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Chemistry 1000 (Fall 2013) Problem Set #4: Metals and Ionic Compounds Solutions Answers to Questions in Silberberg (only those w/out answers at the back of the book) 2.51 The major isotope(s) of tellurium must contain more neutrons than the major isotope(s) of iodine. 2.62 The total positive charge from the cations must be exactly equal in magnitude to the total negative charge from the anions. 2.81 We use roman numerals in the names of ionic compounds containing transition metal cations. 2.85 (a) (b) (c) CsBr BaS CaF2 cesium bromide barium sulfide calcium fluoride 8.3 Main group elements which are metals tend to form cations. These elements can be found toward the bottom left corner of the periodic table. (e.g. all Group 1 metals; Group 2 elements below Be; Group 13 elements below B; Sn; Pb) Main group elements which are nonmetals close to the right side of the periodic table tend to form anions. (e.g. all Group 17 elements, O, S, N) 8.16 Energy is released when cations and anions form ionic bonds. An ionic solid contains many ionic bonds. 8.19 While energy is required to form S2- anions from S atoms and electrons (and also to form K+ ions from K atoms, which release the electrons needed to form S2-), a large amount of energy will be released when ionic bonds are formed between the K+ and S2- ions. This energy released is greater than the energy required to form the initial ions. 8.21 (a) Cs+ [Kr] 5s2 4d10 5p6 (same as Xe) (b) O2[He] 2s2 2p6 (same as Ne) 22 6 2+ S [Ne] 3s 3p (same as Ar) Sr [Ar] 4s2 3d10 4p6 (same as Kr) Cs2S SrO (c) N3[He] 2s2 2p6 (same as Ne) (d) Br[Ne] 3s2 3p6 (same as Ar) 2+ 2 6 + Mg [He] 2s 2p (same as Ne) Li 1s2 (same as He) Mg3N2 LiBr 8.21 was assigned instead of 8.20 because 20d does not give the predicted ionic compound. Instead of Rb2O, you actually get RbO2 (from the Rb+ and O2- ions). This is discussed in Chapter 13. 13.16 The element in Group 1 has the largest atomic radius of all the elements in its period. As such, it will also have the largest atomic volume (since Vatom = 4/3 r3). It follows that the Group 1 element will require a greater volume of space per atom than any other element in the same period in the solid phase. The Group 1 element also has the smallest mass of all the elements in its period. Since density is equal to mass divided by volume, the element with the smallest atomic mass requiring the largest volume (per atom) will have the lowest density of all solids in the period. 13.23 Both Group 1 metals and Group 2 metals react with increasingly vigor with water from the top of the group to the bottom. A Group 2 metal, however, reacts less vigorously with water than the Group 1 metal in the same period. 13.24 (a) Li and Mg; Be and Al (b) Li and Mg Both metals react with N2 to give the corresponding nitride. (Li is the only Group 1 metal that does so.) Both metals react with O2 to give the corresponding oxide. (The other metals in Group 1 react with O2 to give more complex products.) Be and Al Both metals form oxoanions in strong base: [Be(OH)4]2– and [Al(OH)4]– Both have amphoteric oxides (BeO and Al2O3) Both form oxide coatings (passivation layers) so that they do not react with water (c) Li and Mg are similar in size – as are Li+ and Mg2+. Be2+ and Al3+ are similar in charge density, both small and highly charged. 