To determine where dsp² hybridization is found, we need to analyze the given compounds one by one. Let's go through the options step by step.
### Step 1: Understanding dsp² Hybridization
dsp² hybridization involves the mixing of one d orbital, one s orbital, and two p orbitals to form four equivalent dsp² hybrid orbitals. This type of hybridization typically occurs in transition metals that have a coordination number of 4 and can accommodate d orbitals.
### Step 2: Analyzing Each Compound
1. **NiCl₄³⁻**:
- Nickel (Ni) in this compound has a +1 oxidation state (since the overall charge is -3).
- In its neutral state, Ni has 8 electrons in d orbitals and 2 in s orbitals. For Ni⁺, it loses one electron from the s orbital, leaving it with 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸.
- Since there are no vacant d orbitals, NiCl₄³⁻ cannot exhibit dsp² hybridization.
2. **CoCl₄²⁻**:
- Cobalt (Co) in this compound has a +2 oxidation state.
- In its neutral state, Co has 7 electrons in d orbitals and 2 in s orbitals. For Co²⁺, it loses 2 electrons from the s orbital, leaving it with 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷.
- Again, there are no vacant d orbitals, so CoCl₄²⁻ cannot exhibit dsp² hybridization.
3. **CuCl₄³⁻**:
- Copper (Cu) in this compound has a +1 oxidation state.
- In its neutral state, Cu has 10 electrons in d orbitals and 1 in s orbitals. For Cu⁺, it loses one electron from the s orbital, leaving it with 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰.
- All d orbitals are filled, and there are no vacant d orbitals, so CuCl₄³⁻ cannot exhibit dsp² hybridization.
4. **PtCl₄²⁻**:
- Platinum (Pt) in this compound has a +2 oxidation state.
- In its neutral state, Pt has 10 electrons in d orbitals and 2 in s orbitals. For Pt²⁺, it loses 2 electrons from the s orbital, leaving it with 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 4p⁶ 4d¹⁰.
- Since Pt can exhibit rearrangement, one of the d electrons can be promoted to a vacant orbital, allowing for dsp² hybridization.
### Conclusion
The only compound that can exhibit dsp² hybridization is **PtCl₄²⁻**.
### Final Answer
**The correct answer is PtCl₄²⁻.**
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