Animated Solution for Chemistry - Coordination Compounds: Which of the following has an optical isomer?

The quest for optical isomerism in coordination compounds is like a high-stakes game of 3D hide-and-seek. You are not just looking at atoms; you are looking for hidden mirrors and invisible centers. In this epic breakdown, we will dissect a classic JEE problem that tests your spatial visualization skills to the absolute limit.

Before we dive into the options, let's establish the ground rules. For any molecule to exhibit optical isomerism, it must be chiral. What does that mean? It means the molecule must not be superimposable on its mirror image—just like your left and right hands.

In the world of coordination chemistry, the quickest way to fail the chirality test is to possess a Plane of Symmetry (POS) or a Center of Symmetry (COS). If you can imagine a flat sheet of glass slicing through the molecule such that one half perfectly reflects the other, the molecule is symmetric, achiral, and optically inactive.

Let's put our options through the symmetry scanner.

Options (a) and (b): Both $[Co(N{H}_3{)}_{3}Cl{]}^{+}$ and $[Co(en)(N{H}_3{)}_2{]}^{2+}$ have a coordination number of 4. Whether they adopt a tetrahedral or a square planar geometry, the presence of multiple identical monodentate ligands (like the ammonia molecules) guarantees a plane of symmetry. Square planar complexes, in particular, are almost always optically inactive because the molecular plane itself acts as a mirror!

Option (c): The complex $[Co(H_{2}O{)}_4(en){]}^{3+}$ is octahedral, but it's flooded with four identical water molecules. This is a classic $M(AA)a_4$ system. You can easily slice a plane right through the cobalt atom and the bidentate 'en' ring. This plane will perfectly reflect the two water molecules on the left onto the two water molecules on the right. Symmetry achieved; optical activity denied.

Option (d)

Now we arrive at $[Co(en{)}_2(N{H}_3{)}_2{]}^{3+}$. This is an octahedral complex of the famous $M(AA{)}_2{a}_2$ type. It features two bidentate ethylenediamine ('en') rings and two monodentate ammonia ligands.

Because of this specific ligand combination, the complex can arrange itself in two distinct geometrical forms: the trans-isomer and the cis-isomer. Let's pit them against each other.

A Master of Symmetry

Imagine the trans-isomer. The two ammonia ligands are placed exactly opposite to each other, separated by a $180^{\circ }$ angle (one at the top, one at the bottom). The four nitrogen atoms from the two 'en' rings occupy the equatorial plane.

Now, visualize a horizontal plane slicing right through the equator of the molecule. This plane contains the cobalt atom and all four 'en' nitrogen atoms. What happens to the ammonia ligands? The top $N{H}_3$ is perfectly reflected onto the bottom $N{H}_3$. Because of this beautiful equatorial plane of symmetry, the trans-isomer is completely achiral and optically inactive.

The Twisted Champion

Now, let's build the cis-isomer. Here, the two ammonia ligands are placed adjacent to each other at a $90^{\circ }$ angle. This forces the two bulky 'en' rings to occupy the remaining adjacent positions, creating a twisted, propeller-like structure.

Try to find a plane of symmetry now. - If you slice it vertically between the ammonia ligands, the 'en' rings on the sides don't reflect each other properly due to their 3D curvature. - If you slice it horizontally, an ammonia ligand will reflect into an 'en' ring, which is a mismatch.

Because of the spatial constraints and the twisting of the bidentate rings, the cis-isomer completely lacks any plane or center of symmetry. It is a chiral masterpiece!

Because the cis-form of $[Co(en{)}_2(N{H}_3{)}_2{]}^{3+}$ is chiral, it exists as a pair of non-superimposable mirror images—the dextrorotatory ($d$) and levorotatory ($l$) enantiomers. Therefore, this complex exhibits optical isomerism, making Option (d) the undisputed correct answer.

Pro Tip for JEE: Whenever you see an octahedral complex with two or three bidentate rings (like $M(AA{)}_2{a}_2$ or $M(AA{)}_3$), your chirality radar should immediately go off. These twisted structures are the absolute favorites of examiners!