An octahedral complex of Ni(II) must be: | vedclass

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(B) The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.In an octahedral geometry,the coordination number is $6$.For $d^8$ configuration,the $d$-

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outer orbital

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(B) The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.In an octahedral geometry,the coordination number is $6$.For $d^8$ configuration,the $d$-orbitals are $d_{xy}, d_{yz}, d_{zx}, d_{x^2-y^2}, d_{z^2}$.Regardless of whether the ligand is strong field or weak field,the $d^8$ configuration will always have two unpaired electrons in the $e_g$ set ($d_{x^2-y^2}$ and $d_{z^2}$).Since the $3d$ orbitals are not fully available for hybridization,the complex must use the $4d$ orbitals for $sp^3d^2$ hybridization.Therefore,$Ni(II)$ octahedral complexes are always outer orbital complexes.

An octahedral complex of $Ni(II)$ must be:

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