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

Welcome to the fascinating world of coordination chemistry! Today, we are diving deep into a classic problem that tests your spatial visualization skills: identifying optical isomerism in octahedral complexes.

Isomerism occurs when two or more compounds have the same chemical formula but different arrangements of atoms. In coordination compounds, this can manifest as geometrical isomerism (like cis and trans) or optical isomerism.

Optical isomerism is a special type of stereoisomerism where two molecules are non-superimposable mirror images of each other. These mirror-image pairs are called enantiomers, and they have the unique ability to rotate plane-polarized light in opposite directions.

Chirality and Symmetry

For a molecule to exhibit optical isomerism, it must be chiral. But what does chirality mean in the context of a coordination complex?

A molecule is chiral if it lacks any plane of symmetry or center of symmetry. A plane of symmetry is an imaginary flat surface that cuts the molecule exactly in half, such that one half is the perfect mirror reflection of the other.

If you can find even a single plane of symmetry in a molecule, it is achiral and will be optically inactive. Therefore, our mission in this problem is to hunt for symmetry. If we find it, the complex is out of the race!

Why Symmetry Prevails

Let's start by examining options (c) and (d). Option (c) is $[Co(N{H}_3{)}_{4}C{l}_2]Cl$, which is a complex of the type $M{A}_4{B}_2$. Option (d) is $[Co(N{H}_3{)}_{3}C{l}_3]$, a complex of the type $M{A}_3{B}_3$.

Notice that both of these complexes contain only monodentate ligands (ammonia and chloride). Monodentate ligands bind to the central metal atom through a single donor atom.

Because these ligands are simple and don't form rings, complexes of the type $M{A}_4{B}_2$ and $M{A}_3{B}_3$ are highly symmetrical. Whether you arrange them in cis or trans forms, or facial or meridional forms, you will always be able to find at least one plane of symmetry.

Since they possess a plane of symmetry, they are achiral. Thus, we can confidently eliminate options (c) and (d).

Enter Ethylenediamine

Now, let's turn our attention to the remaining options, which feature the ligand ethylenediamine, abbreviated as $en$.

Ethylenediamine is a bidentate ligand. This means it has two nitrogen donor atoms and can bite onto the central cobalt atom at two different positions simultaneously, forming a five-membered chelate ring.

The presence of these bulky chelate rings drastically changes the spatial geometry of the complex and often leads to asymmetry. Let's analyze the trans and cis isomers of $[Co(en{)}_{2}C{l}_2{]}^{+}$ to see how this plays out.

A Case of Perfect Symmetry

Let's visualize the trans isomer of $[Co(en{)}_{2}C{l}_2{]}^{+}$. In the trans configuration, the two identical monodentate ligands (the chloride ions) are positioned exactly opposite to each other, at an angle of $180^{\circ }$.

The two bidentate $en$ ligands occupy the four equatorial positions around the cobalt atom. Now, imagine a flat plane slicing horizontally right through the middle of the molecule, containing the cobalt atom and all four nitrogen atoms of the $en$ ligands.

This equatorial plane acts as a perfect plane of symmetry. The top half of the molecule (containing one chloride ion) is the exact mirror reflection of the bottom half (containing the other chloride ion).

Because of this plane of symmetry, the trans isomer is achiral and optically inactive.

Asymmetry and Optical Activity

Finally, let's examine the cis isomer of $[Co(en{)}_{2}C{l}_2{]}^{+}$. In this configuration, the two chloride ions are adjacent to each other, at an angle of $90^{\circ }$.

Because the chloride ions are adjacent, the two bulky $en$ chelate rings are forced into a spatial arrangement that is highly asymmetric.

Try to visualize slicing this molecule from any angle. You will quickly realize that there is no plane of symmetry that can divide the cis isomer into two identical halves.

Because it completely lacks symmetry, the cis isomer is chiral. It will exist as a pair of non-superimposable mirror images, known as the $d$ (dextrorotatory) and $l$ (levorotatory) enantiomers.

Our structural analysis has led us to a clear conclusion. While the trans isomer possesses a plane of symmetry, the cis isomer is completely asymmetric.

Therefore, the cis isomer is the only complex among the choices that exhibits optical isomerism.

The correct answer is option (a). Always remember that in octahedral complexes, a cis arrangement involving bidentate ligands is a very strong indicator of chirality and optical activity!