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(NH3)3Cl]+ [Pt(NH_3)_3Cl]^+

(b) [Pt(NH3)Cl5] [Pt(NH_3)Cl_5]^-

(c) Pt(NH3)2Cl(NO2)Pt(NH_3)_2Cl(NO_2)

(d) [Pt(NH3)4ClBr]2+ [Pt(NH_3)_4ClBr]^{2+}

STRATEGY

Recall that cis and trans isomers are possible for square planar complexes of the type MA2B2MA_2B_2 and MA2BCMA_2BC and for octahedral complexes of the type MA4B2MA_4B_2 and MA4BCMA_4BC. Cis and trans isomers are not possible when only one ligand differs from the others, as in complexes of the type MA3BMA_3B and MA5BMA_5B.

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(a) No isomers are possible for [Pt(NH3)3Cl]+ [Pt(NH_3)_3Cl]^+, a square planar complex of the type MA3BMA_3B. (first photo below)

(b) No isomers are possible for [Pt(NH3)Cl5] [Pt(NH_3)Cl_5]^-, an octahedral complex of the type MA5BMA_5B. (second photo below)

(c) Cis and trans isomers are possible for Pt(NH3)2Cl(NO2)Pt(NH_3)_2Cl(NO_2), a square planar complex of the type MA2BCMA_2BC. The ClCl^- and NO2NO_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(NH3)4ClBr]2+ [Pt(NH_3)_4ClBr]^{2+}, an octahedral complex of the type MA4BCMA_4BC. The ClCl^- and BrBr^- ligands can be on either adjacent or opposite corners of the octahedron. (the fourth photo below)

worked example 20.6 1
worked example 20.6 2
worked example 20.6 3
worked example 20.6 4

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