Fall Holiday Assignment (08)
Answer Key
(“b/c” stands for “because”)
1.C b/c bond dissociation energy has to do with bonds breaking (they dissociate, they no longer
associate)
2. n/a
3. C
4. E
5. C use P1/T1 = P2/T2
6. C
7. C use Ga: [Ar]4s23d104p1; use the 4p1 electron. Whose HIGHEST values are 4,1,1, ½
8. C
9. E
10. D
11. C
12. A
13. A
14. D use r1/r2 = √M2/√M1
15. A b/c benzene is nonpolar
16. A b/c S has the 3p4 electron, 2 unpaired e- ; i.e.    .
17. A b/c Ga has the 4p1 electron, 1 unpaired e- 
.
18. D
19. B
20. E
21. D b/c CO has same molar mass of N2 (28 g/mol)
22. B
23. A b/c lightest moves fastest
24. A
25. D
26. E
27. E b/c .25/1.25 of 800
28. C
29. E b/c all others are covalent
30. C
31. A b/c 3O2 + CS2  CO2 + 2 SO2
32. E
33. B b/c effective nuclear charge deals with how well the atom’s nuclear charge holds onto its
electrons.
34. B
35. D
36. A
37. C, use PM=dRT; use d= g/L, which is 4g/2L
38. B
39. C, b/c SiO2 is glass
Free Response Answer Key
40. (2004)
Answer the following questions about carbon monoxide, CO(g), and carbon dioxide, CO2(g). Assume that
both gases exhibit ideal behavior.
(a) Draw the complete Lewis structure (electron dot diagram) for the CO molecule and for the CO2
molecule. (a)
(b) Identify the shape of the CO2 molecule . (b)
linear
(c) n/a
(d) A 1.0 mol sample of CO(g) is heated at constant pressure. On the graph below, sketch the expected
plot of volume verses temperature as the gas is heated.
(e) Samples of CO(g) and CO2(g) are placed in 1 L containers at the conditions in the diagram below.
(i) Indicate whether the average kinetic energy of the CO2 is greater than, equal to, or less than the
average kinetic energy of the CO(g) molecules. Justify your answer.
(i) equal to; at the same temperature, all gas molecules have the same kinetic energy
(ii) Indicate whether the root-mean-square speed of the CO2(g) molecules is greater than, equal to
or less than the root-mean-square speed of the CO(g) molecules. Justify your answer.
(ii) less, since CO2 has a molar mass of 44 and CO has a mass of 28, the lighter molecule is
faster at the same temperature
(iii) Indicate whether the number of CO2(g) molecules is greater than, equal, or less than the
number of CO(g) molecules. Justify your answer.
(iii) less; Avogadro’s Hypothesis, equal volumes of gas at the same temperature and pressure
contain equal number of molecules. Since the pressure of CO2 is half the pressure of the CO,
it must contain half as many molecules.
41. (2006)
Answer the following questions about the structures of ions that contain only sulfur and fluorine.
(a) The compounds SF4 and BF3 react to form an ionic compound according to the following equation.
SF4 + BF3  SF3BF4
(i) Draw a complete Lewis structure for the SF3+ cation in SF3BF4.
   
(i)

:S : F:
 :F
  

:F:






(ii) Identify the type of hybridization exhibited by sulfur in the SF3+ cation. (ii)
sp3.
+
(iii) Identify the geometry of the SF3 cation that is consistent with the Lewis structure drawn in
part (a)(i). (iii)
pyramidal
(iv) Predict whether the F—S—F bond angle in the SF3+ cation is larger than, equal to, or smaller
than 109.50˚. Justify your answer. (iv)
smaller than 109.50˚. The lone pair of unbonded
electrons occupies more space than the bonded pairs and, therefore, pushed the bonded
pairs away and hence, a smaller bond angle than a perfect tetrahedron.
(b) The compounds SF4 and CsF react to form an ionic compound according to the following equation.
SF4 + CsF  CsSF5
(i) Draw a complete Lewis structure for the SF5– anion in CsSF5. (i)
(ii) Identify the type of hybridization exhibited by sulfur in the SF5– anion. (ii)
sp3d2
(iii) Identify the geometry of the SF5– anion that is consistent with the Lewis structure drawn in part
(b)(i). (iii) square pyrimidal
(iv) Identify the oxidation number of sulfur in the compound CsSF5. (iv) S = +4
42. (2005)
Use principles of atomic structure, bonding and/or intermolecular forces to respond to each of the
following. Your responses must include specific information about all substances referred to in each
question.
