Download AP CHEMISTRY – Source: 1999 AP Exam CHAPTER 8 TEST

AP CHEMISTRY – Source: 1999 AP Exam
CHAPTER 8 TEST BONDING-KEY
SECTION I
NO CALCULATORS MAY BE USED WITH SECTION I.
Use your time effectively, working as rapidly as you can without losing accuracy. Do not spend too much time on
questions that are too difficult. Go on to the other questions and come back to the difficult ones later if you have time.
It is not expected that everyone will be able to answer all the multiple-choice questions.
Part A
Directions: Each set of lettered choice below refers to the numbered statement immediately following it. Select the
one lettered choice that best fits each statement and then fill in the corresponding oval on the answer sheet. A
choice may be used once, more than once, or not at all in each set.
Use these answers for questions 1 - 4
(A) hydrogen bonding
(B) hybridization
(C) ionic bonding
(D) resonance
(E) van der Waals forces (London dispersion forces)
1. Is used to explain why iodine molecules are held together in the solid state
2. Is used to explain why the boiling point of HF is greater than the boiling point of HBr
3. Is used to explain the fact that the four bonds in methane are equivalent
4. Is used to explain the fact that the carbon-to-carbon bonds in benzene, C6H6, are identical
E
A
B
D
Questions 5-8 refer to the following types of energy.
(A) Activation energy
(D) Kinetic energy
(B) Free energy
(E) Lattice energy
(C) Ionization energy
5.The energy required to convert a ground-state atom in the gas phase to a gaseous positive ion: C
6.The energy change that occurs in the conversion of an ionic solid to widely separated gaseous ions: E
7.The energy in a chemical or physical change that is available to do useful work: B
8.The energy required to form the transition state in a chemical reaction: A
9. The electron-dot structure (Lewis structure) for which of the following molecules would have two unshared pairs
of electrons on the central atom?
(A) H2S
(B) NH3
(C) CH4
(D) HCN
(E) CO2
(D) NH3
(E) H2S
10. Which of the following molecules has a dipole moment of zero?
(A) C6H6 (benzene)
(B) NO
(C) SO2
11. I2(g) + 3 Cl2(g) ---> 2 ICl3(g)
According to the data in the table below, what is the value
of ∆H° for the reaction represented above?
(A) - 870 kJ
(D) + 450 kJ
(B) - 390 kJ
(C) + 180 kJ
(E) + 1,260 kJ
Bond
I---I
Cl---Cl
I---Cl
Average Bond Energy (kj/mole)
150
240
210
12) Given the electronegativities below, which covalent single bond is most polar?
Element:
H
C
N
O
Electronegativity:
2.1
2.5
3.0 3.5
A) C-H
B) N-H
C) O-H
D) O-C
E) O-N
13) The Lewis structure of AsH 3 shows __________ nonbonding electron pair(s) on As.
A) 0
B) 1
C) 2
D) 3
E) Cannot be determined from the data given.
14) The formal charge on carbon in the molecule below is __________.
B) +1
C) +2
D) +3
A) 0
E) -1
15) Using the table of average bond energies below, the H for the reaction is __________ kJ.
H  C  C  H (g) + H  I (g)  H2C  CHI (g)
Bond:
D (kJ/mol):
A) +506
C≡C
839
C=C
614
H-I
299
B) -931
C-I
240
C-H
413
C) -506
D) -129
16) The chloride of which of the following metals should have the greatest lattice energy?
A) potassium
B) rubidium
C) sodium
D) lithium
E) +129
E) cesium
The diagram below is the Born-Haber cycle for the formation of crystalline potassium fluoride.
17) Which energy change corresponds to the first ionization energy of potassium?
A) 2
B) 5
C) 4
D) 3
E) 6
18) In which of the molecules below is the carbon-carbon distance the shortest?
A) H2C  CH2
C) H3C  CH3
E) H3C  CH2  CH3
B) H  C  C  H
D) H2C  C  CH2
19) Which of the following has the bonds correctly arranged in order of increasing polarity?
