1 Pb 1 Pb 8 corners 6 faces = 1+3 = 4 Pb ions 8 corners 2 faces 1 S 1 S 4 edges 1 internal

Transcription

1 1. (Fall 01) The unit cell for lead(ii) sulfide is shown on the right, both as a space-filling model and as a ball-and-stick model. Last year a computational study published in the Journal of Chemical Physics determined that (PbS) is the smallest stable unit that possesses both the same cubic structure and coordination number as the bulk crystal. The merican Institute of Physics issued a press release, and the website ChemistryTimes picked it up adding the movie-like title Baby Crystal is Born.. How many Pb + ions and S ions are contained in one unit cell? (Be sure to show your work.) 1 Pb 1 Pb 8 corners 6 faces = 1+ = 4 Pb ions 8 corners faces 1 S 1 S 1 edges 1 internal = +1 = 4 S ions 4 edges 1 internal B. How many unit cells would it take to form one of the (PbS) baby crystals described above? PbS 1 unit cell 1 (PbS) = 8 unit cells (PbS) 4 PbS C. The ionic radii of Pb + and S for this lattice are 1 pm and 170 pm, respectively. 1. Why is the sulfide ion larger than the lead(ii) ion? What kind of calculation could you do to support your answer? (You do not have to do that calculation here. Instead, describe the calculation you would do.) Even though Pb + has more electrons than S, it s effective nuclear charge must be sufficient greater. s a check, we could calculate the effective nuclear charge of the ions using Slater s rules. Determining the effective nuclear charge of Pb + and S, but that was not required for the exam, so to get a quick answer I used the Excel shielding and effective nuclear charge calculator available at this site: The effective nuclear charge calculated for one of the outer-most electrons in Pb + and S was 8.85 and 4.75, respectively. So the outer-most electrons in Pb + feel an effective nuclear charge almost double that of S, consistent with Pb + being the smaller ion.. Calculate the density of PbS(s). One side of the unit cell is (r Pb+ + r S ) = (1 pm pm) = 606 pm. So the volume of the unit cell is (606 pm) = pm. The unit cell contains four PbS formula units, which would give a mass of 4(m Pb+ + m S ) = 4(07. amu +.06 amu) = amu. Therefore, the density is amu. 10 pm 5 8 = amu/pm 1 6 Converting to more conventional units gives g 10 pm amu/pm = 7.14 g/cm amu 10 cm D. The density of lead(ii) sulfide was determined experimentally to be 7.60 g/cm. 1. How well does your calculated value from part C agree with the experimental value? Very well.

2 Unit I Homework Set, CHM 4610 page. Based on the experimental density, how many of the (PbS) baby crystals would be contained in one cubic centimeter of lead(ii) sulfide? 7.60 g 1 mol PbS units PbS units 1 (PbS) baby crystal 1 cm cm (07..06) g 1 mol PbS units PbS units 0 = baby crystals. (Fall 01) Hey Goldsby, did you make up that last question? The title Baby Crystal is Born is just a little too cute for a physicist. Does lead(ii) sulfide even exist? Good question. Let s take a look.. Draw a thermochemical cycle that would allow you to calculate the enthalpy of formation for lead(ii) sulfide; i.e. H for the reaction Pb(s) + 1 / 8 S 8 (s) PbS(s) Notice that the standard state for sulfur is a solid consisting of cyclic S 8 molecules where the sulfur atoms are connected by single bonds. The enthalpy of sublimation for S 8 (s) is 11.5 kj/mol, and the lattice energy of PbS is 400 kj/mol. Other information is given in the handout. Label each step in your cycle, and make sure that your cycle is based on information provided with the exam. B. Now generate an equation for the enthalpy of formation for lead(ii) sulfide, f H [PbS(s)] using the labels in your thermodynamic cycle, and then plug in the values and calculate the enthalpy of formation. H f [PbS(s)] = H atom [Pb(s)] + IE 1 (Pb) + IE (Pb) + 1 / 8 H sub [S 8 (s)] + 1 / 8 BE(S S) E 1 (S) E (S) + U(PbS) = / 8 (11.5) + 1 / 8 (68) 00 ( 640) + ( 400) = 400 kj/mol So, is PbS(s) stable compared to the elements from which it was formed? No?!? (Correct answer is yes.) I was not able to find an experimentally determined lattice energy for PbS(s). Calculated lattice energies ranged from 400 kj/mol to 00 kj/mol. Using the higher value would have given 400 kj/mol as the enthalpy of formation for PbS(s). The CRC Handbook gives kj/mol as the enthalpy of formation for PbS(s). So the actual lattice energy appears to be around 900 kj/mol. Electrostatic modeling of the lattice energy for PbS(s) may be more due to the polarizable ( soft ) ions having a significant of covalence in their bonding. We will consider hard-soft acid-base theory in Chapter 9.

