Chem 263 Winter 2018 Problem Set #2 Due: February 16

Transcription

1 Chem 263 Winter 2018 Problem Set #2 Due: February Use size considerations to predict the crystal structures of PbF2, CoF2, and BeF2. Do your predictions agree with the actual structures of these three compounds? Why or why not? Use the radius ratio rules (rcation/ranion and compare to 1) to help predict the structures. The radii we have are: Radii found between the slides with the Shannon-Prewitt graphs or the ionic radii periodic table or you can look them up on Wikipedia on the Ionic radius page or found on Web Elements webpage Atom Ion Size CN: 4 Ion Size CN: 6 Ion Size CN: 8 F Å 1.19 Å Pb Å 1.43 Å Be Å 0.59 Å Co Å 0.79 Å low spin Å high spin 1.04 Å You should try the radius ratio rules for all possible CN s. I ll just use the 1.19 Å for Fluorine, since the ionic radii between CN:4 and CN: 6 are almost the same For PbF2, trying 1.19/1.33 = meaning CN: 8, cubic (always make the radius ratio <1, so flip the cation and anion radii if necessary) Or trying 1.19/1.43 = 0.83, still cubic For BeF2, 0.41/1.19 = 0.34 meaning CN: 4, tetrahedral or 0.496/1.19=0.417 which means CN: 6 octahedral (but barely, so the structure more likely has tetrahedral coordination) For CoF2, 0.72/1.19 = or 0.79/1.19=0.664 meaning CN: 6, octahedral or 0.885/1.19=0.744 or 1.04/1.19=0.874 meaning CN: 8, cubic. Actual structures are: PbF2 is fluorite which is cubic, so radius ratio rules agree. BeF2 occurs as cristobalite as well as other SiO2-analogue polymorphs such as α-quartz which is trigonal and has tetrahedral coordination so again they agree. CoF2 is rutile which has an octahedral coordination so they agree. 2. Consider the following thermochemical data for calcium and chlorine in their standard states: S = +201 kj/mol D = +242 kj/mol

2 IE1 = +590 kj/mol IE 2 = kj/mol EA = -349 kj/mol First, use the Born-Landé equation to estimate the lattice energies for CaCl and CaCl2. Specify and justify any assumptions you needed to make to do this. UU = ZZ +ZZ ee 2 NNNN rr ee (1 1 nn ) is the Born-Lande equation where Z+ and Z- are the absolute value of the charges on the cation and anion, respectively, e=1.6022x10-19 C, N=6.023x10 23 atoms/mole, A is Madelung s constant (dependent on the crystal structure), n is the Born-Exponent ACaCl = (treat CaCl as a rock salt structure since Ca + is very similar to K + ) A CaCl2 = (rutile structure) n = [Ar][Ar]=(9+9)/2=9 rca+ is not available, but you can estimate it by knowing that it should be smaller than the atomic radius of Ca (which is 1.97 Ǻ) and should be slightly larger than the radius of K + (1.52 Ǻ), so I m going to use 1.60 Ǻ. rca2+=1.14 Ǻ and rcl-=1.67 Ǻ re= 4πε0r0 = 1.112x10-10 C 2 /J. m * r0 and r0 for CaCl = = 3.27 Ǻ and r0 for CaCl2 is = 2.81 Ǻ So finally, and UU CCCCCCCC = (+1)(1)(1.6022xx10 19 ) 2 (6.023xx10 23 )(1.748) (1.112xx10 10 )(3.27xx ) 9 = JJ 1000 = kkkk/mmmmmm mmmmmm UU CCCCCCCC2 = (+2)(1)(1.6022xx10 19 ) 2 (6.023xx10 23 )(2.408) (1.112xx10 10 )(2.81xx10 10 ) = kkkk/mmmmmm Second, compare your Born-Landé estimates against values from the Kapustinskii equation. UU = VVZZ +ZZ rr cc +rr aa ( rr cc +rr aa ) is the Kapustinskii equation where V is the # of ions in the formula and rr cc, rr aa are the radii of the cation and anion respectively in Angstroms. So we have,

