CHEM 481 Assignent Answers. 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 diensions, and which do not. State your reasoning explicitly: (i) CC... Close-packed CCP. (ii) ACA... Close-packed, but irregular (ixture of CCP and HCP. (iii) BBA... Not close packed - a "B" layer cannot lie directly over a "B" layer.. Draw one layer of close-packed spheres. On this layer ark the positions of the centers of the B layer atos using the sybol and, with the sybol ark the positions of the centers of the C layer atos of an FCC lattice. My graphic is shaded, i.e. white = ; light grey = 4. Deterine the axiu 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 decial fraction of the radii of the close-packed spheres, r s. (Hints: it is valid to think of the T d hole as a sall central ato of a ethane-like olecule; 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 C CP layers.) Consider the pictures at right which set up the trigonoetric paraeters needed to solve the proble. First look at the T d hole. For this syste we can define a right triangle which is half the angle subtended fro 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( 54. 5 ) radius. For the O h hole the geoetry is quite a bit sipler. Shown in the picture is half the face of an FCC unit cell. This right triangle can be solved by pythagorus law, so that: Geoetry of T d hole Geoetry of O h hole ( r+ rh ) = r + r = r = 1414. r so that rh =. 414r. 5. Calciu etal crystallizes in a face-centered cubic unit cell. The density of the solid is 1.54 g/c. What is the radius of a calciu ato? An FCC unit cell contains 4 atos of calciu (just by counting atos at the corners, faces and edges). The radius of an FCC ato 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 volue of the unit cell, and then taking of the volue to get the edge length. This sae principle can be used for any cubic unit cell, adjusting only for the internal geoetry and ato count. 4atos 4. 78g / ol a = V = 8 = = 557. 1 c = 5.57Å. Thus r = /4 5.57 Å = 1.97 Å. d 154. g / c 6. 1 atos / ol
6. Vanadiu etal has a density of 6.11 g/c. Assuing the vanadiu atoic radius is 1. Å, is the vanadiu unit cell siple 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 any atos to ultiply the ass by. This requires us to odify the first equation: 5. 9415g / ol a = V = no atos = no atos.. = no. atos 4. Å. There are only three d 611. g / c 6. 1 atos / ol possibilities to consider: SC, 1 ato and r = a/; BCC, atos and r = a/4; FCC, 4 atos and r = a/4. If SC, a =.4, r = 1. Å, which is not right. If BCC, a = 1.6.4 =. r = 1.1 Å, which is a good fit to the easured radius. If FCC, a = 1.59.4 =.81, and r = 1.5 Å, which does not fit as well as the BCC. So its BCC! 7. Depending on teperature, RbCl can exist in either the rock-salt or cesiu-chloride structure. (a) What is the coordination nuber of the anion and cation in each of these structures? (b) In which of these structures will Rb have the larger apparent radius? Copare 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 sae, this eans that the cation radius will "see" larger in the CsCl structure type. The data tables have Rb: 1.66(6), 1.75(8), which fits. 8. Calculate the axiu fraction of available volue occupied by hard spheres on (a) the siple cubic, (b) the BCC, and (c) the FCC lattices. The general approach is the sae in each case. First study the unit cell diagras of the three cell types: "Sliced" view of Siple Cubic unit cell "Sliced" view of BCC "Sliced" view of FCC We know that the volue of one unit cell is given by V BOX = a for each cube. The volue of the spheres is given by 4πr /. If we can relate the radius to a, we can ratio the two volues and calculate the fraction of volue 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 atos 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 ost 4 dense possible way to pack equal sized spheres into -D space. It is alost 5% ore efficient than SC, so no wonder that the latter is rarely seen for pure eleents. On the other hand, if a suitably sized saller sphere is added to fill the gaps between the large spheres, higher densities are possible. Thus for exaple, the ionic CsCl structure is based on a SC arrangeent of anions (large) with soewhat saller 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 ato on each unit cell edge idway between the Re atos. Sketch this unit cell and deterine (a) the coordination nuber 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 placeent of the neighbouring unit cells. Fro this it is clear that the CN of the sall grey spheres representing the Re atos will be 6 (octahedral), while the coordination nuber of each oxygen is only two (linear). The stoichioetry of this unit cell is 8 1/8 = 1 Re ion; 1 ¼ = oxide ions, hence ReO. 4