3 Unit I Homework Set, CHM 4610 page. (Fall 01) ssign point groups for the following molecules: C v D h C s C h D d B M C 4v B M B D 4h C B M B D h D h C h

4 Unit I Homework Set, CHM 4610 page 4 6. (Fall 014) In class we used a thermodynamic cycle to consider the possibility that MgO(s) is actually composed of Mg and O ions (not Mg and O ions). We found that the additional energy required to remove a second electron from magnesium and give a second electron to oxygen was more than compensated for by the large lattice energy of an ionic compound composed of + and ions. So maybe it s sodium chloride that we ve gotten wrong maybe NaCl is actually composed of Na and Cl ions (not Na and Cl ions).. Make a reasonable estimate of the lattice energy of an ionic compound composed of Na and Cl ions, and explain the reasoning behind your estimate. Lattice energy is proportional to the product charges over the distance between the ions (U Z cat Z anion /d). Na will be significantly smaller than Na, but Cl will be significantly larger than Cl, so let s consider that a wash in terms of the distance between the ions (even through the unit cell will probably be different). But the product of the charges will be x = 4 times greater than that of Na Cl. So a lattice energy 4 times great than that of Na Cl, or 4 x kj/mol ( sig figs, at best) is probably not a horrible estimate. B. Sketch the thermodynamic cycle that would allow you to determine the enthalpy of formation for NaCl assuming it is composed of Na and Cl ions. Label each step in the cycle as to what process is occurring. C. Now use the lattice energy you estimated in part, the cycle you drew in part B, and the data provided in the information packet to calculate the enthalpy of formation for NaCl assuming it is composed of Na and Cl ions. H f (Na+Cl-) = H atom (Na) + IE 1 (Na) + IE (Na) + 1 / H BE (Cl ) + [-E 1 (Cl)] + [-E (Cl)] + [U 0 (Na + Cl - )] = 107 kj/mol kj/mol kj/mol + 1 / (4 kj/mol) + (-49 kj/mol) + [-(-1000* kj/mol)] + (-00 kj/mol) = 600 kj/mol *I had to make up E for Cl because I could not find an experimental value. It could be much larger in magnitude because the nd electron is added to a 4s orbital. D. Does your answer indicate that the salt Na Cl (s) is stable? If so, then why is NaCl(s) composed of Na and Cl ions? If not, than what process in the cycle is most responsible for Na Cl (s) being unstable? ssuming the values provided with the exam are correct, the second ionization energy for sodium is clearly the process most responsible for Na Cl (s) being unstable. no

5 Unit I Homework Set, CHM 4610 page 5 7. (Fall 014) You may have noticed when you were looking at the ionization energy data for sodium that there is a very large difference between IE 1 and IE, compared to the differences between IE and IE and between IE and IE 4. I suppose it s possible that the IE value in the table is incorrect. Or maybe it really is that much larger.. Let s consider the second possibility by using Slater s rules to estimate the second ionization energy for sodium. Be sure to show your work, and draw a box around your final answer. + Na : 1s s p s Na : 1s s p S = = 4.15 Z = Z - S = = 6.85 eff -18 Z eff -17 IE = E -E = 0 - n J = J = J 7 n The ionization energy calculated above is for one atom, but it is necessary to convert the units to kj/mol in order to compare the answer to the value in the table kj 6.0 x10.55 x10 J 7 = 1000 J mole 15,400 kj / mol B. Was the second ionization energy for sodium you calculated considerably larger than the first ionization energy? ccount for the difference you saw in terms of Slater s rules. Yes! It was over three times larger than the second ionization energy given in the table. It is instructive (but not required for the exam) to calculate the first ionization energy using Slater s rules. 6 1 Na : 1s s p s S = = 8.80 Z = Z - S = =.0-18 Z eff -18 IE = E - E = J n = J = J n kj 6.0 x x10 J = 706 kj / mol 1000 J mole which is around 50% higher than the first ionization energy of sodium given in the table (496 kj/mol). We expect IE (Na) to be significantly greater than IE 1 (Na) because removing the second electron cracks the Nobel gas core electrons. The huge difference between IE 1 (Na)and IE (Na) calculated using Slater s rules is due to the shielding factor for the s and p electrons being reduced from 0.85 to 0.5 in the respective calculations. eff

