Chem Practice Exam 2 - 001 (part 1 of 1:] 10 points 1Write...

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Unformatted text preview: 001 (part 1 of 1:] 10 points 1Write the electron—dot notation for the ele ment P. 1.5,. 2':;_I;: 3. g. 5. g. 7. s. P: 9. None of these 002 (part 1 of 1:] 10 points Which property is characteristic of cowalent but NOT ionic compounds? 1. conductor of electricity in molten state 2. insoluble in non—polar solvents 3. aqueous solutions are poor electrical con— ductors =1. Aqueous solutions are good electrical con— ductors. 5. solids with a high melting point 6. formed from elements of quite different electronegativities 7. soluble in polar solvents such as water 003 [part 1 of 1:] 10 points Which property is characteristic of ionic hut NOT of covalent compounds? 1. high melting point 2. low boiling point 3. soft, non—crystalline structure 4. non—conductor of electricity in molten state 5. easily vaporizes 004 [part 1 of 1:] 10 points Ionic compounds involve a (share: transfer) of electrons and typically involve atoms that have [large: similar5 ver different) electroneg— ativities. 1. transfer; similar 2. share; very different 3. transfer; large 4. transfer; very diiferent . share; similar o1 DDS [part 1 of 1:] 10 points How many valence electrons does Ni have? 1.1[11 2.8 3.7 _|"..-‘l Nil 7.6 8. 4 006 [part 1 of 1:] 10 points Which of the following is the best represen— tation of the compound calciuln fluoride? 3— 5. 30a“; 2 [: F :] 6. None is appropriate because calciuni flu— oride is a covalent compound. DDT [part 1 of 1:] 10 points The electronegativity of an atom generally ? going down columns and T going left to right across rows. An element like fluorine would most likely forni aflnj 'E' bond with an element like oxygen. 1. increases; increases; covalent 2. decreases; increases; covaleni 3. decreases; decreases; covalent 4. increases; decreases; covalent 5. increases; decreases; ionic 6. decreases; decreases; ionic ?. increases; increases; ionic 3. decreases; increases; ionic DDS [part 1 of 1:] 10 points Which of the following is the best description of covalent bonding? 1. Bonding that occurs between elements with veljv diiferent electronegativities, and with sharing of electrons. 2. Bonding that occurs between elements with sirnilar electronegativities= and with transfer of electrons. 3. Bonding that occurs between elements with veljv diiferent electronegativities, and with transfer of electrons. 4. Bonding that occurs between elements with sirnilar electronegativities= and with sharing of' electrons. 5. Bonding that occurs between ions and elenients; and with the sharing of electrons. 009 (part 1 of 1:] 10 points Coordinate covalent bond formation is a pro ceas in which -- -. F —— B 1 x \ , 1. both electrons in the covalent bond are F -, F" supplied by the same atom. 2. one atom gives up more than one electron ' F : to another atom. 2 ” I? ' I F—B _ .. \ . 3. one atom glves up an electron to another ' 1: ' atom. ' 4. two atoms Share more than one pair of F_B_F ; electrons. 3- | n : F : 5. one electron in the covalent bond is snp— ” nlied hv each atom. 5 .. .. .. 010 [part1of1j10points : F_B_F : 4 .. l .. How many electrons are around the central 2 F = atom in the expanded valence in I; ? " 1-10' 5- F—FEB—F 2. 8 : F—B—F : 3. 14 '5- .. l .. : F : 4. G 5. 12 "F : . 7. I? 011 (part 1 of 1,] 10 points F—B Which of the following is the correct Lewis \ formula for boron triflnoride (Bng? F F—B—F *3- | : F : F—B—F 9- | F—B—F i 10. l .. : F ' [112 (part 1 of 1:] 10' points Which of the following is the eorreet Lew :H—C—O: formula for formaldehyde [01130:]? 10' I I H l 1' : : 013 (part 1 of 1:] 10 points Which of the following is the correct Lewis | H formula for fluorine [F2]? 0:0" 1' F—F 2 f \ .' H .. H" 2 'FZF' H—C—O 3' F—F 3- | .- 5 : FEF KO : J! . .. 