A) Be  F, Mg  F, N  F, O  F
D) N  F, Be  F, Mg  F, O  F
B) O  F, N  F, Be  F, Mg  F
E) M  gF, Be  F, N  F, O  F
C) O  F, Be  F, Mg  F, N  F
20. CH3CH2OH boils at 78 °C and CH3OCH3 boils at - 24 °C, although both compounds have the same
composition. This difference in boiling points may be attributed to a difference in
(A) molecular mass (B) density (C) specific heat (D) hydrogen bonding (E) heat of combustion
21. What mass of Au is produced when 0.0500 mol of Au2S3 is reduced completely with excess H2?
SOLUTION: 0.0500 mol Au2S3 x 2 mol Au x 197g Au = 19.7 g Au
1 mol Au2S3
1 mol Au
TIP TO SOLVE W/O CALCULATOR: 0.05 x 2 = 0.10 multiply 0.10 x 197 means move decimal one
place to left.
(A)9.85 g
(B) 19.7 g
(C) 24.5 g
(D) 39.4 g
(E) 48.9 g
22. When a solution of potassium chloride is vaporized in a flame, the color of the flame is
(A)blue
(B) yellow
(C) green
(D) violet
(E) white
C10H12O4S(s) + 12O2(g) 10CO2 + SO2(g) + 6H2O(g)
23 . When the equation above is balanced and all coefficients are reduced to their lowest whole-number
terms, the coefficient for O2(g) is
(A) 6
(B) 7
(C) 12
(D) 14
(E) 28
24.The melting point of MgO is higher than that of NaF. Explanations for this observation include which
of the following?
I. Mg 2+ is more positively charged than Na+.
II. O 2- is more negatively charged than F-.
III.The O 2- ion is smaller than the F- ion
(A)II only
(B) I and II only
(C)I and III only
(D)II and III only
(E) I, II, and III
25. H2Se(g) + 4 O2F2(g)  SeF6(g) + 2 HF(g) + 4 O2(g)
Which of the following is true regarding the reaction represented above?
(A) The oxidation number of O does not change.
(B) The oxidation number of H changes from -1 to +1
(C) The oxidation number of F changes from +1 to -1.
(D) The oxidation number of Se changes from -2 to +6
(D) It is a disproportionation reaction for F.
+1 -2
+1 -1
+6 -1
+1 -1
0
H2Se(g) + 4 O2F2(g)  SeF6(g) + 2 HF(g) + 4 O2(g)
26.A 1.0 L sample of an aqueous solution contains 0.10 mol of NaCl and 0.10 mol of CaCl2. What is the
minimum number of moles of AgNO3 that must be added to the solution in order to precipitate all of the
Cl- as AgCl(s)? (Assume that AgCl is insoluble.)
ANALYSIS: 0.10 mol of NaCl contains 0.10 mol Cl- ions. 0.10 mol of CaCl2 contains 0.20 mol Cl- ions. Ag
& Cl combine in a 1 mol:1 mol ratio and 1 mol AgNO3 produces 1 mol Ag+ ions. SO, if you have 0.30 mol
Cl- ions, you need 0.30 mol Ag+ ions, and therefore, 0.30 mol AgNO3.
(A)0.10 mol
(B) 0.20 mol
(C)0.30 mol
(D) 0.40 mol
(E) 0.60 mol
27. Of the following molecules, which has the largest dipole moment?
(A) CO
(B) CO2
(C) O2
(D) HF
(E) F2
ANALYSIS: Immediately rule out B, C and E. There is no dipole moment for any of these molecules! C &
E b/c they are diatomic. There is NO difference in electronegativity, therefore, no dipole moment can
exist! For B, CO2 has a bond dipole for each of the two C=O bonds. However, this is a linear molecule
and the dipoles face in opposite directions, creating no + and – end of the molecule. Of the remaining
choices, A) CO and D) HF, electronegativity differences between the 2 atoms is greater in D. This means
that Q is greater for HF. In addition, a CO triple bond is VERY short, whereas an HF single bond is not as
short. This means r is also greater for HF. Since Dipole moment, μ = Qr, μ is greater for HF than it is for
CO.
28. When hafnium metal is heated in an atmosphere of chlorine gas, the product of the reaction is
found to contain 62.2 percent Hf by mass and 37.4 percent Cl by mass. What is the empirical formula for
this compound?
