The University of British Columbia Okanagan CHEMISTRY. FALL 2009 Final Exam Wednesday, Dec 9, 9AM DO NOT TURN THE PAGE UNTIL INSTRUCTED TO DO SO!

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The University of British Columbia Okanagan C HEM 121 ATOMIC & MOLECULAR HEM 121 CHEMISTRY FALL 2009 Final Exam Wednesday, Dec 9, 9AM DO NOT TURN THE PAGE UNTIL INSTRUCTED TO DO SO!

1 The University of British Columbia Okanagan C HEM 121 ATOMIC & MOLECULAR HEM 121 CHEMISTRY FALL 2009 Final Exam Wednesday, Dec 9, 9AM DO NOT TURN THE PAGE UNTIL INSTRUCTED TO DO SO! Take the time now to fill out the front of the exam booklet. Now! Be sure to indicate clearly your instructor s name. Make sure you have all 3 pages, including this one with some helpful information on the back, and 2 double-sided pages with 10 questions. You should also have a periodic table. Answer all questions in the provided exam booklet, in whatever order and with whatever writing implement brings you the greatest personal and spiritual satisfaction. Make note of the point value of each question, and allocate your time accordingly. Carefully read each question before answering. Where appropriate, you must show your work to receive full credit. Include units and the proper significant figures in all numerical answers. With the exception of a non-programmable calculator, no aids or notes of any kind are permitted or required. Have yourself a Merry Little Christmas, or whatever Generic Winter Solstice Celebration you feel appropriate. The game is afoot! Total Points: 100 Total Time: 3 hours

2 Potentially Helpful Information Constants electron mass = kg proton mass = kg neutron mass = kg h = Js c = m/s R H = J a 0 = m e = C N A = mol 1 k = JK 1 1 ev = 96.5 kj/mol absolute zero = ºC R = Latm/molK = J/molK 1 atm = kpa = 760 torr Equations E = hν E = ½mv 2 E = mc 2 λν = c λ = h/mv hν = hν 0 + KE ΔxΔv h/4πm PV = nrt P = MM = ρrt/p ΔP = X solute P solvent Π = MRT ΔT f = K f m ΔT b = K b m Electronegativity Values H 2.2 Li 1.0 Na 0.9 K 0.8 Be 1.6 Mg 1.3 Ca 1.0 B 2.0 Al 1.6 Ga 1.8 C 2.6 Si 1.9 Ge 2.0 N 3.0 P 2.2 As 2.2 O 3.4 S 2.6 Se 2.6 F 4.0 Cl 3.2 Br 3.0 I 2.7 Page 2/4

3 1. I CLEARLY CANNOT CHOOSE THE WINE IN FRONT OF YOU (10 POINTS, 1 EACH) Write the letter corresponding to the one best answer for each question. a) Which gas has the particles with the highest average molecular speed? a) NH 3 at 25ºC b) CO 2 at 25ºC c) NH 3 at 75ºC d) CO 2 at 75ºC b) CH 3 CN has a higher boiling point than CO 2 because a) CH 3 CN is highly polar, and CO 2 is non-polar b) CH 3 CN has a triple bond, and CO 2 only has double bonds c) CH 3 CN has C-H bonds, and CO 2 does not d) all of the above c) In a boron atom, the 2s orbital is lower energy than the 2p orbital because a) the 2s orbital has a lower value of quantum number n b) the 2s orbital can form a stronger σ-bond, but the 2p can form a weaker π-bond c) the 2s orbital has two electrons, but the 2p has only one d) the 2s orbital penetrates to the nucleus more effectively than a 2p electron d) A S-Cl bond has a) a partial negative charge on Cl and mostly covalent character b) a partial negative charge on S and mostly covalent character c) a partial negative charge on Cl and mostly ionic character d) a partial negative charge on S and mostly ionic character e) no charge separation, because non-metals form covalent bonds that share electrons e) How many electrons are in the 32 S 2 ion? a) 2 b) 14 c) 18 d) 34 f) A methanol solution of a sodium salt burns with a bright orange flame. This orange colour results because a) ground state sodium ions emit photons when heated b) excited state sodium ions emit photons when they relax c) ground state sodium ions are excited when they absorb photons d) sodium salts are orange g) Dipole-dipole forces between molecules are due to a) overlap of atomic orbitals b) permanent charge separation within molecules leading to electrostatic attraction c) electrostatic attraction of protons and electrons d) temporary distortion of electron clouds h) Which material is expected to have the lowest vapour pressure? a) CH 4, because it has the weakest intermolecular forces of attraction b) CH 4, because it has the largest molecules c) HF, because it has the smallest molecules d) HF, because it has the strongest intermolecular forces of attraction Page 1/4

