Head to savemyexams.com for more awesome resources Edexcel A Level Chemistry 6.3 Reactions of Transition Metal Elements Contents 6.3.1 Vanadium 6.3.2 Chromium 6.3.3 Reactions of Ions in Aqueous Solution 6.3.4 Ligand Exchange Page 1 of 12 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers Your notes Head to savemyexams.com for more awesome resources 6.3.1 Vanadium Your notes Colours & Oxidation States Vanadium is a transition metal which has variable oxidation states The table below shows the important ones you need to be aware of Addition of z inc to the vanadium(V) in acidic solution will reduce the vanadium down through each successive oxidation state The colour would successively change from yellow to blue to green to violet The ion with the V at oxidation state +5 exists as a solid compound in the form of a VO 3- ion Usually as NH4VO 3 known as ammonium vanadate(V) It is a reasonably strong oxidising agent Addition of acid to the solid will turn into the yellow solution containing the VO 2 + ion. Page 2 of 12 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers Head to savemyexams.com for more awesome resources Interconversions of Vanadium Ions For vanadium we need to consider the following standard electrode potential values We will use z inc as our chosen oxidising agent The half equations are arranged from high negative EΘ at the top to high positive EΘ at the bottom The best reducing agent is the top right species (V2+) The best oxidising agent is the bottom left species (VO 2+) Reduction from +5 to +4 The two half equations we need to consider are 2 and 5 Vanadium is reduced from an oxidation number of +5 to +4 in half equation 5 The EΘ value for half equation 2 is more negative than the EΘ for half equation 5 Zn is the best reducing agent VO 2 + is the best oxidising agent We can obtain the overall equation by reversing half equation 2 and combining with equation 5 When adding half equations remember to multiply them so each have the same number of electrons 2VO 2 + (aq) + 4H+ (aq) + Zn (s) → 2VO 2+ (aq) + Zn2+ (aq) + 2H2 O (l) Reduction from +4 to +3 The two half equations we need to consider are 2 and 4 Vanadium is reduced from an oxidation number of +4 to +3 in half equation 4 The EΘ value for half equation 2 is more negative than the EΘ for half equation 5 Zn is the best reducing agent VO 2 + is the best oxidising agent We can obtain the overall equation by reversing half equation 2 and combining with equation 4 When adding half equations remember to multiply them so each have the same number of electrons 2VO 2+ (aq)+ 4H+ (aq) + Zn (s) → 2V3+ (aq) + Zn2+ (aq) + 2H2 O (l) Reduction from +3 to +2 The two half equations we need to consider are 2 and 3 Page 3 of 12 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers Your notes Head to savemyexams.com for more awesome resources Vanadium is reduced from an oxidation number of +3 to +2 in half equation 3 The EΘ value for half equation 2 is more negative than the EΘ for half equation 3 Zn is the best reducing agent V3+ is the best oxidising agent We can obtain the overall equation by reversing half equation 2 and combining with equation 3 When adding half equations remember to multiply them so each have the same number of electrons 2V3+ (aq) + Zn (s) → 2V2+ (aq) + Zn2+ (aq) Reduction from +2 to 0 The two half equations we need to consider are 1 and 2 Vanadium is reduced from an oxidation number of +2 to 0 in half equation 1 The EΘ value for half equation 1 is more negative than the EΘ for half equation 2 Zn is not electron releasing with respect to V2+ This means this reaction is not thermodynamically feasible Predicting oxidation reactions The same method can be used to predict whether a given oxidising agent will oxidise a vanadium species to one with a higher oxidation number Exam Tip It is important to not get confused between the two oxo ions of vanadium VO 2 + and VO 2+ Page 4 of 12 