3 and we have UU CCCCCCCC = (2)(1)(1) = kkkk/mmmmmm 3.27 UU CCCCCCCC2 = (3)(2)(1) = kkkk/mmmmmm 2.81 The values agree very well. For CaCl they agree within less than 1% and for CaCl 2 they agree within about 6%. Third, calculate the heats of formation for CaCl and CaCl2. HH ff = SS + DD + IIII + EEEE + UU For this, I just used the calculated lattice energies found by the Kapustinskii equation. So for CaCl, we only need the first IP for Ca and only 1/2D for the single Cl atom and we get HH ff oooo CCCCCCCC = = kkkk/mmmmmm For CaCl2, we need both IP to get to Ca 2+, the full dissociation energy since we have 2 Cl atoms and we need to multiply the EA by 2 to account for both Cl atoms so we get HH ff oooo CCCCCCCC2 = = kkkk/mmmmmm Why are they so different? Based on these numbers, do you expect CaCl to be stable or to disproportionate to CaCl2 and Ca? The heats of formation are different because the lattice energy is more negative for the CaCl2 since the sum of the radii is smaller and the charges on the atoms are larger. Based on these numbers we d expect that CaCl is not stable and would disproportionate to CaCl2 and Ca. 3. Gersten Smith 3.10 parts a and b: aa 2 aa 3 bb 1 aa 1 (aa 2 aa 3 ), bb aa 3 aa 11 2 aa 1 (aa 2 aa 3 ), bb aa 1 aa aa 1 (aa 2 aa 3 ) So,

4 gg 1 ( aaaa 3 2 ( aa 3 2 ii + aa 2 aaaa ( ii kk) + (jj kk)) 2 aaaa jj) (aaaa 3 ( ii kk) (jj kk)) (aaaa 3 aaaa (jj) (ii)) ( aa2 cc 3 + aa2 cc ) Similarly, since the denominator stays the same we get And gg 2 (aaaa 3 2 gg 3 (aa2 3 4 (ii ii) + aa2 3 ( aaaa 3 aaaa (jj) (ii)) ( aa 3 2 ii + aa aaaa jj) (aaaa 3 (jj) (ii)) ( 3(jj) + (ii)) aa 3 aa 3 ii + 2π aa jj aaaa (kk ii) + (kk jj)) 2 ( 3 (jj) + ( ii)) ( aa2 cc 3 2 ) ( 3aa) aa 3 ( ii) + 2π aa jj 4 (ii jj) + aa2 3 4 aa2 (jj ii) + (jj jj)) 4 ( aa2 cc 3 2 ) ( 3 3 kk kk) cc 3 cc kk Part b) The reciprocal lattice of the hexagonal lattice is also a hexagonal lattice. The first Brillouin zone is the Wigner-Seitz cell of the reciprocal lattice. The first Brillouin zone is also a hexagonal prism. For a simple hexagonal Bravais lattice with lattice constants a and c, the reciprocal lattice is a simple hexagonal lattice of lattice constants 2π/c and 4ππ rotated by 30 relative to the direct lattice. Plan View Diagram aa 3 To construct the first Brillouin zone, take perpendicular bisectors and enclose

5 4. A & M 6.1 A (FCC) B (BCC) C (diamond) Θ sin 2 Θ ratio Θ sin 2 Θ ratio Θ sin 2 Θ ratio

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7 5. Using a computer, draw to scale the first five rings of the X-ray ring patterns produced by diffraction from powders of SC, FCC, BCC, and diamond crystal structures of lattice constant a = 5.0 Å. 6. The first four XRD reflections of an unknown cubic alkali halide occur at 2θ = 23.38, 27.07, 38.65, and (Cu Kα radiation). Assign d-spacings and Miller indices to these reflections and calculate the lattice constant. The density of the alkali halide is 2.80 g/cm 3 at room temperature. Identify the substance. To find the d-spacing, just use Bragg s Law and solve for d. nn λλ dd = 2 ssssssθ where n=1 by conventional XRD, λ=1.54 Ǻ which is the length of Cu Kα radiation 2Θ (degrees) Θ (degrees) d- spacing (Ǻ) d 2 Ratios of d 2 Whole integer ratios Miller indices (111) (200) (220) (311) 6.57 Lattice constant values (Ǻ)