ReO With Ba ion at center Conventional Perovskite setting (b) Adding a second cation at the center changes the forula to M O (8 corner plus one central cation = M ; still the sae 1 edge anions for X.) Such structures are known, but are never observed when the two etal ions are of the sae 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 exaple, BaZrO has this structure. The ore conventional unit cell choice for a Perovskite is shown at far right. Iagine starting fro our odified structure with the bariu ion at the iddle, and adding the required 6 oxygen ions to it. Copare the unit at the corner (the growth ) and notice that it is the sae as the central Zr ion in the standard Perovskite. The central Ba + ion becoes then the new corner of a unit cell that has the eight bariu ions at each of the eight corners. Many ionic structures can be described in two alternative fors like this. Another way to view such a coposite structure is the view that the ZrO 6 units are octahedral packed around the large bariu 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 copensate the [ZrO 6 ] unit, which being at the corner is also one per unit cell.. The etal hydride LiH has a density of.77 g/c. The edge of the unit cell is 4.86 Å. If it is assued that the H ions define the lattice points, does the copound have a face-centered cubic or a siple cubic unit cell? This proble, for an ionic solid, is solved uch like #5, except that the LiH unit is counted as if it were the "ato" in the stoichioetric calculation. With this inforation supplied, a slightly shorter approach is to siply calculate the density fro the crystallographic inforation, and copare our results with the known density. For a siple cubic, there is only one LiH per cell, so 795. g / ol d = = 8 = 194. ; whereas for FCC there are four ties as any units V (. 4 86 1 c) 6. 1 ato/ ol in the sae volue, so the density works out to be four ties as great, d =.194 4 =.77 g/c. 1. Confir that in (a) rutile and (b) perovskite the stoichioetry 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
. How any cesiu and chloride ions are there in a single unit cell of CsCl? How any 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 aster, although it is a straightforward application of Hess Law of Heat Suation. Most of the data needed to set-up and solve Born-Haber cycles is contained within the probles. Additional data can be taken fro standard data tables, either in SAL or a General Cheistry book. Make sure you can do these calculations! Be sure that you can construct coherent B-H cycles, as in the exaple provided for (b). The coplete energy calculation forula is provided in each case, but you should derive each one fro its respective cycle: a) Use the following data to calculate the lattice energy of CsCl. The enthalpy of foration of CsC1 is 44 kj/ ol. The enthalpy of subliation of Cs is +78 kj/ol, and the first ionization energy of Cs is +75 kj/ol. The ociation energy of Cl is +4 kj/ol of Cl olecules and the first electron attachent enthalpy of Cl is 49 kj/ol of Cl atos. V = + + + + = + 44+ 78+ 75+ 4 49 =+ 668. 5 kj / ol f subl IE EA b) Use the following data to calculate the lattice energy of CaO. The enthalpy of foration of CaO is 66 kj/ ol. The enthalpy of subliation of Ca is +19 kj/ol, the first ionization energy of Ca is +59 kj/ol, and the second ionization energy of Ca is +1145 kj/ol. Ca + (g) + O - (g) V CaO (c) The enthalpy of ociation of O, is +494 kj/ol of O olecules, the first electron attachent enthalpy of O is 141 kj/ol of O atos, and the second electron attachent enthalpy of O is +845 kj/ol of O nd IE Ca + (g) nd EA O - (g) ions. 1 s t I E 1st EA reverse of of forati V = + + + + + + Ca f subl 1stIE ndie 1stEA ndea (g) O (g) =+ 66 + 19 + 59 + 1145+ 494 141+ 845 = 514 kj / ol subl. ½ BE c) Use the following data to calculate the enthalpy of foration of Ca (s) + ½ O (g) Rb O. The enthalpy of subliation of Rb is +8 kj/ol, and the first ionization energy of Rb is +4 kj/ol. The enthalpy of ociation of O is +494 