Allthe following complexes show a decrease in their weights when placed in a magnetic balance. Then which of the these has square planar geometry ?

Options

A. `Ni(CO)_(4)`
B. `K[AgF_(4)]`
C. `Na_(2)[Zn(CN)_(4)]`
D. None of these

Explanation

To determine which of the given complexes has a square planar geometry, we need to analyze the complexes mentioned in the question and their electronic configurations, hybridization, and the nature of the ligands.

### Step-by-Step Solution:

1. **Identify the complexes**: The complexes given are:
- Ni(CO)₄
- AgF₄⁻
- Ne₂Zn(CN)₄

2. **Analyze each complex**:
- **Ni(CO)₄**: Nickel in this complex is in the zero oxidation state. Nickel (Ni) has an electronic configuration of [Ar] 3d⁸ 4s². In the presence of CO (a strong field ligand), it undergoes sp³ hybridization, leading to a tetrahedral geometry.

- **AgF₄⁻**: Silver (Ag) in this complex is in the +3 oxidation state. The electronic configuration of Ag is [Kr] 4d¹⁰. In the +3 state, Ag has an electronic configuration of 4d⁸. The presence of strong field ligands (F⁻) causes pairing of electrons and results in dsp² hybridization, which leads to a square planar geometry.

- **Ne₂Zn(CN)₄**: Zinc (Zn) is in the +2 oxidation state here. The electronic configuration of Zn is [Ar] 3d¹⁰. Since there are no d electrons available for pairing, Zn will undergo sp³ hybridization, resulting in a tetrahedral geometry.

3. **Determine the geometry**:
- From the analysis:
- Ni(CO)₄: Tetrahedral
- AgF₄⁻: Square planar
- Ne₂Zn(CN)₄: Tetrahedral

4. **Conclusion**: The complex that has square planar geometry is **AgF₄⁻**.

### Final Answer:
The complex with square planar geometry is **AgF₄⁻**.

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