An inorganic red coloured compound (A) on heating gives a compound (B) and a gas (C). (A) on treatment with dil. `HNO_(3)` gives compound (D), brown colour substnace (E) and a neutral oxide (F). Compound (D) on warming gives off again gas (C). Then, (E) will be

Options

A. `Mn_(3)O_(4)`
B. `PbO_(2)`
C. `Pb_(3)O_(4)`
D. `Fe_(2)O_(3)`

Answer

To solve the problem step by step, we will analyze the information given in the question and deduce the identity of the brown-colored substance (E).

### Step 1: Identify the Red-Colored Compound (A)
The problem states that there is an inorganic red-colored compound (A) that, upon heating, produces a compound (B) and a gas (C).

**Hint:** Look for common red-colored inorganic compounds that can decompose upon heating.

### Step 2: Determine the Products of Heating Compound (A)
Upon heating, the red compound (A) gives compound (B) and gas (C). A well-known red inorganic compound is lead(II,IV) oxide, Pb3O4. When Pb3O4 is heated, it decomposes to form lead(II) oxide (PbO) and oxygen gas (O2).

- **Compound (B)**: PbO (Lead(II) oxide)
- **Gas (C)**: O2 (Oxygen gas)

**Hint:** Consider the decomposition reactions of known red inorganic compounds.

### Step 3: Reaction of Compound (A) with Dilute HNO3
When compound (A) (Pb3O4) reacts with dilute nitric acid (HNO3), it produces:
- Compound (D): Lead(II) nitrate (Pb(NO3)2)
- Brown-colored substance (E): Lead(IV) oxide (PbO2)
- Neutral oxide (F): Water (H2O)

The overall reaction can be represented as:
\[ \text{Pb}_3\text{O}_4 + 8 \text{HNO}_3 \rightarrow 3 \text{Pb(NO}_3\text{)}_2 + 2 \text{PbO}_2 + 4 \text{H}_2\text{O} \]

**Hint:** Identify the products of the reaction between lead(II,IV) oxide and dilute nitric acid.

### Step 4: Identify the Brown Color Substance (E)
From the reaction above, we see that the brown-colored substance (E) produced is lead(IV) oxide (PbO2).

**Hint:** Look for the color of the products formed in the reaction to determine which one is brown.

### Step 5: Confirm the Neutral Oxide (F)
The neutral oxide (F) produced in the reaction is water (H2O), which is neutral.

**Hint:** Recall that water is a common neutral oxide.

### Step 6: Analyze Compound (D) on Warming
When compound (D) (lead(II) nitrate, Pb(NO3)2) is warmed, it decomposes to produce:
- Lead(II) oxide (PbO)
- Nitrogen dioxide (NO2)
- Oxygen gas (O2)

This confirms that gas (C) is again released as O2.

**Hint:** Consider the thermal decomposition of lead(II) nitrate.

### Conclusion
The brown-colored substance (E) formed in the reaction is lead(IV) oxide (PbO2).

Thus, the answer is:
**E = PbO2 (Lead(IV) oxide)**