CHEM 481. Solid State. Assignment 4. Answers

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1 CHEM 481. Solid State. Assignment 4. Answers 1. Write a balanced chemical equation for the reaction of an alkali metal with a halogen. Use M to represent the metal and X to represent the halogen. Is the reaction likely to be exothermic or endothermic? Is the product ionic or covalent? M (s) + ½ X (l) MX (s) exothermic reaction, product is an ionic solid. Write a balanced chemical equation for the general reaction of an alkaline earth metal with oxygen. Use the symbol M to represent the alkaline earth element. Is the reaction likely to be exothermic or endothermic? Is the product ionic or covalent? M (s) + ½ O (l) MO (s) exothermic reaction, product is an ionic solid. Would you expect to find calcium naturally occurring in the earth s crust as a free element? Why or why not? Calcium is far too reactive (too easily oxidized) to survive in the environment in the metallic form. Remember the reaction done in the lab where the element is placed in cold water, and produces the hydroxide and hydrogen gas spontaneously. 4. Place the following oxides in order of increasing basicity: CO, SiO, and SnO. See page 49 of the notes; the tin oxide is amphoteric, whereas carbon and silicon oxides are acidic. A reasonable guess from their relative electronegativities is that SnO < SiO < CO. Note that none are strong acid precursors. 5. Complete and balance equations for the following reactions. (a) Li(s) + Cl (g) LiCl (s) (b) Ca(s) + O (g) CaO (s) 6. Write balanced chemical equations for the reaction of hydrogen gas with oxygen, chlorine, nitrogen. H (g) + ½ O (g) H O (l) H (g) + Cl (g) HCl (g) H (g) + N (g) NH (g) (under catalysis) 7. Write an equation for the reaction of sodium and hydrogen. Name the product. Is it ionic or covalent? Predict one physical property and one chemical property of this compound. Na (s) + ½ H NaH (s) sodium hydride is a salt-like hydride, so is ionic; it is a solid and reacts as a strong reducing agent with most oxidizing agents (typically with the re-formation hydrogen gas) 8. One of the pieces of evidence for the hydride ion in metal hydrides comes from electrochemistry. Predict the reactions that occur at each electrode when molten LiH is electrolyzed. Cathode: Li + (l) + e Li (l) Anode: H ½ H (g) + e 9. A method recently suggested for the preparation of hydrogen (and oxygen) proceeds as follows: (a) Sulfuric acid and hydrogen iodide are formed from sulfur dioxide, water, and iodine. (b) The sulfuric acid from the first step is decomposed by heat to water, sulfur dioxide, and oxygen. (c) The hydrogen iodide from the first step is decomposed with heat to hydrogen and iodine. Write a balanced equation for each of these steps and show that their sum is the decomposition of water to hydrogen and oxygen. SO (aq) + H O (l) + I (aq) H SO 4 (l) + HI (g) H SO 4 (l) H O (g) + SO (g) + ½ O (g) HI (g) H (g) + I (g). H O (l) H (g) + O (g) 1. Calcium reacts with hydrogen gas at elevated temperatures to form a hydride. This compound reacts readily with water, so it is an excellent drying agent for organic solvents. (a) Write a balanced equation showing the formation of calcium hydride from Ca and H. Ca (s) + H (g) CaH (s) (b) Write a balanced equation for the reaction of calcium hydride with water. CaH (s) + H O (l) Ca(OH) (aq) + H (g) 1

2 11. Write equations for the reaction of sodium with each of the halogens. Predict several physical properties that are common to all of the alkali metal halides. Na (s) + ½ F (g) NaF (s) Na (s) + ½ Cl (g) NaCl (s) Na (s) + ½ Br (g) NaBr(s) Na (s) + ½ I (g) NaI (s) All are colorless, brittle, ionic solids with intermediate melting points (several hundred, but less than a thousand degrees). All are soluble in water to give solutions of the constituent ions. All conduct electricity in the molten state. 