Among the following species, identify the isostructural pairs `NF_3`, `NO_3^-`, `BF_3`, `H_3O^+`,

Options

A. `[NF_3, NO_3^-]`and, `[BF_3, H_3O^]`
B. `[NF_3, HN_3]`and`[NO_3^-BF_3]`
C. `[NF_3,H_3O^+]`and `[NO_3^-, BF_3]`
D. `[NF_3,H_3O^+]`and`[BF_3, HN_3]`

Answer

To identify the isostructural pairs among the species \( NF_3 \), \( NO_3^- \), \( BF_3 \), and \( H_3O^+ \), we need to analyze the hybridization and geometry of each species.

### Step 1: Analyze \( NF_3 \)
- **Valence Electrons**: Nitrogen (N) has 5 valence electrons, and each Fluorine (F) has 7. Thus, \( NF_3 \) has:
\[ 5 + 3 \times 7 = 26 \text{ electrons} \]
- **Lewis Structure**: Nitrogen forms 3 sigma bonds with 3 fluorine atoms and has 1 lone pair.
- **Steric Number**:
\[ \text{Steric Number} = \text{Number of Sigma Bonds} + \text{Number of Lone Pairs} = 3 + 1 = 4 \]
- **Hybridization**: \( sp^3 \)
- **Geometry**: Tetrahedral

### Step 2: Analyze \( NO_3^- \)
- **Valence Electrons**: Nitrogen has 5, and each Oxygen has 6. The negative charge adds 1 electron. Thus, \( NO_3^- \) has:
\[ 5 + 3 \times 6 + 1 = 24 \text{ electrons} \]
- **Lewis Structure**: Nitrogen forms 3 sigma bonds with 3 oxygen atoms and has no lone pairs.
- **Steric Number**:
\[ \text{Steric Number} = 3 + 0 = 3 \]
- **Hybridization**: \( sp^2 \)
- **Geometry**: Trigonal planar

### Step 3: Analyze \( BF_3 \)
- **Valence Electrons**: Boron has 3 valence electrons, and each Fluorine has 7. Thus, \( BF_3 \) has:
\[ 3 + 3 \times 7 = 24 \text{ electrons} \]
- **Lewis Structure**: Boron forms 3 sigma bonds with 3 fluorine atoms and has no lone pairs.
- **Steric Number**:
\[ \text{Steric Number} = 3 + 0 = 3 \]
- **Hybridization**: \( sp^2 \)
- **Geometry**: Trigonal planar

### Step 4: Analyze \( H_3O^+ \)
- **Valence Electrons**: Oxygen has 6 valence electrons, and each Hydrogen has 1. The positive charge means we lose 1 electron. Thus, \( H_3O^+ \) has:
\[ 6 + 3 \times 1 - 1 = 8 \text{ electrons} \]
- **Lewis Structure**: Oxygen forms 3 sigma bonds with 3 hydrogen atoms and has 1 lone pair.
- **Steric Number**:
\[ \text{Steric Number} = 3 + 1 = 4 \]
- **Hybridization**: \( sp^3 \)
- **Geometry**: Tetrahedral

### Step 5: Identify Isostructural Pairs
- **Isostructural Pairs**:
- \( NF_3 \) and \( H_3O^+ \) both have \( sp^3 \) hybridization and tetrahedral geometry.
- \( NO_3^- \) and \( BF_3 \) both have \( sp^2 \) hybridization and trigonal planar geometry.

### Conclusion
The isostructural pairs are:
- \( NF_3 \) and \( H_3O^+ \) (both tetrahedral, \( sp^3 \))
- \( NO_3^- \) and \( BF_3 \) (both trigonal planar, \( sp^2 \))