When`Hg_(2)Cl_(2)` ionizes, the ins produced and the unpaired electrons present on the cation respectively are

Options

A. `2Hg^(+)` and `2Cl^(-)`, two
B. `Hg_(2)^(2+) ` and `2Cl^(-) `, two
C. `Hg_(2)^(2-)` and `2Cl^(-)` , one
D. `Hg_(2)^(2+)` and `2Cl^(-)`, zero

Explanation

To solve the question regarding the ionization of \( \text{Hg}_2\text{Cl}_2 \) (mercurous chloride) and determine the ions produced along with the number of unpaired electrons in the cation, we can follow these steps:

### Step 1: Understand the Composition of \( \text{Hg}_2\text{Cl}_2 \)
Mercurous chloride, \( \text{Hg}_2\text{Cl}_2 \), consists of two mercury (Hg) atoms and two chloride (Cl) ions.

### Step 2: Ionization of \( \text{Hg}_2\text{Cl}_2 \)
When \( \text{Hg}_2\text{Cl}_2 \) ionizes, it dissociates into:
\[
...