which one of the following ions is not tetrahedral in shape ?

Options

A. `[NiCl_(4)]^(2-)`
B. `NH_4^(+)`
C. `BF_4^(-)`
D. `[Cu(NH_3)_(4)]^(2)`

Explanation

To determine which of the following ions is not tetrahedral in shape, we will analyze the hybridization and geometry of each ion provided in the question: NiCl4^2-, NH4^+, BF4^-, and Cu(NH3)4^2+.

### Step 1: Analyze NiCl4^2-
1. **Determine the oxidation state of Ni**:
- Let the oxidation state of Ni be x.
- The oxidation state of Cl is -1, and there are 4 Cl atoms.
- The overall charge of the ion is -2.
- Thus, the equation becomes: x + 4(-1) = -2
- Solving this gives: x - 4 = -2 → x = +2.

2. **Electronic configuration of Ni**:
- Ni has an atomic number of 28, so its electronic configuration is [Ar] 3d^8 4s^2.
- In the +2 oxidation state, it loses the 4s electrons: [Ar] 3d^8.

3. **Hybridization**:
- Since Cl is a weak field ligand, it does not cause pairing of d electrons.
- The hybridization of Ni in this complex is sp^3 (4 regions of electron density).
- Therefore, NiCl4^2- has a tetrahedral shape.

### Step 2: Analyze NH4^+
1. **Determine the hybridization**:
- Nitrogen (N) has 5 valence electrons, and each hydrogen (H) contributes 1 electron.
- The overall charge is +1, so we have: 5 + 4 - 1 = 8 electrons.
- Hybridization is calculated as: (8 total valence electrons) / 2 = 4.
- This corresponds to sp^3 hybridization.

2. **Geometry**:
- With sp^3 hybridization, NH4^+ has a tetrahedral shape.

### Step 3: Analyze BF4^-
1. **Determine the hybridization**:
- Boron (B) has 3 valence electrons, and each fluorine (F) contributes 1 electron.
- The overall charge is -1, so we have: 3 + 4 - 1 = 6 electrons.
- Hybridization is calculated as: (6 total valence electrons) / 2 = 3.
- This corresponds to sp^3 hybridization.

2. **Geometry**:
- With sp^3 hybridization, BF4^- has a tetrahedral shape.

### Step 4: Analyze Cu(NH3)4^2+
1. **Determine the oxidation state of Cu**:
- Let the oxidation state of Cu be x.
- The oxidation state of NH3 is 0, and there are 4 NH3 molecules.
- The overall charge is +2, so: x + 0 = +2 → x = +2.

2. **Electronic configuration of Cu**:
- Cu has an atomic number of 29, so its electronic configuration is [Ar] 3d^10 4s^1.
- In the +2 oxidation state, it loses one 4s electron and one 3d electron: [Ar] 3d^9.

3. **Hybridization**:
- NH3 is a strong field ligand, which causes pairing of d electrons.
- The hybridization of Cu in this complex is dsp^2 (4 regions of electron density).
- Therefore, Cu(NH3)4^2+ has a square planar shape.

### Conclusion
Among the given ions, **Cu(NH3)4^2+** is not tetrahedral; it has a square planar shape.

### Summary of Shapes:
- NiCl4^2-: Tetrahedral
- NH4^+: Tetrahedral
- BF4^-: Tetrahedral
- Cu(NH3)4^2+: Square planar

### Answer:
**Cu(NH3)4^2+ is not tetrahedral in shape.**