8 By referring to the slides on systematic absences, we can see by looking at the whole integer ratios of the d-spacings that the cubic structure must be FCC, so now we can assign Miller indices based on the structure factors and knowing that for FCC h,k, and l must all be even or all odd. Since we know the structure is cubic, we can calculate the lattice constant (a) by the cubic d- spacing formula aa dd = h 2 + kk 2 + ll 2 The lattice constant values should all be relatively the same as long as the Miller indices were properly assigned to the correct d-spacing values. Now we can find out the substance by dddddddddddddd = mm aaaaaaaaaa = nnnn ( 1 ) VV cccccccc NN AA VV cccccccc Where n is the number of atoms per unit cell which we know to be 4 for an FCC lattice, A is the atomic mass which we will solve for, NA is Avogadro s number and Vcell is equal to since the unit cell is cubic (need to change angstroms to cm). gg 4 aaaaaaaaaa AA ( 2.80gg/cccc3 = mmmmmm ) 6.023xx10 23 aaaaaaaaaa/mmmmmm ( 1 ( cccc) 3) A = g/mol, so by some trial and error and by looking at the alkalis and halides we can see that the mass of RbCl is g/mol which is nearly the same as what we calculated. 7. Derive the simplified structure factor equation for (a) cubic SrTiO3 and (b) CaF2. For what combinations of hkl will systematic absences occur in these two crystals? The total structure factor is φφ kk = ff jj (KK)ee iikk dd jj jj By using the B-cell of the perovskite structure, we can describe SrTiO 3 as simple cubic with a 5- atom basis: Ti (0, 0, 0), Sr ( ½, ½, ½) and O ( ½, 0, 0), (0, ½, 0) and (0, 0, ½) and using KK π (hxx + kkyy + aa llzz)

9 φφ kk = ff TTTT ee ii(2ππ aa (hxx+kkyy+llzz)) (aa(0 x+0 y+0 z)) + ff SSSS ee ii(2ππ aa (hxx+kkyy+llzz)) (aa 1 2 x+1 2 y+1 2 z ) + ff OO [ee ii 2ππ aa (hxx+kkyy+llzz) aa 1 2 x+0 y+0 z + ee ii 2ππ aa (hxx+kkyy+llzz) aa 0 x+1 2 y+0 z + ee ii 2ππ aa (hxx+kkyy+llzz) aa 0 x+0 y+1 2 z Which simplifies down to φφ kk = ff TTTT + ff SSSS ee iiππ(h+kk+ll) + ff OO [ee iiππ(h) + ee iiππ(kk) + ee iiππ(kk) ], where ee iiππ = 1 Without knowing the form factors, no systematic absences will occur for SrTiO 3, but these situations will occur for various values of h, k, and l: All h, k, and l are odd: ff TTTT ff SSSS 3ff OO All h, k, and l are even: ff TTTT + ff SSSS + 3ff OO When 1 is odd/2 are even: ff TTTT ff SSSS + ff OO When 2 are odd/1 is even: ff TTTT + ff SSSS ff OO For CaF 2, you can also see this worked out in Ch. 5 of the Old West book (pg ). We have an FCC lattice of Ca and an 8-atom basis of F, so we should expect to have 12 terms in the structure factor equation. The basis positions are: Ca (0, 0, 0), ( ½, ½, 0), ( ½, 0, ½), (0, ½, ½) and F ( ¼, ¼, ¼), ( ¼, ¼, ¾), ( ¼, ¾, ¼), ( ¾, ¼, ¼), ( ¾, ¾, ¼), ( ¾, ¼, ¾), ( ¼, ¾, ¾), ( ¾, ¾, ¾) φφ kk = ff CCCC 1 + ee iiii(h+kk) + ee iiii(kk+ll) + ee iiii(h+ll) + ff FF [ee iiππ 2 (h+kk+ll) + ee iiππ 2 (h+kk+3ll) + ee iiππ 2 (h+3kk+ll) + ee iiππ 2 (3h+kk+ll) + ee iiππ 2 (3h+3kk+ll) + ee iiππ 2 (3h+kk+3ll) + ee iiππ 2 (h+3kk+3ll) + ee iiππ 2 (3h+3kk+3ll) ] We already know that for FCC, h, k, and l must all be even or all be odd to observe a reflection. Turns out, it s still true for the fluorite structure. The structure factor will be: All odd h,k, and l: 4ff CCCC For h, k, and l all even, with sum 2x an odd number: 4ff CCCC -8ff FF For h, k, and l all even, with sum 2x an even number: 4ff CCCC +8ff FF Otherwise = 0