Consider the following reactions:
(i) `PCl_(3)+3H_(2)O to H_(3)PO_(3) +3HCl`
(ii) `SF_(4)+3H_(2)O to H_(3)SO_(3)+4HF`
(iii) `BCl_(3)+3H_(2)O to H_(3)BO_(3)+3HCl`
(iv) `XeF_(6)+3H_(2)O to XeO_(3)+6HF`
Then according to given information the incorrect statement is:

Options

A. During the (i) reaction the hybridisation of 15th group element does not change
B. During the (ii) reaction the hybridisation of 16th group element does not change
C. During the (iii) reaction the hybridisation of 13th group element does not change
D. During the (iv) reaction the hybridisation of 18th group element does not change

Explanation

To determine the incorrect statement regarding the hybridization changes in the given reactions, let's analyze each reaction step-by-step.

### Step 1: Analyze the First Reaction
**Reaction:** \( PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl \)

1. **Hybridization of \( PCl_3 \):**
- Phosphorus (P) has 5 valence electrons and forms 3 bonds with chlorine (Cl) and has 1 lone pair.
- Therefore, the hybridization of \( PCl_3 \) is \( sp^3 \).

2. **Hybridization of \( H_3PO_3 \):**
- In \( H_3PO_3 \), phosphorus is bonded to three hydroxyl groups (OH) and has one lone pair.
- The hybridization remains \( sp^3 \).

**Conclusion:** No change in hybridization for phosphorus.

### Step 2: Analyze the Second Reaction
**Reaction:** \( SF_4 + 3H_2O \rightarrow H_3SO_3 + 4HF \)

1. **Hybridization of \( SF_4 \):**
- Sulfur (S) has 6 valence electrons, forms 4 bonds with fluorine (F), and has 1 lone pair.
- Therefore, the hybridization of \( SF_4 \) is \( sp^3d \).

2. **Hybridization of \( H_3SO_3 \):**
- In \( H_3SO_3 \), sulfur is bonded to two hydroxyl groups (OH) and has one double bond with oxygen (O).
- The hybridization changes to \( sp^3 \).

**Conclusion:** Change in hybridization for sulfur.

### Step 3: Analyze the Third Reaction
**Reaction:** \( BCl_3 + 3H_2O \rightarrow H_3BO_3 + 3HCl \)

1. **Hybridization of \( BCl_3 \):**
- Boron (B) has 3 valence electrons and forms 3 bonds with chlorine (Cl).
- Therefore, the hybridization of \( BCl_3 \) is \( sp^2 \).

2. **Hybridization of \( H_3BO_3 \):**
- In \( H_3BO_3 \), boron is bonded to three hydroxyl groups (OH).
- The hybridization remains \( sp^2 \).

**Conclusion:** No change in hybridization for boron.

### Step 4: Analyze the Fourth Reaction
**Reaction:** \( XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF \)

1. **Hybridization of \( XeF_6 \):**
- Xenon (Xe) has 8 valence electrons, forms 6 bonds with fluorine (F).
- Therefore, the hybridization of \( XeF_6 \) is \( sp^3d^2 \).

2. **Hybridization of \( XeO_3 \):**
- In \( XeO_3 \), xenon is bonded to three oxygen atoms.
- The hybridization changes to \( sp^3 \).

**Conclusion:** Change in hybridization for xenon.

### Final Conclusion
- **Incorrect Statements:**
- The statement that there is no change in hybridization for the 16th group (sulfur) is incorrect because sulfur does change from \( sp^3d \) to \( sp^3 \).
- The statement regarding xenon also indicates a change in hybridization, which is incorrect if it states otherwise.

### Summary of Incorrect Statements
- The incorrect statement is that there is no change in hybridization for the 16th group elements (sulfur) and possibly for the 18th group elements (xenon).