Chapter 23: Problem 26
Indicate the coordination number and the oxidation number of the metal for each of the following complexes: (a) \(\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) (b) \(\mathrm{Na}_{2}\left[\mathrm{CdBr}_{4}\right]\) (c) \(\left[\mathrm{Pt}(\mathrm{en})_{3}\right]\left(\mathrm{ClO}_{4}\right)_{4}\) (d) \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right]^{+}\) (e) \(\mathrm{NH}_{4}\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{NCS})_{4}\right]\) (f) \(\left[\mathrm{Cu}(\mathrm{bipy})_{2} \mathrm{I}\right] \mathrm{I}\)
Short Answer
Step by step solution
(a) K3[Co(CN)6] Exercise
(a) Coordination Number
(a) Oxidation Number
(b) Na2[CdBr4] Exercise
(b) Coordination Number
(b) Oxidation Number
(c) [Pt(en)3](ClO4)4 Exercise
(c) Coordination Number
(c) Oxidation Number
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Coordination Number
- A coordination number of 4, as seen in \([\text{CdBr}_4]^{2-}\), often leads to tetrahedral or square planar geometries.
- A coordination number of 6, like in the \([\text{Cr}(\text{NH}_3)_2(\text{NCS})_4]^{-}\) complex, usually leads to an octahedral geometry.
Oxidation Number
- For \([\text{CdBr}_4]^{2-}\), with each bromide contributing -1, the oxidation number of cadmium is +2.
- In the complex \([\text{Pt}(\text{en})_3]^{4+}\), the oxidation state of platinum is +4.
Complex Ions
- The geometry of a complex ion like \([\text{Co}(\text{en})_2(\text{C}_2\text{O}_4)]^{+}\) helps dictate its chemical properties.
- Knowing the charge of a complex ion allows for predicting its behavior in reactions.
Ligands
- Bidentate ligands like \(\text{bipy} \) in the \([\text{Cu}(\text{bipy})_2 \text{I}]\) complex help maximize the coordination number while stabilizing the complex.
- Monodentate ligands, such as bromide in \([\text{CdBr}_4]^{2-}\), form only one bond with the central atom.