`(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. Compounds (A) and (B) are respectively:

Options

A. `AgNO_(3) and AgCl`
B. `Pb(NO_(3))_(2) and PbCl_(2)`
C. `Hg_(2)(NO_(3))_(2) and Hg_(2)Cl_(2)`
D. `Cu_(2)(NO_(3))_(2) and Cu_(2)Cl_(2)`

Explanation

To solve the problem, we need to identify compounds (A) and (B) based on the given reactions and the nature of the precipitates formed.

### Step-by-Step Solution:

1. **Identify Compound A:**
- The first reaction states that compound (A) reacts with NaCl to form compound (B), which is a white precipitate.
- The possible compounds given are:
1. AgNO3 and AgCl
2. Pb(NO3)2 and PbCl2
3. Hg2(NO3)2 and Hg2Cl2
4. Cu2(NO3)2 and Cu2Cl2
- Among these, we need to find a compound that forms a white precipitate when reacted with NaCl.

2. **Reactions with NaCl:**
- **AgNO3 + NaCl → AgCl (white ppt.) + NaNO3**
- **Pb(NO3)2 + NaCl → PbCl2 (white ppt.) + NaNO3**
- **Hg2(NO3)2 + NaCl → Hg2Cl2 (white ppt.) + NaNO3**
- **Cu2(NO3)2 + NaCl → Cu2Cl2 (white ppt.) + NaNO3**
- All the options can produce white precipitates, but we need to check the next step to narrow it down.

3. **Identify Compound B:**
- The next reaction states that compound (B) reacts with KI to form compound (C), which is a green precipitate.
- We need to check which of the compounds (B) formed from the previous reactions can react with KI to form a green precipitate.

4. **Reactions with KI:**
- **AgCl + KI → AgI (yellow ppt.)**
- **PbCl2 + KI → PbI2 (yellow ppt.)**
- **Hg2Cl2 + 2KI → Hg2I2 (green ppt.)**
- **Cu2Cl2 + KI → CuI (white ppt.)**
- The only compound that forms a green precipitate is Hg2Cl2 when reacted with KI.

5. **Conclusion:**
- From the reactions, we can conclude that:
- Compound (A) is Hg2(NO3)2 (because it forms Hg2Cl2 when reacted with NaCl).
- Compound (B) is Hg2Cl2 (which forms Hg2I2 when reacted with KI).
- Therefore, the final answer is:
- (A) = Hg2(NO3)2
- (B) = Hg2Cl2

### Final Answer:
- Compounds (A) and (B) are respectively: **Hg2(NO3)2 and Hg2Cl2**.