Black solid `underset(Delta)overset(KOH+Air)to underset(("green"))((A))overset(H_(2)SO_(4))to underset(("purple"))((B))+(C)`
(i) KI on reaction with alkali solution of (B) changes into a compound (D).
(ii) The colour of the compound (B) disappears on treatment with the acidic solution of `FeSO_(4)`
(iii) With cold conc. `H_(2)SO_(4)` compound (B) gives (E), which being explosive decomposes to yield (F) and oxygen. ltBrgt Q. Which of the following options is correct?

Options

A. (C) and (F) are same compounds having same colour.
B. (C) and (F) are different compounds having same colour.
C. Compound (B) forms similar compound (E) with hot and conc.` H_(2)SO_(4)`.
D. Compound (A) does not give same type of reaction in acidic and neutral medium

Explanation

To solve the given question step by step, we will analyze the reactions and compounds mentioned.

### Step 1: Identify the Black Solid
The black solid mentioned in the question is likely manganese dioxide (MnO2). When MnO2 reacts with potassium hydroxide (KOH) in the presence of air and is heated, it forms potassium manganate (K2MnO4), which is a green compound.

**Hint:** Look for common black solids that react with KOH in the presence of air.

### Step 2: Formation of Compound A
The reaction of MnO2 with KOH and air leads to the formation of potassium manganate (K2MnO4), which we denote as compound A. This compound is green in color.

**Hint:** Remember that potassium manganate is formed from manganese compounds in alkaline conditions.

### Step 3: Reaction with H2SO4
When potassium manganate (K2MnO4) is treated with sulfuric acid (H2SO4), it produces potassium permanganate (KMnO4), which is purple in color, along with manganese dioxide (MnO2) as a byproduct.

**Hint:** Identify the changes in oxidation states and colors when reacting with acids.

### Step 4: Identify Compound B and C
From the previous step, we identify:
- Compound B: KMnO4 (purple)
- Compound C: MnO2 (produced as a byproduct)

**Hint:** Pay attention to the color changes associated with different manganese compounds.

### Step 5: Reaction of KI with Compound B
When potassium iodide (KI) reacts with an alkaline solution of KMnO4 (compound B), it forms potassium iodate (KIO3) along with KOH and MnO2. This compound (KIO3) is denoted as compound D.

**Hint:** Consider the reactions of permanganate ions in alkaline solutions.

### Step 6: Color Disappearance with FeSO4
The color of compound B (KMnO4) disappears when treated with an acidic solution of iron(II) sulfate (FeSO4). This is because KMnO4 is reduced to manganese(II) sulfate (MnSO4), which is colorless.

**Hint:** Think about how redox reactions can lead to the loss of color in solutions.

### Step 7: Reaction of Compound B with Cold Concentrated H2SO4
When KMnO4 (compound B) reacts with cold concentrated H2SO4, it forms manganese heptoxide (Mn2O7), which is compound E, along with K2SO4 and water.

**Hint:** Focus on the products formed when KMnO4 reacts with concentrated acids.

### Step 8: Decomposition of Compound E
The compound Mn2O7 (E) is explosive and decomposes to yield manganese dioxide (MnO2, F) and oxygen gas.

**Hint:** Consider the stability of compounds formed from manganese in different oxidation states.

### Conclusion
Now that we have identified all compounds and their reactions:
- A = K2MnO4 (green)
- B = KMnO4 (purple)
- C = MnO2 (black)
- D = KIO3
- E = Mn2O7
- F = MnO2

The correct option based on the analysis is that C and F are the same compounds (both are MnO2).

**Final Answer:** The correct option is that C and F are the same compounds.