`Hg_(2)Cl_(2)` is black ened by `NH_(3)` due to formation of iodide of millon's base

Explanation

To solve the question regarding the reaction of `Hg2Cl2` with `NH3` and whether it leads to the formation of the iodide of Millon's base, we can break down the process step by step.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in this case are Mercury(I) chloride (`Hg2Cl2`) and Ammonia (`NH3`). When ammonia is in aqueous form, it can be represented as Ammonium hydroxide (`NH4OH`).

2. **Reaction of `Hg2Cl2` with `NH3`**:
When `Hg2Cl2` reacts with `NH4OH`, the reaction can be represented as:
\[ Hg_2Cl_2 + 2 NH_4OH \rightarrow 2 NH_2HgCl + 2 H_2O + 2 NH_4Cl \]
Here, `NH2HgCl` is formed, which is a complex of mercury.

3. **Formation of Black Precipitate**:
The black precipitate observed is due to the formation of elemental mercury (`Hg`). This occurs when the mercury(I) ions are reduced during the reaction, leading to the formation of `Hg`, which is black in color.

4. **Clarification on Millon's Base**:
The statement mentions the formation of the iodide of Millon's base. The correct formula for the iodide of Millon's base is `NH2HgI`, which is brown in color, not black. Therefore, the statement that `Hg2Cl2` is blackened by `NH3` due to the formation of the iodide of Millon's base is incorrect.

5. **Conclusion**:
Since the statement is about the formation of the iodide of Millon's base, which is brown, and not related to the black precipitate formed from `Hg`, we conclude that the statement is **false**.

### Final Answer:
The statement is **false**.

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