Question

The geometry and magnetic behaviour of the complex [Ni(CO)₄] are

• A. square planar geometry and diamagnetic
• B. square planar geometry and paramagnetic
• C. tetrahedral geometry and diamagnetic
• D. tetrahedral geometry and paramagnetic

Answer 1

The Correct Option is C

Let's analyze the complex [Ni(CO)₄] to determine its geometry and magnetic behavior.

Step-by-Step Analysis

1. Oxidation State of Nickel:
   • In [Ni(CO)₄], carbon monoxide (CO) is a neutral ligand. Therefore, the oxidation state of nickel in this complex is 0.
2. Electronic Configuration of Nickel:
   • The atomic number of Ni is 28, and the electronic configuration of Ni in the free state is: \( [\mathrm{Ar}]\,3d^8\,4s^2 \).
   • In the Ni(0) state, the electronic configuration becomes: \( [\mathrm{Ar}]\,3d^{10}\,4s^0 \).
3. Hybridization:
   • CO is a strong field ligand and causes the pairing of electrons in the 3d orbitals. This leads to the formation of empty 4p orbitals.
   • The hybridization involved is \( sp^3 \) as four hybrid orbitals are formed to accommodate the CO ligands.
4. Geometry:
   • The \( sp^3 \) hybridization leads to a tetrahedral geometry for the complex \( [\mathrm{Ni(CO)_4}] \).
5. Magnetic Behavior:
   • The 3d-electrons are paired due to CO being a strong field ligand. As a result, there are no unpaired electrons left in the complex.
   • Therefore, the complex is diamagnetic.

Conclusion:

Based on the analysis above, the complex \( [\mathrm{Ni(CO)_4}] \) has a tetrahedral geometry and is diamagnetic.

Thus, the correct answer is:

tetrahedral geometry and diamagnetic