Chapter 9-10 practice test

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1 Class: Date: Chapter 9-10 practice test Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Which one of the following is most likely to be an ionic compound? A. CaCl 2 B. CO 2 C. CS 2 D. SO 2 2. Which one of the following is most likely to be an ionic compound? A. ClF 3 B. FeCl 3 C. NH 3 D. PF 3 3. Which one of the following is most likely to be a covalent compound? A. Rb 2 O B. BaO C. SrO D. SeO 2 4. Which one of the following is most likely to be a covalent compound? A. KF B. CaCl 2 C. SF 4 D. Al 2 O 3 5. Which one of the following compounds utilizes both ionic and covalent bonding? A. CO 2-3 B. Al 2 (SO 4 ) 3 C. CO 2 D. C 6 H 12 O 6 6. The Lewis dot symbol for the chloride ion is A. B. C. D. 1

2 7. Which of the following ionic solids would have the largest lattice energy? A. KF B. KI C. LiF D. LiI 8. Which of the following solids would have the highest melting point? A. NaF B. NaCl C. NaBr D. NaI 9. Which of the following solids would have the lowest melting point? A. KI B. KBr C. KCl D. KF 10. Use the Born-Haber cycle to calculate the lattice energy of KCl(s) given the following data: H(sublimation) K = 79.2 kj/mol I 1 (K) = kj/mol Bond energy (Cl Cl) = kj/mol EA (Cl) = 348 kj/mol H (KCl(s)) = kj/mol A. 165 kj/mol B. 288 kj/mol C. 629 kj/mol D. 707 kj/mol 11. Use the Born-Haber cycle to calculate the standard enthalpy of formation ( H ) for LiCl(s) given the following data: H(sublimation) Li = kj/mol I 1 (Li) = 520 kj/mol Bond energy (Cl Cl) = kj/mol EA (Cl) = 349 kj/mol Lattice energy (LiCl(s)) = 828 kj/mol A. 440 kj/mol B. 320 kj/mol C. 260 kj/mol D. 380 kj/mol 12. Which of the elements listed below would most likely form a polar covalent bond when bonded to oxygen? A. Mg B. H C. Al D. O 2

3 13. Which of the elements listed below would most likely form a nonpolar covalent bond when bonded to bromine? A. Rb B. Br C. C D. O 14. Define electronegativity: A. an atoms ability to attract electrons that are shared in a chemical bond B. an atoms ability to form an ionic bond with another atom C. an atoms ability to donate valence electrons to another atom D. an atoms ability to form a cation 15. Arrange the elements C, O, and H in order of increasing electronegativity A. C < O < H B. H < C < O C. C < H < O D. O < C < H 16. Which one of these polar covalent bonds would have the greatest percent ionic character? A. H Br B. H Cl C. H F D. H I 17. What type of chemical bond holds the atoms together within a water molecule? A. Ionic bond B. Nonpolar covalent bond C. Polar covalent bond D. Coordinate covalent bond 18. List all types of bonding present in the compound CaCO 3 I. ionic bond II. polar covalent bond III. nonpolar covalent bond A. I only B. II only C. III only D. I and II 19. The total number of valence electrons in the compound NH 4 NO 3 is A. 28 B. 30 C. 32 D. 42 3

4 20. The total number of valence electrons in the ion NH + 4 is A. 8 B. 9 C. 10 D The number of lone electron pairs in the N 2 molecule is. A. 1 B. 2 C. 3 D The Lewis structure reveals a triple bond in which of the following molecules? A. Br 2 B. O 2 C. N 2 D. H The Lewis structure reveals an unpaired electron (free radical) in which of the following species? A. NO - 3 B. N 2 O C. NO 2 D. NO The number of lone electron pairs in the NH + 4 ion is. A. 0 B. 1 C. 2 D The number of lone electron pairs in the ClO - 4 ion is. A. 3 B. 4 C. 6 D The number of resonance structures for the sulfur dioxide molecule that satisfy the octet rule is A. 1 B. 2 C. 3 D The number of resonance structures for the nitrate ion that satisfy the octet rule is A. 1 B. 2 C. 3 D. 4 4

