CHE 211 EXAM 4 KEY (CH. 11-­‐13) Answers in BOLD RED NOTE: This is NOT a mastery-­‐based exam. There will be no retakes. It will be curved.
Mark the letter of the best answer to each multiple-­‐choice question clearly on your answer sheet. Answer problems directly on this exam. Show work for any problem that requires calculations and answer thoroughly to ensure maximum partial credit. Each question is worth three points; the Ch. 11 problem is worth 7, while the Ch. 12 & 13 problems are worth 6 points each. CHAPTER 11 (BONDING) 27 question points + 7 problem points = 34 points H
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1. The substance butadiene is used to make synthetic rubber. This is its structure: This molecule has _____ sigma and _____ pi bonds altogether. A. 7 ; 4 B. 9 ; 2 C. 6 ; 4 D. 8 ; 2 2. Based on the structure of butadiene and VSEPR, the C atoms are ____ hybridized A. sp B. dsp2 C. sp3 3. What is NOT part of VB theory? A. antibonding orbitals B. lone pairs 4. How do pi bonds and sigma bonds differ in VB theory? A. sigma bonds only form using s orbitals B. pi bonds are stronger than sigma bonds C. sigma bonds are 2-­‐3 times stronger than pi D. pi bonds do not allow rotation on the bond axis 5. One of the main advantages of hybrid orbitals as opposed to the normal atomic orbital set is that hybrids A. match molecular geometry B. allow resonance C. allow fractional bond orders D. all of these 6. What is the hybridization of the Cl in the molecule ClO2? A. sp B. sp2 7. When a P atom is sp3 hybridized, how many of its hybrid orbitals are half-­‐filled, and thus able to form bonds? A. three B. four C. one D. two 8. The theoretical diatomic molecule OF should have a bond order of _______, based upon its MO diagram. A. 0 B. 0.5 C. 1.0 D. 1.5 9. Which of the following orbitals is most likely to be the σ* MO of F2 ? structure D Problem and answer on the next page for legibility C
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D. sp2 C. a bond = 2 shared electrons D. hybrid orbitals C. sp3 D. d2sp3 CHE 211 Exam 4 KEY PROBLEM SHOW ALL CALCULATIONS in order to assure full credit The acetylide ion consists of two C atoms with a double negative charge, C22– and is well-­‐known, but chemists have an interest in its B + N analog, BN2–. First, GIVE THE LEWIS STRUCTURE of BN2–; then, draw out and fill its MO diagram, using the B2/C2/N2 diagram from your data sheet. Using the filled-­‐in MO diagram, calculate the bond order of BN2– and compare this briefly to the VB (Lewis) structure, noting similarities and differences. EXTRA CREDIT (3 points possible) What uncharged species is BN2– isoelectronic with? Name it and draw its Lewis structure. The species nitrogen, N2, is isoelectronic with BN2–. Its Lewis structure is just A second option would be carbon monoxide
:NΞN: , :CΞO: Many other options do not fit the problem’s criteria – must have 10 electrons + no charge + 2 atoms CHE 211 Exam 4 KEY CHAPTER 12 (STATES) 27 question points + 6 problem points = 33 points 1. An example of an exothermic phase change is A. fusion B. vaporization 2. When a pure substance is going through a phase change, what is NOT changing? A. its temperature B. its density C. its degree of organization D. its volume 3. In a phase diagram, the point beyond which the liquid and gas phases can no longer be observed is the A. triple point B. critical point C. boiling point D. sublimation point 4. Ethanol has a vapor pressure = 100. Torr at 34.9°C and 400. Torr at 63.5°C. What is its ∆HVAP? (all in J/mole) A. 41,700 B. 893 C. 412 D. 18,100 5. Which intermolecular forces are not possible in pure substances? A. London (dispersion) B. ion-­‐dipole C. dipole-­‐dipole 6. In general, the stronger the intermolecular forces in a pure substance, the greater its A. vapor pressure B. density C. capillarity D. surface tension 7. Water is almost unique among substances which are liquids at room temperature in that its A. vapor P is so high B. solid state is less dense than its liquid C. viscosity is so low D heat capacity is low 8. A face-­‐centered cubic unit cell has a total of ______________ within the unit cell. A. 1 atom B. 3 atoms C. 2 atoms 9. In the Band Theory of Conductivity, each band is made up of A. similar molecular orbitals B. covalently bonded electrons C. metal ions D. valence electrons C. sublimation D. freezing D. hydrogen bonding D. 4 atoms PROBLEM DRAW THE LEWIS STRUCTURE with all lone pair electrons Draw the Lewis structure of the compound dimethyl ether (CH3OCH3, with the larger atoms in the bonded order C–O–C and all the H atoms on the two carbons). H
