Subjective Type

Answer the following question: $$[Ni(H_2O)_6]^{2+}$$ is green whereas $$[Ni(CN)_4]^{2-}$$ is colourless. Why? (At no. of $$Ni=28$$)

Solution

In $$[Ni(H_2O)_6]^{2+}, Ni$$is in $$+2$$ oxidation state with the configuration $$3d^8 4s^0$$, i.e., it has two unpaired electrons which do not pair in the presence of weak $$H_2O$$ ligand. The $$d-d$$ transition absorbed red light and the complementary green light is emitted.
On the other hand in $$[Ni(CN)_4]^{2-}, Ni$$ is again in $$+2$$ oxidation state with the electronic configuration $$3d^8$$. In the presence of $$CN^-$$ ligand the two unpaired electrons in the $$3d$$ orbitals pair up. As there is no unpaired electron in $$[Ni(CN)_4]^{2-}$$ therefore the complex is colourless.


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