The hybrid state and oxidation state of S in `SF_4` are respectively

Options

A. `sp^3,+4`
B. `sp^2,+6`
C. `sp^3d,+4`
D. `dsp^3,+6`

Answer

To determine the hybrid state and oxidation state of sulfur (S) in sulfur tetrafluoride (SF₄), we can follow these steps:

### Step 1: Determine the Hybrid State of Sulfur in SF₄

1. **Identify the Valence Electrons of Sulfur**:
Sulfur (S) is in group 16 of the periodic table and has 6 valence electrons.

2. **Count the Number of Surrounding Atoms**:
In SF₄, sulfur is surrounded by 4 fluorine (F) atoms.

3. **Use the Hybridization Formula**:
The formula to determine the hybridization is:
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Valence Electrons} + \text{Number of Monovalent Atoms} \right) \]
Here, the valence electrons of sulfur are 6, and there are 4 fluorine atoms (monovalent).

Plugging in the values:
\[ \text{Hybridization} = \frac{1}{2} \left( 6 + 4 \right) = \frac{10}{2} = 5 \]

4. **Determine the Hybridization Type**:
A hybridization state of 5 corresponds to \( sp^3d \).

### Step 2: Determine the Oxidation State of Sulfur in SF₄

1. **Set Up the Equation for Oxidation State**:
Let the oxidation state of sulfur be \( x \). The oxidation state of fluorine is -1.

2. **Write the Equation**:
Since there are 4 fluorine atoms, the equation can be set up as:
\[ x + 4(-1) = 0 \]
Simplifying this gives:
\[ x - 4 = 0 \]
Thus, \( x = +4 \).

### Final Answer

The hybrid state and oxidation state of sulfur in SF₄ are:
- Hybrid State: \( sp^3d \)
- Oxidation State: +4

### Summary

The correct answer is \( sp^3d \) and +4.