1. Sodium peroxide is the primary product when sodium metal is burned in oxygen. Write a balanced equation for this reaction. Na (s) + O Na O (s) 1. (a) Write balanced equations for the reaction of lithium, sodium, and potassium with O. Specify which metal largely forms oxides, which one forms peroxides, and which one forms superoxides. Li (s) + ½ O (g) Li O (s) a normal oxide Na (s) + O Na O (s) a peroxide K (s) + O KO (s) a superoxide (b) Write Lewis structures for the peroxide and superoxide ion. What point group(s) do these ions belong to? - O O: - : : O O. both are linear, so D h peroxide superoxide 14. A piece of sodium catches on fire in the laboratory! How do you extinguish the fire? What is the worst thing you could do? Fire extinguishers for reactive metals act by smothering the flame. These so-called "class-d" extinguishers may contain either graphite or sodium chloride. The graphite produces a solid coating of metal carbide over the surface of the burning metal, thus smothering the fire. Sodium chloride melts at the temperatures involved, and the liquid forms an inert layer over the surface of the metal, again smothering the combustion. Sand is also useful. Never use water! 15. When magnesium burns in air, it forms both an oxide and a nitride. Write balanced equations for the formation of both compounds. Mg (s) + ½ O MgO (s) Mg (s) + N Mg N (s) 16. Name three uses of limestone. Write a balanced equation for the reaction of limestone with CO in water. Agricultural: to furnish calcium ions to plants and to neutralize acidic soils Bulding: lime (CaO) is used in mortar, and more importantly, in the manufacture of Portland cement Steel-making: In the basic oxygen process, lime (formed in situ from the carbonate) reacts with gangue (silicon dioxide) to form calcium silicate (slag) CaCO (s) + H O (l) + CO (aq) Ca(HCO ) (aq) 17. Explain what is meant by hard water. What causes hard water and what problems are associated with it? Hard water contains metal ions such as Ca + and Mg +. It is formed when ground water seeps through mineral beds of the slightly soluble carbonates of these metals (e.g. the Dolomite in the Rocky Mountains). These ions react with soap forming insoluble soap scum, thus reducing the efficiency of soaps. More problematically, they precipitate the carbonates when the water is heated, causing the plugging of hot water pipes and boilers by "boiler scale". 18. Calcium hydroxide, Ca(OH), has a Ksp of , whereas that for Mg(OH) is Calculate the equilibrium constant for the reaction Ca(OH) (s) + Mg + (aq) Ca + (aq) + Mg(OH) (s) and explain why this reaction can be used in the commercial isolation of magnesium from sea water. Ca(OH) (s) Ca + (aq) + OH (aq) Ksp = Mg + (aq) + OH (aq) Mg(OH) (s) 1/Ksp = 1( ) Ca(OH) (s) + Mg + (aq) Ca + (aq) + Mg(OH) (s) K rxn = Adding calcium hydroxide to sea water will displace magnesium ions by precipitation of magnesium hydroxide and (partial) olving of the calcium ions. The magnesium hydroxide can be recovered and used to produce magnesium. 19. Why is hydrogen bonding important? Give some examples. Hydrogen bonding is responsible for the lower density of the ice-i structure compared to liquid water. This ensures that ice floats, keeping most lakes and oceans from freezing solid when exposed to very cold air masses. If the ice accumulated at the bottom of the lake, in colder climates it would not completely thaw in a summer season, and then progressively more of the lake would freeze until all were frozen, killing all living organisms in the lake. Hydrogen bonding has many important functions in the regulation of biochemical processes. One famous example is the H- bonds that link the two halves of the DNA double helix. ph control of solutions containing living cells is essential, largely

