Which of the following is paramagnetic?

Options

A. `Ni(CO)_(4)`
B. `Fe(CO)_(5)`
C. `V(CO)_(6)`
D. `Cr(CO)_(6)`

Explanation

To determine which of the given compounds is paramagnetic, we need to analyze the electronic configurations of each metal in the coordination complexes and check for the presence of unpaired electrons. Paramagnetic substances have unpaired electrons, while diamagnetic substances have all electrons paired.

### Step-by-Step Solution:

1. **Understanding Paramagnetism**:
- Paramagnetic substances have unpaired electrons in their electronic configuration.
- Diamagnetic substances have all electrons paired.

2. **Identify the Ligand**:
- The ligand mentioned in the question is CO (carbon monoxide), which is a strong field ligand. Strong field ligands tend to cause pairing of electrons in the d-orbitals.

3. **Determine the Oxidation States**:
- For neutral ligands like CO, the oxidation state of the metal remains the same as its elemental state.

4. **Analyze Each Metal**:
- **Nickel (Ni)**:
- Electronic configuration: Ni is in the 0 oxidation state, which means it has a configuration of [Ar] 3d^10 4s^2.
- In the presence of CO, the d-orbitals will be filled completely (d^10), resulting in no unpaired electrons. Therefore, Ni is **diamagnetic**.

- **Iron (Fe)**:
- Electronic configuration: Fe in the 0 oxidation state has a configuration of [Ar] 3d^6 4s^2.
- With CO as a strong field ligand, the d-electrons will pair up, leading to a configuration of d^8. This means all electrons are paired, making Fe **diamagnetic**.

- **Vanadium (V)**:
- Electronic configuration: V in the 0 oxidation state has a configuration of [Ar] 3d^3 4s^2.
- In the presence of CO, the d-electrons will pair, resulting in a configuration of d^5. Since there are unpaired electrons remaining, V is **paramagnetic**.

- **Chromium (Cr)**:
- Electronic configuration: Cr in the 0 oxidation state has a configuration of [Ar] 3d^4 4s^2.
- With CO as a strong field ligand, the d-electrons will pair, resulting in a configuration of d^6. All electrons are paired, making Cr **diamagnetic**.

5. **Conclusion**:
- Among the metals analyzed, only Vanadium (V) has unpaired electrons and is therefore **paramagnetic**.

### Final Answer:
The compound that is paramagnetic is **Vanadium (V)**.