Which chloride of `I^(st)` group basic radicals turns black on treatment with `NH_(3)`?

Answer

To determine which chloride of the first group basic radicals turns black on treatment with NH₃, we can follow these steps:

### Step 1: Identify the First Group Basic Radicals
The first group basic radicals in qualitative analysis are:
- Mercury (Hg²⁺)
- Lead (Pb²⁺)

### Step 2: Examine the Chlorides of the First Group
The chlorides corresponding to these radicals are:
- Mercury(II) chloride (HgCl₂)
- Lead(II) chloride (PbCl₂)

### Step 3: Understand the Reaction with NH₃
When these chlorides are treated with ammonia (NH₃), we need to identify which one forms a black precipitate.

### Step 4: Analyze the Reaction of Mercury(II) Chloride
When mercury(II) chloride (HgCl₂) is treated with ammonia (NH₃), it forms a complex and can lead to the formation of mercury(I) chloride (Hg₂Cl₂) which is a black precipitate:
\[ \text{HgCl}_2 + 2 \text{NH}_3 \rightarrow \text{Hg}_2\text{Cl}_2 + 2 \text{NH}_4\text{Cl} \]
The mercury(I) chloride (Hg₂Cl₂) appears black.

### Step 5: Analyze the Reaction of Lead(II) Chloride
Lead(II) chloride (PbCl₂) does not form a black precipitate when treated with NH₃. Instead, it remains white or forms a white precipitate.

### Conclusion
The chloride of the first group basic radical that turns black on treatment with NH₃ is **Mercury(II) chloride (HgCl₂)**.

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