Question

Which of the following will form octahedral complexes?
A) ${d^4}$ (low spin)
B) ${d^8}$ (high spin)
C) ${d^5}$ (low spin)
D) All of these

Answer 1

We need to know that according to CFT, an octahedral metal complex forms because of the electrostatic interaction of a positively charged metal ion with six negatively charged ligands or with the negative ends of dipoles associated with the six ligands. In addition, the ligands interact with one other electrostatically.

Complete answer:
We have to remember that ${d^1},{d^4},{d^2}$ cannot form octahedral complexes. Octahedral complexes are formed by $s{p^3}{d^2}$ by strong ligands or low spin complexes and by $s{p^3}{d^2}$ by weak field ligands or high spin complexes.
We need to know that an octahedral complex will be formed by ${d^5}$ low spin. Out of 5 d orbitals, 3 will contain electrons and 2 will be empty. These 2 empty d orbitals can undergo $s{p^3}{d^2}$ hybridization (or ${d^2}s{p^3}$ hybridization) to form an octahedral complex. In ${d^6}$ (low spin) complex, two electrons get paired up to make two d orbitals empty. We must know that the hybridization is ${d^2}s{p^3}$ (octahedral and the complex is low spin complex in the absence of half-filled electrons. When two or more types of ligands are coordinated to an octahedral metal center, the complex can exist as isomers. The number of possible isomers can reach 30 for an octahedral complex with six different ligands.

Note:
As we know that octahedral complexes will be favored over tetrahedral ones because: It is more (energetically) favorable to form six bonds rather than four. The CFSE is usually greater for octahedral than tetrahedral complexes. ${d^4}$ forms outer orbital complex in high spin and forms inner orbital complex in low, it cannot form octahedral complex.