Question

Why aqueous cobalt chloride solution when treated with conc. HCl develops a deep blue colored solution?

• A. Due to formation of \([\text{CoCl}_{4}]^{2+}\) having tetrahedral geometry.
• B. Due to formation of \([\text{CoCl}_{4}]^{2-}\) having tetrahedral geometry.
• C. Due to formation of \([\text{CoCl}_{4}]^{2-}\) having octahedral geometry.
• D. Due to formation of \([\text{CoCl}_{4}]^{2+}\) having octahedral geometry.

Hint:

Excess chloride replaces water in the pink octahedral Co(II) ion to give a blue tetrahedral chloro complex.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Aqueous cobalt(II) chloride is pink because of the octahedral ion \([\text{Co}(\text{H}_{2}\text{O})_{6}]^{2+}\).

Step 2: Effect of conc. HCl
Concentrated HCl supplies a large amount of \(\text{Cl}^{-}\), which replaces the water ligands:
\([\text{Co}(\text{H}_{2}\text{O})_{6}]^{2+} + 4\text{Cl}^{-} \rightarrow [\text{CoCl}_{4}]^{2-} + 6\text{H}_{2}\text{O}\)

Step 3: Charge and geometry
The charge is \(+2 + 4(-1) = -2\), so the ion is \([\text{CoCl}_{4}]^{2-}\).
With coordination number 4 and \(sp^{3}\) hybridisation it is tetrahedral and deep blue.

Final Answer:
The blue colour is due to the tetrahedral ion [CoCl4]2-. \[ \boxed{\text{(B) }\text{Due to formation of }[\text{CoCl}_{4}]^{2-}\text{ having tetrahedral geometry.}} \]