The magnetic behaviour or geometry of the complex $\left[ {Ni{{\left( {CO} \right)}_4}} \right]$ is-
A. Square planar geometry and paramagnetic
B. Square planar geometry and diamagnetic
C. Tetrahedral geometry and paramagnetic
D. Tetrahedral geometry and diamagnetic

Hint- In $\left[ {Ni{{\left( {CO} \right)}_4}} \right]$, Ni is in the zero-oxidation state i.e., it has an arrangement of $3{d^8}4{s^2}$. Since CO is a solid field ligand, it causes the matching of unpaired 3d electrons.

Complete answer:
Additionally, it makes the 4s electrons move to the 3d orbital, along these lines offering ascend to $s{p^3}$ hybridization and consequently, tetrahedral shape.
Since no unpaired electrons are available for this situation, $\left[ {Ni{{\left( {CO} \right)}_4}} \right]$ is diamagnetic.

Consequently, option D is the correct option.

Diamagnetic materials have a frail, negative powerlessness to magnetic fields. Diamagnetic materials are somewhat repulsed by a magnetic field and the material doesn't hold the magnetic properties when the outer field is evacuated. In diamagnetic materials all the electrons are paired so there is no lasting net magnetic moment per atom. Diamagnetic properties emerge from the realignment of the electron paths affected by an outside magnetic field. Most of the elements in the periodic table, including copper, silver, and gold, are diamagnetic.

Note: Tetrahedral is a molecular shape that outcomes when there are four bonds and no solitary pairs around the focal atom in the molecule. The atoms attached to the focal atom lie at the edges of a tetrahedron with 109.5° angles between them. Molecules with a tetrahedral electron pair geometry have $s{p^3}$ hybridization at the focal atom.