(A) HfCl
(B) HfCl2
(C) HfCl3
(D) HfCl4
(E) Hf2Cl3
ANALYSIS: Assume 62.2 g Hf & 37.4 g Cl (this gives approx 100 g of product). Use each of these assumed
masses to determine the moles of each element present in the compound.
62.2 g Hf x 1 mol Hf ≈ 1/3 mol Hf
( you can do this in your head, or on paper)
178g Hf
37.4 g Cl x 1 mol Cl
≈ 1 mol Cl
35.5 g Cl
NOW, multiply each of these by the same number to get a whole number ration (instead of 1/3 : 1). If
you multiply by 3 you get a ratio of 1:3, so the formula will be HfCl3).
29. IN the periodic table, as the atomic number increases from 11 to 17, what happens to the atomic
radius?
(A) it remains constant
(D) It decreases only.
(B) It increases only.
(E) It decreases, then increases.
(C) It increases, then decreases.
30. In which of the following processes are covalent bonds broken?
(A) I2(s)  I2(g) (B) CO2(s) CO2(g) (C)NaCl(s)  NaCl(l)
(D) C(diamond)  C(g) (E) Fe(s)Fe(l)
31. In a qualitative analysis for the presence of Pb2+ , Fe2+, and Cu2+ ions in aqueous solution, which of
the following will allow the separation of Pb2+ from the other ions at room temperature?
(A) Adding dilute Na2S(aq) solution
(D) Adding dilute NH3 (aq) solution
(B) Adding dilute HCl(s) solution
(E) Adding dilute HNO3 (aq) solution
(C) Adding dilute NaOH(aq) solution
ANALYSIS: PbCl2 is insoluble, whereas FeCl2 & CuCl2 are soluble. Cl is the only ion that will allow only
the lead compound to precipitate, while leaving the other 2 (FeCl2 & CuCl2) in solution.
32. After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a
student reported a value of 38 percent. The correct value for the percentage of water in the hydrate is
51 percent. Which of the following is the most likely explanation for this difference?
(A) Strong initial heating caused some of the hydrate sample to spatter out of the crucible.
(B) The dehydrated sample absorbed moisture after heating.
(C)The amount of the hydrate sample used was too small.
(D) The crucible was not heated to constant mass before use.
(E) Excess heating caused the dehydrated sample to decompose.
ANALYSIS: (A) if you weighed the hydrate sample before heating (say 2.00g), and then heating caused
part of the sample to be lost, your post-heating mass (say 1.50g) would be lower than the actual (say
1.70g.) When you calculated the % comp that is water, you would divide the difference between the
pre-heating mass (2.00g) and the post heating mass (1.5)by the pre-heating mass (2.00-1.50g/2.00g) x
100= 25%. In fact, the % comp water (if none of the sample was lost)should be (2.00-1.70g/2.00g) x 100
= 15%. Therefore, THE ERROR IS ON THE HIGH END, THE OPPOSITE OF THE SITUATION INDICATED
ABOVE.
(B) Use the 2.00g original mass from (A). Visualize: Heated to release the water in the hydrate. Sample
re-absorbed some moisture. Obtained mass of post-heating sample & it was higher than it should have
been, say 1.80g. Analyze % comp: (2.00-1.80g/2.00g) x 100 = 10%, which is lower than the actual (I used
15% as my “actual” in analysis of choice A, above.) IN THIS CASE, THE ERROR IS ON THE LOW END, AS
THE QUESTION INDICATES.
33. The volume of distilled water that should be added to 10.0 mL of 6.00 M HCl(aq) in order to prepare
a 0.500 M HCl (aq) solution is approximately
ANALYSIS: M1=0.600 mol/L V1=0.010L M2=0.500 mol/L
USE: M1V1=M2V2 to solve for V2.
PRACTICE SOLVING W/O CALCULATOR
(A) 50.0 mL
(B) 60.0 m
(C) 100. mL
(D) 110. mL
(E) 120. mLI think. Check this one
CHEMISTRY-SECTION II PART B
NO CALCULATORS MAY BE USED WITH PART B.
