The correct shape and I-I-I bond angles respectively in `I_(3)^(-)` ion are :

Options

A. Distorted trigonal planar , `135^(@) and 90^(@)`
B. T-shaped, `180^(@) and 90^(@)`
C. Trigonal planar , `120^(@)`
D. Linear , `180^(@)`

Answer

To determine the shape and bond angles of the \( I_3^- \) ion, we can follow these steps:

### Step 1: Determine the Valence Electrons
Iodine (I) is in Group 17 of the periodic table, which means it has 7 valence electrons. In the \( I_3^- \) ion, we have three iodine atoms and one extra electron due to the negative charge.

\[ \text{Total valence electrons} = 3 \times 7 + 1 = 22 \text{ electrons} \]

### Step 2: Draw the Lewis Structure
In the Lewis structure of \( I_3^- \):
- The central iodine atom will form bonds with the two terminal iodine atoms.
- Each bond consists of 2 electrons, which accounts for 4 electrons used in bonding.
- The remaining electrons will be placed as lone pairs on the central iodine atom.

\[ \text{Lone pairs on central I} = 22 - 4 = 18 \text{ electrons} \Rightarrow 9 \text{ lone pairs} \text{ (or 3 lone pairs)} \]

### Step 3: Calculate the Steric Number
The steric number is calculated as the sum of the number of bonded atoms and lone pairs around the central atom.

\[ \text{Steric number} = \text{Number of bond pairs} + \text{Number of lone pairs} = 2 + 3 = 5 \]

### Step 4: Determine the Electron Geometry
With a steric number of 5, the electron geometry is trigonal bipyramidal. In this geometry:
- There are 2 axial positions and 3 equatorial positions.

### Step 5: Consider Lone Pair Repulsion
Lone pairs repel each other more than bond pairs. Therefore, the three lone pairs will occupy the equatorial positions to minimize repulsion. This arrangement leads to the two iodine atoms being positioned axially.

### Step 6: Determine the Molecular Shape
The molecular shape is determined by the positions of the atoms. Since the lone pairs are not counted in the shape, the shape of \( I_3^- \) is linear.

### Step 7: Determine the Bond Angles
In the linear configuration, the bond angle between the iodine atoms is:

\[ \text{Bond angle} = 180^\circ \]

### Final Answer
The correct shape and bond angles of the \( I_3^- \) ion are:
- Shape: Linear
- Bond angle: \( 180^\circ \)

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