Question

Octahedral complex of Ni (II) must be:
A. inner orbital
B. outer orbital
C. inner and outer orbital depending upon the strong or weak field ligand
D. none of these

Answer 1

Hint: In an octahedral complex, the nickel is present in its +2 oxidation state. The electronic configuration of $N{i^{2 + }}$ is \[[Ar]3{{d}^{8}}\]. The d orbital of the ion contains 8 electrons in its outermost shell.

Complete step by step answer:
Inner orbital complex are formed of central metal atom containing hybridization of atomic orbital which have d-orbital containing inner shell electrons and s, p orbital containing outer shell electrons which means that the central metal atom of the complex involve inner shell of d-orbital for hybridization. Here, the d orbitals have lower energy than the s and p orbital.
Outer orbital complexes are formed of central metal atoms containing hybridization of atomic orbital which have s, p, and d orbitals from the outermost shell. In this complex, all the orbitals involved have the same energy level. As the d-orbital is located outside the s and p orbital, the complex formed is known as outer orbital complex.
The given metal ion is \[N{{i}^{2+}}\]. The atomic number of nickel is 28 and the electronic configuration of Ni is $[Ar]3{d^8}4{s^2}$. As nickel is present in its +2 oxidation state therefore two electrons are removed from the electronic configuration forming a new electronic configuration $[Ar]3{d^8}$. In $N{i^{2 + }}$, the d orbital contains 8 electrons which means it has fulfilled its orbital. Thus, the Octahedral complex of Ni (II) will not form an inner orbital complex rather than form an outer orbital complex.
So, the correct answer is “Option B”.

Note:
The most common hybridization formed by inner orbital complex are ${{d}^{2}}s{{p}^{3}}$ and $ds{p^2}$ and the most common hybridization formed by outer orbital complex is $s{p^3}{d^2}$.