Question

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

• A. $X =\left[ Co \left( H _2 O \right)_4 Cl _2\right]^{+}, Y =\left[ CoCl _4\right]^{2-}, Z =$ Tetrahedral
• B. $X =\left[ Co \left( H _2 O \right)_6\right]^{2+}, Y =\left[ CoCl _6\right]^{3-}, Z =$ Octahedral
• C. $X =\left[ Co \left( H _2 O \right)_6\right]^{2+}, Y =\left[ CoCl _4\right]^{2-}, Z =$ Tetrahedral
• D. $X =\left[ Co \left( H _2 O \right)_6\right]^{3+}, Y =\left[ CoCl _6\right]^{3-}, Z =$ Octahedral

Answer 1

The Correct Option is A

The given problem involves understanding the nature of complexes formed by cobalt chloride in the presence of water and hydrochloric acid. We need to identify the complexes \(X\) and \(Y\) and the geometry \(Z\).

1. The pink complex \(X\) is formed when cobalt chloride is dissolved in water. The complex typically formed is \(\left[ Co \left( H _2 O \right)_6\right]^{2+}\), which has an octahedral geometry because six water molecules coordinate with the cobalt ion. However, the presence of chloride ions suggests that some water might be replaced, leading to \(\left[ Co \left( H _2 O \right)_4 Cl _2\right]^{+}\) which also has an octahedral structure due to the coordination number being six.
2. When concentrated hydrochloric acid is added, the pink complex turns deep blue. The reaction leads to the formation of the complex \(\left[ CoCl _4\right]^{2-}\), where chloride ions replace the water molecules. This complex has a tetrahedral geometry since the coordination number is four with chloride ions.
3. Based on these observations, \(X\), \(Y\), and \(Z\) are identified as:
   • \(X = \left[ Co \left( H _2 O \right)_4 Cl _2\right]^{+}\) (Pink color, octahedral)
   • \(Y = \left[ CoCl _4\right]^{2-}\) (Deep blue color, tetrahedral)
   • \(Z =\) Tetrahedral

Thus, the correct answer is:

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

. The presence of different ligands leads to the color change and the change in complex geometry.