The correct statements about [Ni(CO)4] are:

a) Hybridization of [Ni(CO)4] is sp3
b) The complex is diamagnetic.
c) There are no unpaired electrons.
d) The complex is tetrahedral.

Explanation:

a) Hybridization of [Ni(CO)4] is sp3:
In [Ni(CO)4], the central metal atom is nickel (Ni), which belongs to the 3d transition metal series. The electronic configuration of Ni is 1s2 2s2 2p6 3s2 3p6 3d8 4s2. In [Ni(CO)4], the nickel atom forms four sigma bonds with four CO (carbon monoxide) ligands. Each CO ligand donates a lone pair of electrons to form a coordinate bond with the metal atom.

The carbon in CO is sp hybridized, forming a triple bond with oxygen. The oxygen is also sp hybridized, forming a coordinate bond with the metal atom. Therefore, the coordination number of Ni is 4, and the hybridization is sp3.

b) The complex is diamagnetic:
To determine the magnetic nature of [Ni(CO)4], we need to calculate the number of unpaired electrons. In this complex, the nickel atom has 8 valence electrons (3d8). Each CO ligand donates a lone pair of electrons to form a coordinate bond with the metal atom. Hence, all the d orbitals of nickel are occupied, resulting in the absence of unpaired electrons. Therefore, [Ni(CO)4] is diamagnetic.

c) No unpaired electron is there:
As explained above, the nickel atom in [Ni(CO)4] has a completely filled d orbital due to the coordination of four CO ligands. Therefore, there are no unpaired electrons in the complex.

d) The complex is tetrahedral:
In [Ni(CO)4], the nickel atom is surrounded by four CO ligands. Each CO ligand forms a sigma bond with the nickel atom using its lone pair of electrons. The geometry formed by these four ligands is tetrahedral, with the nickel atom at the center and the CO ligands at the four corners of the tetrahedron. Therefore, [Ni(CO)4] is tetrahedral in shape.

In conclusion, the correct statements about [Ni(CO)4] are that its hybridization is sp3, it is diamagnetic, there are no unpaired electrons, and it has a tetrahedral geometry.

In [Ni(CO)₄], the valence shell electronic configuration of ground state Ni atom is 3d⁸4s². All of these 10 electrons are pushed into 3d orbitals and get paired up when strong field CO ligands approach Ni atom. The empty 4s and three 4p orbitals undergo sp³ hybridization and form bonds with four CO ligands to give Ni(CO)₄ with tetrahedral geometry. Since all the electrons are paired, Ni(CO)₄ is diamagnetic.