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers Your notes Head to savemyexams.com for more awesome resources 6.3.2 Chromium Reduction & Oxidation of Chromium Species For chromium we need to consider the following standard electrode potential values We will use z inc and hydrogen peroxide as oxidising agents The half equations are arranged from high negative EΘ at the top to high positive EΘ at the bottom The best reducing agent is the top right species (Zn (s)) The best oxidising agent is the bottom left species (Cr2 O 72-(aq)) Oxidation from +3 to +6 The two half equations we need to consider are 3 and 4 Chromiums oxidation number changes from +6 to +3 in half equation 3 The EΘ value for half equation 3 is more negative than the EΘ for half equation 4 Cr(OH)3 (aq) is the best reducing agent H2 O 2 (aq) is the best oxidising agent We can obtain the overall equation by reversing half equation 3 and combining it with equation 4 When adding half equations remember to multiply them so each has the same number of electrons 2Cr(OH)3 (aq) + 4OH- (aq) + 3H2 O 2 (aq) → 2CrO 4 2- (aq) + 8H2 O (l) This reaction is carried out in alkaline conditions due to the presence of OH- ions in the equation Reduction from +6 to +3 The two half equations we need to consider are 1 and 5 Page 5 of 12 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers Your notes Head to savemyexams.com for more awesome resources Chromiums oxidation number changes from +6 to +3 in half equation 3 The EΘ value for half equation 1 is more negative than the EΘ for half equation 5 Zn is the best reducing agent Cr2 O 72- is the best oxidising agent We can obtain the overall equation by reversing half equation 1 and combining it with equation 5 When adding half equations remember to multiply them so each has the same number of electrons Cr 2 O 7 2- (aq) + 14H+ (aq) + 3Zn (s) → 2Cr 3+ (aq) + 7H2 O (l) + 3Zn2+ (aq) This reaction is carried out under acidic conditions due to presence of H+ in the equation Reduction from +3 to +2 The Cr3+ ion can be further reduced by z inc The two half equations we need to consider are 1 and 2 Chromiums oxidation number changes from +3 to +2 in half equation 3 The EΘ value for half equation 1 is more negative than the EΘ for half equation 2 Zn (s) is the best reducing agent Cr3+ (aq) is the best oxidising agent We can obtain the overall equation by reversing half equation 1 and combining it with equation 2 When adding half equations remember to multiply them so each has the same number of electrons 2Cr 3+ (aq) + Zn (s) → 2Cr 2+ (aq) + Zn2+ (aq) As this reaction is a further step from the previous reduction this reaction is also carried out under acidic conditions The Dichromate(VI) - Chromate(VI) Equilibrium The chromate CrO 42- and dichromate Cr2 O 72- ions can be converted from one to the other by the following equilibrium reaction 2CrO 4 2- (aq) + 2H+ (aq) ⇌ Cr 2 O 7 2- (aq) + H2 O (l) Chromate(VI) ions are stable in alkaline solution, but in acidic conditions the dichromate(VI) ion is more stable Addition of acid will push the equilibrium to the dichromate This results in a colour change from yellow to orange Addition of alkali will remove the H+ ions and push the equilibrium to the chromate This is not a redox reaction as both the chromate and dichromate ions have an oxidation number of +6 This is an acid base reaction Page 6 of 12 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers Your notes Head to savemyexams.com for more awesome resources 6.3.3 Reactions of Ions in Aqueous Solution Reactions with Hydroxide & Ammonia When transition metal ions in aqueous solution react with aqueous sodium hydroxide and aqueous ammonia they form precipitates However some of these precipitates will dissolve in an excess of sodium hydroxide or ammonia to form complex ions in solution The Reactions of Aqueous Transition Metal Ions with Aqueous Sodium Hydroxide Examples of ionic equations for the reactions