Compound 'A' in acidic medium does not give ppt with `H_2S` but in `NH_4OH` medium gives a ppt comp. 'A' is

Options

A. `FeCl_3`
B. `AlCl_3`
C. `ZnCl_2`
D. `SnCl_2`

Answer

To solve the problem of identifying compound 'A', we will analyze the behavior of the compound in different mediums, specifically acidic and basic (ammonium hydroxide) conditions.

### Step-by-Step Solution:

1. **Understanding the Behavior in Acidic Medium:**
- The question states that compound 'A' does not give a precipitate with `H2S` in acidic medium.
- This suggests that compound 'A' is likely a salt that does not form a precipitate of metal sulfide when reacted with `H2S` in the presence of an acid.

2. **Identifying Possible Compounds:**
- One common compound that fits this description is zinc chloride (ZnCl2).
- In acidic conditions, zinc chloride does not produce zinc sulfide (ZnS) precipitate because ZnS is soluble in hydrochloric acid.

3. **Reaction with H2S:**
- When zinc chloride is mixed with `H2S`, the reaction can be represented as follows:
\[ ZnCl_2 + H_2S \rightarrow ZnS \downarrow + 2HCl \]
- However, since the medium is acidic (due to the presence of HCl), the ZnS precipitate dissolves back into the solution, and no visible precipitate is formed.

4. **Understanding the Behavior in Ammonium Hydroxide Medium:**
- The question also states that compound 'A' gives a precipitate in ammonium hydroxide (NH4OH) medium.
- When zinc chloride is added to ammonium hydroxide, the following reaction occurs:
\[ ZnCl_2 + 2NH_4OH \rightarrow Zn(OH)_2 \downarrow + 2NH_4Cl \]
- Here, zinc hydroxide (Zn(OH)2) forms as a white precipitate, which is insoluble in excess ammonium chloride.

5. **Conclusion:**
- Based on the observations from both acidic and basic mediums, we conclude that compound 'A' is zinc chloride (ZnCl2).

### Final Answer:
Compound 'A' is **Zinc Chloride (ZnCl2)**.