The hybridisation of the central atom will change when

Options

A. `NH_(3)` combined with `H^(+)`
B. `H_(3)BO_(2)` combined with `OH`
C. `NH_(3) forms NH_(2)`
D. `H_(2)O` combines with `H^(+)`

Explanation

To determine when the hybridization of the central atom changes, we can analyze several examples of molecules and ions. Here's a step-by-step breakdown of the process:

### Step 1: Understanding Hybridization
Hybridization is the mixing of atomic orbitals to form new hybrid orbitals that can accommodate the bonding pairs of electrons. The type of hybridization depends on the number of electron pairs around the central atom.

### Step 2: Analyzing NH3 and NH4+
1. **NH3 (Ammonia)**:
- Valence electrons of nitrogen = 5
- Number of hydrogen atoms = 3 (monovalent)
- Cationic charge = 0
- Anionic charge = 0
- Calculation:
\[ Z = \frac{1}{2} \left(5 + 3 - 0 + 0\right) = \frac{1}{2} \times 8 = 4 \text{ (electron pairs)} \]
- Hybridization = sp³ (4 electron pairs)

2. **NH4+ (Ammonium ion)**:
- Valence electrons of nitrogen = 5
- Number of hydrogen atoms = 4 (monovalent)
- Cationic charge = 1
- Anionic charge = 0
- Calculation:
\[ Z = \frac{1}{2} \left(5 + 4 - 1 + 0\right) = \frac{1}{2} \times 8 = 4 \text{ (electron pairs)} \]
- Hybridization = sp³ (4 electron pairs)

**Conclusion**: The hybridization does not change from NH3 to NH4+.

### Step 3: Analyzing H3BO3 and H4BO4-
1. **H3BO3 (Boric acid)**:
- Valence electrons of boron = 3
- Number of hydrogen atoms = 3 (monovalent)
- Cationic charge = 0
- Anionic charge = 0
- Calculation:
\[ Z = \frac{1}{2} \left(3 + 3 - 0 + 0\right) = \frac{1}{2} \times 6 = 3 \text{ (electron pairs)} \]
- Hybridization = sp² (3 electron pairs)

2. **H4BO4- (Tetrahydroxyborate ion)**:
- Valence electrons of boron = 3
- Number of hydrogen atoms = 4 (monovalent)
- Cationic charge = 0
- Anionic charge = 1
- Calculation:
\[ Z = \frac{1}{2} \left(3 + 4 - 0 + 1\right) = \frac{1}{2} \times 8 = 4 \text{ (electron pairs)} \]
- Hybridization = sp³ (4 electron pairs)

**Conclusion**: The hybridization changes from sp² in H3BO3 to sp³ in H4BO4-.

### Step 4: Analyzing NH3-2 and NH2-
1. **NH3 (Ammonia)**:
- Hybridization = sp³ (as calculated earlier)

2. **NH2- (Amide ion)**:
- Valence electrons of nitrogen = 5
- Number of hydrogen atoms = 2 (monovalent)
- Cationic charge = 0
- Anionic charge = 1
- Calculation:
\[ Z = \frac{1}{2} \left(5 + 2 - 0 + 1\right) = \frac{1}{2} \times 8 = 4 \text{ (electron pairs)} \]
- Hybridization = sp³ (4 electron pairs)

**Conclusion**: The hybridization does not change from NH3 to NH2-.

### Step 5: Analyzing H2O and H3O+
1. **H2O (Water)**:
- Valence electrons of oxygen = 6
- Number of hydrogen atoms = 2 (monovalent)
- Cationic charge = 0
- Anionic charge = 0
- Calculation:
\[ Z = \frac{1}{2} \left(6 + 2 - 0 + 0\right) = \frac{1}{2} \times 8 = 4 \text{ (electron pairs)} \]
- Hybridization = sp³ (4 electron pairs)

2. **H3O+ (Hydronium ion)**:
- Valence electrons of oxygen = 6
- Number of hydrogen atoms = 3 (monovalent)
- Cationic charge = 1
- Anionic charge = 0
- Calculation:
\[ Z = \frac{1}{2} \left(6 + 3 - 1 + 0\right) = \frac{1}{2} \times 8 = 4 \text{ (electron pairs)} \]
- Hybridization = sp³ (4 electron pairs)

**Conclusion**: The hybridization does not change from H2O to H3O+.

### Final Conclusion
The hybridization of the central atom changes when the number of electron pairs around the atom changes, as seen in the transition from H3BO3 to H4BO4-.