What is not applicable to `TeCl_4` ?

Options

A. The molecule is Sea - saw shaped
B. It is tetrahedral in shape
C. It has 4 bond pair and 1 lone pair of electron
D. The hybrid state of Te is `sp^3d`

Answer

To determine what is not applicable to \( \text{TeCl}_4 \), we will analyze its molecular structure, hybridization, and geometry step by step.

### Step 1: Identify the Group and Valence Electrons
- **TeCl4** contains tellurium (Te) which is in Group 16 of the periodic table.
- The valence electron configuration for Group 16 elements is \( ns^2 np^4 \). Therefore, tellurium has 6 valence electrons.

**Hint:** Remember that the number of valence electrons can be determined by the group number in the periodic table.

### Step 2: Determine the Steric Number
- The steric number is calculated as the number of bond pairs plus the number of lone pairs around the central atom.
- In \( \text{TeCl}_4 \), tellurium forms 4 bonds with chlorine atoms. Therefore, there are 4 bond pairs.
- Since tellurium has 6 valence electrons and 4 are used for bonding, there will be \( 6 - 4 = 2 \) lone electrons, which means there is 1 lone pair.

**Hint:** The steric number helps in determining the hybridization and molecular geometry.

### Step 3: Calculate the Hybridization
- The steric number for \( \text{TeCl}_4 \) is \( 4 \) (bond pairs) + \( 1 \) (lone pair) = \( 5 \).
- A steric number of 5 corresponds to \( sp^3d \) hybridization.

**Hint:** Different steric numbers correspond to different types of hybridization (e.g., 2 = sp, 3 = sp², 4 = sp³, 5 = sp³d).

### Step 4: Determine the Molecular Geometry
- With a steric number of 5 and 1 lone pair, the molecular geometry is described as "seesaw."
- The seesaw shape arises because the lone pair occupies an equatorial position, which minimizes repulsion.

**Hint:** The presence of lone pairs affects the molecular geometry, often leading to shapes that differ from those predicted by simple bond pair counts.

### Step 5: Identify What is Not Applicable
- Based on the analysis:
- The hybridization is \( sp^3d \) (applicable).
- The molecular geometry is seesaw (applicable).
- The bond angles are not tetrahedral (which would be applicable for \( sp^3 \)), but rather they are around 120° and 90° due to the seesaw shape.

Thus, the statement that is **not applicable** to \( \text{TeCl}_4 \) is that it has a tetrahedral shape.

### Final Answer
The statement that is not applicable to \( \text{TeCl}_4 \) is that it has a tetrahedral shape.

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