University:
Amherst College
Course:
CHEM 151 | Introductory Chemistry
Academic year:

2005
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426

Pages:

19
Author:

Preston Ortiz

Fall 2005 Chemistry 11 Name: Solutions Section (circle): 1 2 3 4 5 Exam 3 December 21, 2005 Good morning! This exam packet should contain 19 pages, including 2 pages of reference information and a periodic table. Please write your answers in the spaces provided. If you use the backs of pages, please clearly mark the question number by your work. If you wish, you can remove the reference pages since these pages do not need to be handed in. - There are six multi part questions on this exam. The point values of each question are indicated. Please note that some parts of each question are longer and/or more challenging and thus have greater point values than other parts. It is recommended that you read through the entire exam before commencing work, and then start with the questions that you find most accessible. Do not waste time frustrating yourself over part of one question at the expense of moving on to another. You should have enough time to return to the parts that you find most challenging. As always, your reasoning and the process used in arriving at your answer are the most important aspects to be communicated. Be sure to articulate clearly your thinking as you work through Such question. Full credit cannot be given unless you show all work or reasoning. Pay attention to significant figures when recording your answers. Relax! You have prepared well and you have learned a lot this semester. This is an opportunity to show off the results.of all your hard work. You will have three hours to complete this exam. If you finish early, leave your completed exam on the front desk. Remember: it is a violation of the Statement of Intellectual Responsibility at Amherst College to either give or receive help on this exam. The work you turn in must be yours and yours alone. Good luck! Question Points I 20 II V VI 20 20 20 25 30 Total 135 III IV -1- Score Fall 2005 Chemistry 11 Question I. (20 points) In theory, a diatomic molecule can be formed from one atom of hydrogen and one atom of helium, giving HeH. The ionization energies of hydrogen and helium are 1312 ktf /mol and 2372 kJ/mol, respectively. ' 1. Draw an energy level diagram for the molecular orbitals in the HeH molecule. Use arrows to show how the electrons will fill the MOs. HHe has 3e ~ Is . S closed +0 the, tk ( $ atomic, orhi fa ( energy the. charge distribution based on Coulombc This ' S c &nsisfeni fans ih , h nucleus inkr&chovz Vc He Hat more pro " MO, Mould be fourj neat' 4. What charge on HeH, magnitude and sign, would produce the most stable molecule or ion possible? Explain. What would the bond order be in this molecule or ion? r1 BA ^ , Mod Mk tan h/c u i+L ft - 1 t/ jooid be. I 15 H*e highest you com oyet . l * uf Hk l-lef - ' 3 conft jroibom ^ ^ ) -2- I connbnahdns Fall 2005 Chemistry 11 + ~ 5. Rank the following in order of ionization energy, from lowest to highest: He, HefT, H , He , HeH, H. If there are two or more species that cannot be ordered, indicate as much. Briefly justify your choices. — V is - rFT^ -f On -bkc basis f , / alwAy <> beiWj taster trt> po1! ar\ elecfrtrt ^Oty v < He < 47H+ Me 4 < H from somethin* fa! ts 'Wj neyt cUuryed . 0T s lsU Xn jeheral , Hits rankira is constz -knl i or Me energy teve/ diayon pOsAnie &r e , r>e tomyrhoh luriry*""' 1« pa ** , The hl , ' tors J ll always , heuiral speciesfor be > % ^ cr - ^-hbar^, for * f ' nega-f' ^ e- ' oos . 