The hybridisation number of lone pair of electron and shape of `I_(3)^(Θ)` is .

Options

A. `sp^(3)` d hybridisation
B. It has trigonal bipyramidal shape
C. It is linear
D. It has three lone pair of electrons

Explanation

To determine the hybridization number of lone pairs of electrons and the shape of \( I_3^- \), we can follow these steps:

### Step 1: Determine the Valence Electrons
Iodine (I) has 7 valence electrons. Since \( I_3^- \) has a negative charge, we add one more electron, giving us a total of:
\[ 7 \text{ (from I)} \times 3 + 1 \text{ (for the negative charge)} = 22 \text{ valence electrons} \]

### Step 2: Draw the Lewis Structure
1. Place one iodine atom in the center and the other two iodine atoms on the sides.
2. Connect the central iodine to the two outer iodines with single bonds. This uses 4 electrons (2 for each bond).
3. Subtract the used electrons from the total:
\[ 22 - 4 = 18 \text{ electrons remaining} \]
4. Distribute the remaining electrons to satisfy the octet rule. Each outer iodine will receive 6 electrons (3 lone pairs), using up 12 electrons:
\[ 18 - 12 = 6 \text{ electrons remaining} \]
5. The remaining 6 electrons will be placed as 3 lone pairs on the central iodine atom.

### Step 3: Count Bond Pairs and Lone Pairs
- **Bond pairs**: 2 (the two I-I bonds)
- **Lone pairs**: 3 (on the central iodine)

### Step 4: Determine Hybridization
The hybridization can be calculated using the formula:
\[ \text{Hybridization} = \text{Number of bond pairs} + \text{Number of lone pairs} \]
\[ = 2 + 3 = 5 \]
This corresponds to \( sp^3d \) hybridization.

### Step 5: Determine the Molecular Shape
The arrangement of electron pairs around the central atom (considering lone pairs) leads to a trigonal bipyramidal geometry. However, because there are three lone pairs, they will occupy the equatorial positions to minimize repulsion, resulting in a linear shape for the molecule.

### Conclusion
- **Hybridization**: \( sp^3d \)
- **Shape**: Linear
- **Number of Lone Pairs**: 3