Solution of ` FeCl_(3) " ( yellow )" and NH_(4)SCN " (colourless) "`were mixed in a beaker. Red colour was obtained . On adding `HgCl_(2)` to the solution, the intensity of colour will `………. .`

Answer

To solve the problem, we need to analyze the chemical reactions occurring when the solutions of FeCl₃ and NH₄SCN are mixed, and then consider the effect of adding HgCl₂.

### Step-by-Step Solution:

1. **Identify the Initial Reaction**:
- When FeCl₃ (yellow solution) is mixed with NH₄SCN (colorless solution), they react to form a complex ion, [Fe(SCN)]²⁺, which is red in color.
- The reaction can be represented as:
\[ \text{Fe}^{3+} + \text{SCN}^- \rightarrow [\text{Fe(SCN)}]^{2+} \quad \text{(red color)} \]

2. **Establish the Equilibrium**:
- The formation of the red complex can be considered an equilibrium reaction:
\[ \text{Fe}^{3+} + \text{SCN}^- \rightleftharpoons [\text{Fe(SCN)}]^{2+} \]
- The intensity of the red color is directly related to the concentration of the complex ion [Fe(SCN)]²⁺.

3. **Effect of Adding HgCl₂**:
- When HgCl₂ is added to the solution, it reacts with NH₄SCN. This reaction decreases the concentration of NH₄SCN in the solution.
- The reaction can be represented as:
\[ \text{Hg}^{2+} + 2 \text{SCN}^- \rightarrow \text{Hg(SCN)}_2 \]
- As NH₄SCN is consumed, the concentration of SCN⁻ decreases.

4. **Applying Le Chatelier's Principle**:
- According to Le Chatelier's principle, if the concentration of a reactant (NH₄SCN) decreases, the equilibrium will shift to the left to counteract this change.
- This means that the equilibrium will shift towards the formation of more reactants (Fe³⁺ and SCN⁻), resulting in a decrease in the concentration of the red complex [Fe(SCN)]²⁺.

5. **Conclusion**:
- As a result of the shift in equilibrium, the intensity of the red color will decrease because there is less of the red complex present in the solution.

### Final Answer:
The intensity of the red color will **decrease** upon adding HgCl₂ to the solution.

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