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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

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.

The ferrate ion, \(\mathrm{FeO}_{4}^{2-}\) , is such a powerful oxidizing agent that in acidic solution, aqueous ammonia is reduced to elemental nitrogen along with the formation of the iron(III) ion. a. What is the oxidation state of iron in FeO \(_{4}^{2-},\) and what is the electron configuration of iron in this polyatomic ion? b. If 25.0 \(\mathrm{mL}\) of a 0.243 \(\mathrm{M} \mathrm{FeO}_{4}^{2-}\) solution is allowed to react with 55.0 \(\mathrm{mL}\) of 1.45 \(\mathrm{M}\) aqueous ammonia, what volume of nitrogen gas can form at \(25^{\circ} \mathrm{C}\) and 1.50 \(\mathrm{atm}\) ?

One of the classic methods for the determination of the manganese content in steel involves converting all the manganese to the deeply colored permanganate ion and then measuring the absorption of light. The steel is first dissolved in nitric acid, producing the manganese(II) ion and nitrogen dioxide gas. This solution is then reacted with an acidic solution containing periodate ion; the products are the permanganate and iodate ions. Write balanced chemical equations for both of these steps.

The complex ion \(\mathrm{PdCl}_{4}^{2-}\) is diamagnetic. Propose a structure for \(\mathrm{PdCl}_{4}^{2-}\).

The following statements discuss some coordination compounds. For each coordination compound, give the complex ion and the counterions, the electron configuration of the transition metal, and the geometry of the complex ion. a. \(\mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) is a compound used in novelty devices that predict rain. b. During the developing process of black-and-white film, silver bromide is removed from photographic film by the fixer. The major component of the fixer is sodium thiosul-fate. The equation for the reaction is: $$\operatorname{AgBr}(s)+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(a q) \longrightarrow \\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{Na}_{3}\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right](a q)+\mathrm{NaBr}(a q)$$ c. In the production of printed circuit boards for the electronics industry, a thin layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is: $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)$$ Assume these copper complex ions have tetrahedral geometry.

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