Which of the following are isostructural pairs
A. `SO_4^(-2) and CrO_4^(-2)`
B. `SiCl_4 and TiCl_4`
C. `NH_3 and NO_3^(-)`
D. `BCl_3 and BrCl_3`

Options

A. C and D only
B. A and B only
C. A and C only
D. B and C only

Answer

To determine which pairs are isostructural, we need to analyze the hybridization and geometry of each compound in the options provided. Isostructural compounds have the same type of structure or geometry.

### Step-by-Step Solution:

1. **Option A: `SO_4^(-2)` and `CrO_4^(-2)`**
- **For `SO_4^(-2)` (Sulfate ion)**:
- Sulfur (S) has 6 valence electrons.
- There are 4 oxygen atoms (each contributing 1 electron) and a -2 charge.
- Total = 6 (S) + 4 (O) + 2 (charge) = 12 electrons.
- Hybridization = (12/2) = 6 → SP3 hybridization.
- Geometry: Tetrahedral.

- **For `CrO_4^(-2)` (Chromate ion)**:
- Chromium (Cr) has 6 valence electrons.
- There are 4 oxygen atoms and a -2 charge.
- Total = 6 (Cr) + 4 (O) + 2 (charge) = 12 electrons.
- Hybridization = (12/2) = 6 → SP3 hybridization.
- Geometry: Tetrahedral.

- **Conclusion**: Both `SO_4^(-2)` and `CrO_4^(-2)` are isostructural (both tetrahedral).

2. **Option B: `SiCl_4` and `TiCl_4`**
- **For `SiCl_4` (Silicon tetrachloride)**:
- Silicon (Si) has 4 valence electrons.
- There are 4 chlorine atoms.
- Total = 4 (Si) + 4 (Cl) = 8 electrons.
- Hybridization = (8/2) = 4 → SP3 hybridization.
- Geometry: Tetrahedral.

- **For `TiCl_4` (Titanium tetrachloride)**:
- Titanium (Ti) has 4 valence electrons in its +4 oxidation state.
- Total = 4 (Ti) + 4 (Cl) = 8 electrons.
- Hybridization = (8/2) = 4 → SP3 hybridization.
- Geometry: Tetrahedral.

- **Conclusion**: Both `SiCl_4` and `TiCl_4` are isostructural (both tetrahedral).

3. **Option C: `NH_3` and `NO_3^(-)`**
- **For `NH_3` (Ammonia)**:
- Nitrogen (N) has 5 valence electrons.
- There are 3 hydrogen atoms.
- Total = 5 (N) + 3 (H) = 8 electrons.
- Hybridization = (8/2) = 4 → SP3 hybridization.
- Geometry: Trigonal pyramidal.

- **For `NO_3^(-)` (Nitrate ion)**:
- Nitrogen (N) has 5 valence electrons.
- There are 3 oxygen atoms and a -1 charge.
- Total = 5 (N) + 3 (O) + 1 (charge) = 9 electrons.
- Hybridization = (9/2) = 4.5 → SP2 hybridization.
- Geometry: Trigonal planar.

- **Conclusion**: `NH_3` and `NO_3^(-)` are not isostructural (different geometries).

4. **Option D: `BCl_3` and `BrCl_3`**
- **For `BCl_3` (Boron trichloride)**:
- Boron (B) has 3 valence electrons.
- There are 3 chlorine atoms.
- Total = 3 (B) + 3 (Cl) = 6 electrons.
- Hybridization = (6/2) = 3 → SP2 hybridization.
- Geometry: Trigonal planar.

- **For `BrCl_3` (Bromine trichloride)**:
- Bromine (Br) has 7 valence electrons.
- There are 3 chlorine atoms.
- Total = 7 (Br) + 3 (Cl) = 10 electrons.
- Hybridization = (10/2) = 5 → SP3D hybridization.
- Geometry: Trigonal bipyramidal (with 2 lone pairs, T-shaped).

- **Conclusion**: `BCl_3` and `BrCl_3` are not isostructural (different geometries).

### Final Conclusion:
The isostructural pairs are:
- **Option A: `SO_4^(-2)` and `CrO_4^(-2)`**
- **Option B: `SiCl_4` and `TiCl_4`**