4 H—C. ””” “ ‘3 F—F \H -. u 7 :F—F '0 : 3. _ F .. fr“, F F I x I J : 1.1—0 I314 (part 1 of 1‘] 10 130111tS -- \ -_ “Thieh of the following is the eorreet Lewis -. H. formula for hydrogen sulfide (HES)? H—e—o : 1' H—S—H 6. l .. H 3 H—sEH H—C—O 7- | i H 1 ILTI :m: .4. | E 1 0- H —s — H : 015 [part 1 o1'1:] 10 points How many lone pairs should appear on the central atom in the dot formula. for the chld rate ion 01037? 1. 3 2. 4 3. 1 t 4. 8 5. 2 016 (part 1 011:] 10 points How many double and triple bonds are in the compound CHECCHCHECCH? 1. 4 double, [J triple 2. 0 double, 3 triple 3. 2 double, 1 triple Lu. H.“ 4. 2 double. 2 triple 5. 1 double. 2 triple 017 (part 1 of 1:] 10 points The carbon—nitrogen bond in a. molecule of hy— drogen cyanide (HON) is what kind of bond? 1. douhle 2. single 3. ionic 4. triple c 018 {part 1 of 1:] 10 points Which of the compounds 1. 141013 11. are follow the octet rule? 111. con 1y. XeF4 1. 111 only _ _ _ _ _ _ _ 2. I only 3. 1V only 4. 111 and 1V only 5. l and 111 only U19 (part 1 or 1’] 1U points How many non—hondingr electron pairs are around 1 in IF??? 1.0 2.3t 3.4 4.2 |"..-'l l—n 020 [part 1 of 1:] 10 points ‘Nhich of the following exhibits resonance? 1. C1103? 2. oz 3. 1130+ 4. No; 021 [part 1 of 1:] 10 points How many resonance structures (forms) are there for ozone? 1.4 2.2 3.1 5. none 022 (part 1 of1:] 10 points For the electron dot representation of N203 :0" 3: — N = 0:: 3 / .. 2 :9 / what are the formal charges on each atom going from 1 to 3 in order? 1.11:0= —1 2. —1= —1,o 3.1=D:C 4. o: —1,o (LT! .1= 1:0 at): —1,1 7. o: —1, +2 023 (part 1 of 1:] 10 points What group must atom A come from in the Lewis dot structure for CHgA? The central atom is carbon. It is bonded to two hydrogen atoms and double—bonded to A. Both carbon and A follow the octet rule. 1. VA 2. 11A 3. 1iJIIA =1. IVA 5. 1iJlllA 6. 111A 7. IA 8. VIA 024 [part 1 of1] 10 points The Lewis dot structure of E;er has how many total electrons and how many shared (bonded) pairs? 1. 32; 3 2. 26; 1 3. 26; 3 025 [part 1 of 1:] 10 points Which of the following pairs of elements would he LEAST likely to form a compound by ionic in ending? 1. Al and F 2. C and NI 3. Na and S 4. Cs and O 5. Mg and Br 026 (part 1 of 1:] 10 points N2 has a shorter inter—nuclear distance than 02 because 1. False; 02 has a shorter bond length be— cause oxygen has six valence electrons hut nitrogen has only five. 2. N2 has a linear geometry whereas 02 is angular. 3. nitrogen is less electronegative than oxy— gen. 4. False; N2 and 02 have the same bond lengths because both are covalent diatomic molecules. 5. N2 is more polar than 02. 6. False; 02 has a shorter bond length be— cause oxygen is more electronegative than ni— trogen. 7. N2 is triple—bonded but 02 is double— bonded. 02'? (part 1 of 1:] 10 points Which of the following symbols is used to depict a partial positive charge? 1.0+ 2. 'v+ 6| 028 (part 1 of 1:] 10 points Given that elements C‘- and H are bonded and the electronegativity difierence is 0.4: deter— mine the hond type and the more—negative atom. 1. ionic bond; C 2. None of these 3. ionic bond; same electronegativity =1. nonpolar—covalent hond; C‘- 5. nonpolar—covalent bond; same electroneg— ativity 6. polar—covalent bond; C 7. ionic bond; H 8. nonpolar—covalent bond; H 9. polar—covalent bond; H 10. polar— covalent bond; same electronegativ— ity 029 (part 1 of 1:] 10 points TWhich pair of bonded atoms has the largest dipole moment? 1.C—F 2.0—0 3.C—l\' :1. (3—0 5. 3-1—31 030 (part 1 of 1:] 10 points The concept of “polarity” is due to 1. similarities in attractions for shared elec— trons when the electronic and molecular ge— ometries are the same. 2. the optimal Lewis dot structure for any pair of atoms. 3. atoms that are bonded together but have unequal numbers of lone pair electrons. 4. one atom of a bonded pair of atoms ex— erting a greater attractive force on the shared electrons. 5. the number of electrons shared by any pair of atoms. 001 (part 1 of 1:] 10 points Which of the following molecules has a dipole moment of zero? 1. SE; 2. BClgF 3. NFg 4. HO] (LT! . H30 002 (part 1 of 1:] 10 points Methane (CH4) is NOT a polar compound because 1. it contains no polar bonds. 