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 geometry is paramagnetic.

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.

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.