Animated Solution for Chemistry - Coordination Compounds: Which of the following compounds shows optical isomerism ?

Unlocking Optical Isomerism in Coordination Compounds

Optical isomerism is one of the most fascinating phenomena in coordination chemistry. It occurs when a molecule is chiral—meaning it lacks any plane of symmetry (POS) or center of symmetry (COS). Just like your left and right hands, a chiral molecule and its mirror image are non-superimposable. If you try to place one perfectly over the other, they simply won't match up.

Let's embark on a journey to find which of the given complexes possesses this unique property.

When hunting for optical isomerism, our first step is to weed out the highly symmetric molecules.

Consider option (a), $[Co(CN{)}_6{]}^{3-}$. This is a homoleptic octahedral complex, meaning the central cobalt ion is surrounded by six identical cyanide ligands. Because all the ligands are the same, you can slice this molecule through multiple planes and always get two identical halves. It is perfectly symmetric and therefore achiral.

Similarly, option (c), $[ZnC{l}_4{]}^{2-}$, is a tetrahedral complex with four identical chloride ligands, and option (d), $[Cu(N{H}_3{)}_4{]}^{2+}$, is a square planar complex. Both of these geometries, when bonded to identical ligands, possess multiple planes of symmetry. They are optically inactive.

Now, let's turn our attention to option (b): $[Cr(C_2{O}_4{)}_3{]}^{3-}$.

This is an octahedral complex, but with a twist. The central chromium ion ($C{r}^{3+}$) is surrounded by three oxalate ligands ($C_2{O}_4^{2-}$). Oxalate is a bidentate ligand, which means each oxalate molecule bites onto the central metal at two different positions simultaneously. This forms a complex of the general type $[M(AA{)}_3{]}^{n\pm }$.

When three bidentate ligands coordinate to an octahedral center, they cannot lie flat. Instead, they wrap around the metal ion, creating a 3D structure that resembles the twisted blades of a propeller.

Because of this propeller-like arrangement, the $[Cr(C_2{O}_4{)}_3{]}^{3-}$ complex loses all planes of symmetry. There is no way to slice this molecule into two identical, reflecting halves.

If you place a mirror next to this complex, you generate its mirror image. Try to mentally pick up that mirror image and superimpose it onto the original molecule. You will quickly realize that the "blades" twist in opposite directions! One is a "right-handed" propeller ($d$-form), and the other is a "left-handed" propeller ($l$-form).

Because the mirror images are non-superimposable, the complex exists as a pair of enantiomers. Therefore, $[Cr(C_2{O}_4{)}_3{]}^{3-}$ exhibits optical isomerism.

Pro-Tip: As a general rule in coordination chemistry, octahedral complexes containing three symmetrical bidentate ligands (like $[M(en{)}_3{]}^{3+}$ or $[M(ox{)}_3{]}^{3-}$) are always chiral and will always show optical isomerism!