3 because big changes in the hydronium ion concentration would dramatically affect the hydrogen bonding interactions in the cells, leading to cell death.. Describe the roles that the Group 1 metal play in living systems. See your lecture notes. 1. Discuss the biological properties of the Group elements. See your lecture notes.. In the standard notation of closest-packing of spheres, the letters A, B and C refer to close-packed layers. Which of the following sequences describe closest-packing in dimensions, and which do not. State your reasoning explicitly: (i) ABCABC... Close-packed CCP. (ii) ABACABA... Close-packed, but irregular (mixture of CCP and HCP. (iii) ABBABABBA... Not close packed - a "B" layer cannot lie directly over a "B" layer.. Draw one layer of close-packed spheres. On this layer mark the positions of the centers of the B layer atoms using the symbol and, with the symbol mark the positions of the centers of the C layer atoms of an FCC lattice. My graphic is shaded, i.e. white = ; light grey = 4. Determine the maximum size of (a) the tetrahedral (T d ) hole and (b) the octahedral (O h ) hole in a close-packed structure. Express your result as a decimal fraction of the radii of the close-packed spheres, r s. (Hints: it is valid to think of the T d hole as a small central atom of a methane-like molecule; for the O h hole, consider the four circles obtained by cutting a plane through the hole and the surrounding square array of spheres, i.e. treat is as FCC rather than ABC CP layers.) Consider the pictures at right which set up the trigonometric parameters needed to solve the problem. First look at the T d hole. For this system we can define a right triangle which is half the angle subtended from an apex of the tetrahedron to the center of the hole to another apex. Then: r r+ rh =, so r = r( 15. 1) ) =. 5r, the fraction of h sin( ) radius. For the O h hole the geometry is quite a bit simpler. Shown in the picture is half the face of an FCC unit cell. This right triangle can be solved by pythagorus law, so that: Geometry of T d hole Geometry of O h hole ( r+ rh ) = r + r = r = r so that rh =. 414r. 5. Calcium metal crystallizes in a face-centered cubic unit cell. The density of the solid is 1.54 g/cm. What is the radius of a calcium atom? An FCC unit cell contains 4 atoms of calcium (just by counting atoms at the corners, faces and edges). The radius of an FCC atom is ¼ the length of the face diagonal, and that is of the edge length. We get the edge length by using the density to get the volume of the unit cell, and then taking of the volume to get the edge length. This same principle can be used for any cubic unit cell, adjusting only for the internal geometry and atom count. m a = V = = d 4atoms 4. 78g / mol 154. g / cm 6. 1 atoms / mol 8 = cm = 5.57Å. Thus r = / Å = 1.97 Å.

4 6. Vanadium metal has a density of 6.11 g/cm. Assuming the vanadium atomic radius is 1. Å, is the vanadium unit cell simple cubic, body-centered cubic, or face-centered cubic? This is partly the reverse calculation to #4. Again we can calculate the edge length a, except that we do not know how many atoms to multiply the mass by. This requires us to modify the first equation: m g / mol a = V = no atoms = no atoms.. = no. atoms 4. Å. There are only three d 611. g / cm 6. 1 atoms / mol possibilities to consider: SC, 1 atom and r = a/; BCC, atoms and r = a/4; FCC, 4 atoms and r = a/4. If SC, a =.4, r = 1. Å, which is not right. If BCC, a = =. r = 1.1 Å, which is a good fit to the measured radius. If FCC, a = =.81, and r = 1.5 Å, which does not fit as well as the BCC. So its BCC! 7. Depending on temperature, RbCl can exist in either the rock-salt or cesium-chloride structure. (a) What is the coordination number of the anion and cation in each of these structures? (b) In which of these structures will Rb have the larger apparent radius? Compare your answer with the radii data collected by Shannon (Table on p.77 of the notes). The rock salt (NaCl) structure has CN = 6, while the CsCl has CN = 8. We can answer this by considering the ideal radius ratio's of the two structures, i.e. r + /r =.414 for rock salt and.7 for CsCl. Since the anion is the same, this means that the cation radius will "seem" larger in the CsCl structure type. The data tables have Rb: 1.66(6), 1.75(8), which