6. On the axes below, draw potential energy curves, that is, the bond energy of the molecule as a function of distance between the nuclei, for the following molecules and molecular ions: H2, HefT, HeH, HetT. Keep in mind how properties such as bond strength and bond length might vary among these species. Adesin' lr (or m \ \ We 2 " I tS tkU*Intemuclear Distance -3- oi - cova l-er 4 molecolC 80 0 ' Chemistry 11 Fall 2005 7. Ultimately, all ‘orbitals’ are wavefunctions (or approximations of wavefunctions), which in turn are solutions to the Schrodinger Equation. a. Discuss reasons why we adopt models, such as the localized electron pair model or the molecular orbital model, to describe bonding in molecules, rather than solving the SchrOdinger Equation for the molecule. orh ds and TO Visaah muck easier | p -pricked -to shape and cUar < 3^ _ 1V/-/L ^ 4 localised LPER , b. When solving the Schrodinger Equation for a molecule, one approximation that is often made is that the nuclei are fixed (i.e. don’t move) so that only the motion of the electrons must be considered. Assuming this approximation is valid, and drawing on your knowledge of atoms, for what systems would you predict it is possible to solve such a molecular Schrddinger Equation exactly? Explain. fjoh coles tokicL easier IOLO or oje more T ( yd ( hither bond Ord-er ] are. ' dkosc wkt'cL Are " - flippy 0 - framed tonal bond orders . fo solve.’HaOin .4. or 4 ^ - Sx :. Fall 2005 Chemistry 11 . 24 7 . Question II (20 points) ^^ > , ^ Three salts (A, B, and C) contain octahedral complexes of chromium and have the molecular 3 VS3 formula CrCl3 *6H20. : 5/ f Hofa £r 3 ^ i 3 J 3 1. Draw an energy level diagram for the splitting of the d orbitals in an octahedral field, being sure to label the orbitals. Then, distribute the proper number of electrons for the chromium ion in this complex among the energy levels shown in your drawing. n * > «3 A N . _L J_ f _ L_ o/ y5 c/vy 2. Use the following data to determine how many water molecules are bonded directly to the metal in each complex. Heating 0.270 g of each metal salt results in a mass loss of 0.036, 0.018 g and 0.000 g for salts A, B, and C respectively. M < ®4 HoLis coo mb VIATZZ. &' A A c? >5 — . 0& 7 * mderailo * o ig. QO 0 2 2 r: , 000 / o. ooo ^ 6- I / 0 - =5 0 3. Use the following data to determine how many chlorides are bonded directly to the metal in each complex. Addition of aqueous silver nitrate to 100.0 ml portions of 0.100 M solutions of each salt results in the formation of different masses of silver chloride. Solution A yields 1430 mg, solution B 2870 mg, and solution C 4300 mg. 0 - iOOH iComl - 0 - 0 / 00 males fHd/e 'h 2 Oi /ridles A c mllO lomcally ~ Bonocfi males 0 Complex # * - **M M = ? noofm.323 # * -5- , 03 moles ^ z/ l ~ Qz mole I z /t - s 3 Ci - -z 3 1 I 5's - O 0 /4 Fall 2005 Chemistry 11 4. Draw the three octahedral complexes for salts A, B, and C. a - ^ scr 0 1 C*T H2O f3 + 2- +1 M ' Cr i OH icr ^ £> A 5. Fill in the blank: These three salts are an example of transition metal complexes. Coordination isomerism in 6. In the space below, draw any geometrical (G) or linkage (L) isomers of the structures above that may exist; indicate the type with a (G) or (L). CL a * a4 ^ as\ d 0 O^ l «,0 £ torts (6 ) Qj 0 : no g, 0/ oz I SO, ) LG = j xoHi Cr ! 50 MZH> r i o ( L ) Hornets 04 eis L & &) I same u L t sowers 7. When in solution, the solutions have absorptions at 570 nm, 575 nm, and 620 nm. Identify which complex is which in solution and explain how you made this assignment. of crysf &l field sPhtf / UIQ ~7 C£ ~ St * 5 70 n />o " Ia mssf zphf / m f i strati StforrM 50k° ) 575 lyorJ - £ (PZO ^ Ojc^ d 1 K /K 5 fY )G lU % ( splitffhSj -6 - wrotesd fi ri ^ i ci ci ^ Fall 2005 Chemistry 11 Qnestkra IIL (20 points) 1. Draw an energy level diagram for die molecular orbitals of CO. The diagram should show the energies of the MOs relative to the atomic orbitals and the electron configuration of the molecule 10 veteUfons ; % orloi'hU . d0 - , ir* % 1 ls 9 ^ P H youdhxd / ' ' . ^ ^ -7- Mvndj \ Fall 2005 Chemistry 11 5. Assume the x-axis is the molecule’s intemuclear axis. For each of the o-bonding and antibonding MOs in the molecule, rank