Explain why transition elements are able to form complex ions.
The $\text{Co}^{2+}$ ions make complex ion $G$. Each $G$ ion contains two $\text{Co}^{2+}$ ions, and each metal centre is octahedrally coordinated. One $\text{O}_2$ molecule is present in each $G$ ion and donates one pair of electrons to each $\text{Co}^{2+}$ ion, while one $\text{NH}_2^-$ ion does the same. 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 present in complex ion $G$ are split. State how many d-orbitals are at a higher energy level and how many are at a lower energy level in each $\text{Co}^{2+}$ ion.
The $\text{Co}^{2+}$ ions make a different complex ion, $M$. Each $M$ ion contains two $\text{Co}^{2+}$ ions, both of which are octahedrally coordinated, but the ligands differ from those in $G$. Explain why $G$ and $M$ have different colours.
The $Co^{2+}$ ions make complex ion G. Each G ion contains two Co$^{2+}$ ions, and each one is octahedrally coordinated. One O$_2$ molecule is present in each G ion and donates one pair of electrons to each Co$^{2+}$ ion, and one NH$_2^-$ ion donates one pair of electrons to each Co$^{2+}$ ion. The remaining ligands are NH$_3$ molecules.
Cadmium forms complex ion X, $\text{[Cd(NH}_3)_4\text{]}^{2+}$. When CN$^-$ ions are added to an aqueous solution of X, ligand exchange occurs and complex ion Y is produced. Y contains no NH$_3$ ligands and no H$_2$O ligands. Y is present at a much higher concentration in the mixture than X. The oxidation state and coordination number of cadmium stay the same in this reaction.
Write an ionic equation for this reaction, using the formulae of the complex ions.
Cadmium forms complex ion Z with the same oxidation state and coordination number as in X. Every ligand in Z is Cl$^-$. When NaCl(aq) is added to a solution of X, only a very small amount of Z forms. Write the three cadmium complexes, X, Y and Z, in order of increasing stability constant, $K_{\text{stab}}$.
Ethanedioate ions, C$_2$O$_4^{2-}$, can form complexes with transition element ions. Their concentration can be determined by reacting them with acidified Cr$_2$O$_7^{2-}$ ions. The C$_2$O$_4^{2-}$ ions are protonated to form HOOCCOOH molecules, which are oxidised by Cr$_2$O$_7^{2-}$. The half-equations are shown: Cr$_2$O$_7^{2-} + 14\text{H}^+ + 6e^- \rightleftharpoons 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ and $2\text{CO}_2 + 2\text{H}^+ + 2e^- \rightleftharpoons \text{HOOCCOOH}$.
Construct an equation for the reaction of acidified Cr$_2$O$_7^{2-}$ with HOOCCOOH.
A $25.0\text{ cm}^3$ portion of a Na$_2$C$_2$O$_4$ solution reacts exactly with $16.20\text{ cm}^3$ of $0.0500\text{ mol dm}^{-3}$ K$_2$Cr$_2$O$_7$(aq) after acidification. Calculate the concentration of the Na$_2$C$_2$O$_4$ solution.