`K_(2)HgI_(4)` is `50%` ionised in aqueous solution. Which of the following are correct ?
1. n=7
2. n=3
3. i=2
4. i=4

Explanation

To solve the question regarding the ionization of \( K_2HgI_4 \) in aqueous solution, we will follow these steps:

### Step 1: Understand the Ionization
The compound \( K_2HgI_4 \) dissociates in water. The dissociation can be represented as:
\[ K_2HgI_4 \rightarrow 2K^+ + HgI_4^{2-} \]
From this equation, we can see that one formula unit of \( K_2HgI_4 \) produces 3 ions: 2 potassium ions (\( K^+ \)) and 1 \( HgI_4^{2-} \) ion.

### Step 2: Determine the Degree of Ionization
Given that \( K_2HgI_4 \) is 50% ionized, we can express this as:
\[ \alpha = \frac{50}{100} = 0.5 \]
where \( \alpha \) is the degree of ionization.

### Step 3: Calculate the Number of Dissociated Ions (n)
From the dissociation reaction, we can see that:
- Total number of ions produced from one formula unit = 3 (2 \( K^+ \) and 1 \( HgI_4^{2-} \)).
Thus, \( n = 3 \).

### Step 4: Calculate the Van 't Hoff Factor (i)
The relationship between the degree of ionization (\( \alpha \)), the number of ions produced (\( n \)), and the Van 't Hoff factor (\( i \)) is given by:
\[ \alpha = \frac{i - 1}{n - 1} \]
Substituting the known values:
\[ 0.5 = \frac{i - 1}{3 - 1} \]
This simplifies to:
\[ 0.5 = \frac{i - 1}{2} \]
Multiplying both sides by 2:
\[ 1 = i - 1 \]
Thus, solving for \( i \):
\[ i = 2 \]

### Step 5: Conclusion
Now we can summarize our findings:
- \( n = 3 \) (correct)
- \( i = 2 \) (correct)

### Final Answer
The correct options are:
- 2. \( n = 3 \)
- 3. \( i = 2 \)