Transition metal atoms and transition metal ions form complexes by combining with ligands.

1. (a)Explain why transition elements form complex ions.[1]
2. (b)Co2+ ions form complex ion G. Each G ion contains two Co2+ ions, both of which are octahedrally coordinated. Each G ion contains one O2 molecule, which donates one pair of electrons to each Co2+ ion, and one NH₂- ion, which donates one pair of electrons to each Co2+ ion. The remaining ligands are NH3 molecules.
3. (b)(i)Deduce the formula of complex ion G. Include its overall charge. formula of G[2]
4. (b)(ii)The d-orbitals of the Co2+ ions present in complex ion G are split. State the number of d-orbitals that are at a higher energy level and the number of d-orbitals that are at a lower energy level in each Co2+ ion. number of d-orbitals at a higher energy level number of d-orbitals at a lower energy level[1]
5. (b)(iii)Co2+ ions form a different complex ion, M. Each M ion contains two Co2+ ions, both of which are octahedrally coordinated, but the ligands are different from the ligands in G. Explain why G and M have different colours.[2]
6. (c)Cadmium forms complex ion X, [Cd(NH3)4]2+. When a solution containing CN¯ ions is added to an aqueous solution of X, a ligand exchange reaction takes place, forming complex ion Y. Y contains no NH3 ligands and no H₂O ligands. Y is in a much higher concentration in the mixture than X. The oxidation state and coordination number of cadmium do not change in this reaction.
7. (c)(i)Write an ionic equation for this reaction, using the formulae of the complex ions.[2]
8. (c)(ii)Cadmium forms complex ion Z in the same oxidation state and with the same coordination number as in X. All the ligands in Z are Cl- ions. When NaCl(aq) is added to a solution of X, very little Z forms. Write the three cadmium complexes, X, Y and Z, in order of increasing stability constant, Kstab smallest value of K stab largest value of Kstab[1]
9. (d)Ethanedioate ions, C2O42-, form complexes with transition element ions. The concentration of C2O42- ions can be found by reaction with acidified Cr2O72- ions. C2O42- ions are protonated and form HOOCCOOH molecules which are oxidised by Cr2O72-. The half-equations are shown. Cr2O72- + 14H+ + 6e¯ = 2Cr3+ + 7H2O 2CO2 + 2H+ + 2e¯ HOOCCOOH
10. (d)(i)Construct an equation for the reaction between acidified Cr2O72- and HOOCCOOH.[1]
11. (d)(ii)A 25.0 cm³ sample of a solution of Na2C2O4 reacts with exactly 16.20 cm³ of an acidified solution of 0.0500 mol dm³ K2Cr2O7. Calculate the concentration of the solution of Na2C2O4- [Na2C2O4] = moldm-3[2]