the contribution of the four orbitals that can form retype MOs (C 2s, C 2px, O 2s, O 2p*) from the highest to lowest contribution to that MO. If there is no sp-mixing, then indicate ‘no contribution’ where appropriate. a. d2s b. 0*2s C. 0 2p d. 0*2p Ozs > CiS ^ ^ ^ ^ ^ Pip ycs o ^^ ^ C 2S * > 02 p ^ c2 p 2s 2p 2p 2 2i 25 6. Explain why bonding (or anti-bonding) molecular orbitals cannot be constructed from overlap of the 2s orbital on C and the 2py orbital on O. A sketch might be helpful here. - ^ poW^ - c \ comperefr ! BomdG/ te' f PfO. ta -s mo ( o / e HO does shoW charge (excess eUc4 { Cs\ deisxsrfy) Or> 4ir , Molecule -8- Fall 2005 Chemistry 11 . Question IV (20 points) Shown below is the amino acid histidine. 1. First, complete the drawing by inserting all of the implicit H atoms and draw in the lone pairs. 2 2. Next, indicate on the drawing sp hybridized N and C 3 2 atoms with att and sp hybridized N and C atoms Histtdme - 3. There are four carbon carbon bonds. Give the bond order for each and indicate whether or not there would be a barrier to rotation about that bond. ; - 60 - 2 Ves barrier loarfi r 0 ; BO -1- ^ ^ C BO ; 1 D pyj .£ 00 BAfUtrett Ho ZAZtKtO. 4. Why do some bonds have a barrier to rotation? difficult blc bond mot4 be. ^ IT hoed s malciL roioho\ bToken 4ti roicrfc 5. Circle the H atom shown above that is most easily lost when the amino acid becomes deprotonated. 6. The conjugate base formed upon deprotonation of the histidine is stabilized by delocalization of the negative charge. Draw resonance forms that are consistent with this statement <3• # / > • 0' ! c \ bf © O ’' -9- Chemistry 11 Fall 2005 7. Discuss the hybridization, bond angles, and geometry of the five membered ring that is part of the side group of histidine. Your answer should take into account that the ring is planar and all N C bonds are equivalent. - hybn ' / [ iosorithcj pioofon ^ from ' MoVihfj ** . vAjouid r e*jui*e spin > - + 6. Now consider the complex /rans [Fe(CN)2(NH3)4] . Because CN produces a much stronger crystal field than does NH3, the usual octahedral field splitting is farther distorted in this complex. Using your knowledge of die causes of crystal field splitting and the d orbital geometries, and assuming the CN ligands lie along the z axis, draw a crystal field+ diagram showing the orbital energies and electron configuration in trans [Fe(CN)2(NH3)4] . Label each energy level with the name of the d orbital to which it corresponds (dxy, dz2, etc.). Justifythe energy ordering you choose. “ - " - - - A / jW 1* He CN~ MQ! BcrrH ALonj* d 2r dtf day / - yz 6y * - 12 - ewy d *t Fall 2005 Chemistry 11 2 7. a. Sketch the radial probability distribution (47crzR(r) ) vs. r for die 3, 4, and 5d orbitals on the axes below. Indicate the most probable distance for each, and be sure to label the axes. - tePU Sd b. Explain how these distribution functions are reflected in expected trends in atomic radius. ok-on, ,* " ' ndius h * 4/ c. The atomic radii of Fe, Ru, and Os are 124, 134, and 135 picometers, respectively. Explain this trend (or lack thereof ), especially in light of your response to part 7a. *3 M fe fo\lows fend QkvVc 13V . ^/ > * - ^ I doesn' -f J/ Osf » i is foltouJ m i *n Bef $ i> ^ veer cplomns \>)c, d-z 3 -os d. It is generally observed that as atomic radius increases, ligands in a complex are able to get ‘closer’ to the central metal atom, resulting in more overlap between the ligand electrons and the metal orbitals. Based on this observation, would you expect the d-d transitions in the hexaammine complexes of Ru(lH) to occur at shorter or longer wavelength than in [FefNHj ]3*? What would you expect the magnetic properties of the hexaammine Ru(III) complex to be? ft®* p i&Pjnj ^ Closer fo be Means in ore s; S' y - 13 - phfirntj ) so expect T!OCWL) Fall 2005 Chemistry 11 Question VI. Iron Binding to Heme in Hemoglobin (30 points) / H ' H H 4 . 