Chapter 20: Problem 18
Almost all metals in nature are found as ionic compounds in ores instead of being in the pure state. Why? What must be done to a sample of ore to obtain a metal substance that has desirable properties?
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Chapter 20: Problem 18
Almost all metals in nature are found as ionic compounds in ores instead of being in the pure state. Why? What must be done to a sample of ore to obtain a metal substance that has desirable properties?
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The complex ion \(\mathrm{PdCl}_{4}^{2-}\) is diamagnetic. Propose a structure for \(\mathrm{PdCl}_{4}^{2-}\)
Henry Taube, 1983 Nobel Prize winner in chemistry, has studied the mechanisms of the oxidation-reduction reactions of transition metal complexes. In one experiment he and his students studied the following reaction: $$ \begin{aligned} \operatorname{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(a q) &+\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}(a q) \\ & \longrightarrow \mathrm{Cr}(\mathrm{III}) \text { complexes }+\mathrm{Co}(\mathrm{II}) \text { complexes } \end{aligned} $$ Chromium(III) and cobalt(III) complexes are substitutionally inert (no exchange of ligands) under conditions of the experiment. Chromium(II) and cobalt(II) complexes can exchange ligands very rapidly. One of the products of the reaction is \(\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}^{2+} .\) Is this consistent with the reaction proceeding through formation of \(\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cr}-\mathrm{Cl}-\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\) as an inter- mediate? Explain.
\(\mathrm{CoCl}_{4}^{2-}\) forms a tetrahedral complex ion and \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) forms an octahedral complex ion. What is wrong about the following statements concerning each complex ion and the \(d\) orbital splitting diagrams? a. \(\mathrm{CoCl}_{4}^{2-}\) is an example of a strong-field case having two unpaired electrons. b. Because \(\mathrm{CN}^{-}\) is a weak-field ligand, \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) will be a low-spin case having four unpaired electrons.
A metal ion in a high-spin octahedral complex has two more unpaired electrons than the same ion does in a low-spin octahedral complex. Name some possible metal ions for which this would be true.
Both \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) and \(\mathrm{Ni (\mathrm{SCN})_{4}^{2-}\) have four ligands. The first is paramagnetic, and the second is diamagnetic. Are the complex ions tetrahedral or square planar? Explain.
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