Chemistry 222 Winter 2012 Oregon State University Exam 1 February 2, 2012 Drs. Nafshun, Ferguson, and Watson

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1 Chemistry 222 Winter 2012 Oregon State University Exam 1 February 2, 2012 Drs. Nafshun, Ferguson, and Watson Instructions: You should have with you several number two pencils, an eraser, your 3" x 5" note card, a calculator, and your University ID Card. If you have notes with you, place them in a sealed backpack and place the backpack OUT OF SIGHT or place the notes directly on the table at the front of the room. Fill in the front page of the Scantron answer sheet with your class section number (see below), last name, first name, middle initial, and student identification number. Leave the test form number blank. Section 001 (MWF 8am with Dr. Nafshun) Section 003 (MWF 1pm with Dr. Ferguson) Section 006 (TR 8am with Dr. Nafshun) Section 002 (MWF 11am with Dr. Watson) Section 004 (MWF 3pm with Dr. Ferguson) This exam consists of 22 multiple-choice questions; each has 5 points attached. The last question (Question 23) has 2 points attached. When you finish this exam, proceed to the proctor. Submit your completed Scantron form. You may take your notecard and exam packet with you. PV = nrt R = L atm mol K R = P1 V1 PV 2 2 3RT rms n1t 1 n2t2 Molar Mass 2 kg m 2 s mol K 760 Torr = 1 atm = 760 mm Hg K = C 1 mole = 6.02 x STP = 273 K & 1 atm 1

2 Electron Pair and Molecular Geometries Number of Electron Number of Lone Pairs Electron Pair Geometry Molecular Geometry Groups 2 0 Linear Linear 3 0 Trigonal planar Trigonal planar 1 Trigonal planar Bent 0 Tetrahedral (T d ) Tetrahedral (T d ) 4 1 Tetrahedral (T d ) Trigonal pyramidal 2 Tetrahedral (T d ) Bent 0 Trigonal bipyramidal Trigonal bipyramidal 5 1 Trigonal bipyramidal See-Saw 2 Trigonal bipyramidal T-Shaped 3 Trigonal bipyramidal Linear 0 Octahedral (O h ) Octahedral (O h ) 6 1 Octahedral (O h ) Square pyramidal 2 Octahedral (O h ) Square planar 2

3 1. Determine the electron geometry and molecular geometry of BrF3. The electron geometry is trigonal planar; the molecular geometry is trigonal planar. The electron geometry is trigonal bipyramidal; the molecular geometry is T-shape. The electron geometry is trigonal planar; the molecular geometry is bent. The electron geometry is trigonal bipyramidal; the molecular geometry is see-saw. The electron geometry is tetrahedral; the molecular geometry is trigonal bipyramidal. 2. Consider the molecule below. Determine the molecular geometry at each of the two labeled carbons. Carbon 1 is tetrahedral and Carbon 2 is linear. Carbon 1 is trigonal planar and Carbon 2 is bent. Carbon 1 is bent and Carbon 2 is trigonal planar. Carbon 1 is trigonal planar and Carbon 2 is tetrahedral. Carbon 1 is trigonal pyramidal and Carbon 2 is tetrahedral. 3. Consider the molecule below. Determine the hybridization of the two labeled carbons. Carbon 1 is sp 2 and Carbon 2 is sp 3. Carbon 1 is sp 3 and Carbon 2 is sp 2. Carbon 1 is sp 2 and Carbon 2 is sp. Carbon 1 is sp and Carbon 2 is sp 2. Carbon 1 is sp and Carbon 2 is sp 3. 3

4 4. The H-N-H bond angle in ammonia, NH 3, is: A little greater than A little less than Consider the following five molecules: PCl 5, O 2, CO 2, SO 2, and SF 6. How many of these are polar molecules? One Two Three Four Five 6. Consider tetrafluoroethene, C 2 F 4. Tetrafluoroethene contains: five σ-bonds and no -bonds four σ-bonds and one -bond three σ-bonds and two -bonds five σ-bonds and one -bond four σ-bonds and two -bonds 4

