ExamView - ch 9-10 practice test.tst

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Name: ________________________ 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. CaCl2
B. CO2
C. CS2
D. SO2
____
2. Which one of the following is most likely to be an ionic compound?
A. ClF3
B. FeCl3
C. NH3
D. PF3
____
3. Which one of the following is most likely to be a covalent compound?
A. Rb2O
B. BaO
C. SrO
D. SeO2
____
4. Which one of the following is most likely to be a covalent compound?
A. KF
B. CaCl2
C. SF4
D. Al2O3
____
5. Which one of the following compounds utilizes both ionic and covalent bonding?
A. CO32B. Al2(SO4)3
C. CO2
D. C6H12O6
____
6. The Lewis dot symbol for the chloride ion is
A.
–
B.
C.
D.
–
–
1
ID: A
Name: ________________________
ID: A
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
I1 (K) = 418.7 kJ/mol
Bond energy (Cl–Cl) = 242.8 kJ/mol
EA (Cl) = 348 kJ/mol
H (KCl(s)) = –435.7 kJ/mol
–165 kJ/mol
288 kJ/mol
629 kJ/mol
707 kJ/mol
____ 11. Use the Born-Haber cycle to calculate the standard enthalpy of formation (H ) for LiCl(s) given
A.
B.
C.
D.
the following data:
H(sublimation) Li = 155.2 kJ/mol
I1 (Li) = 520 kJ/mol
Bond energy (Cl–Cl) = 242.7 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
Name: ________________________
ID: A
____ 13. Which of the elements listed below would most likely form a nonpolar covalent bond when bonded
____ 14.
____ 15.
____ 16.
____ 17.
____ 18.
to bromine?
A. Rb
B. Br
C. C
D. O
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
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
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
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
List all types of bonding present in the compound CaCO3
I.
ionic bond
II.
polar covalent bond
III. nonpolar covalent bond
I only
II only
III only
I and II
____ 19. The total number of valence electrons in the compound NH4NO3 is
A. 28
B. 30
C. 32
D. 42
A.
B.
C.
D.
3
Name: ________________________
ID: A
____ 20. The total number of valence electrons in the ion NH4+ is
A. 8
B. 9
C. 10
D. 17
____ 21. The number of lone electron pairs in the N2 molecule is ___.
A. 1
B. 2
C. 3
D. 4
____ 22. The Lewis structure reveals a triple bond in which of the following molecules?
A. Br2
B. O2
C. N2
D. H2
____ 23. The Lewis structure reveals an unpaired electron (free radical) in which of the following species?
A. NO3B. N2O
C. NO2
D. NO2____ 24. The number of lone electron pairs in the NH4+ ion is ___.
A. 0
B. 1
C. 2
D. 3
____ 25. The number of lone electron pairs in the ClO4- ion is ___.
A. 3
B. 4
C. 6
D. 12
____ 26. The number of resonance structures for the sulfur dioxide molecule that satisfy the octet rule is
A. 1
B. 2
C. 3
D. 4
____ 27. The number of resonance structures for the nitrate ion that satisfy the octet rule is
A. 1
B. 2
C. 3
D. 4
4
Name: ________________________
ID: A
____ 28. The azide ion, N3–, is very reactive although it is isoelectronic with the very stable CO 2 molecule.
____ 29.
____ 30.
____ 31.
____ 32.
____ 33.
____ 34.
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.
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. 3
What is the formal charge on the oxygen atom in N2O (the atomic order is N–N–O)?
A. -2
B. -1
C. 0
D. +1
How many covalent bonds will be drawn to bromine in BrO3– 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. 6
How many covalent bonds will be drawn to phosphorous in PO43– 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. 7
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. +1
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
Name: ________________________
ID: A
____ 35. Nitrous oxide, N2O, is sometimes called “laughing gas”. What is the formal charge on the central
____ 36.
____ 37.
____ 38.
____ 39.
____ 40.
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. +1
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. –1
BeF42– is called the fluoberyllate ion. The formal charge on the beryllium atom in this ion is
A. –2
B. –1
C. 0
D. +1
For which of these species does the best Lewis structure have two or more equivalent resonance
structures?