6 Unit I Homework Set, CHM 4610 page 6 8. (Fall 014) ssign point groups for the following molecules. D h D h l Cl 6 BCl T d D h SiCl 4 PCl 5 C 4v SCl 4 BrCl 5 D h C s C s C 1

7 Unit I Homework Set, CHM 4610 page 7 9. (Fall 014) Consider the trigonal bipyrimidal molecule shown on the right. The atoms around the central atom are the same; for example, the molecule might be PCl 5 shown on the previous page. The numbers are just to show where the atoms move following a symmetry operation.. Show the position of the labels after a C rotation about the x axis B. Now, starting with your answer to part (do not start with the original figure) show the position of the labels after a reflection through the xy plane C. What symmetry operation would take the labels directly from their original positions to the positions shown in your answer to part B? In other words, C,x followed by xy = what operation? xz (reflection through the xz plane) D. Do the operations in parts and B commute (i.e. does C,x followed by xy = xy followed by C,x ) for this point group? Show your work by repeating parts and B above, but in reverse order. xy Same structure as obtained in part B, so yes, C,x and xy commute C,x 1 5 4

8 Unit I Homework Set, CHM 4610 page (Fall 014) The unit cell for rutile, TiO, is shown on the right.. How many titanium ions are contained in one unit cell? Show your work. The titanium cations are Ti 4+, so these ions must be the smaller spheres. 8 corners x 1/8 ion/corner = 1 Ti 4+ 1 interior x 1 ion/interior = 1 Ti 4+ Ti 4+ ions total B. How many oxygen ions are contained in one unit cell? Show your work. 4 faces x 1/ ion/face = O - interior x 1 ion/interior = O - 4 O - ions total C. How many TiO formula units are contained in one unit cell? Ti 4+ ions and 4 O - ions TiO formula units D. The density of rutile is 4.g/cm. Calculate the volume of the unit cell in pm. One 100 cm = 10 1 pm. TiO Ti + O = (15.999) = g/mol mass density = volume g 1 mol TiO formula units - g mol formula units g 4. = = cm V cm V cm g - 10 pm V cm cm 4. g/ cm 10 cm E. It would be difficult to calculate the dimensions of the unit cell. Why? pm The unit cell is not a cube, so you cannot get the dimensions by simply taking the cube root. 11. (Fall 015) Orbitals and wave functions.. On the right, sketch the D cross section of a 4p orbital showing all nodal surfaces by dashed lines. Label the nodes surfaces as angular or radial. Indicate the sign of the regions with pluses (+) and minuses ( ), or by shading the negative regions. radial nodes angular node

9 B. Sketch the d z² orbital on the axes provided. Indicate the sign of the regions with pluses (+) and minuses ( ), or by shading the negative regions. You do not need to show any radial nodes. Unit I Homework Set, CHM 4610 page 9 z y x C. Which orbital in a zinc atom is higher in energy? (circle one) d or 4s Which orbital in a zinc atom is better at shielding electrons? d or 4s Which orbital has more radial nodes? d or 4s From which orbital is the electron removed when you ionize zinc? d or 4s Circle the following orbitals that are ungerade: (neither are ungerade) d or 4s Calculate the energy required to remove the last electron from zinc. In other words, calculate IE 0 for a zinc atom. Give your answer in kj/mol. IE 0 = E E 1 = 0 [ J (Z /n )] = J (0 /1 )] = J J (1 kj/1000j) = kj/mol Do you expect your answer to agree well with the experimental value? Explain why or why not in ten words or less. Yes, because Zn 9+ is a one-electron system.

10 Unit I Homework Set, CHM 4610 page (Fall 015) The associative rule holds for all groups, but not so for commutation. group is belian if all of the operations commute with each other, i.e. the order of the operations does not matter. Otherwise the group is said to be non-belian. Some point groups are belian; some are not. Let s take a closer look at the D 4h point group using the hypothetical compound trans-m 4 B. The symbols and B indicate different terminal atoms, and the numbers serve as labels so we can follow the operations. Remember: atoms move, but the axes and operations do not.. Label the second and third structures on the diagram below showing how the atoms move following each operation. B1 1 B 4 4 B 1 B1 What operation would achieve the transition from I to III in a single step? C rotation about the x = -y axis Using the structure on the right to show the point, line, or plane corresponding to that operation. C B. gain, label the second and third structures on the diagram below showing how the atoms, but this time the order of the operations is reversed. B 4 B B1 B1 Following the order in Part B, what operation would achieve the transition from I to III in a single step? C rotation about the x = -y axis Using the structure on the right to show the point, line, or plane corresponding to that operation. C C. Is D 4h belian or non-belian? non-belian, because the operations do not commute

