Animated Solution for Chemistry - Coordination Compounds: Which one of the following complex ions has geometrical isomers?

Geometrical isomerism in coordination compounds is a fascinating phenomenon that arises when ligands can be arranged in different relative spatial positions around the central metal ion. For octahedral complexes (where the coordination number is 6), this typically manifests as cis and trans isomerism.

To quickly determine if a complex can exhibit geometrical isomerism, it is highly effective to classify the complex into general formula types, such as $[M{a}_4{b}_2]$, $[M{a}_3{b}_3]$, or $[M(AA{)}_2{a}_2]$, where 'M' is the central metal, 'a' and 'b' are monodentate ligands, and 'AA' is a symmetrical bidentate ligand.

Let's systematically evaluate each option provided in the question to see if they fit the criteria for geometrical isomerism.

1. The $[Co(en{)}_3{]}^{3+}$ Complex: This complex falls under the $[M(AA{)}_3]$ category. Here, ethylenediamine (en) is a symmetrical bidentate ligand. Because all three ligands are identical and symmetrical, every possible spatial arrangement you can construct will be superimposable on the others. Therefore, it cannot show geometrical isomerism. (Note: It does, however, exhibit optical isomerism because its mirror image is non-superimposable).

2. The $[Ni(N{H}_3{)}_{5}Br{]}^{+}$ Complex: This complex is of the $[M{a}_{5}b]$ type. Imagine placing the single bromide ion at any of the six octahedral vertices. The remaining five positions will always be occupied by identical ammonia molecules. There is no distinct 'adjacent' or 'opposite' relationship that can be formed between two different groups of ligands. Hence, no geometrical isomers exist for this type.

3. The $[Cr(N{H}_3{)}_4(en){]}^{3+}$ Complex: Classified as an $[M{a}_4(AA)]$ type complex, this structure contains one bidentate ligand and four identical monodentate ligands. The bidentate 'en' ligand is constrained to occupy two adjacent (cis) positions. The remaining four positions are filled by the identical ammonia ligands. Once again, no alternative spatial arrangement can be created, meaning no geometrical isomerism.

Finally, we arrive at $[Co(N{H}_3{)}_2(en{)}_2{]}^{3+}$, which is an $[M(AA{)}_2{a}_2]$ type complex. This is where things get interesting! We have two identical monodentate ligands (ammonia) and two symmetrical bidentate ligands (ethylenediamine).

Because we have two distinct ammonia ligands, we have a choice in how we place them relative to each other:

The Cis-Isomer: If we place the two $N{H}_3$ ligands at adjacent positions, meaning the bond angle between them is $90^{\circ }$, we form the cis-isomer. The two 'en' ligands will occupy the remaining four positions. The Trans-Isomer: Alternatively, if we place the two $N{H}_3$ ligands exactly opposite to each other, creating a bond angle of $180^{\circ }$, we form the trans-isomer. In this highly symmetrical arrangement, the two 'en' ligands lie flat in the equatorial plane.

Since $[Co(N{H}_3{)}_2(en{)}_2{]}^{3+}$ can exist in these two distinct spatial arrangements, it is the only complex among the choices that exhibits geometrical isomerism. Therefore, the correct option is (c).