2. it contains no unpaired electrons. 3. The statement is false; methane is a polar molecule. 4. carbon and hydrogen bonds are too simi— lar in electronegativity to form polar bonds. 5. the symmetry of the molecule causes the bond dipoles to cancel one another. 003 (part 1 of 1:] 10 points The smallest h ond angles would he found in a compound with which of the following molec— ular geometry? 1. pyramidal 2. angular 3. trigonal planar 4. square planar 5. linear 004 (part 1 of 1:] 10 points Which represents a false statement? 1. A11 square planar molecules have octaheL dral electronic arrangements. 2. The .5332 hybridization corresponds to trig— onal planar electronic arrangements. 3. All angular molecules are 5332 hybridized. 4. All molecules with four regions of electron density have ago-3 hybridization. What is the name of the molecular geome try shown below? (The dark circle and each white circle represent an atom.) ('N r“. 1. linear 2. trigonal pyramidal 3. trigonal planar 4. tetrahedral U! . square planar 6. quantemar y [I'DIB (part 1 of 10 points 'What is the electronic geometry of IF;r ? 1. trigonal planar 2. octahedral 3. tetrahedral 4. trigonal hipyrainidal 5. T—shaped 00'? [part 2 oi'3) 10 points What is the molecular geometry of IF;r ? 1. cubic 2. T—shaped 3. seesaw 4. square planar 5. tetrahedral 008 [part 3 of 10 points NH: is a cation that is non polar. Is IF;r also non polar T 1. Yes: it’s non polar. 2. No, its polar. 3. ICannot be determined from the structure 009 (part 1 of 1:] 10 points A molecule is known to have tetrahedral elec— tronic arrangement and angular molecular shape. Which of these molecules fits this description? 1. None fits this description. 2. H28 3. 803 4. BeClg s. I; 010 (part 1 of 1:] 10 points A molecule has six areas of high electron den— sity around the central atom and has two lone pairs on the central atom. What is the molec— ular geometry of this molecule? 1. tetrahedral 2. seesaw 3. square pyramidal 4. octahedral 5. square planar 011 (part 1 011:] 1013Oillt3 013 (part 2 of 10 points G1“? the hybridization or 93911 091113131 atom: What is the hybridization of the center atom nitrogenT middle carbon: right carbon, in structure (3? H .. | 1- 519 H —1\ = 0—0 —H | | 2, 5393 HO H 3. .s 3&2 1 5132: 3-133 51.13 p 1. .s 3d 2 5192. 5132. 5103 p 2 5. .s 3 5192. «5‘112 5112 p , 013 (part 2 of 3) 10 points 4 SPI SIDI Sp 1Vl1at is the hybridization of the center atom 111 structure C? 5 5193. 3-102. 5103 1. .sp 6. 3. - 3. 3 5P .51” . "P 2I SP3 7 3 5P 3p “p 3' 5p3d2 2 I ,I 2 3- 519 5310:0111 I I I 1L 513.36; 012 [part 1 oi BI] 10 polnts 5. 51::53 _ A 014 [part 3 of 10 points _ X _ A _ 3“ “‘- "*- “That is the shape of D? A B C- D For structure A, the electronic geometry is 1' 31191131 1. linear. 2. tetrahedral 2. octahedral. 3I cubic 3. cubic. I 4. lmear 4. trigona] hipyramidal, 5I tetmhedmlI 5. trigona] bipyramitlal 6. angular, 6. octahedral G16 (part 1 of 1:] 10 points 019 (part 1 of 1:] 10 points Which of the following has the highest lattice “That is the molecular shape of the 302013 energy? ion? 8 is the central atom. 1. IC-aO 1. tetrahedral 2. MgO 2. trigonal bipyramidal 3. NaCl 3. octahedral 4. K1 4. trigonal planar 5. BaO 5. seesaw 01'? [part 1 of1] 10 points The electronic arrangment is the same as the molecular shape when 6. trigonal pyramidal 020 [part 1 of 1:] 10 points Consider the compound peroxyacetylni— 1. the number of bonding orbitals equals the 1late: an eye “Imam: 11" Smog- munher of anti—bonding orbitals. O O . / \ .c" 2. there are no lone pairs of electrons on the / “NI central atom. CH3 — C N I O 3. there are more shared electrons than non— \O _ 0/ shared electrons. L I I Predict the indicated bond angle. 4. the atoms are joined by sigma bonds. 1. 30°” 5. the molecule is not polar. [118 (part 1 of 1:] 10 points 2 6.0.: Which of the following molecules is incorrectly ' matched with bond angles? 3. 900 1. 5101.. : 109.50 0 D D 4. 109.50 2. ALst : 90 , 120 5 and 18D 3. 