Answer :

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.


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