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Chapter 21: Problem 1

You isolate a compound with the formula \(\mathrm{PtCl}_{4} \cdot 2 \mathrm{KCl}\). From electrical conductance tests of an aqueous solution of the compound, you find that three ions per formula unit are present, and you also notice that addition of \(\mathrm{AgNO}_{3}\) does not cause a precipitate. Give the formula for this compound that shows the complex ion present. Explain your findings. Name this compound.

Short Answer

Expert verified
The compound's formula showing the complex ion present is K₂[PtCl₄K₂], and its name is Potassium hexachloroplatinate(IV). This formula shows that all the chloride ions are included in the complex ion, satisfying the given criteria of three ions per formula unit and no precipitate with AgNO₃.

Step by step solution

01

List Possible Dissociation Reactions

We are given the compound's formula as PtCl₄·2KCl. In water, the possible dissociation reactions for this compound can be: 1. PtCl₄²⁻ + 2K⁺ 2. [PtCl₄K₂]³⁻ + 2K⁺ 3. [PtCl₂K₂]²⁻ + 2K⁺
02

Verify the Number of Ions per Formula Unit

Given that the compound dissociates into three ions per formula unit, we can compare the possible reactions from Step 1 and eliminate the reactions that do not meet this requirement. 1. PtCl₄²⁻ + 2K⁺ → 3 ions ➔ Correct 2. [PtCl₄K₂]³⁻ + 2K⁺ → 3 ions ➔ Correct 3. [PtCl₂K₂]²⁻ + 2K⁺ → 4 ions ➔ Incorrect Since the compound shouldn't cause a reaction with AgNO₃, this suggests silver chloride (AgCl) is not precipitated. To identify which dissociation reaction satisfies this condition, the complex ion must contain all the chloride ions.
03

Check the AgNO₃ Reaction

Comparing candidate reactions from Step 2, we need the one where all chloride ions are included in the complex ion and not free chloride ions are present. 1. PtCl₄²⁻ + 2K⁺ → Free chloride ions available, AgCl can precipitate ➔ Incorrect 2. [PtCl₄K₂]³⁻ + 2K⁺ → No free chloride ions available, AgCl won't precipitate ➔ Correct
04

Identify the Compound's Formula and Name

From the analysis above, we found that the dissociation reaction which agrees with the given criteria (three ions/formula unit and no precipitate with AgNO₃) is: [PtCl₄K₂]³⁻ + 2K⁺ Hence the compound's formula showing the complex ion present is K₂[PtCl₄K₂]. This compound can be named as Potassium hexachloroplatinate(IV).

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Most popular questions from this chapter

a. In the absorption spectrum of the complex ion [Cr(NCS) \(\left._{6}\right]^{3-}\), there is a band corresponding to the absorption of a photon of light with an energy of \(1.75 \times 10^{4} \mathrm{~cm}^{-1}\). Given \(1 \mathrm{~cm}^{-1}=\) \(1.986 \times 10^{-23} \mathrm{~J}\), what is the wavelength of this photon? b. The \(\mathrm{Cr}-\mathrm{N}-\mathrm{C}\) bond angle in \(\left[\mathrm{Cr}(\mathrm{NCS})_{6}\right]^{3-}\) is predicted to be \(180^{\circ}\). What is the hybridization of the \(\mathrm{N}\) atom in the \(\mathrm{NCS}^{-}\) ligand when a Lewis acid-base reaction occurs between \(\mathrm{Cr}^{3+}\) and \(\mathrm{NCS}^{-}\) that would give a \(180^{\circ} \mathrm{Cr}-\mathrm{N}-\mathrm{C}\) bond angle? \(\left[\mathrm{Cr}(\mathrm{NCS})_{6}\right]^{3-}\) undergoes substitution by ethylenediammine (en) according to the equation \(\left[\mathrm{Cr}(\mathrm{NCS})_{6}\right]^{3-}+2 \mathrm{en} \longrightarrow\left[\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}\right]^{+}+4 \mathrm{NCS}^{-}\) Does \(\left[\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}\right]^{+}\) exhibit geometric isomerism? Does \(\left[\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}\right]^{+}\) exhibit optical isomerism?

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.

Draw all geometrical and linkage isomers of square planar \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{SCN})_{2}\right]\).

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. \(\mathrm{Ti}^{3+}\)

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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