5 28. The azide ion, N 3, is very reactive although it is isoelectronic with the very stable CO 2 molecule. This reactivity is reasonable considering that A. a Lewis structure cannot be written for the azide ion that has nitrogen formal charges of zero. B. there is no valid Lewis structure possible for the azide ion. C. there are resonance structures for azide ion but not for carbon dioxide. D. nitrogen cannot form multiple bonds. 29. Assuming the octet rule is obeyed, how many covalent bonds will a nitrogen atom form to give a formal charge of zero? A. 0 B. 1 C. 2 D What is the formal charge on the oxygen atom in N 2 O (the atomic order is N N O)? A. -2 B. -1 C. 0 D How many covalent bonds will be drawn to bromine in BrO 3 for the dot structure that expands the octet to minimize formal charge and if necessary places negative formal charges on the most electronegative atom(s). A. 3 B. 4 C. 5 D How many covalent bonds will be drawn to phosphorous in PO 3 4 for the dot structure that expands the octet to minimize formal charge and if necessary places negative formal charges on the most electronegative atom(s). A. 4 B. 5 C. 6 D The formal charge on the sulfur atom in the resonance structure of sulfur dioxide which has one single bond and one double bond is A. -2 B. -1 C. 0 D What is the formal charge on sulfur in the most favorable Lewis structure for the SCN (thiocyanate) ion based on minimizing formal charge overall? A. -2 B. -1 C. 0 D. +1 5

6 35. Nitrous oxide, N 2 O, is sometimes called laughing gas. What is the formal charge on the central nitrogen atom in the most favorable Lewis structure for nitrous oxide based on minimizing formal charge overall? (The atom connectivity is N N O.) A. 2 B. 1 C. 0 D What is the formal charge on the central nitrogen atom in the most favorable Lewis structure for the fulminate ion, CNO, based on minimizing formal charge overall? A. +2 B. +1 C. 0 D BeF 2 4 is called the fluoberyllate ion. The formal charge on the beryllium atom in this ion is A. 2 B. 1 C. 0 D For which of these species does the best Lewis structure have two or more equivalent resonance structures? A. HCO 2 B. SCN C. CNO D. N Estimate the enthalpy change for the combustion of one mole of acetylene, C 2 H 2, to form carbon dioxide and water vapor. BE(C H) = 456 kj/mol BE(C C) = 962 kj/mol BE(O=O) = 499 kj/mol BE(C=O) = 802 kj/mol BE(O H) = 462 kj/mol A kj/mol B kj/mol C. 155 kj/mol D kj/mol 40. The standard enthalpy of formation of ammonia at 25 C is 46.3 kj/mol. Estimate the N H bond enthalpy at this temperature. (Given: BE(N N)=941.4 kj/mol, BE(H H) = kj/mol) A. 360 kj/mol B. 383 kj/mol C. 391 kj/mol D. 459 kj/mol 6

7 41. Give the number of lone pairs around the central atom and the molecular geometry of IF 5. A. 0 lone pairs, square pyramidal B. 0 lone pairs, trigonal bipyramidal C. 1 lone pair, square pyramidal D. lone pair, octahedral 42. Give the number of lone pairs around the central atom and the geometry of the ion ClO 3. A. 0 lone pairs, trigonal B. 1 lone pair, bent C. 2 lone pairs, T-shaped D. 1 lone pair, trigonal pyramidal 43. According to the VSEPR theory, the geometry of the SO 3 molecule is A. pyramidal. B. tetrahedral. C. seesaw D. trigonal planar. 44. The correct hybridization for CS 2 molecule is best described as A. sp B. sp3 C. sp3d2 D. sp3d 45. The correct hybridization for boron trichloride is A. sp B. sp2 C. sp3 D. sp3d 46. According to the VSEPR theory, the molecular geometry of ammonia is A. linear B. trigonal pyramidal C. bent D. tetrahedral 47. According to VSEPR theory, which one of the following molecules is trigonal bipyramidal? A. PF 5 B. XeF 4 C. NF 3 D. SF Which one of the following molecules has tetrahedral geometry? A. XeF 4 B. BF 3 C. AsF 5 D. CF 4 7

