Explain on the basis of valence bond theory that [Ni(CN)4]2- ion with square planar structure is diamagnetic and [Ni(Cl)4]2-the ion with tetrahedral

According to the valence band theory , the central metal atom or ion under the influence of ligands can use its (n-1)d , ns, np (inner orbital complex) or ns, np, nd (outer orbital complex)orbitals for hybridisation to form equivalent set of orbitals of definite geometry.  In [Ni(CN)4]2- , oxidation state of Ni can be calculated as :

Using overall charge balance as the whole ion has overall -2 charge:

x + 4(-1) = -2   (∵ CN- has -1 negative charge)

x = + 2

Ni is in + 2 oxidation state. 

Electronic configuration of Ni is: [Ar]3d84s2 

Where, [Ar] = 1s22s22p63s23p6

Electronic configuration of Ni+2 = [Ar]3d8

Outer electronic configuration of Ni+2 = 3d8

Since there are 4 CN ions so they can either form tetrahedral or square planar geometry. And CN- is a strong field ligand (according to experimental data of spectro-chemical series) it causes pairing of the 2 unpaired electrons. It undergoes dsp2 (one d orbital, one s and two p orbitals used by the ligands) hybridization and forms square planar structure. Since all the electrons are paired so it is diamagnetic. Paramagnetic compounds- those compounds which have one or more no. of unpaired electrons in their atomic orbitals. Diamagnetic compounds-those compounds in which all the electrons in their atomic orbitals are paired.

Co-ordination numberType of hybridisation Distribution of hybrid orbitals in space

4sp3Tetrahedral

4dsp3Square planar

5sp3dTrigonal bipyramidal

6sp3d2Octahedral

6d2sp3octahedral

In case of [Ni(Cl)4]2- ion, Cl- is a weak field ligand so it will not pair the unpaired electrons of Ni + 2 ion. Therefore it undergoes sp3 hybridization. 

Overall charge balance:

X + 4(-1) = -2

X = + 2.

Since there are 2 unpaired electrons in the d orbital so it is a paramagnetic compound.