Find out number of transformation among following which involves the change of hybridisation of underlined atom.
(a) `H_(2)underline(O)+H^(+) to H_(3)underline(O)^(+) " " (b) NH_(3)+underline(B)F_(3) to NH_(3). Underline(B)F_(3)`
(c ) `underline(X)eF_(6) to underline(X)eF_(5)^(+)+F^(-) " " (d) 2underline(P) Cl_(5) to underline(P)Cl_(4)^+)+PCl_(6)^(-)`
(e ) ` underline(C )H_(3)-CH_(3) to underline(C )H_(3)^(-)+CH_(3)^(+)`

Explanation

To determine the number of transformations that involve a change in hybridization of the underlined atom in the given reactions, we will analyze each reaction step by step.

### Step 1: Analyzing Reaction (a)
**Reaction:** \( H_2O \rightarrow H_3O^+ \)

- **Central Atom:** Oxygen (O)
- **Valence Electrons (V):** 6 (for O)
- **Monovalent Atoms (M):** 2 (2 H atoms in \( H_2O \))
- **Charge:** 0 (for \( H_2O \))

Using the formula for hybridization:
\[ \text{Steric Number} = \frac{V + M + \text{Charge}}{2} = \frac{6 + 2 + 0}{2} = 4 \]
This indicates \( sp^3 \) hybridization.

Now for \( H_3O^+ \):
- **Monovalent Atoms (M):** 3 (3 H atoms in \( H_3O^+ \))
- **Charge:** -1 (for \( H_3O^+ \))

Calculating:
\[ \text{Steric Number} = \frac{6 + 3 - 1}{2} = \frac{8}{2} = 4 \]
This indicates \( sp^3 \) hybridization.

**Conclusion:** No change in hybridization.

### Step 2: Analyzing Reaction (b)
**Reaction:** \( NH_3 + BF_3 \rightarrow NH_3 \cdot BF_3 \)

- **Central Atom in BF3:** Boron (B)
- **Valence Electrons (V):** 3 (for B)
- **Monovalent Atoms (M):** 3 (3 F atoms)
- **Charge:** 0

Calculating:
\[ \text{Steric Number} = \frac{3 + 3 + 0}{2} = 3 \]
This indicates \( sp^2 \) hybridization.

Now for \( NH_3 \cdot BF_3 \):
- **Monovalent Atoms (M):** 4 (3 F + 1 from NH3)
- **Charge:** 0

Calculating:
\[ \text{Steric Number} = \frac{3 + 4 + 0}{2} = \frac{7}{2} = 4 \]
This indicates \( sp^3 \) hybridization.

**Conclusion:** Change from \( sp^2 \) to \( sp^3 \) hybridization.

### Step 3: Analyzing Reaction (c)
**Reaction:** \( XeF_6 \rightarrow XeF_5^+ + F^- \)

- **Central Atom:** Xenon (Xe)
- **Valence Electrons (V):** 8 (for Xe)
- **Monovalent Atoms (M):** 6 (6 F atoms)
- **Charge:** 0

Calculating:
\[ \text{Steric Number} = \frac{8 + 6 + 0}{2} = 7 \]
This indicates \( sp^3d^3 \) hybridization.

Now for \( XeF_5^+ \):
- **Monovalent Atoms (M):** 5 (5 F atoms)
- **Charge:** -1

Calculating:
\[ \text{Steric Number} = \frac{8 + 5 - 1}{2} = \frac{12}{2} = 6 \]
This indicates \( sp^3d^2 \) hybridization.

**Conclusion:** Change from \( sp^3d^3 \) to \( sp^3d^2 \) hybridization.

### Step 4: Analyzing Reaction (d)
**Reaction:** \( PCl_5 \rightarrow PCl_4^+ + PCl_6^- \)

- **Central Atom:** Phosphorus (P)
- **Valence Electrons (V):** 5 (for P)
- **Monovalent Atoms (M):** 5 (5 Cl atoms)
- **Charge:** 0

Calculating:
\[ \text{Steric Number} = \frac{5 + 5 + 0}{2} = 5 \]
This indicates \( sp^3d \) hybridization.

Now for \( PCl_4^+ \):
- **Monovalent Atoms (M):** 4 (4 Cl atoms)
- **Charge:** -1

Calculating:
\[ \text{Steric Number} = \frac{5 + 4 - 1}{2} = \frac{8}{2} = 4 \]
This indicates \( sp^3 \) hybridization.

**Conclusion:** Change from \( sp^3d \) to \( sp^3 \) hybridization.

### Step 5: Analyzing Reaction (e)
**Reaction:** \( C_2H_6 \rightarrow CH_3^- + CH_3^+ \)

- **Central Atom in \( C_2H_6 \):** Carbon (C)
- **Valence Electrons (V):** 4 (for C)
- **Monovalent Atoms (M):** 4 (3 H + 1 C)
- **Charge:** 0

Calculating:
\[ \text{Steric Number} = \frac{4 + 4 + 0}{2} = 4 \]
This indicates \( sp^3 \) hybridization.

Now for \( CH_3^- \):
- **Monovalent Atoms (M):** 3 (3 H atoms)
- **Charge:** -1

Calculating:
\[ \text{Steric Number} = \frac{4 + 3 + 1}{2} = \frac{8}{2} = 4 \]
This indicates \( sp^3 \) hybridization.

**Conclusion:** No change in hybridization.

### Final Summary
The transformations that involve a change in hybridization are:
- Reaction (b): \( sp^2 \) to \( sp^3 \)
- Reaction (c): \( sp^3d^3 \) to \( sp^3d^2 \)
- Reaction (d): \( sp^3d \) to \( sp^3 \)

**Total transformations involving change of hybridization:** **3**