4 i) Consider the molecular orbital of F 2 shown at right. This molecular orbital results from a) bonding overlap of two 2s atomic orbitals b) antibonding overlap of two 2s atomic orbitals c) bonding overlap of two 2p atomic orbitals d) antibonding overlap of two 2p atomic orbitals j) The nucleus of an atom a) occupies very little of the total atomic volume b) has a positive charge due to the neutrons c) has a negative charge due to the electrons d) contains very little of the atomic mass 2. POLAR EXPEDITION (12 POINTS) Consider the three molecules below, in which the central atom is underlined and in boldface type. For each: a) provide a valid Lewis dot structure, clearly showing the position of all electron pairs (2) b) describe the molecular shape as predicted by VSEPR (1) c) indicate whether the molecule should be polar (P) or non-polar (NP) (1) i) [NO 2 ] + ii) HAsCl 2 iii) OSCl 4 3. JUMPING JACK FLASH (6 POINTS) A 3.50L balloon contains 9.93 g of a gas at 73.7 kpa and 36ºC. a) What is the molar mass of the gas? (2) b) What is the volume of the balloon if the temperature is lowered to 10ºC? Assume a constant pressure and ideal behaviour of the gas. (2) c) Explain, in terms of the kinetic molecular theory of gases, why the volume of the gas changes in this way with decreased temperature. (2) 4. AUFBAU BEFORE ME (6 POINTS, 1 EACH) Provide the indicated information for each of the following two atoms: i) 29 Si ii) 81 Br a) the ground-state electronic configuration (using spdf notation), b) the number of unpaired electrons in that configuration, and c) the number of neutrons in the atom. You may employ the noble gas shorthand notation, e.g. [He]2s 1 for lithium. 5. ANOTHER BOHRING QUESTION (6 POINTS) a) In the Bohr model, ionizing an atom may be considered a transition to the n = level. How much energy is required, in Joules, to remove the electron from a hydrogen atom in the n = 2 state? (2) b) If a hydrogen atom in the n = 4 state relaxes to its ground state, what is the wavelength of the emitted photon, in nanometres? (4) Page 2/4

