In which of the following hybridisation of underlined atom changes

Options

A. `underset(-)(N)H_(3) + H^(+) to underline(N) H_(4)^(+)`
B. `underline(B)F_(3) + F^(-) to underline(B)F_(4)^(-)`
C. `underline(C_(2)) H_(2) + H_(2) to underline(C_(2)) H_(6)`
D. All of these

Answer

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

### Step 1: Analyze Option 1 - Ammonia (NH₃) and Ammonium Ion (NH₄⁺)

1. **Identify Valence Electrons:**
- Nitrogen (N) has 5 valence electrons.
- Each Hydrogen (H) has 1 valence electron, and there are 3 H atoms in NH₃:
\[ 1 \times 3 = 3 \]
- Total for NH₃:
\[ 5 + 3 = 8 \text{ valence electrons} \]

2. **Calculate Total Electron Pairs:**
- Divide total valence electrons by 2:
\[ \frac{8}{2} = 4 \text{ total electron pairs} \]

3. **Determine Lone Pairs:**
- NH₃ has 3 bond pairs (N-H bonds), so:
\[ \text{Lone pairs} = 4 - 3 = 1 \]

4. **Determine Hybridization:**
- 4 electron pairs correspond to sp³ hybridization.

5. **For Ammonium Ion (NH₄⁺):**
- Nitrogen still has 5 valence electrons, and there are 4 H atoms:
\[ 1 \times 4 = 4 \]
- Account for the positive charge (-1):
\[ 5 + 4 - 1 = 8 \text{ valence electrons} \]
- Total electron pairs:
\[ \frac{8}{2} = 4 \text{ total electron pairs} \]
- NH₄⁺ has 4 bond pairs and 0 lone pairs, confirming sp³ hybridization.

**Conclusion for Option 1:**
- Hybridization does not change (both are sp³).

### Step 2: Analyze Option 2 - Boron Trifluoride (BF₃) and Tetrafluoroborate Ion (BF₄⁻)

1. **Identify Valence Electrons for BF₃:**
- Boron (B) has 3 valence electrons.
- Each Fluorine (F) has 7 valence electrons, and there are 3 F atoms:
\[ 7 \times 3 = 21 \]
- Total for BF₃:
\[ 3 + 21 = 24 \text{ valence electrons} \]

2. **Calculate Total Electron Pairs:**
- Divide by 2:
\[ \frac{24}{2} = 12 \text{ total electron pairs} \]
- Since we are interested in pairs:
\[ 12 \div 2 = 6 \text{ pairs} \text{ (this is incorrect, it should be 3 pairs)} \]
- Correctly, it should be:
\[ \frac{24}{2} = 12 \text{ electron pairs} \]
- Thus, it is sp² hybridization.

3. **For Tetrafluoroborate Ion (BF₄⁻):**
- Boron still has 3 valence electrons, and there are 4 F atoms:
\[ 7 \times 4 = 28 \]
- Account for the negative charge (+1):
\[ 3 + 28 + 1 = 32 \text{ valence electrons} \]
- Total electron pairs:
\[ \frac{32}{2} = 16 \text{ pairs} \]
- This means sp³ hybridization.

**Conclusion for Option 2:**
- Hybridization changes from sp² (BF₃) to sp³ (BF₄⁻).

### Step 3: Analyze Option 3 - Ethyne (C₂H₂) and Ethane (C₂H₆)

1. **For Ethyne (C₂H₂):**
- Each Carbon (C) has 4 valence electrons, and there are 2 C atoms:
\[ 4 \times 2 = 8 \]
- Each Hydrogen has 1 valence electron, and there are 2 H atoms:
\[ 1 \times 2 = 2 \]
- Total for C₂H₂:
\[ 8 + 2 = 10 \text{ valence electrons} \]
- Total electron pairs:
\[ \frac{10}{2} = 5 \text{ pairs} \]
- Each C in C₂H₂ is sp hybridized.

2. **For Ethane (C₂H₆):**
- Total valence electrons:
\[ 8 + 6 = 14 \]
- Total electron pairs:
\[ \frac{14}{2} = 7 \text{ pairs} \]
- Each C in C₂H₆ is sp³ hybridized.

**Conclusion for Option 3:**
- Hybridization changes from sp (C₂H₂) to sp³ (C₂H₆).

### Final Conclusion:
- The hybridization of the underlined atom changes in options 2 (BF₃ to BF₄⁻) and 3 (C₂H₂ to C₂H₆).