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Chapter 21: Problem 73

When aqueous KI is added gradually to mercury(II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. $\left(\text {Hint} : \mathrm{Hg}^{2+} \text { reacts with } \mathrm{I}^{-} \text { to form } \mathrm{HgI}_{4}^{2-} .\right)\( Would you expect \)\mathrm{Hg} \mathrm{I}_{4}^{2-}$ to form colored solutions? Explain.

Short Answer

Expert verified
The balanced equation for the formation of Mercury(II) iodide is: \[\mathrm{Hg}^{2+} (aq) + 2\mathrm{I}^{-}(aq) \rightarrow \mathrm{HgI}_{2}(s)\]. For the dissolution of the orange precipitate and the formation of Mercury(II) tetrakisiodide complex ion, the equation is: \[\mathrm{HgI}_{2}(s) + 2\mathrm{I}^{-}(aq) \rightarrow \mathrm{HgI}_{4}^{2-}(aq) + 2\mathrm{K}^{+}(aq) \]. Yes, HgI4^2- would be expected to form colored solutions due to a charge transfer process where an electron moves from the iodide ion to the Mercury(II) ion within the complex, resulting in the absorption of certain wavelengths of light and the appearance of color in the solution.

Step by step solution

01

1. Identify ions and charges involved in the reaction

The ions and their charges involved in this reaction are: - Potassium (K^+) - Iodide (I^-) - Mercury(II) (Hg^2+) - Nitrate (NO3^-)
02

2. Write the chemical equation for the formation of the orange precipitate

The formation of the orange precipitate involves a reaction between mercury(II) and iodide ions: \[\mathrm{Hg}^{2+} (aq) + 2\mathrm{I}^{-}(aq) \rightarrow \mathrm{HgI}_{2}(s)\] The balanced equation for the formation of the orange precipitate (Mercury(II) iodide) is given above.
03

3. Write the chemical equation for the dissolution of the orange precipitate

The dissolution of the orange precipitate involves a reaction between Mercury(II) iodide and iodide ions as per the hint: \[\mathrm{HgI}_{2}(s) + 2\mathrm{I}^{-}(aq) \rightarrow \mathrm{HgI}_{4}^{2-}(aq) + 2\mathrm{K}^{+}(aq) \] The balanced equation for the dissolution of the orange precipitate and the formation of Mercury(II) tetrakisiodide complex ion is given above.
04

4. Address whether HgI4^2- is expected to form colored solutions and provide an explanation

Yes, HgI4^2- would be expected to form colored solutions. The color in the solution arises because of the process called "charge transfer." In this case, an electron moves from the iodide ion (I^-) to the Mercury(II) ion (Hg^2+) within the complex, which results in the absorption of certain wavelengths of light. This absorbed light causes the appearance of color in the solutions, with the complementary color being observed. In this case, the orange color is the complementary color produced by the absorption of light by the HgI4^2- complex ion.

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Most popular questions from this chapter

Ethylenediaminetetraacetate (EDTA \(^{4-} )\) is used as a complexing agent in chemical analysis with the structure shown in Fig. \(21.7 .\) Solutions of EDTA \(^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The complex ion virtually prevents the heavy metal ions from reacting with biochemical systems. The reaction of EDTA \(^{4-}\) with \(\mathrm{Pb}^{2+}\) is $$\mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-(a q)} \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad K=1.1 \times 10^{18}$$ Consider a solution with 0.010 mol of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to 1.0 \(\mathrm{L}\) of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing 0.050$M \mathrm{Na}_{4} \mathrm{EDTA} .\( Does \)\mathrm{Pb}(\mathrm{OH})_{2}$ precipitate from this solution? $\left[K_{\mathrm{sp}} \text { for } \mathrm{Pb}(\mathrm{OH})_{2}=1.2 \times 10^{-15}\right]$

Consider aqueous solutions of the following coordination compounds: $\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{I}_{3}, \operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{4}, \mathrm{Na}_{2} \mathrm{Pt}_{6},$ and \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{3}\) . If aqueous AgNO \(_{3}\) is added to separate beakers containing solutions of each coordination compound, how many moles of AgI will precipitate per mole of transition metal present? Assume that each transition metal ion forms an octahedral complex.

Which of the following statement(s) is(are) true? a. The coordination number of a metal ion in an octahedral complex ion is 8. b. All tetrahedral complex ions are low-spin. c. The formula for triaquatriamminechromium(III) sulfate is $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}\left(\mathrm{NH}_{3}\right)_{3}\right]_{2}\left(\mathrm{SO}_{4}\right)_{3}$ d. The electron configuration of \(\mathrm{Hf}^{2+}\) is $[\mathrm{Xe}] 4 f^{12} 6 s^{2}$ e. Hemoglobin contains \(\mathrm{Fe}^{3+}\)

A certain first-row transition metal ion forms many different colored solutions. When four coordination compounds of this metal, each having the same coordination number, are dissolved in water, the colors of the solutions are red, yellow, green, and blue. Further experiments reveal that two of the complex ions are paramagnetic with four unpaired electrons and the other two are diamagnetic. What can be deduced from this information about the four coordination compounds?

The \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) ion is diamagnetic, but \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) is para-magnetic. Explain.
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