(a) At a pressure of 1 atm, the boiling point of NH3(l) is 240 K, whereas the boiling point of NF3(l) is
144 K.
(i) Identify the intermolecular forces(s) in each substance.
force
NH3
NF3
London dispersion
+
+
polar attraction
+
+
hydrogen bonding
+
-
ionic attraction
-
-
(ii) Account for the difference in the boiling points of the substances. (ii) the ability of ammonia
to create intermolecular hydrogen bonds, leads to higher amount of energy to separate
the molecules by boiling them.
(b) The melting point of KCl(s) is 776˚C, whereas the melting point of NaCl(s) is 801˚C.
(i) Identify the type of bonding in each substance. (i) both compounds have ionic bonding
(ii) Account for the difference in the melting points of the substances. (ii) the sodium ion in NaCl
is a smaller size than the corresponding potassium ion in KCl. This smaller size creates a
larger charge density and greater ion Coulombic attraction in the NaCl, making it harder
to melt.
(c) As shown in the table below, the first ionization energies of Si, P, and Cl show a trend.
Element
Si
P
First Ionization Energy
(kJ mol-1)
786
1012
Cl
1251
(i) For each of the three elements, identify the quantum level (e.g., n =1, n = 2, etc.) of the valence
electrons in the atom. (i) Si, n = 3; P, n = 3; Cl, n = 3
(ii) Explain the reasons for the trend in the first ionization energy. (ii) in terms of atomic radius,
Si > P > Cl and nuclear charge Cl > P > Si, the smaller and higher charged chlorine atom
has the greatest attraction for its electrons than the other two. This means that it takes
more energy to remove an electron from chlorine that the other two. The opposite is true
for silicon and it should have the smallest value.
43. (2004)
3
2 Fe(s) + 2 O2(g)  Fe2O3(s) ∆Hf˚ = -824 kJ mol–1
Iron reacts with oxygen to produce iron(III) oxide as represented above. A 75.0 g sample of Fe(s) is
mixed with 11.5 L of O2(g) at 2.66 atm and 298 K.
(a) Calculate the number of moles of each of the following before the reaction occurs.
1 mol
(i) Fe(s) (i) 75.0 g Fe 
= 1.34 mol Fe
55.85 g
PV
RT
(2.66 atm)(11.5 L)
(ii) O2(g) (ii) PV = nRT, n =
(0.0821
L atm
mol K)
(298 K)
= 1.25 mol O2
(b) Identify the limiting reactant when the mixture is heated to produce Fe2O3. Support your answer
with calculations. (b)
Fe; 1.34 mol Fe 
3
2
mol O2
= 1.01 mol O2
2 mol Fe
(c) Calculate the number of moles of Fe2O3 produced when the reaction proceeds to completion.
excess O2, limiting reagent is Fe(c) 1.34 mol Fe 
1 mol Fe2O3
= 0.671 mol Fe2O3
2 mol Fe
44. (2003)
∆H˚vap
Compound
Name
Compound
Formula
(kJ mol-1)
Propane
CH3CH2CH3
19.0
Propanone
CH3COCH3
32.0
1-propanol
CH3CH2CH2OH
47.3
Using the information in the table above, answer the following questions about organic compounds.
(a) For propanone,
(i) draw the complete structural formula (showing all atoms and bonds); (i)
(b)
(i)
(c)
(d)
(ii) predict the approximate carbon-to-carbon-to-carbon bond angle. (ii) 120˚
For each pair of compounds below, explain why they do not have the same value for their standard
heat of vaporization, ∆H˚vap. (You must include specific information about both compounds in each
pair.)