fits. 8. Calculate the maximum fraction of available volume occupied by hard spheres on (a) the simple cubic, (b) the BCC, and (c) the FCC lattices. The general approach is the same in each case. First study the unit cell diagrams of the three cell types: "Sliced" view of Simple Cubic unit cell "Sliced" view of BCC "Sliced" view of FCC We know that the volume of one unit cell is given by V BOX = a for each cube. The volume of the spheres is given by 4πr /. If we can relate the radius to a, we can ratio the two volumes and calculate the fraction of volume occupied. For SC, the radius r = a/, and there is one sphere per unit cell. Thus V SPHERE = 4πa /( 8) =.5 a. For BCC, the radius r = a/4, and there are spheres per unit cell. Thus 4 a V SPHERES = π = 68. a. For FCC the radius r = a/4, and there are four atoms per unit cell. Thus 4 a VSPHERES = 4 4 π = 74. a. We see that the FCC = CCP structure is the densest of the three, and is in fact the most 4 dense possible way to pack equal sized spheres into -D space. It is almost 5% more efficient than SC, so no wonder that the latter is rarely seen for pure elements. On the other hand, if a suitably sized smaller sphere is added to fill the gaps between the large spheres, higher densities are possible. Thus for example, the ionic CsCl structure is based on a SC arrangement of anions (large) with somewhat smaller cation fitting into the big central gap. 9. The ReO structure is cubic with Re at each corner of the unit cell and one O atom on each unit cell edge midway between the Re atoms. Sketch this unit cell and determine (a) the coordination number of the cation and anion and (b) the identity of the structure type that would be generated if a cation were inserted in the center of the ReO structure. (a) The original structure is shown at right. To properly answer the question, you need to carefully consider the placement of the neighbouring unit cells. From this it is clear that the CN of the small grey spheres representing the Re atoms will be 6 (octahedral), while the coordination number of each oxygen is only two (linear). The stoichiometry of this unit cell is 8 1/8 = 1 Re ion; 1 ¼ = oxide ions, hence ReO. 4

5 ReO With Ba ion at center Conventional Perovskite setting (b) Adding a second cation at the center changes the formula to M O (8 corner plus one central cation = M ; still the same 1 edge anions for X.) Such structures are known, but are never observed when the two metal ions are of the same type. However when a different, and indeed larger ion, is placed in the center of the unit cell, we generate a unit cell of Perovskite. For example, BaZrO has this structure. The more conventional unit cell choice for a Perovskite is shown at far right. Imagine starting from our modified structure with the barium ion at the middle, and adding the required 6 oxygen ions to it. Compare the unit at the corner (the growth ) and notice that it is the same as the central Zr ion in the standard Perovskite. The central Ba + ion becomes then the new corner of a unit cell that has the eight barium ions at each of the eight corners. Many ionic structures can be described in two alternative forms like this. Another way to view such a composite structure is the view that the ZrO 6 units are octahedral packed around the large barium ions. Each octahedron consists of a central Zr ion with six oxygen ions associated, thus bearing an overall charge. There is then one Ba + ion per unit cell to compensate the [ZrO 6 ] unit, which being at the corner is also one per unit cell.. The metal hydride LiH has a density of.77 g/cm. The edge of the unit cell is 4.86 Å. If it is assumed that the H ions define the lattice points, does the compound have a face-centered cubic or a simple cubic unit cell? This problem, for an ionic solid, is solved much like #5, except that the LiH unit is counted as if it were the "atom" in the stoichiometric calculation. With this information supplied, a slightly shorter approach is to simply calculate the density from the crystallographic information, and compare our results with the known density. For a simple cubic, there is only one LiH per cell, so m 795. g / mol d = = 8 = 194. ; whereas for FCC there are four times as many units V ( cm) 6. 