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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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Key Concepts

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

Chemical Equations
Chemical equations are a way to represent the reactions that occur between different substances. They show how reactants are transformed into products, using chemical formulas for clarity. In our case, when mercury(II) nitrate reacts with potassium iodide, an orange precipitate forms. This precipitate is Mercury(II) iodide, and its formation can be described using a balanced chemical equation.

A balanced equation ensures that the same number of each type of atom is present on both sides of the equation. For example, when \[\mathrm{Hg}^{2+} (aq) + 2\mathrm{I}^{-}(aq) \rightarrow \mathrm{HgI}_{2}(s),\]two iodide ions neutralize one mercury(II) ion to form solid Mercury(II) iodide.

Complete balancing of chemical equations involves ensuring that both mass and charge are conserved. They help in predicting the quantities of products formed and reactants spent, making them essential tools in chemistry.
Ionic Reactions
Ionic reactions occur when ions in aqueous solutions combine to form new compounds, sometimes resulting in a precipitate. In this context, when KI is added to Hg(NO₃)₂, the ions involved include Hg²⁺, I⁻, K⁺, and NO₃⁻.

Initially, the Hg²⁺ ions react with I⁻ ions to form the orange solid, HgI₂:\[\mathrm{Hg}^{2+} (aq) + 2\mathrm{I}^{-}(aq) \rightarrow \mathrm{HgI}_{2}(s).\]
  • This is a precipitation reaction, where soluble ions form an insoluble solid.

Upon the continued addition of KI, this solid dissolves, forming a new ionic species:\[\mathrm{HgI}_{2}(s) + 2\mathrm{I}^{-}(aq) \rightarrow \mathrm{HgI}_{4}^{2-}(aq),\]where the Mercury(II) tetrakisiodide complex is formed.

During ionic reactions, spectator ions like K⁺ and NO₃⁻ remain dissolved and do not participate in the reaction or alter the precipitate formation.
Complex Ions
Complex ions are formed when a transition metal ion combines with several ligands, such as anions or neutral molecules, to create a more stable configuration. In this exercise, Hg²⁺ forms a complex ion, \(\mathrm{HgI}_{4}^{2-}\), by bonding with four iodide ions (I⁻).

  • In a complex ion like \(\mathrm{HgI}_{4}^{2-}\), the metal ion acts as a central atom.
  • Ligands, like the iodide ions here, donate electron pairs to the central metal ion.

Such complexes are often characterized by intense colors due to charge transfer mechanisms. In \(\mathrm{HgI}_{4}^{2-}\), the electron transfer from I⁻ to Hg²⁺ results in light absorption and emits the complementary orange color. This is due to a process where an electron transitions between orbitals, causing visible color in the solution.

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Most popular questions from this chapter

Ammonia and potassium iodide solutions are added to an aqueous solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} .\) A solid is isolated (compound A), and the following data are collected: i. When 0.105 g of compound A was strongly heated in excess \(\mathrm{O}_{2}, 0.0203 \mathrm{g} \mathrm{CrO}_{3}\) was formed. ii. In a second experiment it took 32.93 \(\mathrm{mL}\) of 0.100\(M \mathrm{HCl}\) to titrate completely the \(\mathrm{NH}_{3}\) present in \(0.341 \mathrm{g} \mathrm{com}-\) pound A. iii. Compound A was found to contain 73.53\(\%\) iodine by mass. iv. The freezing point of water was lowered by \(0.64^{\circ} \mathrm{C}\) when 0.601 \(\mathrm{g}\) compound A was dissolved in 10.00 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O}\left(K_{\mathrm{f}}=\right.\) \(1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol} )\) What is the formula of the compound? What is the structure of the complex ion present? (Hints: \(\mathrm{Cr}^{3+}\) is expected to be six-coordinate, with \(\mathrm{NH}_{3}\) and possibly I- - as ligands. The I- ions will be the counterions if needed.)

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