kj/ol of O olecules, the first electron attachent enthalpy of O is 141 kj/ol of O atos, and the second electron attachent enthalpy of O is +845 kj/ol of O - ions. The lattice energy, V, of Rb O is +5 kj/ol. V = + + + + + =+ 5 f subl 1stIE 1stEA ndea + 5 = f + 8 + 4+ 494 141+ 845; f = 9 kj / ol d) Use the following data to calculate the enthalpy of foration of SrCl. The enthalpy of subliation of Sr is +164 kj/ol, the first ionization energy of Sr is +549 kj/ol, and the second ionization energy of Sr is +164 kj/ol. The enthalpy of ociation of Cl is +4 kj/ol of Cl olecules, and the first electron attachent enthalpy of Cl is 49 kj/ol of Cl atos. The lattice energy, V, of SrCl is +15 kj/ol. V = + + + + + =+ 15 f subl 1stIE ndie 1stEA + 15 = + 164 + 549 + 164 + 4 49; = 88 kj / ol f f 4. Calculate the lattice energy of TlCl by (a) a therocheical cycle and (b) using the Born-Mayer equation. (c) Discuss the relationship between these two nubers. Make use of all the concepts developed in the course which are relevant to this discussion. subliation (Tl) = +18 kj ol 1 f (TlCl) = -4 kj ol 1 Ist I.E. (Tl) = +589 kj ol 1 bond ociation (Cl ) = +4 kj ol 1 electron attachent (Cl) = -49 kj ol 1 a) V H H H = + + + + =+ 4+ 18 + 589 + 4 49 =+ 7475. kj / ol 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 inforation, we ust use the radius ratio rule to predict which structure to 6
use. We need to use the radii data fro 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 assue it is indeed CsCl. The closest past noble gas configuration for thalliu 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 kj ol r 1 1 = 17667 189 1 1 = n 17+ 167 1 1. 65 / (.. ) 1. 5 c) We see that the experiental 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 copounds. 5. Calculate the enthalpy of foration of the hypothetical copound KF assuing a fluorite structure. Use the Born-Meyer equation to obtain the lattice energy and estiate the radius of K + by extrapolation of the data in the table of ionic radii. Both SAL and K&T list the other required therocheical data. Is the lattice energy favorable for this copound? Why then can it not exist? It is hypothetical, so we ust use the Born-Meyer equation to estiate its lattice energy. This estiate is then used in a B- H cycle in order to estiate the enthalpy of foration, using the ethod outlined in 1(d) above. We find that the radius of fluoride is 1.17 for CN4, while that of K + is probably siilar to Ca + with CN8 in the fluorite structure, thus 1.6 Å. Thus: V = Z Z A 189 A B kj ol r 1 1 = 5199. 189 1 = n 117 + 16 1 1 5 / So far this looks great. But we go on. (.. ) 8 V = + + + + + =+ 5 f subl 1stIE ndie 1stEA. + 5 = f + 89 + 419 + 51 + 158 8; f = + 541kJ / ol The foration of KF is endotheric by over 5 kj/ol. Coparison to SrCl in 1(d) shows that the ain contributor to this unfavorable heat of foration is the large size of the nd ionization energy. For potassiu, this second ionization involves the opening of the core orbitals, i.e. those of the argon configuration. This costs a lot of energy, ore 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 copounds is likely to undergo theral decoposition at a lower teperature? Give your reasoning. (a) MgCO and CaCO (decopose to the etal 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 calciu, while for the oxide, the saller Mg will be ore stable than the larger calciu. Both effects contribute to aking the agnesiu carbonate the aterial which decoposes ore easily on heating. (b) CsI and NMe 4 + I (decoposition products are MI + I ; tetraethyl aoniu is a uch larger ion than Cs +. Again, the larger triiodide ion is ore copatible with the large tetraethylaoniu cation, while the saller iodide ion in the product will give a bigger lattice energy with the saller cesiu ion. We predict that cesiu triiodide will be the less therally stable. 7. Which eber of each pair is likely to be the ore soluble in water: (a) SrSO 4 or MgSO 4 We predict a ore stable lattice energy for strontiu sulfate than for agnesiu sulfate, since the larger Sr + is better atched to the large SO 4. Thus the strontiu salt will be less soluble in water, and the agnesiu ore soluble. (b) NaF or NaBF 4? We predict a ore stable lattice with the copatible-sized ions sodiu and fluoride than between sodiu and the large tetrafluoroborate anion. Thus the NaBF 4 is predicted to be ore soluble in water. 7