In `IF_3`. Hybrid state of iodine is `sp^3 d`. There are two lone pair and three bondpairs of electron. Lone pair electorn is positioned on :

Options

A. Axial position only
B. Equatorial position only
C. One is axial and other is equatorial
D. One is axial and other is on two equatorial positions

Answer

To determine the positioning of lone pair electrons in the molecule IF₃ (iodine trifluoride), we can follow these steps:

### Step 1: Identify the Central Atom and Valence Electrons
- The central atom in IF₃ is iodine (I).
- Iodine is in Group 17 of the periodic table and has 7 valence electrons.

### Step 2: Determine Bonds and Lone Pairs
- Iodine forms single bonds with three fluorine (F) atoms.
- Each bond uses one valence electron from iodine, resulting in 3 electrons used for bonding.
- Therefore, the remaining valence electrons for iodine are:
\[ 7 - 3 = 4 \text{ electrons} \]
- These remaining 4 electrons will form 2 lone pairs.

### Step 3: Calculate the Steric Number
- The steric number is calculated as:
\[ \text{Steric Number} = \text{Number of sigma bonds} + \text{Number of lone pairs} \]
- In IF₃, there are 3 sigma bonds (from the three I-F bonds) and 2 lone pairs:
\[ \text{Steric Number} = 3 + 2 = 5 \]

### Step 4: Determine Hybridization
- A steric number of 5 corresponds to an \( sp^3d \) hybridization.

### Step 5: Identify the Geometry
- The geometry for \( sp^3d \) hybridization is trigonal bipyramidal (TBP).

### Step 6: Positioning of Lone Pairs
- In trigonal bipyramidal geometry, the lone pairs are positioned in the equatorial plane to minimize repulsion.
- The more electronegative fluorine atoms will occupy the axial positions.

### Conclusion
- Therefore, the lone pair electrons in IF₃ are positioned in the equatorial positions.

### Final Answer
The lone pair electrons are positioned in the **equatorial positions**.

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