`4I^(-)+Hg^(2+)rarrHgI_(4)^(2-)` , `1` mole each of `Hg^(2+)` and `I^(-)` will form….. Mole `HgI_(4)^(2-)`:

Options

A. `1 "mole"`
B. `0.5 "mole"`
C. `0.25 "mole"`
D. `2 "mole"`

Explanation

To solve the problem, we need to analyze the reaction given and determine how many moles of \( \text{HgI}_4^{2-} \) can be formed from 1 mole each of \( \text{Hg}^{2+} \) and \( \text{I}^- \).

### Step-by-Step Solution:

1. **Write the Balanced Reaction**:
The reaction given is:
\[ 4 \text{I}^- + \text{Hg}^{2+} \rightarrow \text{HgI}_4^{2-} \]
This indicates that 4 moles of iodide ions react with 1 mole of mercury ions to produce 1 mole of tetraiodomercurate(II) ion.

2. **Identify the Stoichiometry**:
From the balanced equation, we see that:
- 4 moles of \( \text{I}^- \) are required for every 1 mole of \( \text{Hg}^{2+} \).
- This means the ratio of \( \text{I}^- \) to \( \text{Hg}^{2+} \) is 4:1.

3. **Determine the Limiting Reagent**:
We are given 1 mole of \( \text{Hg}^{2+} \) and 1 mole of \( \text{I}^- \).
- To fully react with 1 mole of \( \text{Hg}^{2+} \), we would need 4 moles of \( \text{I}^- \).
- Since we only have 1 mole of \( \text{I}^- \), it is the limiting reagent.

4. **Calculate the Amount of Product Formed**:
Since \( \text{I}^- \) is the limiting reagent, we can calculate how much product can be formed:
- From the stoichiometry, 4 moles of \( \text{I}^- \) produce 1 mole of \( \text{HgI}_4^{2-} \).
- Therefore, if we have 1 mole of \( \text{I}^- \), the amount of \( \text{HgI}_4^{2-} \) produced can be calculated as follows:
\[ \text{Moles of } \text{HgI}_4^{2-} = \frac{1 \text{ mole of } \text{I}^-}{4} = 0.25 \text{ moles} \]

5. **Final Answer**:
Thus, from 1 mole each of \( \text{Hg}^{2+} \) and \( \text{I}^- \), we can form **0.25 moles of \( \text{HgI}_4^{2-} \)**.

### Summary:
The answer to the question is:
\[ \text{0.25 moles of } \text{HgI}_4^{2-} \]

---