Question 20.6: Identifying Diastereoisomers for Square Planar and Octahedra......
Identifying Diastereoisomers for Square Planar and Octahedral Complexes
Platinum(II) forms square planar complexes, and platinum(IV) gives octahedral complexes. How many diastereoisomers are possible for each of the following complexes? Draw their structures.
(a) [Pt(NH_3)_3Cl]^+
(b) [Pt(NH_3)Cl_5]^-
(c) Pt(NH_3)_2Cl(NO_2)
(d) [Pt(NH_3)_4ClBr]^{2+}
STRATEGY
Recall that cis and trans isomers are possible for square planar complexes of the type MA_2B_2 and MA_2BC and for octahedral complexes of the type MA_4B_2 and MA_4BC. Cis and trans isomers are not possible when only one ligand differs from the others, as in complexes of the type MA_3B and MA_5B.
(a) No isomers are possible for [Pt(NH_3)_3Cl]^+, a square planar complex of the type MA_3B. (first photo below)
(b) No isomers are possible for [Pt(NH_3)Cl_5]^-, an octahedral complex of the type MA_5B. (second photo below)
(c) Cis and trans isomers are possible for Pt(NH_3)_2Cl(NO_2), a square planar complex of the type MA_2BC. The Cl^- and NO_2^- ligands can be on either adjacent or opposite corners of the square. (the third photo below)
(d) Cis and trans isomers are possible for [Pt(NH_3)_4ClBr]^{2+}, an octahedral complex of the type MA_4BC. The Cl^- and Br^- ligands can be on either adjacent or opposite corners of the octahedron. (the fourth photo below)
