Question

Explain the following:
Low spin octahedral complexes of nickel are not known.

Answer 1

Hint :The oxidation state of the metal likewise decides how little or huge $ \Delta $ is. The higher the oxidation state of the metal, the more grounded the ligand field that is made. If there are two metals with a similar d electron arrangement, the one with the higher oxidation state is bound to be low spin than the one with the lower oxidation state.

Complete Step By Step Answer:
Octahedral complexes have a coordination number of 6, implying that there are six spots around the metal where ligands can attack. Interaction between the electrons of the ligands and those of the metal place produce a crystal field splitting where the $ d{z^2} $ and $ d{x^2} - d{y^2} $ orbitals high in energy, while the other three orbitals of $ dxz,dzy $ , and $ dyz $ , are lower in energy.
Electronic configuration of $ N{i^{2 + }} $ is $ 1{s^2},2{s^2},2{p^6},4{s^0}3{d^8} $ .
Electrons present in $ 3d $ orbital of $ N{i^{2 + }} $ should pair up with each other for the formation of low spin octahedral complexes. So, this will result in one empty $ d $ orbital and $ {d^2}s{p^3} $ hybridization for nickel complexes is not possible.

Note :
remember usually the field strength of the ligand, which is additionally determined by large or small $ \Delta $ , decides if an octahedral complex is high or low spin. This is the place where we utilize the spectro chemical arrangement to decide ligand strength. Strong field ligands, as $ C{N^ - } $ and $ N{O^{2 - }} $ , increment $ \Delta $ which brings about low spin. While feeble field ligands, similar to $ {I^ - } $ and $ C{l^ - } $ , decline the $ \Delta $ which brings about high spin.