The compound which does not show paramagnetism is

Options

A. `[Cu(NH_3)_4]Cl_2`
B. `[Ag(NH_3)_2]Cl`
C. `[NiCl_4]^(2-)`
D. `[CoF_6]^(3-)`

Answer

To determine which compound does not show paramagnetism, we need to analyze the electronic configurations of the central metal ions in the given compounds and check for the presence of unpaired electrons. Paramagnetism arises from the presence of unpaired electrons, while diamagnetism occurs when all electrons are paired.

### Step-by-Step Solution:

1. **Understanding Paramagnetism**:
- Paramagnetism is exhibited by compounds that have unpaired electrons.
- Compounds with all paired electrons exhibit diamagnetism.

2. **Analyzing Compound A: Cu(NH3)4Cl2**:
- Central atom: Copper (Cu)
- Oxidation state: Cu²⁺
- Electronic configuration of Cu²⁺: 3d⁹ 4s⁰
- Since there are unpaired electrons in the 3d subshell, this compound shows paramagnetism.

3. **Analyzing Compound B: Ag(NH3)2Cl**:
- Central atom: Silver (Ag)
- Oxidation state: Ag⁺
- Electronic configuration of Ag⁺: 4d¹⁰
- NH3 is a strong field ligand, which means it will cause pairing of electrons. Since all electrons are paired, this compound does not show paramagnetism (it is diamagnetic).

4. **Analyzing Compound C: NiCl4²⁻**:
- Central atom: Nickel (Ni)
- Oxidation state: Ni²⁺
- Electronic configuration of Ni²⁺: 3d⁸ 4s⁰
- Cl⁻ is a weak field ligand, leading to unpaired electrons in the 3d subshell. Thus, this compound shows paramagnetism.

5. **Analyzing Compound D: CoF6³⁻**:
- Central atom: Cobalt (Co)
- Oxidation state: Co³⁺
- Electronic configuration of Co³⁺: 3d⁶
- F⁻ is also a weak field ligand, resulting in unpaired electrons. Therefore, this compound shows paramagnetism.

### Conclusion:
The compound that does not show paramagnetism is **B: Ag(NH3)2Cl**.