8 49. Predict the geometry around the central atom in SO 2 4. A. trigonal planar B. trigonal pyramidal C. trigonal bipyramidal D. tetrahedral 50. Which of the following substances is/are bent? (i) H 2 S (ii). CO 2 (iii) ClNO (iv) NH 2 (v) O 3 A. only (iii) B. (i) and (v) C. (i), (ii), (iii), and (v) D. (i), (iii), (iv) and (v) 51. A molecule with 3 single bonds and 1 lone pair of electrons around the central atom is predicted to have what type of molecular geometry? A. Tetrahedral B. Trigonal pyramidal C. Trigonal bipyramidal D. Bent 52. A central atom with 4 electron pairs (single bonds and/or lone pairs of electrons) could have which of the following molecular geometries? I. Trigonal bipyramidal II. Tetrahedral III. Trigonal pyramidal IV. Bent A. I and II B. II and III C. II, III, and IV D. I and IV 53. The bond angles in SCl 2 are expected to be A. a little less than B C. a little more than D Which of the following molecules has polar bonds but is a nonpolar molecule? A. PCl 3 B. NCl 3 C. BF 3 D. HF 55. Which of the following molecules has polar bonds but is a nonpolar molecule A. CO B. CO 2 C. CHCl 3 D. Cl 2 8

9 56. Which one of the following molecules is polar? A. PBr 5 B. CCl 4 C. BrF 5 D. XeF Predict the geometry and polarity of the CS 2 molecule. A. linear, polar B. linear, nonpolar C. tetrahedral, nonpolar D. bent, nonpolar 58. Which of the following species has the largest dipole moment (i.e., is the most polar)? A. CH 4 B. CH 3 Br C. CH 3 Cl D. CH 3 F 59. Which of the following species have the same geometries? A. NH 2 and H 2 O B. NH 2 and BeH 2 C. H 2 O and BeH 2 D. NH 2, H 2 O, and BeH Give the number of lone pairs around the central atom and the molecular geometry of CBr 4. A. 0 lone pairs, square planar B. 0 lone pairs, tetrahedral C. 1 lone pair, trigonal bipyramidal D. 1 lone pair, square pyramidal 61. Give the number of lone pairs around the central atom and the molecular geometry of XeF 2. A. 0 lone pairs, linear B. 1 lone pair, bent C. 3 lone pairs, linear D. 2 lone pairs, bent 62. The geometry of the SF 4 molecule is A. seesaw B. trigonal pyramidal. C. square planar. D. trigonal planar. 63. Use VSEPR theory to predict the geometry of the PCl 3 molecule. A. linear B. bent C. trigonal planar D. trigonal pyramidal 9

10 64. According to the VSEPR theory, which one of the following species is linear? A. H 2 S B. HCN C. BF 3 D. H 2 CO 65. According to VSEPR theory, which one of the following species has a tetrahedral geometry? A. IF + 4 B. IF 4 C. PCl + 4 D. PCl Predict the geometry around the central atom in PO 3 4. A. trigonal planar B. trigonal pyramidal C. trigonal bipyramidal D. tetrahedral 67. A central atom with 5 electron pairs (single bonds and/or lone pairs of electrons) could have which of the following molecular geometries? I. Trigonal bipyramidal II. Seesaw III. T-shaped IV. Linear A. I, II, and III B. II, III, and IV C. I, II, III and IV D. I, III, and IV 68. The F Cl F bond angles in ClF 3 are expected to be approximately A. 90 and 180. B only. C. 180 only. D. 120 only. 69. The C N O bond angle in nitromethane, CH 3 NO 2, is expected to by approximately A. 60 B. 90 C D Which one of the following molecules is nonpolar? A. NH 3 B. BeCl 2 C. CH 3 Cl D. H 2 O 10

