Explain why a green solution of potassium manganate $\left(\mathrm{VI}\right)$ ${\mathrm{K}}_2{\mathrm{MnO}}_4$ turns purple and a brown solid is precipitated when carbon dioxide is bubbled into the solution.

When ammonium vanadate is heated with an oxalic acid solution, a compound $\left(\mathrm{Z}\right)$ is formed. A sample of $\left(\mathrm{Z}\right)$ was titrated with ${\mathrm{KMnO}}_4$ solution in hot acidic solution. The resulting liquid was reduced with ${\mathrm{SO}}_2,$ the excess ${\mathrm{SO}}_2$ boiled off, and the liquid again titrated with ${\mathrm{KMnO}}_4.$

The ratio of the volumes of ${\mathrm{KMnO}}_4$ used in the two titrations was $5:1.$ What conclusions can you make regarding the nature of the compound? $({\mathrm{KMnO}}_4$ oxidizes all oxidation states of vanadium to vanadium $(+\mathrm{V}). {\mathrm{SO}}_2$ reduces vanadium $(+\mathrm{V})$ $\left.(+\mathrm{IV}).\right)$ Find the oxidation state of vanadium in $\mathrm{Z}$.