Consider the following three complexes: \(\left(\right.\) Complex 1) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SCN}\right]^{2+}$ \(\left(\right.\) Complex 2) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]^{2+}$ \(\left(\right.\) Complex 3) \(\mathrm{CoClBr} \cdot 5 \mathrm{NH}_{3}\) Which of the three complexes can have (a) geometric isomers, (b) linkage isomers, (c) optical isomers, (d) coordination-sphere isomers?

Geometric isomers occur when two complexes have the same composition but differ in the arrangement of their ligands in space. For a complex to exhibit geometric isomerism, it must have an ambidentate ligand, meaning the ligand can bind through two different atoms. Let's analyze each complex to see if they can form geometric isomers: 1) Complex 1: \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SCN}\right]^{2+}\) - This complex has an ambidentate ligand (SCN^-). The ligand can bind to Co either through N (thiocyanate) or through S (isothiocyanate). Hence, Complex 1 can form geometric isomers. 2) Complex 2: \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]^{2+}\) - The ligands in this complex (NH3 and Cl^-) are not ambidentate. Therefore, Complex 2 cannot have geometric isomers. 3) Complex 3: \(\mathrm{CoClBr} \cdot 5 \mathrm{NH}_{3}\) - The ligands in this complex (Cl^- and Br^-) are not ambidentate. Therefore, Complex 3 cannot have geometric isomers.

Linkage isomers can be observed in a complex with an ambidentate ligand when the ligand alternates between the atoms it can bind with the metal, forming different structures. Let's check for linkage isomers in the given complexes: 1) Complex 1: As mentioned above, this complex has an ambidentate ligand (SCN^-). Hence, Complex 1 can have linkage isomers. 2) Complex 2 and Complex 3: Both the complexes have no ambidentate ligands. Therefore, neither Complex 2 nor Complex 3 can have linkage isomers.

Optical isomers are mirror-image isomers that are non-superimposable. They exhibit chirality, meaning they can rotate plane-polarized light in opposite directions. To have optical isomers, a complex must lack an internal plane of symmetry. Let's look at our three complexes: 1) Complex 1: \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SCN}\right]^{2+}\) - This complex lacks a plane of symmetry. Therefore, Complex 1 can have optical isomers. 2) Complex 2: \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]^{2+}\) - This complex has a plane of symmetry and hence, cannot have optical isomers. 3) Complex 3: \(\mathrm{CoClBr} \cdot 5 \mathrm{NH}_{3}\) - This complex lacks a plane of symmetry. Therefore, Complex 3 can have optical isomers.

Coordination-sphere isomers occur when two complexes have the same composition but differ in the configuration between the central metal atom and its ligands. These isomers arise when a compound contains both coordination and non-coordination groups. Let's check for coordination-sphere isomers in the given complexes: 1) Complex 1: This complex has no non-coordination groups. Hence, Complex 1 cannot have coordination-sphere isomers. 2) Complex 2: This complex has no non-coordination groups. Hence, Complex 2 cannot have coordination-sphere isomers. 3) Complex 3: \(\mathrm{CoClBr} \cdot 5 \mathrm{NH}_{3}\) - This complex can have coordination-sphere isomers because it contains a coordination group (CoClBr) and a non-coordination group (5NH3). In summary: (a) Geometric Isomers: Complex 1 (b) Linkage Isomers: Complex 1 (c) Optical Isomers: Complex 1 and Complex 3 (d) Coordination-Sphere Isomers: Complex 3