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. Type of hybridization in compound (D) is:

Options

A. `sp^(2)`
B. `sp^(3)`
C. `sp^(3)d`
D. No hybridization

Explanation

To solve the question step by step, we need to analyze the reactions and compounds mentioned in the problem.

### Step 1: Identify the Black Solid
The black solid mentioned in the question is likely manganese dioxide (MnO2). This is a common black solid that can react with KOH and air.

**Hint:** Look for common black solids that can form colored compounds upon oxidation.

### Step 2: Reaction with KOH and Air
When MnO2 is heated with KOH and air, it forms potassium manganate (K2MnO4), which is green in color. This compound is denoted as (A).

**Hint:** Remember that manganese can exist in different oxidation states, which can affect the color of the compounds formed.

### Step 3: Reaction with H2SO4
When K2MnO4 reacts with sulfuric acid (H2SO4), it gets oxidized to form potassium permanganate (KMnO4), which is purple in color. This compound is denoted as (B), and along with it, another product (C) is formed.

**Hint:** Consider the role of H2SO4 as an oxidizing agent in this reaction.

### Step 4: Reaction of KI with (B)
The question states that KI reacts with the alkali solution of (B) (KMnO4) to form a compound (D). The reaction produces iodate ion (IO3^-), which is the compound (D).

**Hint:** Look for reactions involving potassium permanganate and iodide ions.

### Step 5: Determine the Structure of (D)
The iodate ion (IO3^-) has a structure where iodine is bonded to three oxygen atoms, with one of the oxygen atoms having a double bond. The lone pair on iodine must also be considered.

**Hint:** Draw the Lewis structure to visualize the bonding and lone pairs.

### Step 6: Calculate Hybridization of (D)
To find the hybridization of the iodate ion (IO3^-), we calculate the steric number:
- Number of sigma bonds: 3 (I-O bonds)
- Number of lone pairs: 1 (on iodine)

Steric number = 3 (sigma bonds) + 1 (lone pair) = 4.

A steric number of 4 corresponds to sp³ hybridization.

**Hint:** Use the formula: Steric number = Number of sigma bonds + Number of lone pairs to find hybridization.

### Conclusion
The type of hybridization in compound (D) (IO3^-) is sp³.

**Final Answer:** The hybridization in compound (D) is sp³.