5 6. ACHEY BREAKY HEAD (13 POINTS) Examine the structure of acetylsalicylic acid (better known as Aspirin), shown at right. Consider the six-membered carbon ring to lie in the xy plane. a) What is the expected approximate COH angle a? (1) b) What is the expected approximate CCO angle b? (1) c) How many π bonds are in Aspirin? (1) d) How many atoms in Aspirin are sp 3 hybridized? (1) e) Which two specific orbitals overlap to form the σ bond(s) between the C and H marked with the symbols? (2) f) Which two specific orbitals overlap to form the π bond(s) between the C and C marked with the symbols? (2) Consider the effects of dissolving 4.50 g of Aspirin into 50.0 g ethanol (C 2 H 6 O). Ethanol has a normal freezing point of 114.6ºC, a density of g/ml, and a K f of 1.99 C/m. g) Will the freezing temperature of the ethanol solution increase, decrease, or stay the same? (1) h) Will the boiling temperature of the ethanol solution increase, decrease, or stay the same? (1) i) Calculate the new freezing temperature of the ethanol solution. (3) 7. NO ZONE LIKE THE O ZONE (10 POINTS) Ozone, O 3, acts as a filter of harmful UV radiation in the stratosphere by absorbing light with a maximum wavelength of 290 nm. Absorption of a photon at 290 nm causes an O-O bond in ozone to break: O 3 O 2 + O a) What is the energy of photons with a wavelength of 290 nm, measured in kj/mol? (2) b) Draw the Lewis resonance structures for the ozone molecule, and use them to determine the O-O bond order in ozone. (3) c) Provide the molecular orbital valence electronic configuration for the O 2 molecule. (For example, the configuration for H 2 is (σ 1s ) 2 ). According to your configuration, what is the bond order in O 2? (3) d) Photons with a wavelength of 290 nm have enough energy to break O-O bonds in O 3, but not those in O 2, with a strength of 498 kj/mol. Are your previous answers consistent with this observation? Explain your answer, briefly. (2) 8. CONSIDER THE FOLLOWING (4 POINTS) Consider the following three atomic orbitals: 6s 5d xy 4f z 3 a) Which of these orbitals has the lowest value of quantum number? (1) b) Which of these orbitals is a member of the largest orbital subshell (i.e. the subshell with the greatest number of other orbitals)? (1) c) Which of these orbitals has the greatest electron density at the atomic nucleus? (1) d) Which orbital is the highest energy in a hydrogen atom? (1) Page 3/4

6 9. TEETH OF THE HYDRAZINE (17 POINTS) Consider the compound hydrazine, N 2 H 4 a) 64.0 g of hydrazine is placed in a 750 ml steel container and heated to 320 C. Hydrazine is a gas under these conditions. The Van der Waals constants for hydrazine are a = 8.35 atml 2 /mol 2 and b = L/mol. Calculate the pressure exhibited by the hydrazine. (2) b) Would you expect equal moles of ethylene (C 2 H 4 ) to exhibit a higher or lower pressure than hydrazine under these conditions? Briefly explain your reasoning. (2) Consider the phase diagram of hydrazine. c) At 180 atm pressure and 0 C, in what phase does hydrazine exist? (1) d) What phase change(s), if any, will occur if the hydrazine in part (c) is heated to 150 C? Explain your reasoning. (2) e) What is the enthalpy of vaporization of hydrazine, in kj/mol? (2) f) What is the name of the point labeled O on the diagram? (1) You can use hydrazine as a rocket fuel, because when combined with N 2 O 4 it reacts exothermically to form N 2 and H 2 O: 2 N 2 H 4 (g) + N 2 O 4 (g) 3 N 2 (g) + 4 H 2 O(g) g) If 6.25 kg of N 2 H 4 reacts with 10.0 kg of N 2 O 4, which material is the limiting reagent? Clearly show all calculations and reasoning. (3) h) Assuming a 100% reaction yield, what volume will the product gases occupy at 250 C under a pressure of 720 torr? (4) 10. LUCY, YOU GOT SOME SPLAININ TO DO! (16 POINTS, 4 EACH) Provide concise rationalizations for the following observations. Note that in each case, there are two observations to explain. a) Removing one electron from N 2 weakens the NN bond, but removing one electron from F 2 strengthens the FF bond. b) Methanol (CH 3 OH) is very soluble in water (H 2 O) but not soluble in carbon tetrachloride (CCl 4 ). c) The boiling point of CF 4 ( 128 C) is higher than that of CH 4 ( 162 C), but lower than that of CH 2 F 2 ( 52 C). d) The ionization energy of Na (5.14 ev) is less than that of Mg (7.65 ev), but the ionization energy of Mg is greater than that of Al (5.99 ev). Page 4/4

DO NOT TURN THE PAGE UNTIL INSTRUCTED TO DO SO!

DO NOT TURN THE PAGE UNTIL INSTRUCTED TO DO SO! The University of British Columbia Okanagan CHEM 121 ATOMIC & MOLECULAR CHEMISTRY Final Exam, Fall 2011 Wednesday, Dec 14, 9AM Name: Circle your Section: Student Number: McNeil / Neeland Signature: DO