# Focus Kimia

## Prediction Or Preparatory Problems Practice IChO 2020 Istanbul Turkey Type 1 (Part 1)

Below we provide training materials to face the 52nd International Chemistry Olympiad (IChO) of 2020 which will be held in Istanbul Turkey.

Problem 1. A frog in a well
The energy levels of an electron in a one-dimensional box are given by:
in which h is the Planck’s constant, m is the mass of the electron, and L is the length of the box.
1)     The π electrons in a linear conjugated neutral molecule are treated as individual particles in a one-dimensional box. Assume that the π electrons are delocalized in the molecular length with the total number of N π electrons and their arrangement is governed by the principles of quantum mechanics.
a.   Derive the general expression for ∆ELUMO–HOMO when an electron is excited from the HOMO to the LUMO.
b.  Determine the wavelength λ of the absorption from the HOMO to the LUMO.
2)     Apply the model of π electrons in a one-dimensional box for three dye molecules with the following structures (see the structural formula). Assume that the π electrons are delocalized in the space between the two phenyl groups with the length L is approximately equal to (2k+1)(0.140) nm, in which k is the number of the double bonds
a.   Calculate the box length L (Å) for each of the dyes.
b.  Determine the wavelength λ (nm) of the absorption for the molecules of the investigated dyes.
3)     Recalculate the box length L (Å) for the three dye molecules, assuming that the π electrons are delocalized over the linear conjugated chain which is presented as a line plotted between the two phenyl groups (see the structural formula). The bond angle C–C–C is 120o and the average length of C–C bond is 0.140 nm.
4)     Give the following experimental data on the wavelength λ of absorption.
a.   Determine the box length L (Å) of the linear conjugated chain for each of the three investigated dyes.
b.  Tabulate the values of the box length L for the dyes calculated above by the three different methods, denoted as 1, 2, and 3. Choose the method which is the most fit to the experimental data.

## Questions and Problems About Reactions in Aqueous Solutions

Questions and Problems About Reactions in Aqueous Solutions

1.      Classify these reactions according to the types discussed in the chapter:
(a) Cl2 + 2OH- → Cl- + ClO- + H2O
(b) Ca2+ + CO32- → CaCO3
(c) NH3 + H+ → NH4+
(d) 2CCl4 + CrO42- → 2COCl2 + CrO2Cl2 + 2Cl-
(e) Ca + F2 → CaF2
(f) 2Li + H2 → 2LiH
(g) Ba(NO3)2 + Na2SO4 → 2NaNO3 + BaSO4
(h) CuO + H2 → Cu + H2O
(i) Zn + 2HCl → ZnCl2 + H2
(j) 2FeCl2 + Cl2 → 2FeCl3
2.      Someone gave you a colorless liquid. Describe three chemical tests you would perform on the liquid to show that it is water.
3.      You are given two colorless solutions, one containing NaCl and the other sucrose (C12H22O11). Suggest a chemical and a physical test that would distinguish between these two solutions. (Physical test: Only the NaCl solution would conduct electricity. Chemical test: Add AgNO3 solution. Only the NaCl solution would give AgCl precipitate.)
4.      Chlorine (Cl2) is used to purify drinking water. Too much chlorine is harmful to humans. The excess chlorine is often removed by treatment with sulfur dioxide (SO2). Balance the following equation that represents this procedure:
Cl2 + SO2 + H2O → Cl- + SO42- + H+
5.      Before aluminum was obtained by electrolytic reduction from its ore (Al2O3), the metal was produced by chemical reduction of AlCl3. Which metals would you use to reduce Al3+ to Al? (Mg, Na, Ca, Ba, K, or Li).
6.      Oxygen (O2) and carbon dioxide (CO2) are colorless and odorless gases. Suggest two chemical tests that would enable you to distinguish between them.
7.      Based on oxidation number, explain why carbon monoxide (CO) is ﬂ ammable but carbon dioxide (CO2) is not. (Oxidation number of C is +2 in CO and +4 (maximum) in CO2.)
8.      Which of these aqueous solutions would you expect to be the best conductor of electricity at 25°C? Explain your answer.
(a) 0.20 M NaCl
(b) 0.60 M CH3COOH
(c) 0.25 M HCl
(d) 0.20 M Mg(NO3)2
9.      A 5.00 x 102 mL sample of 2.00 M HCl solution is treated with 4.47 g of magnesium. Calculate the concentration of the acid solution after all the metal has reacted. Assume that the volume remains unchanged. (1.26 M)