A. HCO2–
B. SCN–
C. CNO–
D. N3–
Estimate the enthalpy change for the combustion of one mole of acetylene, C2H2, 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. +1010 kJ/mol
B. +653 kJ/mol
C. –155 kJ/mol
D. –1010 kJ/mol
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) = 436.4 kJ/mol)
A. 360 kJ/mol
B. 383 kJ/mol
C. 391 kJ/mol
D. 459 kJ/mol
6
Name: ________________________
ID: A
____ 41. Give the number of lone pairs around the central atom and the molecular geometry of IF5.
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 ClO3–.
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 SO3 molecule is
A. pyramidal.
B. tetrahedral.
C. seesaw
D. trigonal planar.
____ 44. The correct hybridization for CS2 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. PF5
B. XeF4
C. NF3
D. SF6
____ 48. Which one of the following molecules has tetrahedral geometry?
A. XeF4
B. BF3
C. AsF5
D. CF4
7
Name: ________________________
ID: A
____ 49. Predict the geometry around the central atom in SO42–.
A. trigonal planar
B. trigonal pyramidal
C. trigonal bipyramidal
D. tetrahedral
____ 50. Which of the following substances is/are bent?
(i) H2S
(ii). CO2
(iii) ClNO (iv) NH2– (v) O3
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
I and II
II and III
II, III, and IV
I and IV
____ 53. The bond angles in SCl2 are expected to be
A. a little less than 109.5°.
B. 109.5°
C. a little more than 109.5°.
D. 120°.
____ 54. Which of the following molecules has polar bonds but is a nonpolar molecule?
A. PCl3
B. NCl3
C. BF3
D. HF
____ 55. Which of the following molecules has polar bonds but is a nonpolar molecule
A. CO
B. CO2
C. CHCl3
D. Cl2
A.
B.
C.
D.
8
Name: ________________________
ID: A
____ 56. Which one of the following molecules is polar?
A. PBr5
B. CCl4
C. BrF5
D. XeF2
____ 57. Predict the geometry and polarity of the CS2 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. CH4
B. CH3Br
C. CH3Cl
D. CH3F
____ 59. Which of the following species have the same geometries?
A. NH2– and H2O
B. NH2– and BeH2
C. H2O and BeH2
D. NH2–, H2O, and BeH2
____ 60. Give the number of lone pairs around the central atom and the molecular geometry of CBr4.
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 XeF2.
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 SF4 molecule is
A. seesaw
B. trigonal pyramidal.
C. square planar.
D. trigonal planar.
____ 63. Use VSEPR theory to predict the geometry of the PCl3 molecule.
A. linear
B. bent
C. trigonal planar
D. trigonal pyramidal
9
Name: ________________________
ID: A
____ 64. According to the VSEPR theory, which one of the following species is linear?
A. H2S
B. HCN
C. BF3
D. H2CO
____ 65. According to VSEPR theory, which one of the following species has a tetrahedral geometry?
A. IF4+
B. IF4–
C. PCl4+
D. PCl4–
____ 66. Predict the geometry around the central atom in PO43–.
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
I, II, and III
II, III, and IV
I, II, III and IV
I, III, and IV
____ 68. The F –Cl –F bond angles in ClF3 are expected to be approximately
A. 90° and 180°.
B. 109.5° only.
C. 180° only.
D. 120° only.
____ 69. The C–N–O bond angle in nitromethane, CH3NO2, is expected to by approximately
A. 60°
B. 90°
C. 109.5°
D. 120°
____ 70. Which one of the following molecules is nonpolar?
A. NH3
B. BeCl2
C. CH3Cl
D. H2O
A.
B.
C.
D.
