Which of the following complexes is(are) paramagnetic ?

Options

A. `[Mn(CN_6)]^(3-)`
B. `[Cr(NH_3)_5]^(3+)`
C. `[Fe(CN)_6]^(4-)`
D. `[Co(CN)_6]^(3-)`

Explanation

To determine which of the given complexes are paramagnetic, we need to analyze each complex for the presence of unpaired electrons. Paramagnetic complexes have unpaired electrons, while diamagnetic complexes have all electrons paired.

### Step-by-Step Solution:

1. **Complex A: Mn(CN)₆³⁻**
- **Oxidation State Calculation**:
- Let the oxidation state of Mn be \( x \).
- The equation is \( x + 6(-1) = -3 \).
- Solving gives \( x = +3 \).
- **Electronic Configuration**:
- Mn has an atomic number of 25, so its electronic configuration is \( [Ar] 3d^5 4s^2 \).
- In the +3 oxidation state, it loses 2 electrons (4s first), resulting in \( [Ar] 3d^4 \).
- **Effect of Ligand**:
- CN⁻ is a strong field ligand, which causes pairing of electrons.
- The 4 electrons in the 3d orbitals will be arranged as: ↑↓ ↑ ↑ ↑ (2 paired, 2 unpaired).
- **Unpaired Electrons**: 2 unpaired electrons.
- **Conclusion**: This complex is **paramagnetic**.

2. **Complex B: Cr(NH₃)₆³⁺**
- **Oxidation State Calculation**:
- Let the oxidation state of Cr be \( x \).
- The equation is \( x + 6(0) = +3 \).
- Thus, \( x = +3 \).
- **Electronic Configuration**:
- Cr has an atomic number of 24, so its configuration is \( [Ar] 3d^5 4s^1 \).
- In the +3 oxidation state, it loses 3 electrons (4s and 2 from 3d), resulting in \( [Ar] 3d^3 \).
- **Effect of Ligand**:
- NH₃ is also a strong field ligand, leading to pairing.
- The 3 electrons will be arranged as: ↑↓ ↑↓ ↑ (3 unpaired).
- **Unpaired Electrons**: 3 unpaired electrons.
- **Conclusion**: This complex is **paramagnetic**.

3. **Complex C: Fe(CN)₆⁴⁻**
- **Oxidation State Calculation**:
- Let the oxidation state of Fe be \( x \).
- The equation is \( x + 6(-1) = -4 \).
- Thus, \( x = +2 \).
- **Electronic Configuration**:
- Fe has an atomic number of 26, so its configuration is \( [Ar] 3d^6 4s^2 \).
- In the +2 oxidation state, it loses 2 electrons (4s first), resulting in \( [Ar] 3d^6 \).
- **Effect of Ligand**:
- CN⁻ is a strong field ligand, leading to pairing.
- The 6 electrons will be arranged as: ↑↓ ↑↓ ↑↓ (all paired).
- **Unpaired Electrons**: 0 unpaired electrons.
- **Conclusion**: This complex is **diamagnetic**.

4. **Complex D: Co(CN)₆³⁻**
- **Oxidation State Calculation**:
- Let the oxidation state of Co be \( x \).
- The equation is \( x + 6(-1) = -3 \).
- Thus, \( x = +3 \).
- **Electronic Configuration**:
- Co has an atomic number of 27, so its configuration is \( [Ar] 3d^7 4s^2 \).
- In the +3 oxidation state, it loses 3 electrons (4s and 2 from 3d), resulting in \( [Ar] 3d^6 \).
- **Effect of Ligand**:
- CN⁻ is a strong field ligand, leading to pairing.
- The 6 electrons will be arranged as: ↑↓ ↑↓ ↑↓ (all paired).
- **Unpaired Electrons**: 0 unpaired electrons.
- **Conclusion**: This complex is **diamagnetic**.

### Final Conclusion:
The complexes that are paramagnetic are:
- **Complex A: Mn(CN)₆³⁻**
- **Complex B: Cr(NH₃)₆³⁺**