Using valence bond theory, predict the hybridisation and magnetic character of the following:

(a) [NiCl₄]^2-

(b) [Co(C₂O₄)₃]^3-

[At. No.: Co=27, Ni=28]

(a) [NiCl₄]^2-:- The atomic number of Ni is 28 and its valence shell electronic configuration is 3d⁸4s².
Ni is in +2 oxidation state in the complex [NiCl₄] ^2-. In this state, there are 8 electrons in the 3d orbital and 4s, 4p and 4d orbitals are vacant.

As Cl^- is a weak field ligand, so it is unable to pair up the electrons of the 3d orbital. Hence, four equivalent hybrid orbitals are formed by one 4s and three 4p orbitals and the complex is sp³ hybridized and it is tetrahedral in shape.

Moreover, it has 2 unpaired electrons in the 3d orbital. So it is paramagnetic in nature.

(b) [Co(C₂O₄)₃]^3-:- The atomic number of Co is 27 and its valence shell electronic configuration is 3d⁷4s².
Co is in +3 oxidation state in the complex [Co(C₂O₄)₃] ^3-. In this state, there are 6 electrons in the 3d orbitals and 4s, 4p and 4d orbitals are vacant.
As (C₂O₄)^2- is a strong field ligand, so it pair up the four unpaired electrons of the 3d orbital. So three 3d orbitals become doubly occupied and two 3d orbital remains vacant.
Hence, six equivalent hybrid orbitals are formed by two 3d, one 4s and three 4p orbitals and the complex is d²sp³ hybridized and it is octahedral in shape.

Moreover, it has no unpaired electron in it. So it is diamagnetic in nature.