Which of the following has the regular tetrahedral structure?

Options

A. `BF_(4)^(-)`
B. `SF_(4)`
C. `XeF_(4)`
D. `[Ni(CN)_(4)]^(2-)`

Explanation

To determine which of the following compounds has a regular tetrahedral structure, we need to analyze the hybridization of each compound based on their valence electrons. Let's go through the compounds step by step.

### Step 1: Analyze BF4⁻
1. **Count Valence Electrons:**
- Boron (B) has 3 valence electrons.
- Each Fluorine (F) has 7 valence electrons, and there are 4 Fluorine atoms: \(4 \times 7 = 28\).
- The negative charge adds 1 electron.
- Total = \(3 + 28 + 1 = 32\) valence electrons.

2. **Determine Electron Pairs:**
- Divide the total by 2 to find the number of electron pairs: \(32 / 2 = 16\) pairs.

3. **Calculate Hybridization:**
- The number of bonding pairs is 4 (from 4 F atoms) and no lone pairs on B.
- Hybridization = \(sp^3\) (4 electron pairs).

4. **Conclusion:**
- BF4⁻ has a tetrahedral structure.

### Step 2: Analyze SF4
1. **Count Valence Electrons:**
- Sulfur (S) has 6 valence electrons.
- Each Fluorine (F) has 7 valence electrons, and there are 4 Fluorine atoms: \(4 \times 7 = 28\).
- Total = \(6 + 28 = 34\) valence electrons.

2. **Determine Electron Pairs:**
- Divide the total by 2: \(34 / 2 = 17\) pairs.

3. **Calculate Hybridization:**
- There are 4 bonding pairs and 1 lone pair on S.
- Hybridization = \(sp^3d\) (5 electron pairs).

4. **Conclusion:**
- SF4 does not have a tetrahedral structure; it has a seesaw shape.

### Step 3: Analyze XeF4
1. **Count Valence Electrons:**
- Xenon (Xe) has 8 valence electrons.
- Each Fluorine (F) has 7 valence electrons, and there are 4 Fluorine atoms: \(4 \times 7 = 28\).
- Total = \(8 + 28 = 36\) valence electrons.

2. **Determine Electron Pairs:**
- Divide the total by 2: \(36 / 2 = 18\) pairs.

3. **Calculate Hybridization:**
- There are 4 bonding pairs and 2 lone pairs on Xe.
- Hybridization = \(sp^3d^2\) (6 electron pairs).

4. **Conclusion:**
- XeF4 does not have a tetrahedral structure; it has a square planar shape.

### Step 4: Analyze Ni(CN)4²⁻
1. **Count Valence Electrons:**
- Nickel (Ni) has 10 valence electrons (as a transition metal).
- Each Cyanide (CN) has 10 valence electrons, and there are 4 CN groups: \(4 \times 10 = 40\).
- Total = \(10 + 40 = 50\) valence electrons.

2. **Determine Electron Pairs:**
- Divide the total by 2: \(50 / 2 = 25\) pairs.

3. **Calculate Hybridization:**
- The coordination number is 4 (4 CN ligands).
- Hybridization = \(sp^3\) (4 electron pairs).

4. **Conclusion:**
- Ni(CN)4²⁻ has a tetrahedral structure.

### Final Conclusion
- The only compound that has a regular tetrahedral structure is **BF4⁻**.