A coloured compound (A) reacts with dilute `H_(2)SO_(4)` to produce a colourless gas (B) and colourless solution (C). The reaction between (B) and the acidified `K_(2)Cr_(2)O_(7)` solution produces a green solution and a slightly yellowise precipitate (D). the substance (D) burns in air to produce a gas (E) which also can change the colour of `K_(2)Cr_(2)O_(7)` solution.
Q. When "B" reacts with "E"

Options

A. A new gas F will be produced
B. It produces D and a colourless liquid
C. There will be no reaction between them
D. It yields B and an acidic oxide

Answer

To solve the problem step by step, we will analyze the reactions and compounds involved based on the information provided.

### Step 1: Identify the colored compound (A)
The colored compound (A) is likely a metal sulfide, such as Manganese(II) sulfide (MnS), which is known to be colored.

**Hint:** Think about common colored compounds that can react with dilute sulfuric acid.

### Step 2: Reaction of A with dilute H2SO4
When the colored compound (A) reacts with dilute sulfuric acid (H2SO4), it produces a colorless gas (B) and a colorless solution (C). The reaction can be represented as:
\[ \text{MnS} + \text{H}_2\text{SO}_4 \rightarrow \text{MnSO}_4 + \text{H}_2\text{S} \]
Here, H2S is the colorless gas (B), and MnSO4 is the colorless solution (C).

**Hint:** Look for reactions where a colored compound produces a gas and a solution.

### Step 3: Reaction of B with acidified K2Cr2O7
Next, we consider the reaction of the colorless gas (B), which is H2S, with acidified potassium dichromate (K2Cr2O7):
\[ 3\text{H}_2\text{S} + \text{K}_2\text{Cr}_2\text{O}_7 + 4\text{H}_2\text{SO}_4 \rightarrow 3\text{S} + 2\text{Cr}_2\text{(SO}_4\text{)}_3 + 7\text{H}_2\text{O} \]
This reaction produces a green solution (due to Cr(III) ions) and a slightly yellowish precipitate (D), which is sulfur (S).

**Hint:** Consider how H2S acts as a reducing agent in the presence of dichromate.

### Step 4: Burning substance D to produce gas E
The yellowish precipitate (D), which is sulfur (S), burns in air to produce a gas (E):
\[ \text{S} + \text{O}_2 \rightarrow \text{SO}_2 \]
Here, SO2 is the gas (E) produced.

**Hint:** Think about the combustion of sulfur and the products formed.

### Step 5: Reaction of B with E
Now, we need to analyze what happens when B (H2S) reacts with E (SO2):
\[ \text{H}_2\text{S} + \text{SO}_2 \rightarrow \text{2H}_2\text{O} + \text{S} \]
In this reaction, hydrogen sulfide (H2S) reacts with sulfur dioxide (SO2) to produce water and sulfur.

**Hint:** Consider the properties of H2S and SO2 and how they might react together.

### Final Summary
In summary, the colored compound (A) reacts with dilute H2SO4 to produce H2S (B) and MnSO4 (C). H2S reacts with acidified K2Cr2O7 to produce a green solution and sulfur (D). Sulfur burns to produce SO2 (E), and when H2S reacts with SO2, it produces water and sulfur.