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What will be the approximate bond angle of the C–O–C bond? Why? The bond angle will be slightly less than 109°; the O will be sp3-­‐hybridized since there are four groups of electrons around it (2 bonds + 2 lone pairs) and the expanded cloud of lone-­‐pair electrons will force the C–O–C bond angle to be a bit less than the ideal 109.5° Which of these intermolecular forces will be present in pure liquid dimethyl ether? EXPLAIN WHY for each. London (dispersion) These will be present because ALL molecular substances have these weak forces which are caused by temporary, fluctuating electron distribution in each molecule. Dipole-­‐dipole The C–O bond is polar, and its bent shape in this molecule will give the ether a dipole moment that will cause there to be dipole-­‐dipole attractions between molecules. Hydrogen bonding This is NOT present because it required H on N, O, or F to produce this intense and fairly unique special type of dipole-­‐dipole force. Ion-­‐Dipole This force CANNOT be present because there are no ions in pure covalent dimethyl ether CHE 211 Exam 4 KEY CHAPTER 13 (MIXTURES) 27 question points + 6 problem points = 33 points 1. What is the strongest type of intermolecular force producing a solution of CsCl in H2O? A. hydrogen bonding B. dispersion forces C. dipole-­‐dipole D. ion-­‐dipole 2. An intermolecular force involving induced dipoles is most likely to be important with which type of solute? A. a solute with a large volume and many electrons B. a solute with a small volume and many electrons C. a solute with a large volume but few electrons D. a solute with a small volume and few electrons 3. When we dissolve an ionic substance in H2O and the solution warms, the process probably has a big negative A. ∆H of solvent separation B. ∆H of solute separation C. ∆H of fusion D. ∆H of hydration 4. To increase the solubility of a gas in a liquid solvent, you can ___________ the T and ______________ the gas’s P. A. lower ; raise B. lower ; lower C. raise ; raise D. raise ; lower 5. A sugar solution is found to contain 35% sugar by weight, with a density of 1.15 g/mL, so a liter of the sugar solution weighs 1150 g. What is the Molarity of this solution? Sugar has a molar mass = 180.0 g/mol. A. 2.11 B. 1.94 C. 0..830 D. 2.24 6. Determine the molality of the sugar solution in question 5. If water has a freezing point depression constant equal to –0.512°/molal, what will the sugar solution’s freezing point be? A. –15°C B. –9°C C. –3°C D. –1.5°C 7. Based on the concentrations you calculated, the osmotic pressure of the sugar solution = _______ atm at 27°C A. 52 B. 5.0 C. 55 D. 12 8. As noted earlier, ethanol has a vapor pressure of 400. Torr at 63.5°. If we have a bottle of wine which has 2.63 moles of ethanol and 47.0 moles H2O per liter, what is the ethanol partial pressure over the wine at 63.5°C? A. 22 Torr B. 21 Torr C. 38 Torr D. 19 Torr 9. A colloid can be detected by the visibility of a beam of light passing through it; this is called A. colloidal illumination B. Brownian motion C. the visibility condition D. the Tyndall effect PROBLEM SHOW ALL WORK ON CALCULATIONS FOR MAXIMUM CREDIT A biochemist isolates a pure protein from a plant source and wishes to estimate the molar mass of the protein so she uses the osmotic pressure of a solution to determine the approximate molar mass. The biochemist dissolves 1.00 g of the protein in enough water to equal 100.0 mL of solution and then tests the osmotic pressure, which she finds to be 9.3 Torr at 25°C. What is the approximate molar mass of this protein? ∏ = MRT, so we can rearrange this to solve for M: M = ∏/RT = ( (9.3 Torr)/760 Torr atm–1)/(0.0821 x 298 K) = 0.00050 M Note that the 9.3 Torr must be converted to atm, and T must be in Kelvin If the solution is thus 5.0 x 10–4 M and we have 0.100 L, then our 1.00 g was = (5.0 x 10–4 mol/L)(.100 L) = 5.0 x 10–5 mole. Thus the weight of a mole is = (1.00 g)(1/5.0 x 10–5 mol) = 20,000 g/mol EXTRA CREDIT (3 Points) s Mark the location of each of the three phases on the following phase diagram: l g