Cobalt chloride when dissolved in water forms pink colored complex...

Cobalt chloride when dissolved in water forms pink coloured is\left[Co{\left({H}_{2}O\right)}_{6}\right]{Cl}_{2}$. This is due to the electronic transition of an electron from${t}_{2g} to {e}_{g}$energy level of${\left[Co{\left({H}_{2}O\right)}_{6}\right]}^{2 +}$complex. When excess of$HCl$is added to the solution${\left[Co{\left({H}_{2}O\right)}_{6}\right]}^{2 +} is transformed into deep blue colour complex $[ {CoCl}_{4}{]}^{2 -}$.

${△}_{t} = \dfrac{4}{9}{△}_{o}$

${△}_{t} < {△}_{o}$

Tetrahedral complexes have smaller crystal field splitting than octahedral complexes.

${CoCl}_{2} + 6{H}_{2}O ⟶ \mathop{\left[Co{\left({H}_{2}O\right)}_{6}\right]{Cl}_{2}}_{}$

${\left[Co{\left({H}_{2}O\right)}_{6}\right]}^{2 +}\overset{Conc . HCl}{\to} [ {CoCl}_{4}{]}^{2 -}$

Hence, $X = {\left[Co{\left({H}_{2}O\right)}_{6}\right]}^{2 +}, Y = {\left[{CoCl}_{4}\right]}^{2 -}, Z = Tetrahedral$