Aqueous solution of a salt 'A', when mixed with NaoH solution and warmed, a black precipitate is formed. Black ppt. is filtered and dissolved in dil. `H_(2)SO_(4)` solution. The resulting solution gives a chocolate coloured precipitate with potassium ferrocyanide solution. the filtrate obtained after filtering off the black precipitate, upon warming with Zn and NaOH evolves a pungent smelling gas. the resulting solution also responds to the ring test. the filtrate does not evolve any gas when it is boiled with urea in the presence of `H_(2)SO_(4)`.
Q. The chocolate coloured precipitate is:

Options

A. `Fe_(2)[Fe(CN)_(6)]`
B. `Cu_(2)[Fe(CN)_(6)]`
C. `HgSO_(4)`
D. `[Fe(H_(2)O)_(5)(NO)]SO_(4)`

Explanation

To solve the question step by step, we will analyze the information provided about the salt 'A' and the reactions it undergoes.

### Step 1: Identify the Black Precipitate
When the aqueous solution of salt 'A' is mixed with NaOH and warmed, a black precipitate is formed. This indicates that the salt likely contains a metal that forms a black hydroxide. The most common metal that fits this description is copper (Cu).

**Reaction:**
\[ \text{Cu(NO}_3\text{)}_2 + 2 \text{NaOH} \rightarrow \text{Cu(OH)}_2 \downarrow + 2 \text{NaNO}_3 \]
Upon warming, copper(II) hydroxide decomposes to form copper(II) oxide, which is black:
\[ 2 \text{Cu(OH)}_2 \xrightarrow{\text{heat}} 2 \text{CuO} + 2 \text{H}_2\text{O} \]

**Hint:** Look for metals that form black precipitates with NaOH.

### Step 2: Dissolve the Black Precipitate
The black precipitate (CuO) is then filtered and dissolved in dilute H₂SO₄. This reaction produces copper(II) sulfate and hydrogen sulfide gas:
\[ \text{CuO} + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{S} \]

**Hint:** Consider what happens when metal oxides react with acids.

### Step 3: Reaction with Potassium Ferrocyanide
The resulting solution (CuSO₄) is treated with potassium ferrocyanide (K₄[Fe(CN)₆]). This reaction leads to the formation of a chocolate-colored precipitate, which is copper(II) ferrocyanide:
\[ \text{CuSO}_4 + K_4[\text{Fe(CN)}_6] \rightarrow \text{Cu}_2[\text{Fe(CN)}_6] \downarrow + 2 \text{K}_2\text{SO}_4 \]

**Hint:** Identify the precipitate formed when copper(II) ions react with ferrocyanide ions.

### Step 4: Analyze the Filtrate
The filtrate obtained after filtering off the black precipitate is warmed with zinc and NaOH, which evolves a pungent-smelling gas (hydrogen sulfide, H₂S). This indicates that the solution contains sulfide ions.

**Hint:** Consider the reactions involving zinc and NaOH that produce gases.

### Step 5: Ring Test and Urea Reaction
The resulting solution responds to the ring test, indicating the presence of ammonium ions (NH₄⁺). The fact that it does not evolve any gas when boiled with urea in the presence of H₂SO₄ suggests that there are no free ammonium ions available for the reaction.

**Hint:** The ring test is a classic test for ammonium ions.

### Conclusion
The chocolate-colored precipitate formed when the solution of CuSO₄ reacts with potassium ferrocyanide is copper(II) ferrocyanide, represented as Cu₂[Fe(CN)₆].

**Final Answer:** The chocolate colored precipitate is Cu₂[Fe(CN)₆].