Explain why transition elements make complex ions.
Complex ion G is formed by $\text{Co}^{2+}$ ions. Each G ion contains two $\text{Co}^{2+}$ ions, and both are octahedrally coordinated. In each G ion, one $\text{O}_2$ molecule donates one pair of electrons to each $\text{Co}^{2+}$ ion, and one $\text{NH}_2^-$ ion donates one pair of electrons to each $\text{Co}^{2+}$ ion. The other ligands are $\text{NH}_3$ molecules. Deduce the formula of complex ion G. Include its overall charge.
The d-orbitals of the $\text{Co}^{2+}$ ions in complex ion G are split. State how many d-orbitals are at the higher energy level and how many are at the lower energy level in each $\text{Co}^{2+}$ ion.
$\text{Co}^{2+}$ ions form a separate complex ion, M. Each M ion contains two $\text{Co}^{2+}$ ions, and both are octahedrally coordinated, but the ligands are not the same as those in G. Explain why G and M have different colours.
Co$^{2+}$ ions form complex ion G. Each G ion contains two Co$^{2+}$ ions, both of which are octahedrally coordinated. Each G ion contains one O$_2$ molecule, which donates one pair of electrons to each Co$^{2+}$ ion, and one NH$_2^-$ ion, which donates one pair of electrons to each Co$^{2+}$ ion. The other ligands are NH$_3$ molecules.
Cadmium forms complex ion X, $[\text{Cd(NH}_3)_4]^{2+}$. When a solution containing CN$^-$ ions is added to an aqueous solution of X, a ligand exchange reaction occurs, producing complex ion Y. Y has no NH$_3$ ligands and no H$_2$O ligands. Y is present in a much higher concentration in the mixture than X. The oxidation state and coordination number of cadmium do \textbf{not} change in this reaction.
Using the formulae of the complex ions, write the ionic equation for this reaction.
Cadmium forms complex ion Z in the same oxidation state and with the same coordination number as in X. All of the ligands in Z are Cl$^-$ ions. When NaCl(aq) is added to a solution of X, only a very small amount of Z is formed. Write the three cadmium complexes, X, Y and Z, in order of increasing stability constant, $K_{\text{stab}}$.
Ethanedioate ions, $\text{C}_2\text{O}_4^{2-}$, form complexes with transition element ions. The concentration of $\text{C}_2\text{O}_4^{2-}$ ions can be determined by reaction with acidified $\text{Cr}_2\text{O}_7^{2-}$ ions. $\text{C}_2\text{O}_4^{2-}$ ions are protonated and form HOOCCOOH molecules which are oxidised by $\text{Cr}_2\text{O}_7^{2-}$. The half-equations are shown. $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightleftharpoons 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ $2\text{CO}_2 + 2\text{H}^+ + 2e^- \rightleftharpoons \text{HOOCCOOH}$
Construct an equation for the reaction of acidified $\text{Cr}_2\text{O}_7^{2-}$ with HOOCCOOH.
A $25.0\text{ cm}^3$ sample of a solution of Na$_2$C$_2$O$_4$ reacts with exactly $16.20\text{ cm}^3$ of an acidified solution of $0.0500\text{ mol dm}^{-3}$ K$_2$Cr$_2$O$_7$. Calculate the concentration of the solution of Na$_2$C$_2$O$_4$.