1. Oxidation states This is a method of interpreting redox reactions This method, also called oxidation number, assigns numbers to atoms to show how many electrons they have used in bonding This method can be applied to covalent bonding as well as ionic bonding

2. Used to...tell if oxidation or reduction has taken place work out what has been oxidised and/or reduced construct half equations and balance redox equations ATOMS AND SIMPLE IONS The number of electrons which must be added or removed to become neutral atomsNa in Na= 0neutral already... no need to add any electrons cationsNa in Na + = +1need to add 1 electron to make Na + neutral anionsCl in Cl¯ = -1 need to take 1 electron away to make Cl¯ neutral OXIDATION STATES

3. Q. What are the oxidation states of the elements in the following? a) C b) Fe 3+ c) Fe 2+ d) O 2- e) He f) Al 3+ Q. What are the oxidation states of the elements in the following? a) C b) Fe 3+ c) Fe 2+ d) O 2- e) He f) Al 3+ Used to...tell if oxidation or reduction has taken place work out what has been oxidised and/or reduced construct half equations and balance redox equations ATOMS AND SIMPLE IONS The number of electrons which must be added or removed to become neutral atomsNa in Na= 0neutral already... no need to add any electrons cationsNa in Na + = +1need to add 1 electron to make Na + neutral anionsCl in Cl¯ = -1 need to take 1 electron away to make Cl¯ neutral

4. OXIDATION STATES Q. What are the oxidation states of the elements in the following? a) C (0)b) Fe 3+ (+3)c) Fe 2+ (+2) d) O 2- (-2) e) He (0) f) Al 3+ (+3) Q. What are the oxidation states of the elements in the following? a) C (0)b) Fe 3+ (+3)c) Fe 2+ (+2) d) O 2- (-2) e) He (0) f) Al 3+ (+3) Used to...tell if oxidation or reduction has taken place work out what has been oxidised and/or reduced construct half equations and balance redox equations ATOMS AND SIMPLE IONS The number of electrons which must be added or removed to become neutral atomsNa in Na= 0neutral already... no need to add any electrons cationsNa in Na + = +1need to add 1 electron to make Na + neutral anionsCl in Cl¯ = -1 need to take 1 electron away to make Cl¯ neutral

5. OXIDATION STATES MOLECULES The SUM of the oxidation states adds up to ZERO ELEMENTSH in H 2 = 0both are the same and must add up to Zero COMPOUNDSC in CO 2 = +4 O in CO 2 = -21 x +4 and 2 x -2 = Zero

6. because CO 2 is a neutral molecule, the sum of the oxidation states must be zero for this, one element must have a positive OS and the other must be negative OXIDATION STATES Explanation MOLECULES The SUM of the oxidation states adds up to ZERO ELEMENTSH in H 2 = 0both are the same and must add up to Zero COMPOUNDSC in CO 2 = +4 O in CO 2 = -21 x +4 and 2 x -2 = Zero

7. HOW DO YOU DETERMINE WHICH IS THE POSITIVE ONE? the more electronegative species will have the negative value electronegativity increases across a period and decreases down a group O is further to the right than C in the periodic table so it has the negative value OXIDATION STATES MOLECULES The SUM of the oxidation states adds up to ZERO ELEMENTSH in H 2 = 0both are the same and must add up to Zero COMPOUNDSC in CO 2 = +4 O in CO 2 = -21 x +4 and 2 x -2 = Zero

8. HOW DO YOU DETERMINE THE VALUE OF AN ELEMENT’S OXIDATION STATE? from its position in the periodic table and/or the other element(s) present in the formula OXIDATION STATES MOLECULES The SUM of the oxidation states adds up to ZERO ELEMENTSH in H 2 = 0both are the same and must add up to Zero COMPOUNDSC in CO 2 = +4 O in CO 2 = -21 x +4 and 2 x -2 = Zero

9. OXIDATION STATES in SO 4 2- the oxidation state of S = +6there is ONE S O = -2there are FOUR O’s +6 + 4(-2) = -2 so the ion has a 2- charge COMPLEX IONS The SUM of the oxidation states adds up to THE CHARGE e.g.NO 3 - sum of the oxidation states = - 1 SO 4 2- sum of the oxidation states = - 2 NH 4 + sum of the oxidation states = +1 Examples

10. OXIDATION STATES What is the oxidation state (OS) of Mn in MnO 4 ¯ ? the oxidation state of oxygen in most compounds is - 2 there are 4 O’s so the sum of its oxidation states- 8 overall charge on the ion is - 1 therefore the sum of all the oxidation states must add up to - 1 the oxidation states of Mn four O’s must therefore equal - 1 therefore the oxidation state of Mn in MnO 4 ¯is +7 +7 + 4(-2) = - 1 COMPLEX IONS The SUM of the oxidation states adds up to THE CHARGE e.g.NO 3 - sum of the oxidation states = - 1 SO 4 2- sum of the oxidation states = - 2 NH 4 + sum of the oxidation states = +1 Examples