Question #4: For each of the following reactions, in part (i) write a balanced equation and in part
(ii) answer the question about the reaction. In part (i), coefficients should be in terms of lowest
whole numbers. Assume that solutions are aqueous unless otherwise indicated. Represent
substances in solutions as ions if the substances are extensively ionized. Omit formulas for
any ions or molecules that are unchanged by the reaction.
General Scoring Note for Question 4:
Five points are earned for each reaction and are distributed as follows:
Four points are earned for part i.
one point for correct reactants,
two points for correct products,
one point for the correct coefficients in the balanced equation.
One point is earned for the correct answer in part ii.
(a) Calcium oxide powder is added to distilled water.
CaO(s) + H2O(l)  Ca(OH)2
OR
CaO(s) + H2O(l)  Ca2+(aq) + OH-(aq)
(b)Liquid bromine is shaken with a 0.5 M sodium iodide solution.
Br2(l) + 2Na+(aq) + 2 I- (aq)  2Na+(aq) + 2Br- (aq) + I2(s)
Br2(l) + 2 I- (aq)  2Br- (aq) + I2(s)
(c)Solid lead (II) carbonate is added to a 0.5 M sulfuric acid solution.
PbCO3(s) + H2SO4(aq)  PbSO4 (s) + H2CO3(aq)
(Carbonic acid breaks down into H2O & CO2)
PbCO3(s) + 2H+(aq) + SO4 2- (aq)  PbSO4 (s) + H2O (l) + CO2(g) OR
PbCO3(s) + H+(aq) + HSO4 - (aq)  PbSO4 (s) + H2O (l) + CO2(g)
(d) A mixture of powdered iron(III) oxide and powdered aluminum metal is heated strongly.
heat
Fe2O3 (s) + Al(s)  Al2O3(s) + Fe(s)
(e) Carbon dioxide gas is passed over hot, solid sodium oxide.
CO2 (g) + Na2O (s)  Na2CO3
Your responses to the rest of the questions in this part of the examination will be graded on the basis of
the accuracy and relevance of the information cited. Explanations should be clear and well organized.
Examples and equations may be included in your responses where appropriate. Specific answers are
preferable to broad, diffuse responses.
Question 6:
1a. Compare the structural models used to describe solids, liquids, and gases. Include
(1) distance between particles, (2) relative strength of intermolecular forces, and (3) the
extent of entropy.
1b.
(i) Is vaporization an endothermic or exothermic process? Support your answer
with an explanation.
(ii) In what ways is the high heat of vaporization of water important to each of us
personally?
2006 Question6:Answer each in terms of principles of molecular behavior & chemical
concepts.
(a) The structures for glucose, C6H12O6, and cyclohexane, C6H12, are shown below.
Identify the type(s) of intermolecular attractive forces in:
(i) pure glucose
Answer: Hydrogen bonding OR dipole-dipole interactions OR van de Waals interactions.
(London dispersion forces may also be mentioned.)
(ii) pure cyclohexane
Answer: London dispersion forces
(b) Glucose is soluble in water but cyclohexane is not soluble in water. Explain.
Answer:
Glucose is polar and cyclohexane is nonpolar.
• Polar solutes (such as glucose) are generally soluble in polar solvents such as water.
• Nonpolar solutes (such as cyclohexane) are not soluble in the polar solvent.
Students generally answered Part a of this question correctly. Students seemed to have trouble
differentiating between intermolecular and intramolecular forces throughout this question.
Question 8: Answer the following questions using principles of chemical bonding and
molecular structure.
ANSWER: http://apcentral.collegeboard.com/apc/public/repository/chemistry-released-exam1999.pdf The “correct” answers are shown. In addition, actual student answers are shown,
with analyses of their answers.
(a) Consider the carbon dioxide molecule, CO2, and the carbonate ion, CO3 2-.
(i) Draw the complete Lewis electron-dot structure for each species.
(ii) Account for the fact that the carbon-oxygen bond length in CO3 2-is greater
than the carbon-oxygen bond length in CO2.
(b) Consider the molecules CF4 and SF4.
(i) Draw the complete Lewis electron-dot structure for each molecule.
(ii) In terms of molecular geometry, account for the fact that the CF 4 molecule is
nonpolar, whereas the SF4 molecule is polar.