The hybridization and magnetic nature of `[Mn(CN)_6]^(4-) and [Fe(CN)_6]^(3-)` respectively are

Options

A. `d^2sp^3` and diamagnetic
B. `sp^3d^2` and diamagnetic
C. `d^2sp^3` and paramagnetic
D. `sp^3d^2` and paramagnetic

Answer

To determine the hybridization and magnetic nature of the complexes \([Mn(CN)_6]^{4-}\) and \([Fe(CN)_6]^{3-}\), we will follow these steps:

### Step 1: Determine the oxidation state of the central metal ion

1. **For \([Mn(CN)_6]^{4-}\)**:
- Let the oxidation state of Mn be \(x\).
- The CN ligand has a charge of \(-1\), and there are 6 CN ligands.
- The overall charge of the complex is \(-4\).
- Therefore, we can set up the equation:
\[ x + 6(-1) = -4 \]
- Solving for \(x\):
\[ x - 6 = -4 \implies x = +2 \]
- Thus, the oxidation state of Mn is \(+2\).

2. **For \([Fe(CN)_6]^{3-}\)**:
- Let the oxidation state of Fe be \(y\).
- Again, the CN ligand has a charge of \(-1\), and there are 6 CN ligands.
- The overall charge of the complex is \(-3\).
- Setting up the equation:
\[ y + 6(-1) = -3 \]
- Solving for \(y\):
\[ y - 6 = -3 \implies y = +3 \]
- Thus, the oxidation state of Fe is \(+3\).

### Step 2: Determine the electron configuration of the metal ions

1. **For \(Mn^{2+}\)**:
- The electron configuration of Mn is \([Ar] 3d^5 4s^2\).
- For \(Mn^{2+}\), we remove 2 electrons (from the 4s orbital):
\[ Mn^{2+}: [Ar] 3d^5 \]

2. **For \(Fe^{3+}\)**:
- The electron configuration of Fe is \([Ar] 3d^6 4s^2\).
- For \(Fe^{3+}\), we remove 3 electrons (2 from 4s and 1 from 3d):
\[ Fe^{3+}: [Ar] 3d^5 \]

### Step 3: Determine the hybridization

1. **For \([Mn(CN)_6]^{4-}\)**:
- The CN ligand is a strong field ligand, which causes pairing of electrons in the 3d orbitals.
- The electron configuration after pairing is:
\[ 3d^6 \text{ (two pairs and two unpaired)} \]
- The hybridization involves 6 orbitals (3d, 4s, and 4p):
- Therefore, the hybridization is \(d^2sp^3\).

2. **For \([Fe(CN)_6]^{3-}\)**:
- Similarly, CN being a strong field ligand causes pairing.
- The electron configuration after pairing is:
\[ 3d^6 \text{ (two pairs and one unpaired)} \]
- The hybridization also involves 6 orbitals:
- Therefore, the hybridization is \(d^2sp^3\).

### Step 4: Determine the magnetic nature

1. **For \([Mn(CN)_6]^{4-}\)**:
- After pairing, there are no unpaired electrons.
- Thus, it is **diamagnetic**.

2. **For \([Fe(CN)_6]^{3-}\)**:
- After pairing, there is one unpaired electron.
- Thus, it is **paramagnetic**.

### Final Answer
- The hybridization of \([Mn(CN)_6]^{4-}\) is \(d^2sp^3\) and it is diamagnetic.
- The hybridization of \([Fe(CN)_6]^{3-}\) is \(d^2sp^3\) and it is paramagnetic.