The correct answer is option: 3. \([Pt(NH_3)_6]^{3+}\).
Geometrical isomerism arises in coordination compounds when there are different spatial arrangements possible for ligands around a metal center. This typically occurs in complexes with coordination numbers greater than 4, and it is a result of the different ways the ligands can be arranged in the coordination sphere.
Let us analyze each complex ion mentioned:
1. \([Cr(H_2O)_4Cl_2]^{+}\):
- This complex has a coordination number of 6, with two different ligands: water (H₂O) and chloride (Cl⁻).
- Geometrical isomerism is possible here. The cis and trans isomers can be formed based on the arrangement of water and chloride ligands around the central chromium ion.
2. \([Pt(NH_3)_2Cl_2]\):
- This complex has a coordination number of 4, with two different ligands: ammonia (NH₃) and chloride (Cl⁻).
- Geometrical isomerism is possible, and the isomers can be classified as cis and trans based on the arrangement of ammonia and chloride ligands.
3. \([Pt(NH_3)_6]^{3+}\):
- This complex has a coordination number of 6, with only one type of ligand: ammonia (NH₃).
- No geometrical isomerism is possible here because all six positions around the platinum ion are occupied by the same ligand (NH₃). The spatial arrangement is the same in all cases.
4. \([Co(CN)_4(NC)_2]^{3-}\):
- This complex has a coordination number of 6, with two different ligands: cyanide (CN⁻) and isocyanide (NC⁻).
- Geometrical isomerism is possible here. The cis and trans isomers can be formed based on the arrangement of cyanide and isocyanide ligands.
In summary, only the complex ion \([Pt(NH_3)_6]^{3+}\) does not show geometrical isomerism, as all the ligands around the platinum ion are identical. The other three complexes have different ligands, leading to the possibility of geometrical isomerism.
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