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Chapter 21: Problem 40

Give formulas for the following complex ions. a. tetrachloroferrate(III) ion b. pentaammineaquaruthenium(III) ion c. tetracarbonyldihydroxochromium(III) ion d. amminetrichloroplatinate(II) ion

Short Answer

Expert verified
The formulas for the given complex ions are as follows: a. Tetrachloroferrate(III) ion: \(\ce{[Fe(Cl)4]^{3-}}\) b. Pentaammineaquaruthenium(III) ion: \(\ce{[Ru(NH3)5(H2O)]^{3+}}\) c. Tetracarbonyldihydroxochromium(III) ion: \(\ce{[Cr(CO)4(OH)2]^{3-}}\) d. Amminetrichloroplatinate(II) ion: \(\ce{[Pt(NH3)Cl3]^{2-}}\)

Step by step solution

01

a. Tetrachloroferrate(III) ion

We can break down the name into three parts - the counterion (chloride), the central ion (iron), and its oxidation state (III, meaning a charge of +3). The Latin prefix "tetra" indicates that there are four chloride ions. Thus, our complex ion is as follows: 1. Chloride: Cl^- 2. Iron: Fe^3+ 3. Oxidation state of central ion: +3 4. Number of chloride ions: 4 Now we can combine these to generate the formula for tetrachloroferrate(III) ion: \[ \ce{[Fe(Cl)4]^{3-}} \]
02

b. Pentaammineaquaruthenium(III) ion

Here, the name of the complex ion has the following parts: 1. Central ion: ruthenium, Ru^3+ 2. Prefix penta indicates five ammine ligands: \( \ce{NH3} \) 3. One aqua ligand: \( \ce{H2O} \) 4. Oxidation state of central ion: +3 With this information, the formula for pentaammineaquaruthenium(III) ion can be written as: \[ \ce{[Ru(NH3)5(H2O)]^{3+}} \]
03

c. Tetracarbonyldihydroxochromium(III) ion

We can break down the name as follows: 1. Central ion: chromium, Cr^3+ 2. Prefix tetra indicates four carbonyl ligands: \( \ce{CO} \) 3. Prefix di indicates two hydroxo ligands: \( \ce{OH} \) 4. Oxidation state of central ion: +3 The formula for tetracarbonyldihydroxochromium(III) ion can then be written as: \[ \ce{[Cr(CO)4(OH)2]^{3-}} \]
04

d. Amminetrichloroplatinate(II) ion

Lastly, we break down the name of this complex ion: 1. Central ion: platinum, Pt^2+ 2. One ammonium ligand: \( \ce{NH3} \) 3. Prefix tri indicates three chloro ligands: Cl^- 4. Oxidation state of central ion: +2 So, the formula for amminetrichloroplatinate(II) ion is: \[ \ce{[Pt(NH3)Cl3]^{2-}} \]

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Most popular questions from this chapter

What is the lanthanide contraction? How does the lanthanide contraction affect the properties of the 4\(d\) and 5\(d\) transition metals?

Draw the \(d\) -orbital splitting diagrams for the octahedral complex ions of each of the following. a. \(\mathrm{Zn}^{2+}\) b. \(\mathrm{Co}^{2+}\) (high and low spin) c. \(\mathrm{Ti}^{3+}\)

\(\mathrm{CoCl}_{4}^{2-}\) 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.

Use standard reduction potentials to calculate $\mathscr{C}^{\circ}, \Delta G^{\circ},\( and \)K$ (at 298 K) for the reaction that is used in production of gold: $$2 \mathrm{Au}(\mathrm{CN})_{2}-(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}(a q)$$ The relevant half-reactions are $$\begin{aligned} \operatorname{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-0.60 \mathrm{V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-1.26 \mathrm{V} \end{aligned}$$

Sketch a d-orbital energy diagram for the following. a. a linear complex ion with ligands on the \(x\) axis b. a linear complex ion with ligands on the \(y\) axis
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