Question

\([Co(H_{2}O)_{6}]^{2+}\) (Pink) \(+4Cl^{-}\rightleftharpoons[CoCl_{4}]^{2-}\) (Blue) \(+6H_{2}O\) (Endothermic reaction). If the mixture is transferred from room temperature to a freezing ice bath, what is expected to happen?

• A. Colour will remain the same
• B. Colour will become deeper blue
• C. It will become colourless
• D. Colour will become pink

Hint:

Remember this important equilibrium: \[ \boxed{ [Co(H_2O)_6]^{2+} \;\rightleftharpoons\; [CoCl_4]^{2-} } \] \[ \boxed{ \begin{aligned} \text{Pink} &\rightarrow [Co(H_2O)_6]^{2+}\\ \text{Blue} &\rightarrow [CoCl_4]^{2-} \end{aligned} } \] Heating favours the blue complex, whereas cooling favours the pink complex.

Answer 1

The Correct Option is D

Concept: The given equilibrium is \[ [Co(H_2O)_6]^{2+}+4Cl^- \rightleftharpoons [CoCl_4]^{2-}+6H_2O. \] The forward reaction is stated to be endothermic. According to Le Chatelier's principle,

• Increasing temperature favours the endothermic direction.

• Decreasing temperature favours the exothermic (reverse) direction.
Therefore, cooling shifts the equilibrium towards the pink hexaaquacobalt(II) complex.

Step 1: Identify the colours of the complexes.
\[ \boxed{ \begin{aligned} [Co(H_2O)_6]^{2+} &\rightarrow \text{Pink}\\ [CoCl_4]^{2-} &\rightarrow \text{Blue} \end{aligned} } \]

Step 2: Apply Le Chatelier's principle.
Since the forward reaction is endothermic, \[ \boxed{\text{Cooling favours the reverse reaction}.} \] Hence, the equilibrium shifts towards \[ [Co(H_2O)_6]^{2+}. \]

Step 3: Predict the observed colour.
As more pink complex is formed, the solution gradually changes from blue to \[ \boxed{\text{Pink}.} \] Therefore, \[ \boxed{\textbf{Option (D)}} \] is the correct answer.