10
Name: ________________________
ID: A
____ 71. Which one of the following molecules has a non-zero dipole moment?
A. BeCl2
B. Br2
C. BF3
D. IBr
____ 72. Which one of the following molecules has a zero dipole moment?
A. CO
B. CH2Cl2
C. SO3
D. SO2
____ 73. Which of the following species has the largest dipole moment (i.e., is the most polar)?
A. H2
B. H2O
C. H2S
D. H2Se
Short Answer
1. Use VSEPR theory to predict the molecular geometry of CO32–.
2. According to VSEPR theory, which of the following triatomic ions should be linear: N 3–, I3–, NO2–,
ClO2–, SCN–.
3. Using periodic trends, arrange the following molecules in order of increasing dipole moment: NH 3,
PH3, AsH3.
4. Explain why CO2 is nonpolar, but OCS is polar.
5. Which of the following molecules has polar bonds but is a nonpolar molecule? PCl5, PCl3, NCl3 and
CO2
6. According to the VSEPR theory, the geometrical structure of PF5 is
7. Draw a Lewis structure for PF5 that shows the correct atom arrangement predicted by VSEPR
theory.
8. How does the geometrical structure of PF5 differ from that of IF5?
9. Ozone (O3) is an allotropic form of oxygen. Use VSEPR theory to predict the shape of the ozone
molecule.
11
ID: A
Chapter 9-10 practice test
Answer Section
MULTIPLE CHOICE
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A
EK.2.C.2
B
EK.2.C.2
D
EK.2.C.1
C
EK.2.C.1
B
EK.2.C.1
B
EK.2.C.4
C
EK.2.C.2
A
EK.2.C.1
A
EK.2.C.1
D
EK.5.C.2
D
EK.5.C.2
B
EK.2.C.1
B
EK.2.C.1
A
EK.2.C.1
B
EK.2.C.1
C
EK.2.C.1
C
EK.2.C.1
D
EK.2.C.1
C
EK.2.C.4
A
EK.2.C.4
B
EK.2.C.4
PTS: 1
DIF: Easy
REF: Section: 9.2
PTS: 1
DIF: Easy
REF: Section: 9.2
PTS: 1
DIF: Easy
REF: Section: 9.4
PTS: 1
DIF: Easy
REF: Section: 9.4
PTS: 1
DIF: Medium
REF: Section: 9.4
PTS: 1
DIF: Medium
REF: Section: 9.2
PTS: 1
DIF: Medium
REF: Section: 9.3
PTS: 1
DIF: Medium
REF: Section: 9.3
PTS: 1
DIF: Medium
REF: Section: 9.3
PTS: 1
DIF: Difficult
REF: Section: 9.3
PTS: 1
DIF: Difficult
REF: Section: 9.3
PTS: 1
DIF: Medium
REF: Section: 9.5
PTS: 1
DIF: Medium
REF: Section: 9.5
PTS: 1
DIF: Easy
REF: Section: 9.5
PTS: 1
DIF: Medium
REF: Section: 9.5
PTS: 1
DIF: Medium
REF: Section: 9.5
PTS: 1
DIF: Medium
REF: Section: 9.5
PTS: 1
DIF: Medium
REF: Section: 9.5
PTS: 1
DIF: Medium
REF: Section: 9.6
PTS: 1
DIF: Medium
REF: Section: 9.6
PTS: 1
DIF: Medium
REF: Section: 9.6
1
ID: A
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C
EK.2.C.4
C
EK.2.C.4
A
EK.2.C.4
D
EK.2.C.4
B
EK.2.C.4
C
EK.2.C.4
A
EK.2.C.4
D
EK.2.C.4
B
EK.2.C.4
C
EK.2.C.4
B
EK.2.C.4
D
EK.2.C.4
C
EK.2.C.4
D
EK.2.C.4
B
EK.2.C.4
A
EK.2.C.4
A
EK.2.C.4
D
EK.5.C.2
C
EK.5.C.2
C
EK.2.C.4
D
EK.2.C.4
