Which of the following pairs is isostructural (i.e having the same shape and hybridization ?
`[NF_(3) " and " BF_(3)]`
`[BF_(4)^(-) and NH_(4)^(+)]`

Options

A. `[NF_(3) " and " BF_(3)]`
B. `[BF_(4)^(-) and NH_(4)^(+)]`
C. `[BCl_(3) and BrCl_(3)]`
D. `[NH_(3) and NO_(3)^(-)]`

Explanation

To determine which of the given pairs is isostructural (having the same shape and hybridization), we will analyze each pair step by step.

### Step 1: Analyze the first pair - NF3 and BF3

1. **Determine the steric number for NF3:**
- Nitrogen (N) has 5 valence electrons.
- Fluorine (F) has 7 valence electrons, and there are 3 fluorine atoms.
- Total valence electrons = 5 (from N) + 3 × 7 (from F) = 5 + 21 = 26.
- Steric number = (Number of valence electrons + Number of surrounding atoms) / 2 = (5 + 3) / 2 = 4.
- Hybridization = sp³.

2. **Geometry of NF3:**
- NF3 has 3 sigma bonds and 1 lone pair on nitrogen.
- The molecular geometry is trigonal pyramidal.

3. **Determine the steric number for BF3:**
- Boron (B) has 3 valence electrons.
- Fluorine (F) has 7 valence electrons, and there are 3 fluorine atoms.
- Total valence electrons = 3 (from B) + 3 × 7 (from F) = 3 + 21 = 24.
- Steric number = (3 + 3) / 2 = 3.
- Hybridization = sp².

4. **Geometry of BF3:**
- BF3 has 3 sigma bonds and no lone pairs on boron.
- The molecular geometry is trigonal planar.

### Conclusion for the first pair:
- NF3 is trigonal pyramidal (sp³) and BF3 is trigonal planar (sp²).
- **They are not isostructural.**

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### Step 2: Analyze the second pair - BF4⁻ and NH4⁺

1. **Determine the steric number for BF4⁻:**
- Boron (B) has 3 valence electrons.
- Fluorine (F) has 7 valence electrons, and there are 4 fluorine atoms.
- Total valence electrons = 3 (from B) + 4 × 7 (from F) + 1 (for the negative charge) = 3 + 28 + 1 = 32.
- Steric number = (3 + 4 + 1) / 2 = 4.
- Hybridization = sp³.

2. **Geometry of BF4⁻:**
- BF4⁻ has 4 sigma bonds and no lone pairs on boron.
- The molecular geometry is tetrahedral.

3. **Determine the steric number for NH4⁺:**
- Nitrogen (N) has 5 valence electrons.
- Hydrogen (H) has 1 valence electron, and there are 4 hydrogen atoms.
- Total valence electrons = 5 (from N) + 4 × 1 (from H) - 1 (for the positive charge) = 5 + 4 - 1 = 8.
- Steric number = (5 + 4 - 1) / 2 = 4.
- Hybridization = sp³.

4. **Geometry of NH4⁺:**
- NH4⁺ has 4 sigma bonds and no lone pairs on nitrogen.
- The molecular geometry is tetrahedral.

### Conclusion for the second pair:
- Both BF4⁻ and NH4⁺ are tetrahedral (sp³).
- **They are isostructural.**

### Final Answer:
The correct answer is the second pair: **[BF4⁻ and NH4⁺]**.

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