In which of the following reaction, Hybridisation of underline atom gets changed

Options

A. `XeF_4 + SbF_5 rarr`
B. `H_3PO_2 overset(disproportionation)rarr`
C. `H_2SO_4 +NaCl rarr`
D. `NH_3 + BF_3 rarr`

Answer

To determine in which reaction the hybridization of the underlined atom changes, we will analyze each option step by step.

### Step 1: Analyze the first reaction (Xenon Tetrafluoride with Antimony Pentafluoride)

1. **Identify the reactants**: Xenon tetrafluoride (XeF4) and Antimony pentafluoride (SbF5).
2. **Determine the hybridization of Xenon in XeF4**:
- Valence electrons of Xenon = 8 (group 18 element).
- Number of monovalent fluorine atoms = 4.
- Steric number = (8 + 4) / 2 = 12 / 2 = 6.
- Hybridization = sp³d².
3. **Determine the hybridization of Xenon in the product (XeF3⁺)**:
- After losing one fluorine, the effective valence electrons = 8 - 1 = 7.
- Number of monovalent fluorine atoms = 3.
- Steric number = (7 + 3) / 2 = 10 / 2 = 5.
- Hybridization = sp³d.
4. **Conclusion**: The hybridization of Xenon changes from sp³d² to sp³d.

### Step 2: Analyze the second reaction (H3PO2 dissociating to H2PO2⁻)

1. **Identify the reactants**: H3PO2 dissociates to H2PO2⁻.
2. **Determine the hybridization of Phosphorus in H3PO2**:
- Valence electrons of Phosphorus = 5.
- Number of monovalent atoms (3 Hydrogens) = 3.
- Steric number = (5 + 3) / 2 = 8 / 2 = 4.
- Hybridization = sp³.
3. **Determine the hybridization of Phosphorus in H2PO2⁻**:
- Valence electrons = 5.
- Number of monovalent atoms (2 Hydrogens and 1 negative charge) = 3.
- Steric number = (5 + 3) / 2 = 8 / 2 = 4.
- Hybridization = sp³.
4. **Conclusion**: The hybridization of Phosphorus remains sp³ in both cases.

### Step 3: Analyze the third reaction (H2SO4 reacting to form Sodium Sulphate and HCl)

1. **Identify the reactants**: H2SO4 and the products Sodium Sulphate and HCl.
2. **Determine the hybridization of Sulfur in H2SO4**:
- Valence electrons of Sulfur = 6.
- Number of monovalent atoms (4 Oxygens) = 4.
- Steric number = (6 + 4) / 2 = 10 / 2 = 5.
- Hybridization = sp³d.
3. **Determine the hybridization of Sulfur in the product**:
- The structure of Sodium Sulphate does not change the hybridization of Sulfur.
- It remains sp³d.
4. **Conclusion**: The hybridization of Sulfur does not change.

### Step 4: Analyze the fourth reaction (Ammonia reacting with BF3)

1. **Identify the reactants**: Ammonia (NH3) and Boron trifluoride (BF3).
2. **Determine the hybridization of Nitrogen in NH3**:
- Valence electrons of Nitrogen = 5.
- Number of monovalent atoms (3 Hydrogens) = 3.
- Steric number = (5 + 3) / 2 = 8 / 2 = 4.
- Hybridization = sp³.
3. **Determine the hybridization of Boron in BF3**:
- Valence electrons of Boron = 3.
- Number of monovalent atoms (3 Fluorines) = 3.
- Steric number = (3 + 3) / 2 = 6 / 2 = 3.
- Hybridization = sp².
4. **Conclusion**: The hybridization of Nitrogen remains sp³, and that of Boron remains sp².

### Final Conclusion:

The only reaction where the hybridization of the underlined atom changes is in the first option (Xenon in XeF4 reacting with SbF5).