`HgCl_(2)` gives orange ppt , with (a) ___, which dissolves in excess of it forming (b) ____ called (c ) ____.

Answer

To solve the question, we need to fill in the blanks based on the reaction of mercuric chloride (HgCl₂) with potassium iodide (KI). Here’s the step-by-step solution:

### Step 1: Identify the Reactants
- The question states that HgCl₂ gives an orange precipitate. We know that HgCl₂ reacts with potassium iodide (KI) to produce an orange precipitate.

### Step 2: Write the Reaction
- The reaction can be written as:
\[ \text{HgCl}_2 + 2 \text{KI} \rightarrow \text{HgI}_2 \downarrow + 2 \text{KCl} \]
- Here, HgI₂ (mercuric iodide) is the orange precipitate formed.

### Step 3: Identify the Excess Reactant
- The question mentions that the orange precipitate dissolves in excess of the reactant. When excess KI is added to HgI₂, it forms a soluble complex.

### Step 4: Write the Complex Formation Reaction
- The reaction with excess KI can be written as:
\[ \text{HgI}_2 + 2 \text{KI} \rightarrow \text{K}_2\text{HgI}_4 \]
- K₂HgI₄ is the soluble complex formed.

### Step 5: Name the Complex
- The complex K₂HgI₄ is commonly known as Nessler's reagent.

### Final Answer
- Therefore, the completed statement is:
- HgCl₂ gives orange ppt with (a) **KI**, which dissolves in excess of it forming (b) **K₂HgI₄** called (c) **Nessler's reagent**.