Cobalt chloride when dissolved in water forms pink coloured is$\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]{\mathrm{Cl}}_2$ . This is due to the electronic transition of an electron from ${\mathrm{t}}_{2\mathrm{g}}$ to ${\mathrm{e}}_{\mathrm{g}}$ energy level of ${\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$ complex. When excess of $\mathrm{HCl}$ is added to the solution ${\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$ is transformed into deep blue colour complex $[{\mathrm{CoCl}}_4{]}^{2-}$.

${△}_{\mathrm{t}}= \frac{4}{9} {△}_{\mathrm{o}}$

${△}_{\mathrm{t}} < {△}_{\mathrm{o}}$

Tetrahedral complexes have smaller crystal field splitting than octahedral complexes.

$$\begin{gathered} {\mathrm{CoCl}}_2+6{\mathrm{H}}_2\mathrm{O}⟶\underset{ \\ }{\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]{\mathrm{Cl}}_2 } \end{gathered}$$

${\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+} \stackrel{\mathrm{Conc}. \mathrm{HCl}}{\to } [{\mathrm{CoCl}}_4{]}^{2-}$

Hence, $\mathrm{X}={\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}, \mathrm{Y}={\left[{\mathrm{CoCl}}_4\right]}^{2-}, \mathrm{Z}=\mathrm{Tetrahedral}$