The correct statement with respect to the complexes `Ni(CO)_4 and [Ni(CN)_4]^(2-)` is

Options

A. Nickel is in the same oxidation state in both
B. Both have tetrahedral geometry
C. Both have square planar geometry
D. Have tetrahedral and square planar geometry respectively

Explanation

To analyze the complexes \( Ni(CO)_4 \) and \( [Ni(CN)_4]^{2-} \), we need to determine their hybridization, geometry, and the nature of the ligands involved.

### Step 1: Analyze \( Ni(CO)_4 \)

1. **Identify the oxidation state of Nickel (Ni)**: In \( Ni(CO)_4 \), carbon monoxide (CO) is a neutral ligand. Therefore, the oxidation state of Ni is 0.

2. **Electron configuration of Ni**: The electron configuration of Ni (atomic number 28) is \( [Ar] 3d^8 4s^2 \).

3. **Determine the number of valence electrons**: Since Ni is in the 0 oxidation state, it has all its electrons available for bonding, which totals to 10 valence electrons (8 from \( 3d \) and 2 from \( 4s \)).

4. **Identify the ligand field strength**: CO is a strong field ligand, which means it will cause pairing of the \( 3d \) electrons.

5. **Hybridization**: The \( 3d \) electrons will pair up, and the \( 4s \) electrons will also be used for bonding. The hybridization will be \( sp^3 \) since there are 4 ligands (tetrahedral geometry).

6. **Geometry**: The geometry of \( Ni(CO)_4 \) is tetrahedral.

### Step 2: Analyze \( [Ni(CN)_4]^{2-} \)

1. **Identify the oxidation state of Nickel (Ni)**: In \( [Ni(CN)_4]^{2-} \), cyanide (CN) is a negatively charged ligand. Therefore, to balance the charge of -2 from four cyanide ligands, the oxidation state of Ni must be +2.

2. **Electron configuration of Ni in +2 state**: The electron configuration of Ni in the +2 oxidation state is \( [Ar] 3d^8 \) (removing 2 electrons from the \( 4s \) orbital).

3. **Determine the number of valence electrons**: In the +2 state, Ni has 8 valence electrons available for bonding.

4. **Identify the ligand field strength**: CN is also a strong field ligand, which will cause pairing of the \( 3d \) electrons.

5. **Hybridization**: With 4 ligands, the hybridization will be \( dsp^2 \) since it involves \( 3d \), \( 4s \), and \( 4p \) orbitals.

6. **Geometry**: The geometry of \( [Ni(CN)_4]^{2-} \) is square planar.

### Conclusion

- \( Ni(CO)_4 \) has tetrahedral geometry with \( sp^3 \) hybridization.
- \( [Ni(CN)_4]^{2-} \) has square planar geometry with \( dsp^2 \) hybridization.

### Final Statement

The correct statement with respect to the complexes \( Ni(CO)_4 \) and \( [Ni(CN)_4]^{2-} \) is that \( Ni(CO)_4 \) has tetrahedral geometry and \( [Ni(CN)_4]^{2-} \) has square planar geometry.

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