Variable oxidation states

2. Variable oxidation states Transition metals can form a number of stable ions, each with the metal in a different oxidation state. Variable oxidation states are possible because the 4s and 3d sub-levels are very close in energy. It is relatively easy to lose electrons from either of these sub-levels. The oxidation state of the metal in a compound is noted as a Roman numeral after the name of the metal. Iron(II) means that the compound contains Fe2+, whereas iron(III) contains Fe3+.

3. Identify the oxidation states

4. Vanadium Ammonium Vanadate (V) (white solid) can be added to dilute HCl to give the orange Dioxovanadium (V) ion VO3- + 2H+ VO2+ + H2O If granulated zinc is now added it reduces the vanadium over a period of several minutes and gives several colour changes Blue [VO(H2O)5]2+ Vanadium (IV) Green [VCl2(H2O)4] + Vanadium (III) Violet [V(H2O)6]2+ Vanadium (II)

5. Chromium Chromium in yellow Chromate (VI) is in its highest oxidation state. If we add acid the solution goes orange 2CrO42- + 2H+ Cr2O72- + H2O The Dichromate ion will also be reduced by zinc and HCl Orange Cr2O72- Cr (VI) Green [CrCl2(H2O)4]+ Cr (III) Blue [Cr(H2O)6]2+ Cr (II)

6. Oxidation states of vanadium

7. Oxidation states and bonding At lower oxidation states, transition metals form ionic bonds. At higher oxidation states, transition metals cannot form monatomic ions. Instead they bond covalently to form compounds or molecular ions. For example, manganese forms ionic bonds in manganese(II) chloride and covalent bonds in the MnO4– oxoanion. Transition metals are often in their highest oxidation state when bonded to very electronegative elements like oxygen and fluorine.

8. Common oxidation states

9. rally with your partner:describe everything you can about the diagram below and write 2 half equations MnO4- Mn2+ Fe2+ Fe3+

10. Construct an overall redox equation MnO4- + 8H+ + 5Fe2+  Mn2+ + 4H2O + 5Fe3+

12. Redox titrations In a titration, the concentration of a solution is determined by titrating with a solution of known concentration. In redox titrations, an oxidizing agent is titrated against a reducing agent. Electrons are transferred from one species to the other. Indicators are sometimes used to show the endpoint of the titration. However, most transition metal ions naturally change colour when changing oxidation state.

13. Oxidation states of manganese

14. Potassium manganate(VII) titration

15. Potassium manganate(VII) calculation

16. Using redox titrations to work out % of iron in iron tablets • Example question and answer

17. rally with your partner:describe everything you can about the diagram below and write 2 half equations Cr2O72- 2Cr3+ Fe2+ Fe3+

18. Construct an overall redox equation Cr2O72- + 14H+ + 6Fe2+  2Cr3+ + 7H2O + 6Fe3+

19. Potassium dichromate(VI) titrations Potassium dichromate(VI) (K2Cr2O7) is an oxidizing agent used in titrations. The oxidation state of the chromium ion is reduced from +6 to +3. The colour change in the titration is not very visible, so an indicator of sodium diphenylaminesulfonate is used. This turns from colourless to purple at the endpoint. The solution being titrated against must be acidified with excess dilute sulfuric acid.

20. Potassium dichromate(VI) equation

21. Using redox titrations to work out % of iron in lawn sand • Example question and answer

22. Potassium dichromate(VI) calculation

23. Oxidation of transition metal ions in alkaline solutions

24. Preparation of Ions in solution In redox titrations we saw a reduction of metal in a high oxidation state to a low one. This occurs in acidic solutions MnO4- Mn2+ H+ Cr2O72- 2Cr3+ H+ In alkaline solutions the opposite is possible (oxidation of metals in low oxidation states). This is due to the tendency to form –ve ions If we wish to store solutions such as Fe2+ ions we must store them in acid conditions . In alkaline conditions the ion will precipitate out as the hydroxide. Preparation of metal complexes with high oxidation numbers is done by :- i) adding an alkali ii) adding an oxidising agent

26. Redox reactions Transition metals are able to exist in many different oxidation states, which is why they often undergo redoxreactions. Oxidation of transition metals occurs most easily in alkaline solution. This is because negative ions tend to form in alkaline solution and it is easier to lose electrons from a negatively-charged species. Reduction of transition metals occurs most easily in acidic solution.

27. [Co(H2O)6]2+ + 2OH– [Co(H2O)4(OH)2] + 2H2O [Co(H2O)4(OH)2] + 6NH3 [Co(NH3)6]2+ + 2OH– + 4H2O 4[Co(NH3)6]2+ + O2 + 2H2O 4[Co(NH3)6]3+ + 4OH– 2[Co(OH)6]4– + H2O2 2[Co(OH)6]3– + 2OH– Oxidation of cobalt(II) In ammoniacal solution, Co2+ is oxidized to Co3+ by oxygen in the air. Several reactions occur because ammonia acts as both a base and a ligand: Co2+ can also be oxidized by hydrogen peroxide (H2O2) after adding an alkali such as sodium hydroxide.

28. 2CrO42– + 2H+ Cr2O72– + H2O Cr2O72– + 14H+ + 3Zn 2Cr3+ + 7H2O + 3Zn2+ Zn + 2Cr3+ Zn2+ + 2Cr2+ Reduction of chromium(VI) In aqueous solution, chromium has an oxidation state of 6+. It exists in alkaline solution as CrO42– and as Cr2O72– in acidic solution. Chromium(VI) can be reduced to Cr3+ and Cr2+ by zinc in acid solution . Cr2+ is easily oxidized to Cr3+ in the presence of oxygen, but hydrogen is produced during the reduction, which excludes air.

29. Preparation of Chromate (VI) in Alkaline Conditions We know addition of NaOH to a Chromium(III) salt will give a green precipitate. Precipitate Cr(H2O)63+ [Cr(H2O)3(OH)3] Excess NaOH cause this green precipitate to dissolve to give a green solution of [Cr(OH)6]3- If H2O2 is added to this a yellow solution of Chromate (VI) is produced 2 [Cr(OH)6]3- + 3 H2O2 2CrO42- + 2OH- + 8H2O Needs :- alkaline conditions & oxidising agent H2O2

30. Identify the reaction