13.106 E photon hc 6.626 10 34 J Hz 2.9979 10 10 nm 8 m 9 s 589.2nm 1m 4 sig. fig. E photon 3.371 1019 J This is the energy for a single photon (and the answer to this question, as phrased). To calculate the energy in J/mol, multiply by Avogadro’s number: E 3.371 1019 J photon 6.022141 10 23 photons mol J E 2.030 105 mol 13.107 Be(s) + 2OH–(aq) + 2 H2O(l) → [Be(OH)4]2–(aq) + H2(g) Zn(s) + 2OH–(aq) + 2 H2O(l) → [Zn(OH)4]2–(aq) + H2(g) 2 Al(s) + 2 OH–(aq) + 6 H2O(l) → 2 [Al(OH)4]–(aq) + 3 H2(g) 13.129 Step 1: Write a balanced chemical equation for this reaction Cl2(g) + 2 NaOH(aq) → NaOCl(aq) + NaCl(aq) + 2 H2O(l) Step 2: Calculate the mass of NaOCl to be produced First, you must calculate the mass of 1000. L of bleach solution. g 10001LmL 1.07 106 g msolution 1000.L 1.07 mL mNaOCl 5.25% 100% .25% 1.07 106 g 5.62 104 g msolution 5100 % Step 3: Calculate the moles of NaOCl to be produced nNaOCl 5.62 104 g 1mol 755mol 74.4419g Step 4: Calculate the moles of Cl2 to be consumed nCl 755molNaOCl 1molCl 2 755molCl 2 1molNaOCl Step 5: Calculate the volume of Cl2 to be consumed 2 PV nRT Pa m 755mol 8.3145 mol 1kPa 16.9m 3 nRT K 273.15K V 101.3kPa P 1000Pa 3 Note that 1 m3 = 1000 L, so this is 16,900 L (3 sig. fig.) of chlorine gas. 23.28 Addition of cryolite lowers the melting point of Al2O3. The electrolysis (which requires that the Al2O3 be in liquid state) can therefore be performed at a lower temperature. 23.54 (a) If they are not kept separate, the Cl2 and NaOH produced in the chlor-alkali process will react to give NaOCl, NaCl and H2O. Additional Practice Problems 1. Name each of the following compounds: (a) CuI copper(I) iodide (b) CuS copper(II) sulfide (c) BaF2 barium fluoride (d) NiO nickel(II) oxide 2. Give the formula for each of the following compounds: (a) iron (III) chloride FeCl3 (b) aluminium oxide Al2O3 (c) cobalt (III) oxide Co2O3 (d) magnesium fluoride MgF2 (e) lithium sulfide Li2S (f) copper(II) sulfide CuS 3. Suppose that you want to make 12 g of lithium oxide. What are the minimum masses of lithium and oxygen you will need? 4 Li(s) + O2(g) → 2 Li2O(s) 6.941 g/mol 31.9988 g/mol 29.8814 g/mol n Li2O 12 g n Li n Li2O 1mol 0.40mol 29.8814 g 4molLi 0.80mol 2molLi2 O m Li 0.80mol nO2 n Li2O 2 significant figures for all three answers! 6.941g 5.6 g 1mol 1molO2 0.20mol 2molLi2 O mO2 0.20mol 31.9988 g 6.4 g 1mol 4. Write a balanced equation for each of the following reactions and name all products: (a) Na and Br2 2 Na(s) + Br2(l) → 2 NaBr(s) sodium bromide (b) Li and N2 6 Li(s) + N2(g) → 2 Li3N(s) lithium nitride (c) Li and O2 4 Li(s) + O2(g) → 2 Li2O(s) lithium oxide (d) K and H2O 2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2(g) or 2 K(s) + 2 H2O(l) → 2 K+(aq) + 2 OH-(aq) + H2(g) potassium hydroxide and hydrogen gas 5. The group 13 metals react readily with oxygen. This would seem to imply that aluminium products should rust easily; however, they do not. Why don’t aluminium products rust? Pure aluminium reacts quickly with oxygen to form a thin layer of aluminium oxide on its surface. This layer of aluminium oxide does not react further with oxygen. As such, aluminium products do not rust. (This effect can be enhanced by “anodizing” the aluminium which deposits a slightly thicker film of aluminium oxide on the surface of the aluminium.) 6. (a) The bonding in BeCl2 is different from the bonding in MgCl2 or CaCl2. What is different about BeCl2 and why is this the case? Beryllium chloride is not an ionic compound (unlike MgCl2 or CaCl2). For an ionic compound to exist, both the cation and anion must be relatively stable. A beryllium cation (Be2+) would only have one shell of electrons (electron configuration 1s2). As such, it would be very small (smaller than helium!) – too small to fully stabilize a +2 charge. So, BeCl2 is a covalent compound while MgCl2 and CaCl2 are ionic. (b) This is not the only case of beryllium behaving differently from the other alkaline earth metals. Outside of Group 2, what element has the most similar chemical properties to beryllium? This is an example of what kind of relationship? The chemistry of beryllium and aluminium are often similar. This is an example of a diagonal relationship. 