## Precipitation Reactions in Aqueous Solution

One common type of reaction that occurs in aqueous solution is the precipitation reaction, which results in the formation of an insoluble product, or precipitate. A precipitate is an insoluble solid that separates from the solution. Precipitation reactions usually involve ionic compounds. For example, when an aqueous solution of lead(II) nitrate [Pb(NO3)2] is added to an aqueous solution of potassium iodide (KI), a yellow precipitate of lead iodide (PbI2) is  formed:

Pb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2KNO3(aq)

Potassium nitrate remains in solution. Figure 1 shows this reaction in progress. The preceding reaction is an example of a metathesis reaction (also called a double displacement reaction), a reaction that involves the exchange of parts between two compounds. (In this case, the compounds exchange the NO3- and I- ions.) As we will see, the precipitation reactions discussed in this chapter are examples of metathesis reactions.

Figure 1 Formation of yellow PbI2 precipitate as a solution of Pb(NO3)2 is added to a solution of KI.

Solubility
How can we predict whether a precipitate will form when a compound is added to a solution or when two solutions are mixed? It depends on the solubility of the solute, which is defned as the maximum amount of solute that will dissolve in a given quantity of solvent at a specif  c temperature. Chemists refer to substances as soluble, slightly soluble, or insoluble in a qualitative sense. A substance is said to be soluble if a fair amount of it visibly dissolves when added to water. If not, the substance is described as slightly soluble or insoluble. All ionic compounds are strong electrolytes, but they are not equally soluble. Table 1 classif es a number of common ionic compounds as soluble or insoluble. Keep in mind, however, that even insoluble compounds dissolve to a certain extent. Figure 2 shows several precipitates.

Table 1 Solubility Rules for Common Ionic Compounds in Water at 25˚C

Molecular Equations, Ionic Equations, and Net Ionic Equations
The equation describing the precipitation of lead iodide on page 100 is called a molecular equation because the formulas of the compounds are written as though all species existed as molecules or whole units. A molecular equation is useful because it identifes the reagents (that is, lead nitrate and potassium iodide). If we wanted to bring about this reaction in the laboratory, we would use the molecular equation. However, a molecular equation does not describe in detail what actually is happening in solution.

Figure 2 Appearance of several precipitates. From left to right: CdS, PbS, Ni(OH)2, Al(OH)3.

As pointed out earlier, when ionic compounds dissolve in water, they break apart into their component cations and anions. To be more realistic, the equations should show the dissociation of dissolved ionic compounds into ions. Therefore, returning to the reaction between potassium iodide and lead nitrate, we would write

## Explanation About Electrolytes versus Nonelectrolytes

All solutes that dissolve in water fit into one of two categories: electrolytes and nonelectrolytes. An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. A nonelectrolyte does not conduct electricity when dissolved in water. Figure 1 shows an easy and straightforward method of distinguishing between electrolytes and nonelectrolytes. A pair of platinum electrodes is immersed in a beaker of water. To light the bulb, electric current must ﬂow from one electrode to the other, thus completing the circuit. Pure water is a very poor conductor of electricity. However, if we add a small amount of sodium chloride (NaCl), the bulb will glow as soon as the salt dissolves in the water. Solid NaCl, an ionic compound, breaks up into Na+ and Cl- ions when it dissolves in water. The Na+ ions are attracted to the negative electrode and the Cl- ions to the positive electrode. This movement sets up an electrical current that is equivalent to the ﬂow of electrons along a metal wire. Because the NaCl solution conducts electricity, we say that NaCl is an electrolyte. Pure water contains very few ions, so it cannot conduct electricity. Comparing the lightbulb’s brightness for the same molar amounts of dissolved substances helps us distinguish between strong and weak electrolytes. A characteristic of strong electrolytes is that the solute is assumed to be 100 percent dissociated into ions in solution. (By dissociation we mean the breaking up of the compound into cations and anions.) Thus, we can represent sodium chloride dissolving in water as