11 71. Which one of the following molecules has a non-zero dipole moment? A. BeCl 2 B. Br 2 C. BF 3 D. IBr 72. Which one of the following molecules has a zero dipole moment? A. CO B. CH 2 Cl 2 C. SO 3 D. SO Which of the following species has the largest dipole moment (i.e., is the most polar)? A. H 2 B. H 2 O C. H 2 S D. H 2 Se Short Answer 1. Use VSEPR theory to predict the molecular geometry of CO According to VSEPR theory, which of the following triatomic ions should be linear: N 3, I 3, NO 2, ClO 2, SCN. 3. Using periodic trends, arrange the following molecules in order of increasing dipole moment: NH 3, PH 3, AsH Explain why CO 2 is nonpolar, but OCS is polar. 5. Which of the following molecules has polar bonds but is a nonpolar molecule? PCl 5, PCl 3, NCl 3 and CO 2 6. According to the VSEPR theory, the geometrical structure of PF 5 is 7. Draw a Lewis structure for PF 5 that shows the correct atom arrangement predicted by VSEPR theory. 8. How does the geometrical structure of PF 5 differ from that of IF 5? 9. Ozone (O 3 ) is an allotropic form of oxygen. Use VSEPR theory to predict the shape of the ozone molecule. 11

12 Chapter 9-10 practice test Answer Section MULTIPLE CHOICE 1. ANS: A PTS: 1 DIF: Easy REF: Section: 9.2 OBJ: EK.2.C.2 2. ANS: B PTS: 1 DIF: Easy REF: Section: 9.2 OBJ: EK.2.C.2 3. ANS: D PTS: 1 DIF: Easy REF: Section: ANS: C PTS: 1 DIF: Easy REF: Section: ANS: B PTS: 1 DIF: Medium REF: Section: ANS: B PTS: 1 DIF: Medium REF: Section: ANS: C PTS: 1 DIF: Medium REF: Section: 9.3 OBJ: EK.2.C.2 8. ANS: A PTS: 1 DIF: Medium REF: Section: ANS: A PTS: 1 DIF: Medium REF: Section: ANS: D PTS: 1 DIF: Difficult REF: Section: 9.3 OBJ: EK.5.C ANS: D PTS: 1 DIF: Difficult REF: Section: 9.3 OBJ: EK.5.C ANS: B PTS: 1 DIF: Medium REF: Section: ANS: B PTS: 1 DIF: Medium REF: Section: ANS: A PTS: 1 DIF: Easy REF: Section: ANS: B PTS: 1 DIF: Medium REF: Section: ANS: C PTS: 1 DIF: Medium REF: Section: ANS: C PTS: 1 DIF: Medium REF: Section: ANS: D PTS: 1 DIF: Medium REF: Section: ANS: C PTS: 1 DIF: Medium REF: Section: ANS: A PTS: 1 DIF: Medium REF: Section: ANS: B PTS: 1 DIF: Medium REF: Section: 9.6 1

