Which ligands are in the inner coordination sphere for [PtBr(NH3)3]Br∙H2O ?

Question

BlackTom AI · Accepted Answer

Question restated: Which ligands are in the inner coordination sphere for [PtBr(NH3)3]Br·H2O?

Option a: 3 NH3, 1 Br−, and 1 H2O. This would imply four ligands in the inner sphere (three NH3 and one Br−) plus a coordinated water molecule, totaling five entities around Pt. In a typical square-planar Pt(II) complex, the coordination number is 4. Adding an outer-sphere Br−/H2O as counterion or lattice water is more consistent with a five-coordinate description only in special cases. Thus, this option incorrectly assigns both Br− and H2O inside the inner sphere.

Option b: 3 NH3 and 2 Br−. This would yield a total of five ligands (three NH3 and two Br−) in the inner sphere, which is inconsistent with the common coordination geometry of Pt(II) complexes like [PtBr(NH3)3]+ that are typically square-planar with four ligands. Therefore, this option is unlikely.

Option c: 3 NH3, 2 Br−, and 1 H2O. This describes five ligands in the inner sphere (three NH3, two Br−, and a water). As with option a, this contradicts the standard four-coordinate geometry of this complex and misplaces a water molecule as a coordinated ligand. It also overcounts ligands around Pt.

Option d: 3 NH3. This would place only three ligands around Pt, which would be insufficient for a stable Pt(II) coordination environment in a typical complex of this type; a fourth ligand is expected in the inner sphere. Hence, this cannot be correct.

Option e: 3 NH3 and 1 Br−. This arrangement yields four ligands around Pt: three ammine ligands and one bromide, which matches the expected square-planar coordination of [PtBr(NH3)3]+. In the given formula [PtBr(NH3)3]Br·H2O, the extra Br− and H2O are typically external (outer-sphere) counterion and lattice/solvent waters, not ligands bound to Pt. Therefore, the inner coordination sphere correctly comprises three NH3 and one Br−, aligning with standard chemistry for this complex.

Summary of reasoning:
- The inner coordination sphere of a typical Pt(II) ammine complex with a single halide ligand is four-coordinate (square-planar).
- Counting ligands shows that options a, c, and b add too many ligands to Pt or misplace solvent as an inner-sphere ligand.
- Option d provides too few ligands for a stable square-planar Pt(II) center.
- Option e provides four ligands total: three NH3 and one Br−, matching the expected geometry and the given formula where Br and H2O are outside the coordination sphere.

Note on common misconceptions:
- It might be tempting to include H2O as a coordinated ligand when the formula is written with ·H2O; however, in many crystal structures, water acts as a lattice solvent rather than a bound ligand, so it does not populate the inner sphere.
- Counting outer-sphere counterions separately from inner-sphere ligands is crucial; confusing the two leads to incorrect conclusions about the ligand set.

End of analysis for each option. The reasoning above demonstrates why each alternative is incorrect or correct based on typical coordination chemistry of Pt(II) complexes and the given formula.

Which ligands are in the inner coordination sphere for [PtBr(NH3)3]Br∙H2O ?单项选择题

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