`(A)+NaCl to (B)` (white ppt.)
`(B)+Kito(C)` (green ppt).
`(C)+underset(("excess"))(KI) to (D)+(E)` (colourless solution)
`(E)+NH_(3)+KOH to (F)`
Q. Type of hybridization in compound (E) is:

Options

A. `d^(2)sp^(3)`
B. `sp^(3)d^(2)`
C. `sp^(3)`
D. `dsp^(2)`

Explanation

To determine the type of hybridization in compound (E), we will follow the reactions step by step:

### Step 1: Identify Compound A
- **Reaction**: (A) + NaCl → (B) (white ppt.)
- **Analysis**: A white precipitate indicates the formation of a compound that is insoluble in water. The likely candidate for compound (A) is **Hg₂(NO₃)₂** (mercury(I) nitrate), which reacts with NaCl to form **Hg₂Cl₂** (mercury(I) chloride) as the white precipitate (B) and NaNO₃.

### Step 2: Identify Compound B
- **Compound B**: Hg₂Cl₂ (white ppt.)

### Step 3: Identify Compound C
- **Reaction**: (B) + KI → (C) (green ppt.)
- **Analysis**: When Hg₂Cl₂ reacts with potassium iodide (KI), it forms **Hg₂I₂** (mercury(I) iodide), which is a green precipitate (C).

### Step 4: Identify Compound D and E
- **Reaction**: (C) + excess KI → (D) + (E) (colourless solution)
- **Analysis**: When Hg₂I₂ reacts with excess KI, it forms **K₂HgI₄** (potassium tetraiodomercurate(II)) which is soluble in water, and elemental mercury (Hg). The compound (E) is K₂HgI₄, which is a colorless solution.

### Step 5: Determine the Hybridization of Compound E
- **Compound E**: K₂HgI₄
- **Analysis**: In K₂HgI₄, the central mercury atom is surrounded by four iodide ions. The geometry around the mercury atom is tetrahedral, which corresponds to **sp³ hybridization**.

### Conclusion
The type of hybridization in compound (E) is **sp³ hybridization**.

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