1. Only two hydrogen atoms are shown explicitly in the Rawing. Draw in the remaining H atoms and lode pairs on the picture, and determine the * ' total number of a j Zfcand n 11 bonds in porphyrin. fN N* Shown at left is the porphyrin ligand that is similar to the one that binds to Fe(II) in hemoglobin. The Questions below seek to understand the properties of mis system in terms of the concepts introduced in Chemistry 11 . Restrict your answers to the space provided , ^ H H 2. Question about the single bonds to H a. In generalfriot just in this molecule), are single bonds between carbon and hydrogen atom? ALWAYS the same length? Explain. ' ho 5p C - rt longs* c - rt * b. Compare the bond lengths in water (H2O), ammonia (NH3), and methane (CH4). Explain. Sp d z -H -- UH u sUo'kd 0 m* It ley ri C 3. Consider the two nitrogen atoms shown above with one single and one double bond to carbon. Sketch the orbitals that contribute to a and 71 bonding and show their overlap. Use a localized electron pair model to completely describe the orbitals used to make the bonds between the N and C atoms as well as to contain any left over elections on N. ^ foTTf 1 4. Predict the geometry and bond angles about the N for the two C-N-C systems in #3. 2 I ~LO -piOAfi ^ & jT' / - H - /£ _ /2 Chemistry 11 Fall 2005 5 . In fact, all 8 N-C bonds in the entire molecule are identical in length and strength and the molecule is planar. How can we understand this? (words or pictures) 0Ut\ [op 7, of LP c — & * ' *3* 1 e 6. Iron binding occurs with the simultaneous loss of two protons. Circle the protons lost and put a square around the atoms that bind to the iron. Heow many porphyrin lone pairs are { Ml/ involved in metal bonding? ^ 3L pt 7. Some metals can form square planar complexes with ligands like porphyrin. a. What orbitals does the metal provide for the ligands’ electrons in square planar complexes? Include n and 1 orbital designations in your answer. — b. What happens to the energy of the metal’s d orbitals in square planar complexes? Be specific and include d orbital labels. a - "l ? 8. In fact, when Fe(II) binds to the porphyrin in hemoglobin, it is octahedral. The fifth coordination site is occupied by a histidine ligand. The histidine free amino acid is shown below, though the only important part for metal bonding is the five-membered ring. Which atom in the ring would most likely bond to the metal and why? SPV A) L?*JL pair auJCcy (row If ^^ 5/ e Histidine | y - 15 - I Q . vx fclF-Zj for , r\ n Chemistry 11 Fall 2005 9. It has been mentioned in class and in your text that O2 bonds to the protein hemoglobin reversibly without oxidation, whereas O2 binding to just the porphyrin by itself in solution without die protein is accompanied by an irreversible oxidation of the metal and reduction of the O2. The presence of the protein around the metal-porphyrin complex is said to prevent A oxidation by preventing the from bonding perpendicular to the porphyrin plane. When O2 binds to the Fe(II) in solution, it binds perpendicular. How can you understand this 62 ^ observation? HINT: You might consider which empty molecular orbitals on the O2 might (in the perpendicular arrangement in solution) overlap with filled atomic orbitals on the Fe(II) in porphyrin to allow the electron to be transferred from one to the other. - V7)") IT?