Which of the following is not expected to show paramagnetism?

Options

A. `[Nl(H_2O)_6]^(2+)`
B. `Nl(CO)_4`
C. `[Nl(NH_3)_4]^(2+)`
D. `[Co(NH_3)_6]^(2+)`

Answer

To determine which of the given options is not expected to show paramagnetism, we need to analyze the electronic configurations of the complexes provided and check for the presence of unpaired electrons. Paramagnetism is exhibited by species that have unpaired electrons.

### Step-by-Step Solution:

1. **Identify the Complexes**: We need to analyze the complexes mentioned in the question, which include nickel and cobalt complexes with different ligands.

2. **Nickel (II) Complex with Water (Ni(H2O)2+)**:
- Nickel in +2 oxidation state has the electronic configuration of [Ar] 3d^8.
- Water (H2O) is a weak field ligand, which means it does not cause pairing of electrons.
- The 3d orbitals will have 2 unpaired electrons.
- **Conclusion**: Ni(H2O)2+ is paramagnetic.

3. **Nickel (0) Complex with Carbonyl (Ni(CO)4)**:
- Nickel in 0 oxidation state has the electronic configuration of [Ar] 4s^2 3d^8.
- Carbonyl (CO) is a strong field ligand, which causes pairing of electrons.
- All electrons will pair up in the 3d orbitals.
- **Conclusion**: Ni(CO)4 is diamagnetic (not paramagnetic).

4. **Cobalt (II) Complex with Ammonia (Co(NH3)6^2+)**:
- Cobalt in +2 oxidation state has the electronic configuration of [Ar] 3d^7.
- Ammonia (NH3) is a weak field ligand, which does not cause pairing.
- The 3d orbitals will have 3 unpaired electrons.
- **Conclusion**: Co(NH3)6^2+ is paramagnetic.

5. **Final Analysis**:
- From the analysis, we find that Ni(H2O)2+ and Co(NH3)6^2+ are paramagnetic due to unpaired electrons.
- Ni(CO)4 is the only complex that does not show paramagnetism as it has all paired electrons.

### Final Answer:
**Ni(CO)4 is not expected to show paramagnetism.**

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