To determine which ion is not tetrahedral in shape, we will analyze the geometry of each given option step by step.
### Step 1: Analyze BF4⁻
- **Boron (B)** has an atomic number of 5, with an electronic configuration of 1s² 2s² 2p¹.
- Boron typically forms BF3, and upon accepting one fluorine atom, it becomes BF4⁻.
- The hybridization for BF4⁻ is sp³, which leads to a tetrahedral geometry.
### Step 2: Analyze NH4⁺
- **Nitrogen (N)** has an atomic number of 7, with an electronic configuration of 1s² 2s² 2p³.
- When nitrogen forms NH4⁺, it has four bonding pairs and no lone pairs.
- The hybridization is also sp³, resulting in a tetrahedral shape.
### Step 3: Analyze [Cu(NH3)4]²⁺
- **Copper (Cu)** has an atomic number of 29, with an electronic configuration of [Ar] 3d¹⁰ 4s¹.
- In the +2 oxidation state, copper loses two electrons (one from 4s and one from 3d), resulting in a configuration of 3d⁹.
- Ammonia (NH3) is a strong field ligand, causing electron pairing and leading to dsp² hybridization.
- The geometry of [Cu(NH3)4]²⁺ is square planar, not tetrahedral.
### Step 4: Analyze NiCl4²⁻
- **Nickel (Ni)** has an atomic number of 28, with an electronic configuration of [Ar] 3d⁸ 4s².
- In NiCl4²⁻, nickel is in the +2 oxidation state, losing two electrons from the 4s orbital.
- Chlorine is a weak field ligand, which does not cause electron pairing, leading to sp³ hybridization.
- Therefore, NiCl4²⁻ has a tetrahedral geometry.
### Conclusion
After analyzing all the options, the ion that is not tetrahedral in shape is **[Cu(NH3)4]²⁺**, which has a square planar geometry.
### Final Answer
**[Cu(NH3)4]²⁺ is the ion that is not tetrahedral in shape.**
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