* ') 7T *0 ) - So 4 -' C & fra) d* IOUJSP 1 ' ' ^ •u 4 ff - > u TT* tU « = hv - hv continued on next page - 17 - i k\- c —X E kinetic (ejected e — —V ZA = A V Collision rate with container wall (collisions per second): 0 Z=4 N Fall 2005 Chemistry 11 Equations, Constants, Conversion Factors, and Miscellaneous Information Kinetic energy: — 1 E kinetic = mv2 2 deBroglie wavelength: X= — -p = mv Heisenberg’s uncertainty principle: - Energy levels of a one electron atom: XPS: Ap x Ax > — 4 rc or mAvx Ax > ^ E„ = -2.178 x 10 18 J ^ ^ - j — h Ait Ex NA = 6.022 x 1023 mol-1 Ideal gas constant: R = 0.08206 L atm mol 1 K-1 = 8.314 J mol-1 K-1 Fundamental charge: e = 1.60218 x 10-19 C Avogadro’s number: ' Proton mass: mp = 1.673 x 10~27 kg mn = 1.675 x 10-27 kg Electron mass: me = 9.109 x 10-31 kg Speed of light: c = 2.9979 x 108 m s-1 Planck's constant: h = 6.626 x 10-34 J s Neutron mass: 1 atm = 760 torr = 101.325 kPa = 101325 Pa 1 M = 1 mol LTI 1 L = 103 mL = 103 cm3 1 nm = 10-9 m 1 A = 0.1 nm = 10~ 1 kg = 103 g 1 mg = 103 g TK = TX + 273 1 J = 1 N m = 1 kg m2 s~2 STP = 1.000 atm, 273.15 K ,om 1 kJ = 103 J Spectrochemicalseries: T < Br < C1 < F < OH < H2Q < NH3 < en < CN < CO " / - 18 - Full 2005 11 ( ' hemistry* PERIODIC CHART OF THE ELEMENTS IA mg IVB IIA VB VIB IB VIU VtIB tIB !VA |UA INERT VA VIA VIIA GASES T 1 2 H He 1.80797 4 0028 H 1.0 D 797 4 3 ^ Li Be 6 939 ».01» TT i? 57 50 14 « Si S Cl Al P Ar 26 9815 29.088 38.9719 92.864 9 S.4S3 99.949 ?1 22 IT 24 25 7H 27 7F 29 411 TT T? T5 TW 4% 4K TT /5 76 77 16 Tj 14 15 30 Tl 32 TT 411 TT 50 TT Na Mjg 22.9698 20 W " e O F Ne B Cr,n; N 0067 15 SS 34 I» 0994 JO 103 loan 12 T9 T T 6 ' 1 17 As Se Br Kr K Ca Sc5» Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge 36 102 47.90 50.942 51 698 51.9990 55 »41 59.1992 59.71 99.54 85.97 89.72 12 56 74.9218 78.98 79.666 81.90 44 > 50 £s Sr Y Zr Nb Mo Tc Rb Ru Rb Pd 95.47 97 £ 2 88.905 91.22 92.906 65 94 1991 tfl1.07 102 605 108 4 , MS 55 72 7T 74 TUI 10 b 70 70 Cs Ba La Hf Ta W Re Os ir Pt Au 192.685 117.94 118.91 178.49 188.949 199.85 161.2 190.2 192.2 195.09 199.997 TOB MO Fr Bh Hs Mt) Ra) Ac Rf) Db) Sg | [ 262] ( 261| [ 26 [ 229) PH 17 U 13 |B9 1229 104 13271 PEL hturfein h Wt MMDM W ITWIt S Wfl sfrvll WMnOnfcMtt ** ** $htf l <$ winfwwi lo llx: 1461 -dm: ol th« . CMWHffiMi on At bn * The pp>jp dut ittkins urcd httn are tfw foirwChMnrad Afcflnrf mAb«n ^ Actinium. Aluminum Americium Antimony Argon Arsenic Astatine Barium Rcrkclium Beryllium Bismuth Bohrium Boron Bromine BJ 1362 1299 Ua TT? ? ?~ m or 1ST Tl Pb Bi Po At Rn 204.97 207.19 286.960 [ 210 ( 21111 ( 2221 J http:jVchjcnllab.pt .marM.iiJpa.isdui'pc-riodk 'pTinlabtr.j'if 'i uH ( 2 72| DT 02 iff SO '>1 4 Aitmide SortM W 91 92 93 9T 94 W W 97 Too 99 1U1 TU? TUT U Np Pu Es Fm Md No AmCm Lr Th Pa Bk Cf ( 211) ( 231) ( 242) 253| ) 256) 257] ( 24 f | ( 247 ( 2491 ( 294 212 241) 2 S 6) 8691 1218.03 At Gallium Ra Bk Germanium Gold Hafnium Hassimu Gadolinium | Fm F Fr Gd Ga | Ge Au Nitrogen Hf Hs Osmium Oxygen Palladium Phosphorus Bh Helium He B Br Cd Hoimium Ho H In I Hydrogen | Mercury Molybdenum Neodymium Neon Neptunium Nickel Niobium FT Eu Nobelium Platinum | | | | Mo Scandium Seaborgnnn Nd Selenium Se Ne Silicon Silver Sodium Strontium Si Ag Np Ni Nb Sc Sg Na Sr S Ta N No Os O Sulfur Tantalum Technetium Tellurium Tc Te Pd Terbium ThaiIium Tb Tl P PI Pu Po Thorium Th Thulium Tin Titanium Tungsten Uranium Vanadium Tm Sn Ti ir Plutonium Polonium Potassium Fe Praesodymium Promethium Lanthanum Kr La Protactinium Pr Pit! Pa Ijjwrcncium Lr Radium Ra Xenon Xe Lead Lilhium Lutetium Pb Li Radon Rhenium Yllcrbium Yttrium Y Magnesium Mg Rn Rc Rh Rb Zinc Zirconium Zr Indium Iodine Iridium Iron Cadmium Calcium Californium Carbon Cerium Cesium Ca Cf C Cc Cs Chlorine Cl Chromium Cobalt Copper Curium Cr Co Cu Cm Dubnium Db Manganese Mn Dy Es Meitnerium Mt Mendelevium Md Dysprosium Einsteinium 54 ' Erbium Europium Fermiurn Fluorine Francium Be ? TTT TT LantHa rids Senas * nr la HD 7N 71 m I 61 fli rill 114 Ii 5 117 Bn Sm Lu Tb Ho Ce Gd Yb Pr Ell Er Pm Dy Tm | ( 147) 150 35 15198 I 57 25 159 9M 192 59 194.930 IG 3.26 188 1164 17J. 174.97 I 40 12 140.987 Ac A1 Am Sb Ar As Bi TW TOT 55 Cd In Sn Sb Te I Xe 112.40 114.02 116.89 12171 127 68 126.904 691 ID Krypton Lu Rhodium Rubidium Ruthenium Rutherfordium Samarium - 19 - K Ru Rf Sm W u V Yb Zn

Chemistry Exam 3 Fall 2005 Solutions

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