Cobalt chloride when dissolved in water forms pink colored complex X which has octahedral geometry. This solution on treating with conc HCl forms deep blue complex, Y which has a Z geometry. X,Y and Z, respectively, are

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$X = {[Co {(H_{2} O)}_{4} {Cl}_{2}]}^{+}, Y = {[{CoCl}_{4}]}^{2 -}, Z = Tetrahedral$

$X = {[Co {(H_{2} O)}_{6}]}^{2 +}, Y = {[{CoCl}_{4}]}^{2 -}, Z = Tetrahedral$

$X = {[Co {(H_{2} O)}_{6}]}^{2 +}, Y = {[{CoCl}_{6}]}^{3 -}, Z = Octahedral$

$X = {[Co {(H_{2} O)}_{6}]}^{3 +}, Y = {[{CoCl}_{6}]}^{3 -}, Z = Octahedral$

answer is C.

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Detailed Solution

$\begin{array}{l} {CoCl}_{2} + 6 H_{2} O \to {[Co {(H_{2} O)}_{6}]}^{2 +} + 2 {Cl}^{-} \\ {[Co {(H_{2} O)}_{6}]}^{2 +} + 4 HCl \to {[{CoCl}_{4}]}^{2 -} + 4 H_{3} O^{+} + 2 H_{2} O \end{array}$

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