7. Write a balanced equation for each of the following reactions and name all products: (a) Ca and O2 2 Ca(s) + O2(g) → 2 CaO(s) calcium oxide (b) Mg and H2O Mg(s) + 2 H2O(l) → Mg(OH)2(s) + H2(g) magnesium hydroxide and hydrogen gas (c) Ga and Br2(l) 2 Ga(s) + 3 Br2(l) → 2 GaBr3(s) gallium bromide (d) CaCO3 and heat CaCO3(s) → CaO(s) + CO2(g) calcium oxide and carbon dioxide (e) MgCO3 and HCl(aq) MgCO3(s) + 2 HCl(aq) → MgCl2(aq) + CO2(g) + H2O(l) magnesium chloride and carbon dioxide and water or MgCO3(s) + 2 H+(aq) → Mg2+(aq) + CO2(g) + H2O(l) magnesium cation and carbon dioxide and water 8. (a) (b) What is hard water? Hard water is water containing Ca2+ and/or Mg2+ cations. It may also contain other cations with large positive charges. e.g. Fe3+ Explain how/why water becomes hard. Your answer should involve at least one balanced chemical equation. Water dissolves minerals such as MgCO3 or CaCO3 as it passes through limestone, etc. This is possible because CO2 from the air dissolves in water, making the water acidic: H2O(l) + CO2(g) H2CO3(aq) The acidic water dissolves carbonates, making soluble bicarbonates: CaCO3(s) + H2CO3(aq) Ca(HCO3)2(aq) (c) (d) 9. (a) (b) Why do deposits build up in pots and kettles which are regularly used to heat hard water? Heating water reduces the solubility of gases such as CO2. This makes the water less acidic and some of the soluble bicarbonates are converted back to carbonates and precipitate out. This is an example of Le Châtelier’s principle as removal of CO2(g) drives both equilibria shown in part (b) toward the reactant side. What can be done to avoid these deposits? Explain the underlying chemistry. Water can be softened. In other words, remove the Ca2+, Mg2+, etc. cations. One way to do this is to use an ion exchange column in which anionic beads are initially saturated with Na+ cations. When hard water passes through the ion exchange column, the Ca2+/Mg2+/etc. cations are more strongly attracted to the anionic beads than the Na+ cations were. So, the Ca2+/Mg2+/etc. cations adsorb to the beads and Na+ cations are released into the water. Since Na2CO3 is more soluble in water than CaCO3 or MgCO3, it does not precipitate the to the same extent when the water is boiled and the CO2 evaporates out of solution. Write a balanced chemical equation for the industrial production of sodium metal. Include all states of matter. 2NaCl(l ) 2Na(l ) Cl2(g) Write a balanced chemical equation for the industrial production of sodium hydroxide. Include all states of matter. 2NaCl(aq) 2H2O(l ) 2NaOH(aq) H2(g) Cl2(g) (c) (d) or 2Cl(aq) 2H2O(l ) 2OH(aq) H2(g) Cl2(g) These two industrial processes are based on the same method for making an otherwise unfavourable reaction occur. What is this method? electrolysis Despite the fact that the two processes have reagents in common and employ the same method, they lead to different products. Why? Water is present in the production of sodium hydroxide. When pure NaCl is electrolyzed the best electron acceptors are the Na+ cations, so neutral Na is produced. When water is present, the best electron acceptors are the H+ cations, so hydrogen gas is produced and the Na+ cations do not react.