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

Options

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

Answer

To determine which of the following sets do not have `sp^3d` hybridization, we will calculate the hybridization of each compound in the given options using the concept of steric number. The steric number can be calculated using the formula:

\[ \text{Steric Number} = \frac{1}{2} \left( V + M - C + A \right) \]

Where:
- \( V \) = number of valence electrons on the central atom
- \( M \) = number of monovalent atoms attached to the central atom
- \( C \) = positive charge on the compound
- \( A \) = negative charge on the compound

### Step 1: Analyze the first pair - \( \text{PF}_4^- \) and \( \text{BrF}_3 \)

1. **For \( \text{PF}_4^- \)**:
- Valence electrons of phosphorus (P) = 5
- Monovalent atoms (F) = 4
- Positive charge (C) = 0
- Negative charge (A) = 1

\[ \text{Steric Number} = \frac{1}{2} \left( 5 + 4 - 0 + 1 \right) = \frac{1}{2} \times 10 = 5 \]
- Hybridization = \( sp^3d \)

2. **For \( \text{BrF}_3 \)**:
- Valence electrons of bromine (Br) = 7
- Monovalent atoms (F) = 3
- Positive charge (C) = 0
- Negative charge (A) = 0

\[ \text{Steric Number} = \frac{1}{2} \left( 7 + 3 - 0 + 0 \right) = \frac{1}{2} \times 10 = 5 \]
- Hybridization = \( sp^3d \)

### Step 2: Analyze the second pair - \( \text{ICl}_2^+ \) and \( \text{SF}_4 \)

1. **For \( \text{ICl}_2^+ \)**:
- Valence electrons of iodine (I) = 7
- Monovalent atoms (Cl) = 2
- Positive charge (C) = 1
- Negative charge (A) = 0

\[ \text{Steric Number} = \frac{1}{2} \left( 7 + 2 - 1 + 0 \right) = \frac{1}{2} \times 8 = 4 \]
- Hybridization = \( sp^3 \)

2. **For \( \text{SF}_4 \)**:
- Valence electrons of sulfur (S) = 6
- Monovalent atoms (F) = 4
- Positive charge (C) = 0
- Negative charge (A) = 0

\[ \text{Steric Number} = \frac{1}{2} \left( 6 + 4 - 0 + 0 \right) = \frac{1}{2} \times 10 = 5 \]
- Hybridization = \( sp^3d \)

### Step 3: Analyze the third pair - \( \text{XF}_2 \) and \( \text{I}_3^- \)

1. **For \( \text{XF}_2 \)** (assuming X is a halogen):
- Valence electrons of X = 7
- Monovalent atoms (F) = 2
- Positive charge (C) = 0
- Negative charge (A) = 0

\[ \text{Steric Number} = \frac{1}{2} \left( 7 + 2 - 0 + 0 \right) = \frac{1}{2} \times 9 = 4.5 \]
- Hybridization = Not applicable (not a valid steric number)

2. **For \( \text{I}_3^- \)**:
- Valence electrons of iodine (I) = 7
- Monovalent atoms (I) = 2
- Positive charge (C) = 0
- Negative charge (A) = 1

\[ \text{Steric Number} = \frac{1}{2} \left( 7 + 2 - 0 + 1 \right) = \frac{1}{2} \times 10 = 5 \]
- Hybridization = \( sp^3d \)

### Step 4: Analyze the fourth pair - \( \text{SF}_4^- \) and \( \text{SCl}_4 \)

1. **For \( \text{SF}_4^- \)**:
- Valence electrons of sulfur (S) = 6
- Monovalent atoms (F) = 4
- Positive charge (C) = 0
- Negative charge (A) = 1

\[ \text{Steric Number} = \frac{1}{2} \left( 6 + 4 - 0 + 1 \right) = \frac{1}{2} \times 11 = 5.5 \]
- Hybridization = Not applicable (not a valid steric number)

2. **For \( \text{SCl}_4 \)**:
- Valence electrons of sulfur (S) = 6
- Monovalent atoms (Cl) = 4
- Positive charge (C) = 0
- Negative charge (A) = 0

\[ \text{Steric Number} = \frac{1}{2} \left( 6 + 4 - 0 + 0 \right) = \frac{1}{2} \times 10 = 5 \]
- Hybridization = \( sp^3d \)

### Conclusion:
From the analysis, we find that:
- The first pair \( \text{PF}_4^- \) and \( \text{BrF}_3 \) both have \( sp^3d \) hybridization.
- The second pair \( \text{ICl}_2^+ \) has \( sp^3 \) hybridization and \( \text{SF}_4 \) has \( sp^3d \) hybridization.
- The third pair \( \text{XF}_2 \) has an invalid steric number, while \( \text{I}_3^- \) has \( sp^3d \) hybridization.
- The fourth pair \( \text{SF}_4^- \) has an invalid steric number, while \( \text{SCl}_4 \) has \( sp^3d \) hybridization.

Thus, the pair that does not have both compounds with \( sp^3d \) hybridization is **Option B: \( \text{ICl}_2^+ \) and \( \text{SF}_4 \)**.