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Chapter 19: Problem 75

How could you determine experimentally whether the compound \(\mathrm{Ga}_{2} \mathrm{Cl}_{4}\) contains two gallium(II) ions or one gallium(I) and one gallium(III) ion? (Hint: Consider the electron configurations of the three possible ions.)

Short Answer

Expert verified
To determine experimentally whether the compound \(\mathrm{Ga}_{2}\mathrm{Cl}_{4}\) contains two gallium(II) ions or one gallium(I) and one gallium(III) ion, perform a redox reaction with potassium permanganate (\(\mathrm{KMnO}_{4}\)) in an acidic solution. Prepare a solution of the compound and add the potassium permanganate in a dilute sulfuric acid solution. If the deep purple color of the potassium permanganate disappears, it indicates the presence of gallium(II) ions, meaning the compound contains two gallium(II) ions. If the deep purple color remains unchanged, it suggests the compound consists of one gallium(I) and one gallium(III) ion.

Step by step solution

01

Determine the electron configurations of the possible ions

Gallium is found in group 13 of the periodic table and has an atomic number of 31. Its electron configuration can be represented as [Ar]4s²3d¹04p¹. To find the electron configurations of the three possible ions, we need to remove one, two, or three electrons from a neutral gallium atom. For gallium(I), one electron is removed: [Ar]4s²3d¹04p⁰. For gallium(II), two electrons are removed: [Ar]4s⁰3d¹04p¹. For gallium(III), three electrons are removed: [Ar]4s⁰3d⁹4p⁰.
02

Consider the reactivity of the ions

By looking at the electron configurations of the three possible ions, we can predict their reactivity. Gallium(II) has a partially filled 3d orbital which makes it less stable (and therefore more reactive) than gallium(I) and gallium(III), which have completely filled or empty orbitals. As a result, gallium(II) ions are more likely to undergo redox reactions.
03

Design an experiment to identify the ions

One possible way to experimentally determine the composition of the compound is to react it with a suitable reagent that distinguishes between gallium(I), gallium(II), and gallium(III) ions based on their reactivity. For this experiment, we can use a redox reagent like potassium permanganate (\(\mathrm{KMnO}_{4}\)) in an acidic solution. Potassium permanganate is a strong oxidizing agent which will react with gallium(II) ions to form gallium(III) ions, while the permanganate ions (\(\mathrm{MnO}_{4}^-\)) will be reduced to manganese(II) ions (\(\mathrm{Mn}^{2+}\)). This reaction can be monitored by the disappearance of the deep purple color of the potassium permanganate solution.
04

Carry out the experiment and analyze the results

To perform the experiment, prepare a solution of the compound \(\mathrm{Ga}_{2}\mathrm{Cl}_{4}\) and add potassium permanganate in an acidic solution (e.g., a dilute solution of sulfuric acid). If the deep purple color of potassium permanganate disappears, it indicates the presence of gallium(II) ions in the compound, which means the compound contains two gallium(II) ions. If the deep purple color remains unchanged, it suggests that the compound consists of one gallium(I) and one gallium(III) ion.

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Most popular questions from this chapter

Tin forms compounds in the +2 and +4 oxidation states. Therefore, when tin reacts with fluorine, two products are possible. Write balanced equations for the production of the two tin halide compounds and name them.

Write balanced equations describing the reaction of lithium metal with each of the following: \(\mathrm{O}_{2}, \mathrm{S}, \mathrm{Cl}_{2}, \mathrm{P}_{4}, \mathrm{H}_{2}, \mathrm{H}_{2} \mathrm{O},\) and HCl.

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While selenic acid has the formula \(\mathrm{H}_{2} \mathrm{SeO}_{4}\) and thus is directly related to sulfuric acid, telluric acid is best visualized as \(\mathrm{H}_{6} \mathrm{TeO}_{6}\) or \(\mathrm{Te}(\mathrm{OH})_{6}\) a. What is the oxidation state of tellurium in \(\operatorname{Te}(\mathrm{OH})_{6} ?\) b. Despite its structural differences with sulfuric and selenic acid, telluric acid is a diprotic acid with \(\mathrm{p} K_{\mathrm{a}_{1}}=7.68\) and \(\mathrm{p} K_{\mathrm{a}_{2}}=11.29 .\) Telluric acid can be prepared by hydrolysis of tellurium hexafluoride according to the equation $$\operatorname{TeF}_{6}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\operatorname{Te}(\mathrm{OH})_{6}(a q)+6 \mathrm{HF}(a q)$$ Tellurium hexafluoride can be prepared by the reaction of elemental tellurium with fluorine gas:$$\operatorname{Te}(s)+3 \mathrm{F}_{2}(g) \longrightarrow \operatorname{Te} \mathrm{F}_{6}(g)$$.If a cubic block of tellurium (density \(=6.240 \mathrm{g} / \mathrm{cm}^{3}\) ) measuring \(0.545 \mathrm{cm}\) on edge is allowed to react with 2.34 L fluorine gas at 1.06 atm and \(25^{\circ} \mathrm{C}\), what is the \(\mathrm{pH}\) of a solution of \(\mathrm{Te}(\mathrm{OH})_{6}\) formed by dissolving the isolated \(\operatorname{Te} \mathrm{F}_{6}(g)\) in \(115 \mathrm{mL}\) solution? Assume \(100 \%\) yield in all reactions.

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