Propane and propanone(i) propane, 26 electrons, molar mass = 44
propanone, 32 electrons, molar mass = 58
higher # electrons means larger van der Waal forces, larger molar mass means a slower
molecule, the oxygen creates a polar molecule and dipol–dipole interactions
(ii) Propanone and 1-propanol(ii) 1-propanol has an –OH which creates a site for hydrogen
bonding with other –OH on adjacent molecules increasing intermolecular forces that
must be overcome in order to vaporize the liquid.
n/a
Given the structural formula for propyne below,
H
| 
H—C—CC—H
|
H
(i) indicate the hybridization of the carbon atom indicated by the arrow in the structure above;
(i) sp
(ii) indicate the total number of sigma () bands and the total number of pi (π) bonds in the
molecule(ii) 6 sigma, 2 pi
45. (1997)
Explain each of the following observations using principles of atomic structure and/or bonding.
(a) Potassium has a lower first-ionization energy than lithium. (a) potassium’s valence electron is 4s1
while lithium’s is 2s1. potassium’s electron is shielded by more electrons than lithium and is
therefore more easily removed at a lower energy.
(b) The ionic radius of N3- is larger than that of O2-. (b)
The addition of electrons to a neutral
atom produces an anion that is significantly larger than its parent atom. Even though both
ions are isoelectronic, there is a greater nuclear positive charge in the oxide ion causing its
electrons to be more tightly pulled toward the nucleus.
(c) A calcium atom is larger than a zinc atom. (c) Even though a zinc atom contains 10 more
electrons than calcium, these are all 3d electrons, filling an inner shell, not adding another
larger one. There is a corresponding increase of 10 more protons for the zinc and this increase
in nuclear charge pulls the electrons in more tightly and reducing its size.
(d) Boron has a lower first-ionization energy than beryllium. (d) Boron’s last electron is 2p1 and it
receives the benefit of effective shielding by its completed 2s electrons. Thus it is easier to
remove this electron.
46. (2006)
Suppose that a stable element with atomic number 119, symbol Q, has been discovered.
(a) Write the ground-state electron configuration for Q, showing only the valence-shell electrons.
(a) 8 s1
(b) Would Q be a metal or a nonmetal? Explain in terms of electron configuration. (b) metal. It has
one valence electron that it easily looses, consistent with the other alkali metals in its family.
(c) On the basis of periodic trends, would Q have the largest atomic radius in its group or would it have
the smallest? Explain in terms of electronic structure. (c)
Q would have the largest atomic
radius in its group. Every time another s-orbital is added to an element, it creates a bigger
electronic shell.
(d) What would be the most likely charge of the Q ion in stable ionic compounds? (d)
+1
47. (2006)
Answer the following questions that relate to the analysis of chemical compounds.
(a) A compound containing the elements C, H, N, and O is analyzed. When a 1.2359 g sample is
burned in excess oxygen, 2.241 g of CO2(g) is formed. The combustion analysis also showed that
the sample contained 0.0648 g of H.
(i) Determine the mass, in grams, of C in the 1.2359 g sample of the compound.
12.011
(i) C = 2.241 g 
= 0.6116 g C
(12.011 2 16.00)
(ii) When the compound is analyzed for N content only, the mass percent of N is found to be 28.84
percent. Determine the mass, in grams, of N in the original 1.2359 g sample of the compound.
(ii) N = 1.2359 g  28.84% = 0.3564 g N
(iii)
Determine the mass, in grams, of O in the original 1.2359 g sample of the compound.
(iii) O = 1.2359 g – (0.6116 g C + 0.3564 g N + 0.0648 g H) = 0.2031 g O
(iv)
Determine the empirical formula of the compound.
0.0648 g H
0.3564 g N
0.6116 g C
(iv)
-1
g
1.0079 g mol
14.007 g mol
12.011 mol
0.2031 g O
16.00 g mol
mmol: 50.92 C
64.29 H
25.45 N
12.69 O
ratio: 4
5
2
1
empirical formula: C4H5N2O
(b) A different compound, which has the empirical formula CH2Br, has a vapor density of 6.00 g L-1 at
375 K and 0.983 atm. Using these data, determine the following.
(i) The molar mass of the compound
Lgatm
)(375K)
(mass)(RT)
(mass)RT (6.00 g)(0.0821 molgK
=
(i) PV = nRT =
; molar mass =
= 188 g/mol
molar mass
PV
(0.983 atm)(1 L)
(ii) The molecular formula of the compound(ii) CH2Br has a molar mass of 93.9; 188/93.9 = 2;
therefore, the molecular formula is 2(CH2Br) = C2H4Br2