Animated Solution for Chemistry - Coordination Compounds: The complex that can show optical activity is (ox = oxalate)

Welcome to a fascinating journey into the 3D world of coordination chemistry! Today, we are tackling a classic problem: identifying which of the given octahedral complexes exhibits optical activity.

To solve this, we must first understand what makes a molecule optically active. The golden rule is chirality. A molecule is chiral, and thus optically active, if it lacks any plane of symmetry ($\sigma$) and center of symmetry ($i$). In simpler terms, its mirror image must be non-superimposable on the original molecule, much like your left and right hands.

Let's begin by examining the $trans$ isomers given in options (a) and (b): $trans\text{-}[Cr(C{l}_2)(ox{)}_2{]}^{3-}$ and $trans\text{-}[Fe(N{H}_3{)}_2(CN{)}_4{]}^{-}$.

In a $trans$ octahedral complex, identical ligands are positioned exactly opposite to each other (at a $180^{\circ }$ angle). Because of this highly symmetrical arrangement, you can easily slice the molecule in half with an imaginary plane.

For instance, in $trans\text{-}[Cr(C{l}_2)(ox{)}_2{]}^{3-}$, the plane containing the central Chromium atom and the four oxygen atoms of the oxalate ligands perfectly bisects the two Chlorine atoms. Since a plane of symmetry exists, the molecule is achiral and optically inactive.

Next, we look at option (c), which is $cis\text{-}[Fe(N{H}_3{)}_2(CN{)}_4{]}^{-}$.

While $cis$ isomers are generally less symmetrical than their $trans$ counterparts, this specific complex only contains monodentate ligands (ammonia and cyanide). In such cases, a plane of symmetry can still be found passing through the central metal and bisecting the angle between the identical $cis$ ligands. Therefore, this complex is also optically inactive.

Finally, we arrive at option (d): $cis\text{-}[CrC{l}_2(ox{)}_2{]}^{3-}$.

This complex is special because it contains two bidentate oxalate ($ox$) ligands arranged in a $cis$ configuration. Bidentate ligands form ring-like structures (chelates) with the central metal. When two such rings are forced into adjacent ($cis$) positions, they create a twisted, propeller-like geometry.

This unique 3D arrangement completely destroys any potential plane of symmetry. If you draw the mirror image of $cis\text{-}[CrC{l}_2(ox{)}_2{]}^{3-}$, you will find that no matter how you rotate it, it will never perfectly superimpose onto the original molecule.

These non-superimposable mirror images are called enantiomers. Because it exists as a pair of enantiomers, $cis\text{-}[CrC{l}_2(ox{)}_2{]}^{3-}$ is optically active.

Thus, the correct answer is option (d). Always remember: a $cis$ octahedral complex with at least two bidentate ligands is a prime candidate for optical activity!