The complex having square planar geometry is

Options

A. `[Ni(CO)_(4)]`
B. `[MnCl_(4)]^(2-)`
C. `[CuCl_(4)]^(2-)`
D. `[Cu(NH_(3))_(4)]^(2+)`

Answer

To determine which complex has a square planar geometry, we can analyze the given complexes one by one based on their electronic configurations and the nature of their ligands.

### Step-by-Step Solution:

1. **Identify the complexes:**
- The complexes given are Ni(CO)4, MnCl4^2-, CoCl4^2-, and Cu(NH3)4^2+.

2. **Analyze Ni(CO)4:**
- Nickel (Ni) in Ni(CO)4 has an oxidation state of 0. The electronic configuration of Ni is [Ar] 3d^8 4s^2.
- CO is a strong field ligand, which causes pairing of electrons in the d-orbitals.
- The hybridization will be sp^3, leading to a tetrahedral geometry. **(Not square planar)**

3. **Analyze MnCl4^2-:**
- Manganese (Mn) has an oxidation state of +2 in MnCl4^2-. The electronic configuration is [Ar] 3d^5.
- Chloride (Cl) is a weak field ligand, which does not cause pairing of electrons.
- The hybridization will be sp^3, resulting in a tetrahedral geometry. **(Not square planar)**

4. **Analyze CoCl4^2-:**
- Cobalt (Co) in CoCl4^2- has an oxidation state of +2. The electronic configuration is [Ar] 3d^7.
- Chloride (Cl) is a weak field ligand, leading to unpaired electrons.
- The hybridization will be sp^3, resulting in a tetrahedral geometry. **(Not square planar)**

5. **Analyze Cu(NH3)4^2+:**
- Copper (Cu) in Cu(NH3)4^2+ has an oxidation state of +2. The electronic configuration is [Ar] 3d^9.
- Ammonia (NH3) is a strong field ligand, which causes pairing of electrons.
- The hybridization will be dsp^2, leading to a square planar geometry. **(This is square planar)**

### Conclusion:
The complex that has square planar geometry is **Cu(NH3)4^2+**.