in the table above [Fe(H2 O)6]2+ (aq) + 2OH- (aq) → [Fe(H2 O)4(OH)2 ] (s) + 2H2 O (l) [Cu(H2 O)6]2+ (aq) + 2OH- (aq) → [Cu(H2 O)4(OH)2 ] (s) + 2H2 O (l) [Fe(H2 O)6]3+ (aq) + 3OH- (aq) → [Fe(H2 O)3(OH)3] (s) + 3H2 O (l) The Reactions of Aqueous Transition Metal Ions with Ammonia Page 7 of 12 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers Your notes Head to savemyexams.com for more awesome resources Your notes Examples of ionic equations for the reactions in the table above [Fe(H2 O)6]2+ (aq) + 2NH3 (aq) → [Fe(H2 O)4(OH)2 ] (s) + 2NH4+ (aq) [Cu(H2 O)6]2+ (aq) + 2NH3 (aq) → [Cu(H2 O)4(OH)2 ] (s) +2NH4+ (aq) [Fe(H2 O)6]3+ (aq) + 3NH3 (aq) → [Fe(H2 O)3(OH)3] (s) +3NH4+ (aq) [Cu(H2 O)4(OH)2 ]2+ (aq) + 4NH3 (aq) → [Cu(H2 O)2 (OH)4] (aq) + 2H2 O (l) + 2OH- (aq) Solutions of metal aqua ions react asacids with aqueous ammonia, whilst some react further with excess ammonia Initially, ammonia acts as abaseto remove one H+ ion per ammonia molecule used With excess ammonia, some metal ions undergo ligand substitution with NH3 Exam Tip It is easiest to remember the formulas of the precipitates by remembering that the number of OHions substituted is the same as the value of the charge on the initial ion Page 8 of 12 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers Head to savemyexams.com for more awesome resources Ionic Equations Reaction with limited OH- and limited NH3 Your notes The bases OH- and ammonia when in limited amounts form the same hydroxide precipitates. They form in deprotonation acid-base reactions For example, consider the reaction that occurs when aqueous sodium hydroxide is added to copper(II) sulfate solution [Cu(H2 O)6 ]2+ (aq) + 2OH- (aq) → [Cu(H2 O)4 (OH)2 ] (s) + 2H2 O (l) This seems like a ligand substitution reaction - two hydroxide ions replacing two water molecules However this is actually a deprotonation reaction - two hydroxide ions removing hydrogen ions from two of the water ligands converting them into water molecules The two ligands that have lost hydrogen ions are now hydroxide ligands Reaction with excess OH- From above, we have seen how hydrated transition metal ions can be deprotonated by adding a base such as aqueous sodium hydroxide to form a metal hydroxide precipitate For example [Cr(H2 O)6 ]3+ (aq) + 3OH- (aq) → [Cr(H2 O)3 (OH)3 ] (s) +3H2 O (l) When an excess of sodium hydroxide is added further deprotonation takes place [Cr(H2 O)3 (OH)3 ] (s) + 3OH- (aq) → [Cr(OH)6 ]3- (aq) + 3H2 O (l) In this reaction, chromium(III) hydroxide acts as an acid, as it is reacting with a base Chromium(III) hydroxide can also act as a base because it can react with acids as follows [Cr(H2 O)3 (OH)3 ] (s) + 3H+ (aq) → [Cr(H2 O)6 ]3+ (aq) A metal hydroxide that can act as both an acid and a base is called an amphoteric hydroxide This is an example of amphoteric behaviour Reaction with excess NH3 With excess NH3 ligand substitution reactions occur with Cu, Co and Cr and their precipitates dissolve The ligands NH3 and H2 O are similar in siz e and are uncharged Ligand exchange occurs without a change of co-ordination number for Co and Cr For example, when excess aqueous ammonia is added to a copper(II) hydroxide precipitate it dissolves forming a deep blue solution [Cu(H2 O)4 (OH)2 ] (s) + 4NH3 (aq) → [Cu(NH3 )4 (H2 O)2 ]2+ (aq) + 2H2 O (l) + 2OH- (aq) This is a ligand substitution - four ammonia molecules replace two water molecules and two hydroxide ions In these reactions, NH3 is acting as a Lewis base donating an electron pair Reaction with Cl - Page 9 of 12 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers Head to savemyexams.com for more awesome resources The chloride ligand undergoes substitution reactions with Cu, Co and Fe(III) The Cl– ligand is much larger in siz e than H2 O and is charged This means that there is a change in: Coordination number from 6 to 4 Shape from octahedral to tetrahedral The overall charge on the complex For