1 atom/ mol in the same volume, so the density works out to be four times as great, d = =.77 g/cm. 1. Confirm that in (a) rutile and (b) perovskite the stoichiometry is consistent with the structure. The unit cells are shown below: Rutile Perovskite Rutile: 8 Ti 1/8 th + 1 full Ti = Ti; 4 O ½ + full O = 4 O. Thus TiO. Perovskite: 1 Ti = Ti at centre; 8 Ca 1/8 th = 1 Ca; 6 O ½ = O. Thus CaTiO. 5

6 . How many cesium and chloride ions are there in a single unit cell of CsCl? How many zinc and sulfide ions are there in a single unit cell of zinc blende? CsCl has 1 Cs at centre; 8 1/8 th = 1 Cl ZnS has 4 Zn in centre, 8 1/8 th + 6 1/ = 4 S.. Born-Haber Cycles require a lot of practice to master, although it is a straightforward application of Hess Law of Heat Summation. Most of the data needed to set-up and solve Born-Haber cycles is contained within the problems. Additional data can be taken from standard data tables, either in SAL or a General Chemistry book. Make sure you can do these calculations! Be sure that you can construct coherent B-H cycles, as in the example provided for (b). The complete energy calculation formula is provided in each case, but you should derive each one from its respective cycle: a) Use the following data to calculate the lattice energy of CsCl. The enthalpy of formation of CsC1 is 44 kj/ mol. The enthalpy of sublimation of Cs is +78 kj/mol, and the first ionization energy of Cs is +75 kj/mol. The ociation energy of Cl is +4 kj/mol of Cl molecules and the first electron attachment enthalpy of Cl is 49 kj/mol of Cl atoms. V = = = kj / mol f subl IE EA b) Use the following data to calculate the lattice energy of CaO. The enthalpy of formation of CaO is 66 kj/ mol. The enthalpy of sublimation of Ca is +19 kj/mol, the first ionization energy of Ca is +59 kj/mol, and the second ionization energy of Ca is kj/mol. Ca + (g) + O - (g) V CaO (c) The enthalpy of ociation of O, is +494 kj/mol of O molecules, the first electron attachment enthalpy of O is 141 kj/mol of O atoms, and the second electron attachment enthalpy of O is +845 kj/mol of O nd IE Ca + (g) nd EA O - (g) ions. 1 s t I E 1st EA reverse of of formati V = Ca f subl 1stIE ndie 1stEA ndea (g) O (g) = = 514 kj / mol subl. ½ BE c) Use the following data to calculate the enthalpy of formation of Ca (s) + ½ O (g) Rb O. The enthalpy of sublimation of Rb is +8 kj/mol, and the first ionization energy of Rb is +4 kj/mol. The enthalpy of ociation of O is +494 kj/mol of O molecules, the first electron attachment enthalpy of O is 141 kj/mol of O atoms, and the second electron attachment enthalpy of O is +845 kj/mol of O - ions. The lattice energy, V, of Rb O is +5 kj/mol. V = =+ 5 f subl 1stIE 1stEA ndea + 5 = f ; f = 9 kj / mol d) Use the following data to calculate the enthalpy of formation of SrCl. The enthalpy of sublimation of Sr is +164 kj/mol, the first ionization energy of Sr is +549 kj/mol, and the second ionization energy of Sr is +164 kj/mol. The enthalpy of ociation of Cl is +4 kj/mol of Cl molecules, and the first electron attachment enthalpy of Cl is 49 kj/mol of Cl atoms. The lattice energy, V, of SrCl is +15 kj/mol. V = =+ 15 f subl 1stIE ndie 1stEA + 15 = ; = 88 kj / mol f f 4. Calculate the lattice energy of TlCl by (a) a thermochemical cycle and (b) using the Born-Mayer equation. (c) Discuss the relationship between these two numbers. Make use of all the concepts developed in the course which are relevant to this discussion. sublimation (Tl) = +18 kj mol 1 f (TlCl) = -4 kj mol 1 Ist I.E. (Tl) = +589 kj mol 1 bond ociation (Cl ) = +4 kj mol 1 electron attachment (Cl) = -49 kj mol 1 a) V H H H = = = kj / mol f subl IE EA b) To use the Born-Meyer equation, we need to know the inter-ionic distance, but also what value of the Madelung constant to use. In the absence of any better information, we must use the radius ratio rule to predict which structure to 6