D
EK.2.C.4
A
EK.2.C.4
B
EK.2.C.4
PTS: 1
DIF: Medium
REF: Section: 9.6
PTS: 1
DIF: Medium
REF: Section: 9.6
PTS: 1
DIF: Medium
REF: Section: 9.6
PTS: 1
DIF: Medium
REF: Section: 9.6
PTS: 1
DIF: Difficult
REF: Section: 9.8
PTS: 1
DIF: Difficult
REF: Section: 9.8
PTS: 1
DIF: Difficult
REF: Section: 9.7
PTS: 1
DIF: Medium
REF: Section: 9.7
PTS: 1
DIF: Difficult
REF: Section: 9.7
PTS: 1
DIF: Difficult
REF: Section: 9.9
PTS: 1
DIF: Difficult
REF: Section: 9.9
PTS: 1
DIF: Difficult
REF: Section: 9.7
PTS: 1
DIF: Difficult
REF: Section: 9.7
PTS: 1
DIF: Difficult
REF: Section: 9.7
PTS: 1
DIF: Difficult
REF: Section: 9.7
PTS: 1
DIF: Medium
REF: Section: 9.7
PTS: 1
DIF: Medium
REF: Section: 9.8
PTS: 1
DIF: Difficult
REF: Section: 9.10
PTS: 1
DIF: Difficult
REF: Section: 9.10
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
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ID: A
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EK.2.C.4
A
EK.2.C.4
D
EK.2.C.4
D
EK.2.C.4
D
EK.2.C.4
B
EK.2.C.4
C
EK.2.C.4
A
EK.2.C.4
C
EK.2.C.1
B
EK.2.C.1
C
EK.2.C.1
B
EK.2.C.1
D
EK.2.C.1
A
EK.2.C.4
B
EK.2.C.4
C
EK.2.C.4
A
EK.2.C.4
D
EK.2.C.4
B
EK.2.C.4
C
EK.2.C.4
D
EK.2.C.4
C
EK.2.C.4
A
EK.2.C.4
D
EK.2.C.4
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.2
PTS: 1
DIF: Medium
REF: Section: 10.2
PTS: 1
DIF: Medium
REF: Section: 10.2
PTS: 1
DIF: Medium
REF: Section: 10.2
PTS: 1
DIF: Medium
REF: Section: 10.2
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
PTS: 1
DIF: Medium
REF: Section: 10.1
3
ID: A
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B
EK.2.C.1
D
EK.2.C.1
C
EK.2.C.1
B
EK.2.C.1
PTS: 1
DIF: Medium
REF: Section: 10.2
PTS: 1
DIF: Medium
REF: Section: 10.2
PTS: 1
DIF: Medium
REF: Section: 10.2
PTS: 1
DIF: Medium
REF: Section: 10.2
DIF: Medium
REF: Section: 10.1
SHORT ANSWER
1. ANS:
trigonal planar
PTS: 1
OBJ: EK.2.C.4
2. ANS:
N3–, I3–, and SCN– are linear
PTS: 1
OBJ: EK.2.C.4
3. ANS:
DIF: Medium
REF: Section: 10.1
DIF: Medium
REF: Section: 10.2
AsH3 < PH3 < NH3
PTS: 1
OBJ: EK.2.C.1
4. ANS:
In CO2 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
OBJ: EK.2.C.1
5. ANS:
DIF: Medium
REF: Section: 10.2
DIF: Medium
REF: Section: 10.2
PCl5 and CO2
PTS: 1
OBJ: EK.2.C.1
6. ANS:
trigonal bipyramidal
PTS: 1
OBJ: EK.2.C.4
DIF: Medium
REF: Section: 10.1
4
ID: A
7. ANS:
PTS: 1
OBJ: EK.2.C.4
8. ANS:
DIF: Medium
REF: Section: 10.1
PF5 is trigonal bipyramidal, whereas IF5 is square pyramidal
PTS: 1
OBJ: EK.2.C.4
9. ANS:
DIF: Medium
REF: Section: 10.1
DIF: Medium
REF: Section: 10.1
Bent
PTS: 1
OBJ: EK.2.C.4
5
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