`(A)+NaCl to (B)` (white ppt.)
`(B)+KIto(C)` (green ppt).
`(C)+underset(("excess"))(KI) to (D)+(E)` (colourless solution)
`(E)+NH_(3)+KOH to (F)`
Q. Colour of the compound (F) is:

Options

A. Yellow
B. Blue
C. White
D. Brown

Explanation

To solve the question regarding the color of compound F, we will analyze the reactions step by step:

### Step 1: Identify Compound A
- **Reaction**: (A) + NaCl → (B) (white ppt.)
- **Analysis**: The white precipitate formed when compound A reacts with NaCl suggests that compound A is likely to be mercurous nitrate, \( \text{Hg}_2(\text{NO}_3)_2 \). When it reacts with NaCl, it forms mercurous chloride, \( \text{Hg}_2\text{Cl}_2 \), which is a white precipitate.

### Step 2: Identify Compound B
- **Compound B**: \( \text{Hg}_2\text{Cl}_2 \) (white ppt.)

### Step 3: Identify Compound C
- **Reaction**: (B) + KI → (C) (green ppt.)
- **Analysis**: When compound B \( \text{Hg}_2\text{Cl}_2 \) reacts with potassium iodide (KI), it forms mercurous iodide, \( \text{Hg}_2\text{I}_2 \), which is a green precipitate.

### Step 4: Identify Compound D and E
- **Reaction**: (C) + excess KI → (D) + (E) (colorless solution)
- **Analysis**: When excess KI is added to \( \text{Hg}_2\text{I}_2 \), it forms a colorless solution of the Nessler reagent, \( \text{K}_2\text{HgI}_4 \) (compound E) and elemental mercury (D).

### Step 5: Identify Compound F
- **Reaction**: (E) + NH₃ + KOH → (F)
- **Analysis**: The Nessler reagent \( \text{K}_2\text{HgI}_4 \) reacts with ammonia (NH₃) and potassium hydroxide (KOH) to form a brown precipitate of \( \text{Hg}_2\text{O} \cdot \text{NH}_2 \) (compound F) along with mercury ions.

### Conclusion
- **Color of Compound F**: The color of compound F is brown.

### Final Answer
The color of the compound (F) is **brown**.

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