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Write balanced chemical equations to represent the following observations. (In some instances the complex involved has been discussed previously in the text.) (a) Solid silver chloride dissolves in an excess of aqueous ammonia. (b) The green complex \(\left[\mathrm{Cr}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}\), on treatment with water over a long time, converts to a brown-orange complex. Reaction of \(\mathrm{AgNO}_{3}\) with a solution of the product precipitates \(3 \mathrm{~mol}\) of \(\mathrm{AgCl}\) per mole of Cr present. (Write two chemical equations.) (c) When an \(\mathrm{NaOH}\) solution is added to a solution of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{a}\) precipitate forms. Addition of excess \(\mathrm{NaOH}\) solution causes the precipitate to dissolve. (Write two chemical equations.) (d) A pink solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) turns deep blue on addition of concentrated hydrochloric acid.

Short Answer

Expert verified
a) AgCl(s) + 2 NH₃(aq) → [Ag(NH₃)₂]⁺(aq) + Cl⁻(aq) b) 1. [Cr(en)₂Cl₂]Cl(s) + H₂O(l) → [Cr(en)₂Cl(OH)]Cl (Orange-brown complex) 2. [Cr(en)₂Cl(OH)]Cl + 3 AgNO₃ → Cr(en)₂(OH)Cl + 3 AgCl + 3 NO₃⁻ c) 1. Zn(NO₃)₂(aq) + 2 NaOH(aq) → Zn(OH)₂(s) + 2 NaNO₃(aq) 2. Zn(OH)₂(s) + 2 NaOH(aq) → Na₂[Zn(OH)₄](aq) d) Co(NO₃)₂(aq) + 6 HCl(aq) → [CoCl₆]⁴⁻(aq) + 2 NO₃⁻(aq) + 6 H⁺(aq)

Step by step solution

01

a) Silver chloride dissolving in aqueous ammonia

A solid silver chloride (AgCl) is dissolving in an excess of aqueous ammonia (NH₃). The silver chloride ionizes to form a complex ion with ammonia, known as diamminesilver(I) complex ion, [Ag(NH₃)₂]⁺. Ammonia acts as a ligand, forming a complex ion with the silver ion. The balanced chemical equation is as follows: AgCl(s) + 2 NH₃(aq) → [Ag(NH₃)₂]⁺(aq) + Cl⁻(aq)
02

b) Conversion of green complex to a brown-orange complex

The green complex [Cr(en)₂Cl₂]Cl is converted to a brown-orange complex by reacting with water over time. When treated with AgNO₃, the product precipitates 3 moles of AgCl per mole of Cr present. The first and second chemical equations are given below: 1. [Cr(en)₂Cl₂]Cl(s) + H₂O(l) → [Cr(en)₂Cl(OH)]Cl (Orange-brown complex) 2. [Cr(en)₂Cl(OH)]Cl + 3 AgNO₃ → Cr(en)₂(OH)Cl + 3 AgCl + 3 NO₃⁻
03

c) Precipitation of a product and then dissolution with excess NaOH

A sodium hydroxide (NaOH) solution is added to a solution of zinc nitrate, Zn(NO₃)₂. This reaction produces a precipitate, which dissolves when an excess of NaOH solution is added. The two chemical equations are written below: 1. Zn(NO₃)₂(aq) + 2 NaOH(aq) → Zn(OH)₂(s) + 2 NaNO₃(aq) (Formation of precipitate) 2. Zn(OH)₂(s) + 2 NaOH(aq) → Na₂[Zn(OH)₄](aq) (Dissolution of precipitate)
04

d) Color change of Co(NO₃)₂ solution on addition of concentrated HCl

A pink solution of cobalt nitrate, Co(NO₃)₂, turns deep blue upon the addition of concentrated hydrochloric acid (HCl). This color change is due to the formation of the hexachlorocobaltate complex ion, [CoCl₆]⁴⁻. The balanced chemical equation for this reaction is as follows: Co(NO₃)₂(aq) + 6 HCl(aq) → [CoCl₆]⁴⁻(aq) + 2 NO₃⁻(aq) + 6 H⁺(aq)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Complexation
Chemical complexation involves the combination of metal ions with ligands to form complex ions. Ligands are atoms, ions, or molecules that can donate a pair of electrons to form a coordinate bond with a central metal ion. These complex ions often exhibit unique properties such as distinct colors, solubility behaviors, and reactivities that differ from those of their individual components.

In the original exercise, the concept of chemical complexation is observed when solid silver chloride (AgCl) reacts with an aqueous ammonia solution. The ammonia molecules act as ligands, coordinating with the silver ions to form a soluble [Ag(NH3)2]+ complex. This is a classic example of how ligands can stabilize metal ions in solution, leading to increased solubility of an otherwise insoluble substance.
Precipitation Reactions
A precipitation reaction occurs when two soluble substances react to form an insoluble product, known as a precipitate. Such reactions are essential for understanding solubility rules and predicting which combinations of ionic compounds can lead to the formation of a solid.