13 22. ANS: C PTS: 1 DIF: Medium REF: Section: ANS: C PTS: 1 DIF: Medium REF: Section: ANS: A PTS: 1 DIF: Medium REF: Section: ANS: D PTS: 1 DIF: Medium REF: Section: ANS: B PTS: 1 DIF: Difficult REF: Section: ANS: C PTS: 1 DIF: Difficult REF: Section: ANS: A PTS: 1 DIF: Difficult REF: Section: ANS: D PTS: 1 DIF: Medium REF: Section: ANS: B PTS: 1 DIF: Difficult REF: Section: ANS: C PTS: 1 DIF: Difficult REF: Section: ANS: B PTS: 1 DIF: Difficult REF: Section: ANS: D PTS: 1 DIF: Difficult REF: Section: ANS: C PTS: 1 DIF: Difficult REF: Section: ANS: D PTS: 1 DIF: Difficult REF: Section: ANS: B PTS: 1 DIF: Difficult REF: Section: ANS: A PTS: 1 DIF: Medium REF: Section: ANS: A PTS: 1 DIF: Medium REF: Section: ANS: D PTS: 1 DIF: Difficult REF: Section: 9.10 OBJ: EK.5.C ANS: C PTS: 1 DIF: Difficult REF: Section: 9.10 OBJ: EK.5.C ANS: C PTS: 1 DIF: Medium REF: Section: ANS: D PTS: 1 DIF: Medium REF: Section: ANS: D PTS: 1 DIF: Medium REF: Section: ANS: A PTS: 1 DIF: Medium REF: Section: ANS: B PTS: 1 DIF: Medium REF: Section:

14 46. ANS: B PTS: 1 DIF: Medium REF: Section: ANS: A PTS: 1 DIF: Medium REF: Section: ANS: D PTS: 1 DIF: Medium REF: Section: ANS: D PTS: 1 DIF: Medium REF: Section: ANS: D PTS: 1 DIF: Medium REF: Section: ANS: B PTS: 1 DIF: Medium REF: Section: ANS: C PTS: 1 DIF: Medium REF: Section: ANS: A PTS: 1 DIF: Medium REF: Section: ANS: C PTS: 1 DIF: Medium REF: Section: ANS: B PTS: 1 DIF: Medium REF: Section: ANS: C PTS: 1 DIF: Medium REF: Section: ANS: B PTS: 1 DIF: Medium REF: Section: ANS: D PTS: 1 DIF: Medium REF: Section: ANS: A PTS: 1 DIF: Medium REF: Section: ANS: B PTS: 1 DIF: Medium REF: Section: ANS: C PTS: 1 DIF: Medium REF: Section: ANS: A PTS: 1 DIF: Medium REF: Section: ANS: D PTS: 1 DIF: Medium REF: Section: ANS: B PTS: 1 DIF: Medium REF: Section: ANS: C PTS: 1 DIF: Medium REF: Section: ANS: D PTS: 1 DIF: Medium REF: Section: ANS: C PTS: 1 DIF: Medium REF: Section: ANS: A PTS: 1 DIF: Medium REF: Section: ANS: D PTS: 1 DIF: Medium REF: Section:

15 70. ANS: B PTS: 1 DIF: Medium REF: Section: ANS: D PTS: 1 DIF: Medium REF: Section: ANS: C PTS: 1 DIF: Medium REF: Section: ANS: B PTS: 1 DIF: Medium REF: Section: 10.2 SHORT ANSWER 1. ANS: trigonal planar PTS: 1 DIF: Medium REF: Section: ANS: N 3, I 3, and SCN are linear PTS: 1 DIF: Medium REF: Section: ANS: AsH 3 < PH 3 < NH 3 PTS: 1 DIF: Medium REF: Section: ANS: In CO 2 the two bond moments point in opposite directions and are of equal magnitude. Therefore, they cancel. In OCS, even though the two bond moments point in opposite directions, they are not of the same magnitude and do not cancel. PTS: 1 DIF: Medium REF: Section: ANS: PCl 5 and CO 2 PTS: 1 DIF: Medium REF: Section: ANS: trigonal bipyramidal PTS: 1 DIF: Medium REF: Section:

16 7. ANS: PTS: 1 DIF: Medium REF: Section: ANS: PF 5 is trigonal bipyramidal, whereas IF 5 is square pyramidal PTS: 1 DIF: Medium REF: Section: ANS: Bent PTS: 1 DIF: Medium REF: Section:

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