Animated Solution for Chemistry - Coordination Compounds: Which of the following complex species is not expected to exhibit optical isomerism?

When we talk about optical isomerism in coordination compounds, we are essentially hunting for chirality. A molecule is chiral if it cannot be superimposed on its mirror image. The quickest way to check for chirality is to look for a Plane of Symmetry (POS) or a Center of Symmetry (COS). If either is present, the molecule is achiral and optically inactive.

Let's evaluate the given options. Options (a), (b), and (d) all contain the bidentate ligand ethylenediamine ($en$). Bidentate ligands form chelate rings, which often lead to chiral structures, especially in their $cis$ configurations. For instance, $[Co(en{)}_3{]}^{3+}$ is a classic example of a chiral complex that exists as non-superimposable enantiomers.

Now, let's turn our attention to option (c), $[Co(N{H}_3{)}_{3}C{l}_3]$. This is an octahedral complex of the type $M{A}_3{B}_3$. Complexes of this type exhibit a special kind of geometrical isomerism, existing in two forms:

1. Facial (fac) isomer: Three identical ligands occupy the corners of one triangular face of the octahedron. 2. Meridional (mer) isomer: Three identical ligands lie in a single plane that bisects the octahedron, much like the meridian of a globe.

Let's check these isomers for symmetry.

In the fac-isomer, if you imagine a plane passing through the central Cobalt atom, one $N{H}_3$ ligand, and one $Cl$ ligand, it perfectly bisects the angle between the remaining two $N{H}_3$ ligands and the remaining two $Cl$ ligands. Because this plane divides the molecule into two identical mirror-image halves, the fac-isomer possesses a plane of symmetry and is optically inactive.

In the mer-isomer, the three $N{H}_3$ ligands and the Cobalt atom all lie in the same plane. This plane also contains one of the $Cl$ ligands and perfectly reflects the other two $Cl$ ligands onto each other. Thus, the mer-isomer also possesses a plane of symmetry and is optically inactive.

Since both possible geometrical isomers of $[Co(N{H}_3{)}_{3}C{l}_3]$ contain a plane of symmetry, the complex as a whole cannot exhibit optical isomerism.

Key Takeaway: It is a universal rule in coordination chemistry that $M{A}_3{B}_3$ type octahedral complexes never show optical isomerism. Keep this in your arsenal for quick problem-solving!