`SF_6` and `SF_4` involves hybridisation of the type ____ and ____ respectively.

Options

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

Answer

To determine the hybridization types for SF₆ (sulfur hexafluoride) and SF₄ (sulfur tetrafluoride), we can follow these steps:

### Step 1: Identify the Central Atom and its Valency
- The central atom in both SF₆ and SF₄ is sulfur (S).
- The valency of sulfur is 6, as it can form six bonds.

### Step 2: Count the Monovalent Elements
- In SF₆, there are 6 fluorine (F) atoms.
- In SF₄, there are 4 fluorine (F) atoms.

### Step 3: Apply the Hybridization Formula
The formula for calculating the number of hybrid orbitals (H) is:
\[ H = \frac{V + M - C + A}{2} \]
Where:
- \( V \) = Valency of the central atom
- \( M \) = Number of monovalent elements
- \( C \) = Charge of cation (if any)
- \( A \) = Charge of anion (if any)

### Step 4: Calculate Hybridization for SF₆
For SF₆:
- \( V = 6 \) (valency of sulfur)
- \( M = 6 \) (number of fluorine atoms)
- \( C = 0 \) (no cationic charge)
- \( A = 0 \) (no anionic charge)

Substituting into the formula:
\[ H = \frac{6 + 6 - 0 + 0}{2} = \frac{12}{2} = 6 \]
This indicates that SF₆ has 6 hybrid orbitals, which corresponds to sp³d² hybridization.

### Step 5: Calculate Hybridization for SF₄
For SF₄:
- \( V = 6 \) (valency of sulfur)
- \( M = 4 \) (number of fluorine atoms)
- \( C = 0 \) (no cationic charge)
- \( A = 0 \) (no anionic charge)

Substituting into the formula:
\[ H = \frac{6 + 4 - 0 + 0}{2} = \frac{10}{2} = 5 \]
This indicates that SF₄ has 5 hybrid orbitals, which corresponds to sp³d hybridization.

### Final Answer
Thus, the hybridization types for SF₆ and SF₄ are:
- SF₆: sp³d²
- SF₄: sp³d