Which of the following complex shows `sp ^(3) d ^(2)` hybridization?

Options

A. `[Cr (NO _(2)) _(6) ] ^(3-)`
B. `[Fe (CN) _(6) ] ^(4-)`
C. `[CoF_(6) ] ^(3-)`
D. `[Ni(CO) _(4) ]`

Answer

To determine which of the given complexes shows \( sp^3d^2 \) hybridization, we will follow these steps:

### Step 1: Identify the oxidation state of the central metal atom in each complex.

1. **For the first complex (assumed to be Cr(NO2)6)**:
- Let the oxidation state of Cr be \( x \).
- The oxidation state of \( NO2 \) is -1, and since there are 6 \( NO2 \) ligands, the equation becomes:
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]

2. **For the second complex (assumed to be Fe(CN)6)**:
- Let the oxidation state of Fe be \( x \).
- The oxidation state of \( CN \) is -1, and since there are 6 \( CN \) ligands:
\[ x + 6(-1) = -4 \implies x - 6 = -4 \implies x = +2 \]

3. **For the third complex (assumed to be CoF6)**:
- Let the oxidation state of Co be \( x \).
- The oxidation state of \( F \) is -1, and since there are 6 \( F \) ligands:
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]

4. **For the fourth complex (assumed to be Ni(CO)4)**:
- The carbonyl \( CO \) is a neutral ligand, so:
\[ x = 0 \]

### Step 2: Determine the electronic configuration of the central metal atom in its oxidation state.

1. **For Cr in +3 state**:
- Ground state: \( [Ar] 3d^5 4s^1 \)
- In +3 state: \( [Ar] 3d^3 \)

2. **For Fe in +2 state**:
- Ground state: \( [Ar] 3d^6 4s^2 \)
- In +2 state: \( [Ar] 3d^6 \)

3. **For Co in +3 state**:
- Ground state: \( [Ar] 3d^7 4s^2 \)
- In +3 state: \( [Ar] 3d^6 \)

4. **For Ni in 0 state**:
- Ground state: \( [Ar] 3d^8 4s^2 \)

### Step 3: Analyze the hybridization based on the ligands and the number of unpaired electrons.

1. **For Cr(NO2)6**:
- \( Cr^{3+} \) has 3 unpaired electrons in \( 3d \) and uses \( 4s \) and \( 4p \) orbitals for hybridization.
- Hybridization: \( d^2sp^3 \)

2. **For Fe(CN)6**:
- \( Fe^{2+} \) has 6 electrons in \( 3d \) and due to the strong field ligand \( CN^- \), pairing occurs.
- Hybridization: \( d^2sp^3 \)

3. **For CoF6**:
- \( Co^{3+} \) has 6 electrons in \( 3d \) and \( F^- \) is a weak field ligand, so no pairing occurs.
- Hybridization: \( sp^3d^2 \)

4. **For Ni(CO)4**:
- \( Ni^0 \) has 10 electrons in \( 3d \) and \( CO \) is a strong field ligand, leading to pairing.
- Hybridization: \( sp^3 \)

### Conclusion:
The complex that shows \( sp^3d^2 \) hybridization is **CoF6**.