5 7. Describe a pi bond. Side by side overlap of p orbitals End to end overlap of p orbitals s Orbital overlapping with the end of a p orbital Overlap of two s orbitals p Orbital overlapping with a d orbital 8. Which molecule's bonding scheme can be described as sigma bonds formed by the overlap of hydrogen 1s orbitals and carbon sp hybridized orbitals, a sigma bond formed by the overlap of carbon sp hybridized orbitals, and pi bonds formed by the overlap of carbon unhybridized 2p orbitals? C 2 C 2 H 2 C 2 H 4 C 2 H 6 C 2 H 8 9. Consider MO (Molecular Orbital Theory). For the O 2 molecule, there are electrons in the π 2p* anti-bonding orbitals?

6 10. Consider MO (Molecular Orbital Theory). The F 2 2- ion is: paramagnetic diamagnetic trimagnetic tetramagnetic jay-z-magnetic 11. Molecular orbital theory predicts the N 2 + ion has a bond order of: A student measures the pressure inside their home to be 1.05 atm. How high would the column of mercury be in a barometer? mm 724 mm 746 mm 760 mm 798 mm 6

7 13. A gas at 35.0 C occupies 4.50 L. At what temperature will the gas occupy 9.00 L? 343 C 70.0 C 616 C 1.16 C 17.5 C 14. Using the graph below, determine the gas that has the lowest density at STP. A B C D All of the gases have the same density at STP. 15. A container containing 1 mole of gas, originally at a pressure of P 1, a volume of V 1, and at a temperature of 373 ºC undergoes two processes. The first occurs at constant temperature with the pressure being doubled. The second takes place at constant pressure with the temperature being raised to 473 ºC. What is the final volume? 22.4 L V 1 2V 1 0.5V 1 4V 1 7

8 16. The density of a gas is 1.43 g/l at STP. What is the gas? Cl 2 CH 4 O 2 Ne He 17. A compound is found to be 30.45% N and % O by mass. If 1.63 g of this compound occupy 389 ml at 0.00 C and 775 mm Hg, what is the molecular formula of the compound? NO 2 N 2 O N 4 O 2 N 2 O 5 N 2 O A mixture of moles of N 2, moles of O 2, and unknown number of moles of Ar gases are placed into a L flask. The temperature and pressure of the flask and gases are measured to be K and 734 mm Hg. How many moles of argon gas are present? mol Ar (g) mol Ar (g) mol Ar (g) mol Ar (g) mol Ar (g) 8

9 19. The root-mean-square speed of N 2 (g) at 760 mm Hg and 297 K is: 3874 m/s x 10 6 m/s 514 m/s 264 m/s 1040 m/s 20. Consider the following five gases: H 2 (g) O 2 (g) He (g) Cl 2 (g) Ar (g) Of these, the gas with the highest velocity at room temperature is: H 2 (g) O 2 (g) He (g) Cl 2 (g) Ar (g) 21. A student combusts grams of propane, C 3 H 8 (g). How many liters of CO 2 (g) are produced at 300 K and 1.20 atm? C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) 334 L L 299 L 200 L 861 L 9

10 22. The van der Waals equation for real gases recognizes that: the non-zero volumes of gas particles effectively decrease the amount of "empty space" between them the molecular attractions between particles of gas decreases the pressure exerted by the gas molar volumes of gases are different gas particles have non-zero volumes and interact with each other All of the above statements are true. 23. Because of Chemistry I like pie. My new favorite movie character is Bond, James Bond. I have attained a magnetic personality. I may purchase a hybrid vehicle. I ve been getting a whole lot of dates by using pick-up lines that include the words polar, dipole, lobes, 180 degrees, see-saw, tetrahedral, and orbitals. 10