Valence shell electron pair repulsion theory `(VSEPR)` can be used to predict the approximate shape of molecules. Electrons in bonds and in lone pairs can be thought of as `"charge cloud"` that repel one another stay as far apart possible, thus causing molecules to assume specific shapes.
The repulsive interactions of electro pair is decrease in the order :
Lone pair-lone pair `gt` Lone pair -bond pair `gt` Bond pair -bond pair.
These repulsions effect result in deviations from idealized shapes and alteration in bond angles in molecules
Which of the following is linear ?

Options

A. `I_(3^(-))`
B. `ICl_(2^(+))`
C. `HgCl_(2)`
D. `XeF_(2)`

Answer

To determine which of the given molecules is linear using the Valence Shell Electron Pair Repulsion (VSEPR) theory, we will analyze each molecule step by step.

### Step 1: Analyze I3⁻
1. **Identify the central atom**: Iodine (I).
2. **Count valence electrons**: Iodine has 7 valence electrons. The negative charge adds 1 more electron, giving us a total of 8 electrons.
3. **Determine bond pairs and lone pairs**: I3⁻ forms 2 bonds with other iodine atoms, using 2 electrons. This leaves 6 electrons, which means there are 3 lone pairs.
4. **Calculate steric number**: Steric number = number of bond pairs + number of lone pairs = 2 (bond pairs) + 3 (lone pairs) = 5.
5. **Determine hybridization and shape**: A steric number of 5 corresponds to sp³d hybridization, which has a trigonal bipyramidal geometry. However, with 3 lone pairs, the molecular shape is linear due to the arrangement of lone pairs minimizing repulsion.

**Conclusion**: I3⁻ is linear.

### Step 2: Analyze ICl2⁺
1. **Identify the central atom**: Iodine (I).
2. **Count valence electrons**: Iodine has 7 valence electrons. The positive charge removes 1 electron, giving us 6 electrons.
3. **Determine bond pairs and lone pairs**: ICl2⁺ forms 2 bonds with chlorine atoms, using 2 electrons. This leaves 4 electrons, which means there are 2 lone pairs.
4. **Calculate steric number**: Steric number = 2 (bond pairs) + 2 (lone pairs) = 4.
5. **Determine hybridization and shape**: A steric number of 4 corresponds to sp³ hybridization, which has a tetrahedral geometry. With 2 lone pairs, the molecular shape is bent.

**Conclusion**: ICl2⁺ is not linear.

### Step 3: Analyze HgCl2
1. **Identify the central atom**: Mercury (Hg).
2. **Count valence electrons**: Mercury has 2 valence electrons from the 6s orbital.
3. **Determine bond pairs and lone pairs**: HgCl2 forms 2 bonds with chlorine atoms, using both electrons. There are no lone pairs left.
4. **Calculate steric number**: Steric number = 2 (bond pairs) + 0 (lone pairs) = 2.
5. **Determine hybridization and shape**: A steric number of 2 corresponds to sp hybridization, which is linear.

**Conclusion**: HgCl2 is linear.

### Step 4: Analyze XeF2
1. **Identify the central atom**: Xenon (Xe).
2. **Count valence electrons**: Xenon has 8 valence electrons.
3. **Determine bond pairs and lone pairs**: XeF2 forms 2 bonds with fluorine atoms, using 2 electrons. This leaves 6 electrons, which means there are 3 lone pairs.
4. **Calculate steric number**: Steric number = 2 (bond pairs) + 3 (lone pairs) = 5.
5. **Determine hybridization and shape**: A steric number of 5 corresponds to sp³d hybridization. The shape is linear due to the arrangement of lone pairs minimizing repulsion.

**Conclusion**: XeF2 is linear.

### Final Conclusion:
The linear molecules among the options provided are:
- I3⁻
- HgCl2
- XeF2