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explain why octahedral complexes of Ni+2 must be outer orbital complexes?
Formation of [Ni (CN4)] 2- is outer orbital complexes:
Since the coordination number of Ni in the complex is 4, the configuration of Ni+2 , at first sight, shows that the complex is formed by sp 3 hybridization and it is paramagnetic and it has two unpaired electrons. However, experiments show that the complex is diamagnetic, i.e. it do not have unpaired electrons, this is possible when 3d electrons rearrange against the hund’s rule as shown in the below configuration. This is also in accordance with the fact that the ligand involved here is a strong i.e. CN- ion.
Hence now dsp2 hybridization, involving one 3d, one 4s and two 4p orbitals, takes place leading to four dsp2 hybrid orbitals, each of which accepts four electron pairs from CN- ion forming [Ni (CN4)] 2 ion. Thus the resulting complex is square planar and is diamagnetic as it has no unpaired electrons.
The electronic configuration of inner orbital complex chromium atom and Cr3+ ion may be shown as below.
the coordination number of Ni in the complex is 4, the configuration of Ni+2 , at first sight, shows that the complex is formed by sp 3 hybridization and it is paramagnetic and it has two unpaired electrons. However, experiments show that the complex is diamagnetic, i.e. it do not have unpaired electrons, this is possible when 3d electrons rearrange against the hund’s rule as shown in the below configuration. This is also in accordance with the fact that the ligand involved here is a strong i.e. CN- ion.Hence now dsp2 hybridization, involving one 3d, one 4s and two 4p orbitals, takes place leading to four dsp2 hybrid orbitals, each of which accepts four electron pairs from CN- ion forming [Ni (CN4)] 2 ion. Thus the resulting complex is square planar and is diamagnetic as it has no unpaired electrons.
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