Which of the following set do not have `sp^(3)d` hybridization :-

Options

A. `PF_(5)& BrF_(3)`
B. `I Cl_(2)^(+) & SF_(4)`
C. `XeF_(2) & ICl_(2)^(-)`
D. `AsF_(4)^(-) & SCl_(4)`

Answer

To determine which of the given sets do not have `sp^3d` hybridization, we will use the formula for calculating hybridization:

**Hybridization = 1/2 * (Number of valence electrons in the central atom + Number of monovalent ligands + Oxidation state)**

Let's analyze each set step by step.

### Step 1: Analyze the first set - PCl5 and BrF3

1. **PCl5**:
- Central atom: Phosphorus (P)
- Valence electrons in P: 5
- Monovalent ligands (Cl): 5
- Oxidation state: 0 (neutral molecule)

Calculation:
\[ \text{Hybridization} = \frac{1}{2} \times (5 + 5 + 0) = \frac{1}{2} \times 10 = 5 \]
This indicates `sp^3d` hybridization.

2. **BrF3**:
- Central atom: Bromine (Br)
- Valence electrons in Br: 7
- Monovalent ligands (F): 3
- Oxidation state: 0 (neutral molecule)

Calculation:
\[ \text{Hybridization} = \frac{1}{2} \times (7 + 3 + 0) = \frac{1}{2} \times 10 = 5 \]
This also indicates `sp^3d` hybridization.

### Step 2: Analyze the second set - ICl2+ and SF4

1. **ICl2+**:
- Central atom: Iodine (I)
- Valence electrons in I: 7
- Monovalent ligands (Cl): 2
- Oxidation state: +1 (due to the positive charge)

Calculation:
\[ \text{Hybridization} = \frac{1}{2} \times (7 + 2 - 1) = \frac{1}{2} \times 8 = 4 \]
This indicates `sp^3` hybridization (not `sp^3d`).

2. **SF4**:
- Central atom: Sulfur (S)
- Valence electrons in S: 6
- Monovalent ligands (F): 4
- Oxidation state: 0 (neutral molecule)

Calculation:
\[ \text{Hybridization} = \frac{1}{2} \times (6 + 4 + 0) = \frac{1}{2} \times 10 = 5 \]
This indicates `sp^3d` hybridization.

### Step 3: Analyze the third set - XeF2 and ICl2-

1. **XeF2**:
- Central atom: Xenon (Xe)
- Valence electrons in Xe: 8
- Monovalent ligands (F): 2
- Oxidation state: 0 (neutral molecule)

Calculation:
\[ \text{Hybridization} = \frac{1}{2} \times (8 + 2 + 0) = \frac{1}{2} \times 10 = 5 \]
This indicates `sp^3d` hybridization.

2. **ICl2-**:
- Central atom: Iodine (I)
- Valence electrons in I: 7
- Monovalent ligands (Cl): 2
- Oxidation state: -1 (due to the negative charge)

Calculation:
\[ \text{Hybridization} = \frac{1}{2} \times (7 + 2 + 1) = \frac{1}{2} \times 10 = 5 \]
This indicates `sp^3d` hybridization.

### Step 4: Analyze the fourth set - AsF4- and SCl4

1. **AsF4-**:
- Central atom: Arsenic (As)
- Valence electrons in As: 5
- Monovalent ligands (F): 4
- Oxidation state: -1 (due to the negative charge)

Calculation:
\[ \text{Hybridization} = \frac{1}{2} \times (5 + 4 + 1) = \frac{1}{2} \times 10 = 5 \]
This indicates `sp^3d` hybridization.

2. **SCl4**:
- Central atom: Sulfur (S)
- Valence electrons in S: 6
- Monovalent ligands (Cl): 4
- Oxidation state: 0 (neutral molecule)

Calculation:
\[ \text{Hybridization} = \frac{1}{2} \times (6 + 4 + 0) = \frac{1}{2} \times 10 = 5 \]
This indicates `sp^3d` hybridization.

### Conclusion:
The only set that does not have `sp^3d` hybridization is the second set (ICl2+ and SF4).

### Final Answer:
**The set that does not have `sp^3d` hybridization is: ICl2+ and SF4.**