11. HYDROGEN +1 except 0atom (H) and molecule (H 2 ) -1hydride ion, H¯ in sodium hydride NaH OXYGEN -2 except 0atom (O) and molecule (O 2 ) -1in hydrogen peroxide, H 2 O 2 +2in F 2 O FLUORINE -1 except 0atom (F) and molecule (F 2 ) OXIDATION STATES CALCULATING OXIDATION STATE - 1 Many elements can exist in more than one oxidation state In compounds, certain elements are used as benchmarks to work out other values

12. HYDROGEN +1 except 0atom (H) and molecule (H 2 ) -1hydride ion, H¯ in sodium hydride NaH OXYGEN -2 except 0atom (O) and molecule (O 2 ) -1in hydrogen peroxide, H 2 O 2 +2in F 2 O FLUORINE -1 except 0atom (F) and molecule (F 2 ) OXIDATION STATES Q. Give the oxidation state of the element other than O, H or F in... SO 2 NH 3 NO 2 NH 4 + IF 7 Cl 2 O 7 NO 3 ¯NO 2 ¯SO 3 2- S 2 O 3 2- S 4 O 6 2- MnO 4 2- What is odd about the value of the oxidation state of S in S 4 O 6 2- ? Q. Give the oxidation state of the element other than O, H or F in... SO 2 NH 3 NO 2 NH 4 + IF 7 Cl 2 O 7 NO 3 ¯NO 2 ¯SO 3 2- S 2 O 3 2- S 4 O 6 2- MnO 4 2- What is odd about the value of the oxidation state of S in S 4 O 6 2- ? CALCULATING OXIDATION STATE - 1 Many elements can exist in more than one oxidation state In compounds, certain elements are used as benchmarks to work out other values

13. OXIDATION STATES A. The oxidation states of the elements other than O, H or F are SO 2 O = -22 x -2 = - 4overall neutralS = +4 NH 3 H = +13 x +1 = +3overall neutralN = - 3 NO 2 O = -2 2 x -2 = - 4 overall neutralN = +4 NH 4 + H = +14 x +1 = +4overall +1N = - 3 IF 7 F = -17 x -1 = - 7overall neutralI = +7 Cl 2 O 7 O = -2 7 x -2 = -14 overall neutralCl = +7 (14/2) NO 3 ¯ O = -2 3 x -2 = - 6 overall -1N = +5 NO 2 ¯ O = -2 2 x -2 = - 4 overall -1N = +3 SO 3 2- O = -2 3 x -2 = - 6 overall -2S = +4 S 2 O 3 2- O = -2 3 x -2 = - 6 overall -2S = +2 (4/2) S 4 O 6 2- O = -2 6 x -2 = -12 overall -2S = +2½ ! (10/4) MnO 4 2- O = -2 4 x -2 = - 8overall -2Mn = +6 What is odd about the value of the oxidation state of S in S 4 O 6 2- ? An oxidation state must be a whole number (+2½ is the average value)

14. METALS have positive values in compounds value is usually that of the Group Number Al is +3 where there are several possibilities the values go no higher than the Group No.Sn can be +2 or +4 Mn can be +2,+4,+6,+7 NON-METALS mostly negative based on their usual ionCl usually -1 can have values up to their Group No. Cl +1 +3 +5 or +7 OXIDATION STATES CALCULATING OXIDATION STATE - 2 The position of an element in the periodic table can act as a guide

15. OXIDATION STATES Q. What is the theoretical maximum oxidation state of the following elements? NaPBaPbSMnCr What will be the usual and the maximum oxidation state in compounds of? LiBrSrOBN+1 Q. What is the theoretical maximum oxidation state of the following elements? NaPBaPbSMnCr What will be the usual and the maximum oxidation state in compounds of? LiBrSrOBN+1 METALS have positive values in compounds value is usually that of the Group Number Al is +3 where there are several possibilities the values go no higher than the Group No.Sn can be +2 or +4 Mn can be +2,+4,+6,+7 NON-METALS mostly negative based on their usual ionCl usually -1 can have values up to their Group No. Cl +1 +3 +5 or +7 CALCULATING OXIDATION STATE - 2 The position of an element in the periodic table can act as a guide

16. OXIDATION STATES CALCULATING OXIDATION STATE - 2 The position of an element in the periodic table can act as a guide A. What is the theoretical maximum oxidation state of the following elements? NaPBaPbSMnCr +1+5+2+4+6+7+6 What will be the usual and the maximum oxidation state in compounds of? LiBrSrOBN USUAL +1-1+2-2+3 -3 or +5 MAXIMUM +1+7+2+6+3 +5