In the exercise, we see a precipitation reaction when solutions of zinc nitrate, Zn(NO3)2, and sodium hydroxide, NaOH, are mixed, resulting in the formation of zinc hydroxide, Zn(OH)2. This insoluble substance falls out of solution as a precipitate. Conversely, excess NaOH can result in the re-dissolution of the precipitate, indicating the formation of a soluble complex ion or a change in the precipitate's overall charge. These steps demonstrate the reversibility of precipitation reactions under certain conditions and how a change in the chemical environment can affect solubility.
Ligand Exchange Reactions
Ligand exchange reactions involve the replacement of one or more ligands in a complex ion with other ligands. These reactions can significantly impact the properties of the complex, including its color, shape, and reactivity. This can be seen when additional ligands interact with the central metal ion, leading to a rearrangement of its coordination sphere.

This is exemplified in the given exercise when concentrated hydrochloric acid is added to a pink cobalt nitrate solution, Co(NO3)2, resulting in a deep blue complex. The chloride ions from the HCl displace the water molecules from the coordination sphere of cobalt, forming a [CoCl6]4− complex. Ligand exchange reactions like this are often employed in both analytical chemistry to test for the presence of certain metal ions and in various industrial applications.

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Most popular questions from this chapter

A manganese complex formed from a solution containing potassium bromide and oxalate ion is purified and analyzed. It contains \(10.0 \% \mathrm{Mn}, 28.6 \%\) potassium, \(8.8 \%\) carbon, and \(29.2 \%\) bromine by mass. The remainder of the compound is oxygen. An aqueous solution of the complex has about the same electrical conductivity as an equimolar solution of \(\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\). Write the formula of the compound, using brackets to denote the manganese and its coordination sphere.

(a) A compound with formula \(\mathrm{RuCl}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) is dissolved in water, forming a solution that is approximately the same color as the solid. Immediately after forming the solution, the addition of excess \(\mathrm{AgNO}_{3}(a q)\) forms \(2 \mathrm{~mol}\) of solid \(\mathrm{AgCl}\) per mole of complex. Write the formula for the compound, showing which ligands are likely to be present in the coordination sphere. (b) After a solution of \(\mathrm{RuCl}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) has stood for about a year, addition of \(\mathrm{AgNO}_{3}(a q)\) precipitates \(3 \mathrm{~mol}\) of \(\mathrm{AgCl}\) per mole of complex. What has happened in the ensuing time?

Two different compounds have the formulation \(\mathrm{CoBr}\left(\mathrm{SO}_{4}\right) \cdot 5 \mathrm{NH}_{3}\). Compound \(\mathrm{A}\) is dark violet, and compound \(\mathrm{B}\) is red-violet. When compound \(\mathrm{A}\) is treated with \(\mathrm{AgNO}_{3}(a q)\), no reaction occurs, whereas compound \(\mathrm{B}\) reacts with \(\mathrm{AgNO}_{3}(a q)\) to form a white precipitate. When compound \(\mathrm{A}\) is treated with \(\mathrm{BaCl}_{2}(a q)\), a white precipitate is formed, whereas compound \(\mathrm{B}\) has no reaction with \(\mathrm{BaCl}_{2}(a q)\). (a) Is Co in the same oxidation state in these complexes? (b) Explain the reactivity of compounds \(\mathrm{A}\) and \(\mathrm{B}\) with \(\mathrm{AgNO}_{3}(a q)\) and \(\mathrm{BaCl}_{2}(a q)\). (c) Are compounds \(A\) and \(B\) isomers of one another? If so, which category from Figure \(24.17\) best describes the isomerism observed for these complexes? (d) Would compounds \(\mathrm{A}\) and \(\mathrm{B}\) be expected to be strong electrolytes, weak electrolytes, or nonelectrolytes?

(a) What is the difference between Werner's concepts of primary valence and secondary valence? What terms do we now use for these concepts? (b) Why can the \(\mathrm{NH}_{3}\) molecule serve as a ligand but the \(\mathrm{BH}_{3}\) molecule cannot?

The complex \(\left[\mathrm{Ru}(\mathrm{EDTA})\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{-}\) undergoes substitution reactions with several ligands, replacing the water molecule with the ligand. \(\left[\mathrm{Ru}(\mathrm{EDTA})\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{-}+\mathrm{L} \longrightarrow[\mathrm{Ru}(\mathrm{EDTA}) \mathrm{L}]^{-}+\mathrm{H}_{2} \mathrm{O}\) The rate constants for several ligands are as follows: $$ \begin{array}{ll} \hline \text { Ligand, } \mathrm{L} & k\left(M^{-1} s^{-1}\right) \\ \hline \text { Pyridine } & 6.3 \times 10^{3} \\ \text { SCN }^{-} & 2.7 \times 10^{2} \\ \mathrm{CH}_{3} \mathrm{CN} & 3.0 \times 10 \\ \hline \end{array} $$ (a) One possible mechanism for this substitution reaction is that the water molecule dissociates from the complex in the rate-determining step, and then the ligand \(\mathrm{L}\) fills the void in a rapid second step. A second possible mechanism is that \(L\) approaches the complex, begins to form a new bond to the metal, and displaces the water molecule, all in a single concerted step. Which of these two mechanisms is more consistent with the data? Explain. (b) What do the results suggest about the relative basicities of the three ligands toward Ru(III)? (c) Assuming that the complexes are all low spin, how many unpaired electrons are in each?

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