The `I_(3)^(Θ)` ion has .

Options

A. Five equatorial lone pairs on the central I atom and two axial bonding pairs in a trigonal bipyramidal arrangement.
B. Five equatorial lone pairs on the central I atom and two axial bonding pairs in a pentagonal bipyramidal arrangement
C. Three equatorial lone pairs on the central I atom and two axial bonding pairs in a trigonal bipyramidal arrangement .
D. Two equatorial lone pairs on the central I atom and three axial bonding pairs in a trigonal bipyramidal arrangement .

Explanation

To determine the geometry and shape of the \( I_3^- \) ion using VSEPR theory, we can follow these steps:

### Step 1: Identify the central atom and count the valence electrons
The central atom in the \( I_3^- \) ion is iodine (I). Iodine is in group 17 of the periodic table and has 7 valence electrons. Since there are three iodine atoms in the ion and it carries a negative charge, we need to account for the additional electron from the charge.

- Total valence electrons = \( 3 \times 7 + 1 = 22 \) electrons.

### Step 2: Determine the number of bonding pairs and lone pairs
In the \( I_3^- \) ion, there are two iodine atoms bonded to the central iodine atom. This means there are 2 bonding pairs. The remaining electrons will be lone pairs.

- Total electrons used in bonding = 2 (for the two I-I bonds).
- Remaining electrons = \( 22 - 4 = 18 \) electrons.
- Lone pairs = \( \frac{18}{2} = 9 \) lone pairs.

### Step 3: Arrange the electron pairs
According to VSEPR theory, we need to arrange the electron pairs around the central atom to minimize repulsion.

- The total number of electron pairs (bonding + lone pairs) is \( 2 \) (bonding pairs) + \( 3 \) (lone pairs) = \( 5 \) electron pairs.

### Step 4: Determine the hybridization
With 5 electron pairs, the hybridization can be determined as follows:

- Hybridization = \( sp^3d \).

### Step 5: Identify the geometry
The arrangement of 5 electron pairs corresponds to a trigonal bipyramidal geometry.

### Step 6: Determine the molecular shape
In the case of \( I_3^- \), the three lone pairs will occupy the equatorial positions to minimize repulsion, while the two iodine atoms will occupy the axial positions.

- Therefore, the molecular shape is linear.

### Conclusion
The \( I_3^- \) ion has a trigonal bipyramidal electron geometry with a linear molecular shape.

### Summary of the solution:
1. Central atom: Iodine (I).
2. Total valence electrons: 22.
3. Bonding pairs: 2, Lone pairs: 3.
4. Hybridization: \( sp^3d \).
5. Geometry: Trigonal bipyramidal.
6. Molecular shape: Linear.