In `TeCl_4`, the central tellurium involves the hybridization

Options

A. `sp^3`
B. `sp^3d`
C. `sp^3d^2`
D. `dsp^2`

Explanation

To determine the hybridization of the central tellurium atom in TeCl₄ (tellurium tetrachloride), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Central Atom**:
- The central atom in TeCl₄ is tellurium (Te).

2. **Determine the Group of Tellurium**:
- Tellurium is in Group 16 of the periodic table.

3. **Count Valence Electrons**:
- Elements in Group 16 have 6 valence electrons. Therefore, tellurium has 6 valence electrons.

4. **Count the Number of Chlorine Atoms**:
- TeCl₄ has 4 chlorine (Cl) atoms bonded to the tellurium atom.

5. **Determine Electrons Used for Bonding**:
- Each chlorine atom forms a single covalent bond with tellurium, which means tellurium uses 4 of its valence electrons to bond with the 4 chlorine atoms.

6. **Calculate Remaining Valence Electrons**:
- After forming 4 bonds, tellurium has 6 - 4 = 2 valence electrons left.

7. **Identify Lone Pairs**:
- The remaining 2 valence electrons will form a lone pair on the tellurium atom.

8. **Determine the Steric Number**:
- The steric number is calculated by adding the number of bonded atoms to the number of lone pairs.
- Here, there are 4 bonded chlorine atoms and 1 lone pair:
- Steric Number = 4 (bonded atoms) + 1 (lone pair) = 5.

9. **Determine Hybridization**:
- The hybridization corresponding to a steric number of 5 is sp³d.

10. **Identify the Geometry**:
- The geometry of a molecule with sp³d hybridization and a steric number of 5 is trigonal bipyramidal.

### Final Answer:
The central tellurium in TeCl₄ involves **sp³d hybridization**.

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