Isostructural group of molecules are

Options

A. `NH_(3) , NF_(3) , BF_(3)`
B. `NO_(3)^(-) , NO_(2)^(-) , SF_(4)`
C. `XeO_(4) , NH_(4)^(+) , CH_(4)`
D. `CH_(3)^(-) , NH_(3) , NF_(3)`

Answer

To determine the isostructural group of molecules, we need to analyze the hybridization and molecular geometry of each option provided in the question. Isostructural molecules have the same shape and hybridization.

### Step-by-Step Solution:

1. **Understanding Hybridization**:
- The hybridization can be calculated using the formula:
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Valence Electrons of Central Atom} + \text{Monovalent Atoms} + \text{Charge} \right) \]
- We will apply this formula to each molecule in the options.

2. **Analyzing Option 1: NH3, NF3, BF3**:
- **NH3**:
- Nitrogen has 5 valence electrons, 3 hydrogen atoms (monovalent).
- Hybridization:
\[ \frac{1}{2} (5 + 3 + 0) = 4 \quad \Rightarrow \quad \text{sp}^3 \]
- Shape: Pyramidal (3 bond pairs, 1 lone pair).
- **NF3**:
- Similar to NH3, nitrogen has 5 valence electrons, 3 fluorine atoms.
- Hybridization:
\[ \frac{1}{2} (5 + 3 + 0) = 4 \quad \Rightarrow \quad \text{sp}^3 \]
- Shape: Pyramidal (3 bond pairs, 1 lone pair).
- **BF3**:
- Boron has 3 valence electrons, 3 fluorine atoms.
- Hybridization:
\[ \frac{1}{2} (3 + 3 + 0) = 3 \quad \Rightarrow \quad \text{sp}^2 \]
- Shape: Trigonal planar (3 bond pairs, 0 lone pairs).
- **Conclusion**: Not isostructural.

3. **Analyzing Option 2: NO3-, NO2-, SF4**:
- **NO3-**:
- Nitrogen has 5 valence electrons, 3 oxygen atoms, and a -1 charge.
- Hybridization:
\[ \frac{1}{2} (5 + 3 + 1) = 4 \quad \Rightarrow \quad \text{sp}^2 \]
- Shape: Trigonal planar (3 bond pairs, 0 lone pairs).
- **NO2-**:
- Nitrogen has 5 valence electrons, 2 oxygen atoms, and a -1 charge.
- Hybridization:
\[ \frac{1}{2} (5 + 2 + 1) = 4 \quad \Rightarrow \quad \text{sp}^2 \]
- Shape: Bent (2 bond pairs, 1 lone pair).
- **SF4**:
- Sulfur has 6 valence electrons, 4 fluorine atoms.
- Hybridization:
\[ \frac{1}{2} (6 + 4 + 0) = 5 \quad \Rightarrow \quad \text{sp}^3d \]
- Shape: Seesaw (4 bond pairs, 1 lone pair).
- **Conclusion**: Not isostructural.

4. **Analyzing Option 3: XeO4, NH4+, CH4**:
- **XeO4**:
- Xenon has 8 valence electrons, 4 oxygen atoms.
- Hybridization:
\[ \frac{1}{2} (8 + 0 + 0) = 4 \quad \Rightarrow \quad \text{sp}^3 \]
- Shape: Tetrahedral (4 bond pairs, 0 lone pairs).
- **NH4+**:
- Nitrogen has 5 valence electrons, 4 hydrogen atoms, and a +1 charge.
- Hybridization:
\[ \frac{1}{2} (5 + 4 - 1) = 4 \quad \Rightarrow \quad \text{sp}^3 \]
- Shape: Tetrahedral (4 bond pairs, 0 lone pairs).
- **CH4**:
- Carbon has 4 valence electrons, 4 hydrogen atoms.
- Hybridization:
\[ \frac{1}{2} (4 + 4 + 0) = 4 \quad \Rightarrow \quad \text{sp}^3 \]
- Shape: Tetrahedral (4 bond pairs, 0 lone pairs).
- **Conclusion**: All are isostructural (tetrahedral).

5. **Analyzing Option 4: CH3-, NH3, NF3**:
- **CH3-**:
- Carbon has 4 valence electrons, 3 hydrogen atoms, and a -1 charge.
- Hybridization:
\[ \frac{1}{2} (4 + 3 + 1) = 4 \quad \Rightarrow \quad \text{sp}^3 \]
- Shape: Pyramidal (3 bond pairs, 1 lone pair).
- **NH3**:
- Already calculated as sp3, pyramidal.
- **NF3**:
- Already calculated as sp3, pyramidal.
- **Conclusion**: All are isostructural (pyramidal).

### Final Conclusion:
- **Isostructural Groups**:
- Option 3 (XeO4, NH4+, CH4) and Option 4 (CH3-, NH3, NF3) are isostructural.