Animated Solution for Chemistry - Coordination Compounds: The number of optical isomers possible for $[Cr(C_2{O}_4{)}_3{]}^{3-}$ is ...

When we dive into the world of coordination compounds, we aren't just looking at flat, two-dimensional drawings on a piece of paper. We are exploring intricate, three-dimensional architectures that behave in fascinating ways. One of the most beautiful phenomena in this 3D realm is optical isomerism—the ability of a molecule to exist as two non-superimposable mirror images, much like your left and right hands.

In this problem, we are asked to find the number of optical isomers for the complex ion $[Cr(C_2{O}_4{)}_3{]}^{3-}$.

First, let's break down the components of our complex. The central metal ion is Chromium in a $+3$ oxidation state ($C{r}^{3+}$). Surrounding it are three oxalate ligands, denoted as $C_2{O}_4^{2-}$ or simply 'ox'.

The crucial piece of information here is the nature of the oxalate ligand. It is a symmetrical bidentate ligand. "Bidentate" means it has two "teeth" or donor atoms (in this case, two oxygen atoms) that can simultaneously bite onto the central chromium ion.

Since we have $3$ of these bidentate ligands, the total number of coordinate bonds formed with the central metal is $3\times 2=6$. A coordination number of $6$ dictates that the complex will adopt an octahedral geometry.

To determine if a molecule is optically active (and thus has optical isomers), we must act as molecular detectives hunting for symmetry. Specifically, we are looking for a plane of symmetry or a center of symmetry.

Imagine trying to slice the $[Cr(C_2{O}_4{)}_3{]}^{3-}$ octahedron perfectly in half so that one side is the exact mirror reflection of the other. Because the three bidentate oxalate ligands wrap around the central chromium ion like the blades of a propeller, they create a helical twist. This "propeller" arrangement completely destroys any internal symmetry.

There is no plane you can cut through, and no center point you can invert through, that will leave the molecule looking identical. Because it lacks all elements of symmetry, the molecule is chiral.

Because the molecule is chiral, its mirror image is a completely distinct entity. If you build a 3D model of this complex and then build its exact mirror reflection, you will find that no matter how you twist, turn, or rotate the reflection, you can never perfectly superimpose it onto the original model.

These two non-superimposable mirror images are called enantiomers. In coordination chemistry, they are often referred to as the dextro ($d$ or $\Delta$) and laevo ($l$ or $\Lambda$) forms, depending on how they rotate plane-polarized light.

Therefore, the complex $[Cr(C_2{O}_4{)}_3{]}^{3-}$ exists as exactly two optical isomers.

This specific structural motif is a classic in competitive exams. You can save precious time by remembering this golden rule: Any octahedral complex of the general formula $[M(AA{)}_3{]}^{n\pm }$ (where AA is a symmetrical bidentate ligand) will always be optically active and will always exist as exactly $2$ optical isomers. Furthermore, because all three ligands are identical and symmetrically arranged, this type of complex can never exhibit geometrical (cis/trans) isomerism.