11 Unit I Homework Set, CHM 4610 page (Fall 015) ssign point groups for the following molecules. O h D h D h C v D h C v C 4v C v D 4h

12 Unit I Homework Set, CHM 4610 page (Fall 015) Silver(I) chloride must have a pretty stable lattice because gcl(s) is one of the few exceptions to that rule that all chloride salts are soluble in water except.... Draw and label a thermodynamic cycle that would allow you to calculate the lattice energy, U 0, of gcl(s) from the enthalpy of formation and any other necessary parameters. One of the steps in your cycle is given below. Complete the cycle and label all of the steps. g (g) + Cl (g) IE 1 (g) H o atom(g) g(g) Cl(g) E 1 (Cl) H o BE(Cl Cl)/ U 0 [gcl(s)] g(s) + ½Cl (g) H o f[gcl(s)] gcl(s) B. Given ΔH f [gcl(s)] = 17 kj/mol and the other information contained in the handout, calculate the lattice energy for gcl(s). First write an equation using the labels from your thermodynamic cycle ΔH f [gcl(s)] = ΔH atom (g) + IE 1 (g) + ΔH BE (Cl-Cl)/ + [-E 1 (Cl)] + U 0 [gcl(s)] and then calculate the lattice energy, U 0, in kj/mol. U 0 [gcl(s)] = ΔH f [gcl(s)] {ΔH atom (g) + IE 1 (g) + ΔH BE (Cl-Cl)/ + [-E 1 (Cl)]} = -17 {( / +(-49)} = -915 kj/mol The lattice energy of gcl(s) is kj/mol. C. g + is larger than Na + but smaller than K +. Does the lattice energy you calculated in Part B make sense compared to the lattice energies of NaCl(s) and KCl(s)? Based on ion size, we expect the magnitude of the lattice to increase as the size of the cation decreases. So the magnitude of the lattice energies should go as follows: KCl < gcl < NaCl. The lattice energies for KCl and NaCl are -699 kj/mol and -788 kj/mol, respectively, but the magnitude of the lattice energy for gcl (-915 kj/mol) is larger than expected. D. Given that the ionic radii of g + and Cl are 11 pm and 181 pm, respectively, predict the structure of the gcl(s) unit cell. Show your work. 11/181 = 0.64, so the NaCl (ccp) or Nis (hcp) structure is expected.

13 15. (Fall 015) The unit cell for a well-known superconductor is shown below Unit I Homework Set, CHM 4610 page 1 Ba Y Ba. How many Ba ions are contained in one unit cell? Show your work. interior B. How many O ions are contained in one unit cell? Show your work. 1 edge x 1/4 edge + 8 face x 1/ face = + 4 = 7 C. How many Y ions are contained in one unit cell? Show your work. 1 interior D. How many Cu ions are contained in one unit cell? Show your work. 8 edge x 1/4 edge + 8 corner x 1/8 corner = + 1 = E. What is the empirical formula of the compound? YBa Cu O 7

14 15. (continued from previous page) Unit I Homework Set, CHM 4610 page 14 Write the empirical formula of the compound (your answer to Part E on the previous page) in the space below. YBa Cu O 7 F. t first glance, it might see that it would be impossible to determine the dimensions of the unit cell from ionic radii; however, with a little geometry, you should be able to estimate some reasonable values. Using the ionic radii provided from Table 4.4 in Huheey 4e, estimate the dimensions of the unit cell and calculate the volume. Several approaches, here s one: Estimate the volume by assuming the three sections of the unit cell are cubes. For the top and bottom sections, the sides would be r Cu(+) + r O(-). V top section = V bottom section = (r Cu(+) + r O(-) ) = [(87 pm) + (11 pm)] = 7. x 10 7 pm For the middle section, the closest cation-anion contact is between the edge oxygens and the central yttrium ion. So s + s = diagonal = r Y(+) + r O(-). s = [r Y(+) + r O(-) ] => s = [r Y(+) + r O(-) ] = 1.414(115.9 pm + 11 pm) = 5 pm V middle section = s = (5 pm) =.8 x 10 7 pm V unit cell = V top section + V middle section + V bottom section = 7. x 10 7 pm +.8 x 10 7 pm + 7. x 10 7 pm = 1.8 x 10 8 pm x (10 cm/10 1 pm) = 1.8 x 10 - cm G. Now calculate the density. YBa Cu O 7 => 666 g/mol x (1 mol / 6.0 x 10 unit cells) = 1.1 x 10-1 g 1.1 x 10-1 g / 1.8 x 10 - cm = 6.1 g/cm (pretty good agreement with 6. g/cm from Wikipedia*) *