Which of the following complex(es) is paramagnetic as well as low spin ?

Options

A. `[Cr(CN)_(6)]^(4-)`
B. `[Ni(NH_(3))_(6)]^(2+)`
C. `[Fe(CN)_(6)]^(4-)`
D. `[Cr(NH_(3))_(6)]^(2+)`

Answer

To determine which of the following complexes is paramagnetic as well as low-spin, we need to analyze the oxidation states, electronic configurations, and the nature of the ligands involved in each complex.

### Step-by-Step Solution:

1. **Identify the Complexes**:
We need to analyze the given complexes. In this case, we have:
- A: [Cr(CN)6]4-
- B: [Ni(NH3)6]2+
- C: [Fe(CN)6]4-
- D: [Cr(NH3)6]2+

2. **Determine the Oxidation States**:
- For **[Cr(CN)6]4-**:
- CN is a neutral ligand, so the oxidation state of Cr is +2.
- For **[Ni(NH3)6]2+**:
- NH3 is also a neutral ligand, so the oxidation state of Ni is +2.
- For **[Fe(CN)6]4-**:
- CN is a neutral ligand, so the oxidation state of Fe is +2.
- For **[Cr(NH3)6]2+**:
- NH3 is a neutral ligand, so the oxidation state of Cr is +2.

3. **Electronic Configurations**:
- **Cr2+** (3d4): The electronic configuration is [Ar] 3d4.
- **Ni2+** (3d8): The electronic configuration is [Ar] 3d8.
- **Fe2+** (3d6): The electronic configuration is [Ar] 3d6.

4. **Identify the Nature of the Ligands**:
- CN- is a strong field ligand, which causes pairing of electrons.
- NH3 is a weaker field ligand compared to CN-.

5. **Determine Spin States**:
- For **[Cr(CN)6]4-**:
- With Cr in +2 oxidation state (3d4) and CN- being a strong field ligand, the electrons will pair up, leading to a low-spin state. Therefore, it is paramagnetic (due to unpaired electrons) and low-spin.
- For **[Ni(NH3)6]2+**:
- Ni in +2 oxidation state (3d8) with NH3 being a weak field ligand will not cause pairing, leading to a high-spin state. It is paramagnetic but not low-spin.
- For **[Fe(CN)6]4-**:
- Fe in +2 oxidation state (3d6) with CN- being a strong field ligand will also cause pairing, leading to a low-spin state. However, it is diamagnetic as all electrons are paired.
- For **[Cr(NH3)6]2+**:
- Cr in +2 oxidation state (3d4) with NH3 being a weak field ligand will not cause pairing, leading to a high-spin state. It is paramagnetic but not low-spin.

6. **Conclusion**:
The only complex that is both paramagnetic and low-spin is **[Cr(CN)6]4-**.

### Final Answer:
**The complex that is paramagnetic as well as low-spin is [Cr(CN)6]4-.**