`HgCI_(2)+ "excess of" KI rarr (A) underset(NH_(3)) to (B)` ,(A) and (B) respectively are

Options

A.
B. `(Y),(X)`
C. both (X)
D. both (Y)

Answer

To solve the question, we need to determine the compounds A and B formed from the reaction of mercury(II) chloride (HgCl₂) with excess potassium iodide (KI) and then with ammonia (NH₃) and sodium hydroxide (NaOH).

### Step-by-Step Solution:

1. **Reaction of HgCl₂ with KI**:
- When mercury(II) chloride (HgCl₂) reacts with excess potassium iodide (KI), it undergoes a double displacement reaction.
- The reaction can be represented as:
\[ \text{HgCl}_2 + 2 \text{KI} \rightarrow \text{HgI}_2 + 2 \text{KCl} \]
- Here, mercuric iodide (HgI₂) is formed, which is a scarlet precipitate. Thus, we identify **A** as:
\[ A = \text{HgI}_2 \quad (\text{scarlet precipitate of mercuric iodide}) \]

2. **Further Reaction of HgI₂ with NH₃ and NaOH**:
- The next step involves treating the mercuric iodide (HgI₂) with ammonia (NH₃) and sodium hydroxide (NaOH).
- The reaction can be represented as:
\[ \text{HgI}_2 + 2 \text{KI} + 2 \text{NH}_3 + 2 \text{NaOH} \rightarrow \text{K}_2\text{HgI}_4 + 2 \text{NaCl} + 2 \text{NH}_4\text{I} \]
- The product formed in this reaction is potassium tetraiodomercurate(II) (K₂HgI₄), which is also known as Nessler's reagent. Thus, we identify **B** as:
\[ B = \text{K}_2\text{HgI}_4 \quad (\text{Nessler's reagent}) \]

### Final Answer:
- **A** = HgI₂ (scarlet precipitate of mercuric iodide)
- **B** = K₂HgI₄ (Nessler's reagent)