IIT JEE Advanced 2006 Paper 1

In $K_2[Ni(CN)_4]$ ; $Ni^{2+}(3d^8)$ The electronic configuration is

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$CN^{-}$ being a strong-field ligand causes pairing of electrons

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So, no unpaired electrons, hence the complex is diamagnetic. In $K_2[Ni(Cl)_4]$ ; $Ni^{2+}(3d^8)$ The electronic configuration is

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$Cl^{-}$ being a weak-field ligand does not cause pairing of electrons, so there are two unpaired electrons and the complex is paramagnetic.