Animated Solution for Chemistry - Coordination Compounds: Which one of the following has largest number of isomers? ($R=\text{alkyl group}$, $\text{en}=\text{ethylenediamine}$)

Isomerism in coordination chemistry is like a molecular puzzle. You are given a set of atoms, and your job is to figure out how many distinct ways they can be arranged in 3D space. In this problem, we are tasked with finding which of the given complexes exhibits the highest number of isomers. To do this, we must systematically evaluate both geometrical (cis/trans) and optical (d/l) isomerism for each option.

The $M{A}_4{B}_2$ Case

Let's start with $[Ru(N{H}_3{)}_{4}C{l}_2{]}^{+}$. This is a classic octahedral complex of the type $M{A}_4{B}_2$.

Imagine the central Ruthenium atom surrounded by four ammonia ligands and two chloride ligands. The two chlorine atoms can either be placed adjacent to each other at a $90^{\circ }$ angle (the cis form) or directly opposite each other at a $180^{\circ }$ angle (the trans form).

Do these forms show optical isomerism? No. Both the cis and trans forms possess a plane of symmetry, meaning their mirror images are perfectly superimposable. Therefore, this complex yields exactly 2 isomers.

The $M{A}_{5}B$ Case

Next up is $[Co(N{H}_3{)}_{5}Cl{]}^{2+}$. This is an $M{A}_{5}B$ type complex.

With five identical ammonia ligands and only one unique chloride ligand, there is no relative positioning that can create a distinct geometrical arrangement. No matter where you place the chlorine atom, the molecule looks exactly the same. Furthermore, it has multiple planes of symmetry, ruling out any optical activity. Thus, this complex has only 1 isomer.

The Square Planar Case

Option C presents $[Ir(P{R}_3{)}_{2}H(CO){]}^{2+}$. This is a square planar complex of the type $M{A}_{2}BC$.

In a square planar geometry, the two identical phosphine ligands can be placed adjacent to each other (cis) or opposite to each other (trans). This gives us 2 geometrical isomers. However, square planar complexes inherently possess a molecular plane of symmetry (the plane containing all the atoms). Because of this, they almost never exhibit optical isomerism. So, we are left with 2 isomers.

The Bidentate Champion

Finally, we arrive at $[Co(en{)}_{2}C{l}_2{]}^{+}$. This is an octahedral complex featuring a symmetrical bidentate ligand, ethylenediamine (en), making it an $M(AA{)}_2{B}_2$ type complex.

Like the $M{A}_4{B}_2$ case, it can form cis and trans geometrical isomers. The trans isomer has the two chlorine atoms opposite each other. This arrangement is highly symmetrical and possesses a plane of symmetry, rendering it optically inactive.

But the magic happens in the cis isomer! When the two chlorine atoms are adjacent, the bulky bidentate 'en' rings wrap around the remaining positions in a way that completely destroys any plane of symmetry. Because it lacks symmetry, the cis isomer is chiral. Its mirror image is non-superimposable, meaning it exists as a pair of enantiomers: the dextrorotatory ($d$) and levorotatory ($l$) forms.

Let's tally up the isomers for Option D: 1. The trans isomer (optically inactive) 2. The cis-$d$ isomer (optically active) 3. The cis-$l$ isomer (optically active)

This gives a grand total of 3 isomers. Comparing all the options, $[Co(en{)}_{2}C{l}_2{]}^{+}$ clearly takes the crown. Always remember: when dealing with octahedral complexes containing bidentate ligands, the cis form is a prime suspect for optical activity!