULE. Ni(NO 3 ) 3 THIS SALT CONTAINS 2 ELEMENTS WITH MULTIPLE OXIDATION STATES (Ni AND N). YOU HAVE TWO WAYS TO SOLVE. 1)USE THE DISSOCIATION REACTION AND BALANCE FOR CHARGE TO GET THE METAL (Ni) OXIDATION STATE. THEN SOLVE FOR NITROGEN. 2)OR SOLVE FOR NITROGEN IN THE NITRATE ION FIRST, THEN SOLVE FOR Ni. SEE SOLUTIONS IN NEXT TWO SLIDES…

16. Ni HAS A OXIDATION STATE OF 3+ FROM THE DISSOCIATION ABOVE NITROGEN HAS MULTIPLE OXIDATION STATES OXYGEN IS 2- IN MOST COMPOUNDS …OXYGEN RULE. THIS SALT CONTAINS 2 ELEMENTS WITH MULTIPLE OXIDATION STATES (Ni AND N). 1)WRITE THE DISSOCIATION REACTION AND BALANCE FOR CHARGE TO GET THE METAL (Ni) OXIDATION STATE. THEN SOLVE FOR NITROGEN. Ni(NO 3 ) 3  Ni + 3 NO 3 - 0 = X + 3 (-1) X = Ni = +3 CHARGE FROM TABLE E Ni(NO 3 ) 3 +3 + 3(X) + 9(-2) = 0 X = OXIDATION STATE OF N = 5+

17. Ni HAS MULTIPLE OXIDATION STATES, SOLVE FOR IT. NITROGEN HAS AN OXIDATION STATE OF 5+ FROM THE DISSOCIATION CALCULATION OXYGEN IS 2- IN MOST COMPOUNDS …OXYGEN RULE. THIS SALT CONTAINS 2 ELEMENTS WITH MULTIPLE OXIDATION STATES (Ni AND N). 1)OR SOLVE FOR NITROGEN IN THE NITRATE ION FIRST, THEN SOLVE FOR Ni. SEE SOLUTIONS IN NEXT TWO SLIDES… NO 3 - CHARGE FROM TABLE E Ni(NO 3 ) 3 X + 3(+5) + 9(-2) = 0 X = OXIDATION STATE OF NI = 3+ X + 3(-2) = -1 X = OXIDATION STATE OF N = 5+