31:3: 120° 5- 1200 4. SF5 : 90D and 180D 5. Belg : slightly less than lflgc 021 (part 1 of 1:] 10 points Which of the following has bond angles slightly LESS than 1200? 1. N03? 2. S03 3. SF2 4. I; 5. 03 022 [part 1 of 1:] 10 points A molecule has one lone non—bonded pair of electrons on the central atom and three atoms bonded to the central atom. What is its electronic arrangement and its hybridization? 1. trigonal planar; sp2 2. pyramidal; .sp2 3. tetrahedral; spa 4. trigonal planar; sp3 5. pyramidal; .3333 6. angular; 5333 Y. tetrahedral; 5393 023 (part 1 of 1:] 10 points NF; has (2, 3; 4) regions of high electron density and (2, 4; I3) bonded electrons. 1.4;4‘ 2.3;4 3.2;2 4.3;2 5.2;6 6.2;4 8.4;6 9. 4; 2 024 (part 1 of 1:] 10 points SC):— is moreyf'less polar than S03: because 1. more; the tetrahedral geometry increases the polarity of S — 0 bonds 2. less; it is a resonance structure. 3. more; it has four polar S —0 bonds rather than three polar S — 0 bonds. 4. less; the four polar S—O bonds in 803;— are symmetrical and cancel the dipole moments. 5. more; the four polar S —0 bonds in 803‘ are arranged non—symmetrically leaving a net dipole moment. 025 [part 1 of 1:] 10 points Which of the following molecules is incorrectly matched with molecular geometry? 1. ALng : trigonal bipyramidal 2. 1"er : octahedral 3. H20 : angular =1. SiF4 : tetrahedral . NHg : tetrahedral 026 (part 1 of 1:] 10 points How many or bonds and how many rr bonds are there in propanol (CflchgCHQOHj? “.7! 1.11;e. 2.9;2 3.10;1 4.9;1 5. 11; 1 [127 [part 1 of 1:] 10 points Which of the following is TRUE about anti— b ending orbitals of the molecular orbital the my? 1. Antibonding orbitals are higher in energy than their corresponding bonding orbitals. 2. Unshared electrons are placed in anti— bonding orbitals. 3. Antibonding orbitals are made from the overlap of sigma and pi orbitals, 4. The strongest bonds between atoms have no electrons in antibonding orbitals. 5. Although antibonding orbitals may ac— cept electrons; the electrons never remain for long. D28 (part 1 012) 10 points What is the expected bond order for the diatomic species BE? 1 1 ' 2 3 2. _ 2 3. 2 4. =1 5. 3 6. 1 029 (part 2 of 2:] 10 points What is the magnetism and number of un— paired electrons of Bg? 1. paramagnetic; 3 2. paramagnetic; 2 3. diamagnetic 4. paramagnetic; 1 5. paramagnetic; i1 030 [part 1 of 1:] 10 points Which species has the highest bond order? 1. N3 2.31; LL"! 0 m G31 (part 1 of1j 10 points TWrite out the molecular orbital configurations for 03. 02—, 03—. What is the bond order of each of the species [in the order 02: 02—, 03‘)? 1. 2; 1; U 1 2.1;—;U 3.1;1_;1 :3 =1. 2; —; 1 2. 2: 1 032 (part 1 of 1:] 10 points TWhich of the following statements is true about a molecule with a bond order of one? 5. 3; 1. The molecule has a single bond. 2. The molecule has no electrons in anti— bonding orbitals. 3. Two side—by—side p orbitals combine to form pi bond and pi antibond orbitals; there fore the bond order is 1. 4. The molecule is as stable as molecules with bond orders of two and three. 033 [part 1 of 1:] 10 points What is the bond order for 02‘? 1. 2 2.1.5 3.1 [134 [part 1 of 1:] 10 points The molecular orbital configuration of a di— atomic molecule is on} of; org“.2 0552 033,2 @331. The bond order for this molecule is 1.1. 2. 3. U! MID? Mill—l 035 (part 1 of 1:] 10 points When constructing an energy level diagram for a heteronuclear diatomic molecule the di— agram should be skewed because of the 1. di__erence in the number of electrons. 2. di__erence in atomic weights. 3. di‘Ierence in electronegativity. =1. diference in atomic sizes. ...
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This note was uploaded on 10/31/2009 for the course CH 56435 taught by Professor Flowers during the Spring '08 term at University of Texas at Austin.

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Chem Practice Exam 2 - 001 (part 1 of 1:] 10 points 1Write...

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