Molecular Geometry

The question pertains to the geometry and magnetic behavior of the complex [Ni(CO)₄]. Let's break down the problem step-by-step.

1. Firstly, understand the electronic configuration of the central metal atom, Nickel (Ni). The ground state electronic configuration of Ni is [Ar] 3d⁸ 4s².
2. In the complex [Ni(CO)₄], nickel is in the zero oxidation state, which means it retains its ground state electronic configuration.
3. When forming the complex, the CO ligands, which are strong field ligands, cause the pairing of electrons. The 4s and 3d electrons pair up and facilitate the hybridization of 4s, 4p, and one of the 3d electrons to form four sp³ hybrid orbitals. This results in a tetrahedral geometry.
4. Tetrahedral geometry is explained by the sp³ hybridization of the orbitals involved in chemical bonding.
5. CO is a strong field ligand according to the spectrochemical series, causing the electrons to pair up, resulting in no unpaired electrons in the nickel atom.
6. The absence of unpaired electrons makes [Ni(CO)₄] diamagnetic.
7. Thus, the geometry of this complex is tetrahedral, and its magnetic behavior is diamagnetic.

Let's review the options:

• Square planar geometry and diamagnetic: Incorrect, as square planar geometry is not typical for Ni(CO)₄; tetrahedral is more likely due to the hybridization involving sp³.
• Square planar geometry and paramagnetic: Incorrect for the same reasons above; CO ligands cause electron pairing.
• Tetrahedral geometry and diamagnetic: Correct, as explained above.
• Tetrahedral geometry and paramagnetic: Incorrect, as the strong field CO ligands cause electron pairing, resulting in diamagnetic behavior.

The correct answer is: tetrahedral geometry and diamagnetic.