To determine which complex has a square planar geometry, we will analyze the given complexes one by one, focusing on their electronic configurations, oxidation states, and hybridization.
### Step-by-Step Solution:
1. **Analyze Ni(CO)₄:**
- Nickel (Ni) has an outer electronic configuration of 3d⁸ 4s².
- In Ni(CO)₄, CO is a strong field ligand and causes pairing of electrons.
- After pairing, the configuration becomes 3d¹⁰.
- The hybridization is sp³, leading to a tetrahedral geometry.
- **Conclusion:** Ni(CO)₄ does not have square planar geometry.
2. **Analyze MnCl₄²⁻:**
- Manganese (Mn) has an outer electronic configuration of 3d⁵ 4s².
- In MnCl₄²⁻, we determine the oxidation state of Mn:
- Let x be the oxidation state: x - 4 = -2 → x = +2.
- Thus, Mn is in the +2 oxidation state with a configuration of 3d⁵.
- Chlorine is a weak field ligand, so no pairing occurs.
- The hybridization is sp³, resulting in tetrahedral geometry.
- **Conclusion:** MnCl₄²⁻ does not have square planar geometry.
3. **Analyze CuCl₄²⁻:**
- Copper (Cu) has an outer electronic configuration of 3d¹⁰ 4s¹.
- In CuCl₄²⁻, we find the oxidation state of Cu:
- Let x be the oxidation state: x - 4 = -2 → x = +2.
- Thus, Cu is in the +2 oxidation state with a configuration of 3d⁹.
- Chlorine is a weak field ligand, so no pairing occurs.
- The hybridization is sp³, leading to tetrahedral geometry.
- **Conclusion:** CuCl₄²⁻ does not have square planar geometry.
4. **Analyze Cu(NH₃)₄²⁺:**
- Copper (Cu) has an outer electronic configuration of 3d¹⁰ 4s¹.
- In Cu(NH₃)₄²⁺, we find the oxidation state of Cu:
- Let x be the oxidation state: x - 4 = -2 → x = +2.
- Thus, Cu is in the +2 oxidation state with a configuration of 3d⁹.
- NH₃ is a strong field ligand, which can cause pairing.
- The electron from the 3d orbital can move to the empty 4p orbital.
- The hybridization becomes dsp², leading to square planar geometry.
- **Conclusion:** Cu(NH₃)₄²⁺ has square planar geometry.
### Final Answer:
The complex that has square planar geometry is **Cu(NH₃)₄²⁺**.