7 use. We need to use the radii data from the table of ionic radii. When we don t know the correct structure type, a reasonable procedure is to first try the CN6 radii, such that: + r 164. = = 98. This predicts the CsCl structure. Ideally we should now check the answer using CN8 radii, but there r 167. is no data available for chloride with this radius, so we will assume it is indeed CsCl. The closest past noble gas configuration for thallium is Xe. For Cl it is Ar. We take the average of the n values of these two configurations, or ½ (9+1) = 1.5 V = Z Z A 189 A B AB kj mol r 1 1 = = n / (.. ) 1. 5 AB c) We see that the experimental lattice energy is 14% higher than the calculated value. This is a large discrepancy, and suggests additional covalent bonding beyond the ionic lattice forces. This is not terribly surprising, since Tl in Group 1 is a very soft acid, and chloride is a borderline base. So we expect a lot of covalency in their compounds. 5. Calculate the enthalpy of formation of the hypothetical compound KF assuming a fluorite structure. Use the Born-Meyer equation to obtain the lattice energy and estimate the radius of K + by extrapolation of the data in the table of ionic radii. Both SAL and K&T list the other required thermochemical data. Is the lattice energy favorable for this compound? Why then can it not exist? It is hypothetical, so we must use the Born-Meyer equation to estimate its lattice energy. This estimate is then used in a B- H cycle in order to estimate the enthalpy of formation, using the method outlined in 1(d) above. We find that the radius of fluoride is 1.17 for CN4, while that of K + is probably similar to Ca + with CN8 in the fluorite structure, thus 1.6 Å. Thus: V = Z Z A 189 A B AB kj mol r 1 1 = = AB n / So far this looks great. But we go on. (.. ) 8 V = =+ 5 f subl 1stIE ndie 1stEA. + 5 = f ; f = + 541kJ / mol The formation of KF is endothermic by over 5 kj/mol. Comparison to SrCl in 1(d) shows that the main contributor to this unfavorable heat of formation is the large size of the nd ionization energy. For potassium, this second ionization involves the opening of the core orbitals, i.e. those of the argon configuration. This costs a lot of energy, more than the lattice energy of a typical MX salt can supply. Thus KF cannot exist. 6. Which one of each of the following pairs of isostructural compounds is likely to undergo thermal decomposition at a lower temperature? Give your reasoning. (a) MgCO and CaCO (decompose to the metal oxide and carbon dioxide). The answer hinges on the relative size of the carbonate (reactant) and oxide (product) lattice energies. For Mg, the lattice energy with the big carbonate is going to be less favorable than for calcium, while for the oxide, the smaller Mg will be more stable than the larger calcium. Both effects contribute to making the magnesium carbonate the material which decomposes more easily on heating. (b) CsI and NMe 4 + I (decomposition products are MI + I ; tetramethyl ammonium is a much larger ion than Cs +. Again, the larger triiodide ion is more compatible with the large tetramethylammonium cation, while the smaller iodide ion in the product will give a bigger lattice energy with the smaller cesium ion. We predict that cesium triiodide will be the less thermally stable. 7. Which member of each pair is likely to be the more soluble in water: (a) SrSO 4 or MgSO 4 We predict a more stable lattice energy for strontium sulfate than for magnesium sulfate, since the larger Sr + is better matched to the large SO 4. Thus the strontium salt will be less soluble in water, and the magnesium more soluble. (b) NaF or NaBF 4? We predict a more stable lattice with the compatible-sized ions sodium and fluoride than between sodium and the large tetrafluoroborate anion. Thus the NaBF 4 is predicted to be more soluble in water. 7