When concentrated HCl is added to aqueous solution of $$CoCl_2$$, its colour changes from reddish pink to deep blue. Which complex ion gives blue colour in reaction?

Let us begin by recalling that an aqueous solution of $$CoCl_2$$ does not remain as simple ions. The $$Co^{2+}$$ ion is immediately surrounded by six water molecules acting as ligands. Therefore, in water the principal species present is the octahedral hexaaquacobalt(II) complex

$$[Co(H_2O)_6]^{2+}$$

This complex ion possesses a characteristic reddish-pink colour. Now we add concentrated hydrochloric acid. Concentrated HCl supplies a very high concentration of chloride ions $$Cl^-$$. According to the principles of coordination chemistry, if a ligand present in large excess can bind more strongly (or comparably strongly) than the original ligand, ligand substitution will occur.

The relevant equilibrium that describes the possible ligand exchange is

$$[Co(H_2O)_6]^{2+} + 4\,Cl^- \;\rightleftharpoons\; [CoCl_4]^{2-} + 6\,H_2O$$

We are using the idea of Le-Chatelier’s principle. The addition of a large excess of $$Cl^-$$ ions from concentrated HCl pushes the equilibrium to the right. Consequently, water molecules are displaced and four chloride ions coordinate to the cobalt(II) centre, forming the tetrahedral tetrachlorocobaltate(II) ion.

This newly formed complex, $$[CoCl_4]^{2-}$$, is deep blue in colour. Therefore, the observed change from reddish pink to deep blue can be directly attributed to the formation of $$[CoCl_4]^{2-}$$.

None of the other complexes listed fit the experimental facts: $$[CoCl_6]^{4-}$$ and $$[CoCl_6]^{3-}$$ would be hexachloro complexes whose existence under these conditions is not favoured, and $$[Co(H_2O)_6]^{2+}$$ is the original pink species, not blue.

Hence, the correct answer is Option D.