Explain on the basis of valence bond theory that [Ni(CN)4]2− ion with square planar structure is diamagnetic and the [NiCl4]2− ion with tetrahedral geometry is paramagnetic.
Solution 1
Ni is in the +2 oxidation state i.e., in d8 configuration.
d8 Configuration 
There are 4 CN− ions. Thus, it can either have a tetrahedral geometry or square planar geometry. Since CN− ion is a strong field ligand, it causes the pairing of unpaired 3d electrons.
It now undergoes dsp2 hybridization. Since all electrons are paired, it is diamagnetic.
In case of [NiCl4]2−, Cl− ion is a weak field ligand. Therefore, it does not lead to the pairing of unpaired 3delectrons. Therefore, it undergoes sp3 hybridization.
Since there are 2 unpaired electrons in this case, it is paramagnetic in nature.
Solution 2
`[Ni(CN)_4]^(2-) : Ni(28): 1s^22s^2 2p^6 3s^2 3p^6 4s^2 3d^8`
or last shell EC is in `[Ni(CN)_4]^(2-)` , Ni is in +2 state
`:.Ni^2:3d^8 4s^0`
CN being strong field ligand pairs up the `e^(-1)`s thus
`[Ni(CN)_4]^(2-)`: 
No unpaired e-, hence `[Ni(CN)_4]^(2-)` is diamagnetic
`[NiCl_4]^(2-)`: Ni is in +2 state thus
`Ni^(2+): 3d^8 4s^2`
Cl- being a weak field ligand cannot pair up the `e^(-1)s`
`:.[NiCl_4]^(2-)` 
Due to unpaired e-1s, `[NiCl4]^(2-)` is paramagnetic
