`I_(2)` has less solubility in water and its solubility increase on adding KI solution. When KI and `I_(2)` react then a species 'X' is formed by which solubility of `I_(2)` increases.
Geometry of anionic part of X is

Options

A. Trigonal bipyramidal
B. Square pyramidal
C. Pentagon
D. Liner

Explanation

To solve the question regarding the solubility of iodine in water and the geometry of the anionic part of the species formed when potassium iodide (KI) is added, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Iodine's Solubility**:
- Iodine (I₂) is a non-polar molecule, which makes it less soluble in polar solvents like water.

2. **Reaction with Potassium Iodide (KI)**:
- When KI is added to a solution containing I₂, a reaction occurs that forms a new species. This species is known to be \( KI_3 \), where I₂ reacts with iodide ions (I⁻) from KI.

3. **Identifying the Anionic Part**:
- In the compound \( KI_3 \), the anionic part is \( I_3^- \). This can also be represented as \( I_2I^- \), indicating that it consists of two iodine atoms bonded to one iodide ion.

4. **Determining the Hybridization**:
- To find the geometry of the anionic part \( I_3^- \), we need to determine its hybridization.
- The formula for hybridization is:
\[ \text{Hybridization} = \frac{1}{2}(V + M - C) \]
where:
- \( V \) = number of valence electrons of the central atom
- \( M \) = number of monovalent atoms surrounding the central atom
- \( C \) = charge of the species (with sign)

5. **Calculating for \( I_3^- \)**:
- For \( I_3^- \):
- The central atom is iodine (I).
- Iodine has 7 valence electrons (V = 7).
- There are 2 monovalent iodide ions surrounding it (M = 2).
- The charge of the species is -1 (C = -1).
- Plugging in the values:
\[ \text{Hybridization} = \frac{1}{2}(7 + 2 - (-1)) = \frac{1}{2}(7 + 2 + 1) = \frac{1}{2}(10) = 5 \]
- This indicates that the hybridization is \( sp^3d \).

6. **Determining the Geometry**:
- The hybridization \( sp^3d \) corresponds to a trigonal bipyramidal geometry. This means that the arrangement of atoms around the central iodine in \( I_3^- \) will be in a trigonal bipyramidal shape.

### Conclusion:
The geometry of the anionic part \( I_3^- \) is **trigonal bipyramidal**.

---