example, when excess / concentrated hydrochloric acid is added to a copper(II) hydroxide precipitate it forms a yellow solution [Cu(H2 O)6 ]2+ (aq) + 4Cl– (aq) → [CuCl4 ]2– (aq) + 6H2 O (l) This is a ligand substitution - four chloride ions replace six water molecules In these reactions, Cl– is acting as a Lewis base donating an electron pair Similar reactions occur with: Pink [Co(H2 O)6]2+ (aq) forming blue [CoCl4]2– (aq) Yellow [Fe(H2 O)6]3+ (aq) forming orange [FeCl4]– (aq) Page 10 of 12 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers Your notes Head to savemyexams.com for more awesome resources 6.3.4 Ligand Exchange Your notes Ligand Exchange Ligand exchange (or ligand substitution) is when one ligand in a complex is replaced by another Ligand exchange forms a new complex that is more stable than the original one The ligands in the original complex can be partially or entirely substituted by others The complex ion can change its charge or remain the same depending on the ligand involved There are no changes in coordination number, or the geometry of the complex, if the ligands are of a similar size But, if the ligands are of a different size, for example water ligands and chloride ligands, then a change in coordination number and the geometry of the complex will occur Addition of a high concentration of chloride ions (from conc HCl or saturated NaCl) to an aqueous ion leads to a ligand substitution reaction. The Cl- ligand is larger than the uncharged H2 O and NH3 ligands so therefore ligand exchange can involve a change of co-ordination number For example when concentrated hydrochloric acid is added slowly and continuously to a copper(II) sulfate solution the colour changes from blue to green then finally yellow The equation for this reaction is [Cu(H2 O)6 ]2+ (aq) + 4Cl- (aq) ⇌ [CuCl4 ]2- (aq) + 6H2 O (l) We can see that all six water ligands have been replaced by four chloride ions This reaction involves a change in coordination number from 6 to 4 Note that despite the charge on the complex changing from +2 to -2, there has been no change in oxidation number of the copper We can also see that this reaction is reversible, which helps to explain the observed colour change The hexaaquacopper(II) ion is blue The tetrachlorocuprate(II) ion is yellow The green colour is due to a mixture of the blue and yellow complex ions A similar reaction also takes place with cobalt resulting in a blue solution and a change in coordination number from 6 to 4 [Co(H2 O)6 ]2+ (aq) + 4Cl- (aq) ⇌ [CoCl4 ]2- (aq) + 6H2 O (l) Exam Tip Be careful: If solid copper chloride (or any other metal) is dissolved in water it forms the aqueous [Cu(H2 O)6]2+ complex and not the chloride [CuCl4 ]2- complex Page 11 of 12 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers Head to savemyexams.com for more awesome resources The Chelate Effect & Stability The replacement of monodentate ligands with bidentate and multidentate ligands in complex ions is called the chelate effect It is an energetically favourable reaction, meaning that ΔGꝋ is negative The driving force behind the reaction is entropy The Gibbs equation reminds us of the link between enthalpy and entropy: ΔGꝋ = ΔHreactionꝋ – T ΔSsystemꝋ Reactions in solution between aqueous ions usually come with relatively small enthalpy changes However, the entropy changes are always positive in chelation because the reactions produce a net increase in the number of particles A small enthalpy change and relative large positive entropy change generally ensures that the overall free energy change is negative For example, when EDTA chelates with aqueous cobalt(II) two reactants becomes seven product species [Co(H2 O)6 ]2+ (aq)+ EDTA4- (aq) → [CoEDTA]2- (aq) + 6H2 O (l) The ligand EDTA readily chelates with aqueous transition metal ions in an energetically favourable reaction Page 12 of 12 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers Your notes
