Question

Henry Taube, the 1983 Nobel Prize winner in chemistry has studied the mechanisms of oxidation-reduction reactions involving transition metal complexes. In one experiment he and his students studied the following reaction:

$\mathrm{Cr}{{\left({\mathrm{H}}_2\mathrm{O}\right)}_6}^{2+}\left(\mathrm{aq}\right)+\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{Cl}}^{2+}\left(\mathrm{aq}\right)\to \mathrm{Cr}\left(\mathrm{III}\right)\mathrm{complexes}+\mathrm{Co}\left(\mathrm{II}\right)\mathrm{complexes}$

Chromium (III) and cobalt(III) complexes are substitutionally inert (no exchange of ligands) under conditions of the experiment. However, chromium(II) and cobalt(II) complexes can exchange ligands very rapidly. One of the products of the reaction is .$\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5{\mathrm{Cl}}^{2+}$

Is this consistent with the reaction proceeding through formation of ${\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{Cr}-\mathrm{Cl}-\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5$as an intermediate? Explain your answer.

Short answer

Yes, the reaction proceeds through the intermediate${\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{Cr}-\mathrm{Cl}-\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5$ to give the product$\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5{\mathrm{Cl}}^{2+}$

Step-by-step solution

The complex Cr(H2O)62+ has chromium in +2 oxidation state

Here chromium in +2 oxidation state can readily interacts with the ligand - chloride ion of cobalt complex.

The electron is transferred through chloride ion from chromium (+2) to cobalt (+3) complex (Co(NH3)5Cl2-) thereby reducing cobalt (+3) to cobalt (+2) state

Hence chromium (+3) complex (Co(NH₃)₅Cl^2- ) is formed through the single step intermediate where chloride ion is dissociated from cobalt complex and attached to chromium complex.