17. OXIDATION STATES CALCULATING OXIDATION STATE - 2 Q. What is the oxidation state of each element in the following compounds/ions ? CH 4 PCl 3 NCl 3 CS 2 ICl 5 BrF 3 PCl 4 + H 3 PO 4 NH 4 Cl H 2 SO 4 MgCO 3 SOCl 2

18. OXIDATION STATES CALCULATING OXIDATION STATE - 2 Q. What is the oxidation state of each element in the following compounds/ions ? CH 4 C = - 4H = +1 PCl 3 P = +3Cl = -1 NCl 3 N = +3Cl = -1 CS 2 C = +4S = -2 ICl 5 I = +5Cl = -1 BrF 3 Br = +3F = -1 PCl 4 + P = +4Cl = -1 H 3 PO 4 P = +5H = +1O = -2 NH 4 ClN = -3H = +1Cl = -1 H 2 SO 4 S = +6H = +1O = -2 MgCO 3 Mg = +2H = +4O = -2 SOCl 2 S = +4Cl = -1O = -2

19. manganese(IV) oxide shows that Mn is in the +4 oxidation state in MnO 2 sulphur(VI) oxide for SO 3 S is in the +6 oxidation state dichromate(VI) for Cr 2 O 7 2- Cr is in the +6 oxidation state phosphorus(V) chloride for PCl 5 P is in the +5 oxidation state phosphorus(III) chloride for PCl 3 P is in the +3 oxidation state OXIDATION STATES THE ROLE OF OXIDATION STATE IN NAMING SPECIES To avoid ambiguity, the oxidation state is often included in the name of a species Q. Name the following...PbO 2 SnCl 2 SbCl 3 TiCl 4 BrF 5

20. OXIDATION STATES Q. Name the following...PbO 2 lead(IV) oxide SnCl 2 tin(II) chloride SbCl 3 antimony(III) chloride TiCl 4 titanium(IV) chloride BrF 5 bromine(V) fluoride manganese(IV) oxide shows that Mn is in the +4 oxidation state in MnO 2 sulphur(VI) oxide for SO 3 S is in the +6 oxidation state dichromate(VI) for Cr 2 O 7 2- Cr is in the +6 oxidation state phosphorus(V) chloride for PCl 5 P is in the +5 oxidation state phosphorus(III) chloride for PCl 3 P is in the +3 oxidation state THE ROLE OF OXIDATION STATE IN NAMING SPECIES To avoid ambiguity, the oxidation state is often included in the name of a species

21. REDOXWhen reduction and oxidation take place OXIDATION Removal (loss) of electrons ‘OIL’ species will get less negative or more positive REDUCTIONGain of electrons ‘RIG’ species will become more negative or less positive REDOX REACTIONS OXIDATION AND REDUCTION IN TERMS OF ELECTRONS Oxidation and reduction are not only defined as changes in O and H

22. REDOXWhen reduction and oxidation take place OXIDATION Removal (loss) of electrons ‘OIL’ species will get less negative or more positive REDUCTIONGain of electrons ‘RIG’ species will become more negative or less positive REDUCTION in O.S. Species has been REDUCED e.g. Cl is reduced to Cl¯ (0 to -1) INCREASE in O.S. Species has been OXIDISED e.g. Na is oxidised to Na + (0 to +1) REDOX REACTIONS OXIDATION AND REDUCTION IN TERMS OF ELECTRONS Oxidation and reduction are not only defined as changes in O and H

23. REDUCTION in O.S. INCREASE in O.S. Species has been REDUCED Species has been OXIDISED REDOX REACTIONS OXIDATION AND REDUCTION IN TERMS OF ELECTRONS Q. State if the changes involve oxidation (O) or reduction (R) or neither (N) Fe 2+ —>Fe 3+ I 2 —>I¯ F 2 —> F 2 O C 2 O 4 2- —> CO 2 H 2 O 2 —> O 2 H 2 O 2 —> H 2 O Cr 2 O 7 2- —> Cr 3+ Cr 2 O 7 2- —> CrO 4 2- SO 4 2- —> SO 2

24. REDOX REACTIONS REDUCTION in O.S. INCREASE in O.S. Species has been REDUCED Species has been OXIDISED OXIDATION AND REDUCTION IN TERMS OF ELECTRONS Q. State if the changes involve oxidation (O) or reduction (R) or neither (N) Fe 2+ —>Fe 3+ O +2 to +3 I 2 —>I¯ R 0 to -1 F 2 —> F 2 OR 0 to -1 C 2 O 4 2- —> CO 2 O +3 to +4 H 2 O 2 —> O 2 O -1 to 0 H 2 O 2 —> H 2 O R -1 to -2 Cr 2 O 7 2- —> Cr 3+ R +6 to +3 Cr 2 O 7 2- —> CrO 4 2- N +6 to +6 SO 4 2- —> SO 2 R +6 to +4