Question

The complex that can show fac- and mer- isomers is:
(A) $\left[Co{\left(N{H}_3\right)}_3{\left(N{O}_2\right)}_3\right]$
(B) $\left[Pt{\left(N{H}_3\right)}_{2}C{l}_2\right]$
(C) $\left[CoC{l}_2{\left(en\right)}_2\right]$
(D) ${\left[Co{\left(N{H}_3\right)}_{4}C{l}_2\right]}^{+}$

Answer 1

Hint: Octahedral Molecular Geometry describes the form of compounds composed of six or symmetrically arranged atoms or groups of ligands around a central atom, describing the vertices of an octahedron.

Complete step by step solution:
> The eight faces of the octahedron, hence, the octa prefix. Octahedron is one of the platonic solids, while octahedral molecules typically have a center atom and no ligand relation.
Only octahedral complexes of general formulation $\left[M{A}_3{B}_3\right]$ can have facial isomerism. If one kind is situated adjacently at the corners of an octahedral face and forms its own face, it is known as face isomerism (facial isomerism). Two equatorial ligands and one axial. There are two ligands. In the Equatorial Position of the octahedral all three of the same ligands are present, this means the meridional isomerism (Mer-).
> Facial isomer or cis isomer is also recognized. In one triangular side there are three A ligands and on the other triangular side there are three B ligands. The mer isomer is also referred to as a meridional isomer or trans isomer.
In order to explain Stoichiometries and Isomerism in Coordination Compounds, Alfred Werner developed the principle of octahedral coordination geometry.

From the above information says that option A is the correct option.

Note: The term "octahedral" is used by chemists somewhat loosely, focusing on the structure of the bonds with the central atom and not taking into account variations between ligands. In the mathematical sense , for example, ${\left[Co{\left(N{H}_3\right)}_6\right]}^{3+}$ is not octahedral, due to orientation of N-H bonds.