Chemistry 1C Review Problems, Chapters 13, 14 and 16 ...

Chemistry 1C Review Problems, Chapters 13, 14 and 16 vanKoppen Spring 2002
Exam 1: Friday, April 26 9 – 9:50 PM Chem. 1179
In studying for the exam do homework problems, the quiz, examples in the text and in lecture again!!! Do the assigned
problems and additional problems if needed. After you finish all the assigned problems, work the following review problems
using one 8.5 x 11 inch page of notes which you are allowed to have during the exam (use only two thirds of one side of the
page so you will have room for more notes for the second midterm and final).
Check your quiz and exam scores on the web: http://www.chem.ucsb.edu/
Go to Undergrads —> Course Web Pages –> Chem 1C-1CL (Van Koppen) You will need to set up a new student account by
entering your perm, a 4-digit pin and confirm the pin. Then use your perm and pin number to check your scores.
Structure and Bonding
Chemical bonds between metals and non-metals tend to be ionic. Metals tend to lose the number of electrons needed to attain
a noble gas configuration and the non-metal atoms tend to gain the number of electrons needed to attain a noble gas
configuration. e.g. Na + – Cl –
Chemical bonds between non-metal atoms tend to be covalent. In covalent compounds, pairs of shared electrons make the
bonds between the atoms. e.g. H – H
Atom and ion size strongly influence bonding and chemistry:
Covalent bonds: a smaller atomic radius => shorter bond distance => better orbital overlap => stronger bond
Ionic bonds: a smaller ionic radius => shorter bond distance => greater V(r) => stronger bond
NOTE: The Coulomb force of attraction = V(r) = Q 1 Q 2 /4πε o r, where Q 1 and Q 2 are the numerical charges on the ions
and r is the distance between the centers of the ions. For Na + – Cl – , Q 1 = +1 Q 2 = –1, and r is the distance from the
center of Na + to the center of Cl – (See text pg. 568).
Cohesion of crystalline solids: Molecular, Ionic, Metallic and Covalent Crystals
1) Molecular Crystals: Molecules are held together in their crystal lattice by intermolecular forces, including hydrogenbonding,
dipole-dipole forces, or London Dispersion Forces (LDF). Intermolecular forces are generally weaker than those in
ionic, metallic and covalent crystals. As a result, molecular crystals typically have low melting points, are soft, and have poor
conductivity. Examples of molecular crystals include noble gases, oxygen, nitrogen, halogens, compounds such as CO 2 ,
organic compounds, proteins, and metal halides with low ionicity.
Molecular crystals are of great scientific value. If proteins and other macromolecules are obtained in the crystalline state,
their structures can be determined by x-ray diffraction. Knowing the three-dimensional structures of biological molecules is
the starting point for understanding their functions.
2) Ionic Crystals: Compounds formed by atoms with significantly different electronegativities are ionic. e.g. NaCl, CsCl,
ZnS, CaF 2 , etc. The strong and long range electrostatic forces make ionic crystals hard, high-melting, brittle solids that are
electrical insulators. Melting an ionic crystal, however, disrupts the lattice and sets the ions free to move, and so ionic liquids
are good electrical conductors.
3) Metallic Crystals: The characteristic property of metals is their good conductivity of electricity and heat. How do metals
conduct electricity and heat (see lecture notes and pg 745-6 text) What is the bonding in metallic crystals The valence
electrons in a metal are delocalized in molecular orbitals that extend over the entire crystal. The “sea” of electrons in metals
accounts for strong bonding and high boiling points. Metals have a large range in melting points (e.g. Gallium melts in your
hand at 29.78 o C and Tungsten, the highest melting elemental metal, melts at 3410 o C). Note that Gallium, like most metals,
does have a high boiling point, 2400 o C.
4) Covalent Crystals: In covalent crystals atoms are linked by covalent bonds rather than by either the electrostatic
attractions of molecular and ionic crystals or the sea of valence-electrons (“the glue”) in metals. Covalent crystals, also known
as network atomic solids, have high melting and boiling points due to the strong covalent bonds between the atoms. They are
typically hard and brittle. e.g. diamond, graphite, black and red phosphorous. Diamond is an insulator whereas graphite is an
electrical conductor. Explain why. (see pg. 751).
APPROACH TO PROBLEM SOLVING:
When you are asked to rationalize or predict relative boiling points, melting points, heats of vaporization, surface tension, etc.,
what is the first question you should ask yourself What are the INTERMOLECULAR FORCES Or what type of
crystalline solid is this Metallic, Covalent, Ionic or Molecular
The general trend in strength of intermolecular forces from strongest to weakest are:
Ionic > Hydrogen–bonding > dipole – dipole > LDF (or vanderWaal forces)
(H bound to N,O,F) (polar molecules) (induced dipole forces)
Note: LDF is considered a weak force. However, for large molecules, with large surface areas, LDF can be quite large giving
rise to relatively high boiling points. For example, a non-polar hydrocarbon, C 25 H 52 , is solid at room temperature. Its boiling
point is 400 o C. Compare the intermolecular forces for C 25 H 52 and H 2 O.
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Problem: Predict the trend in boiling points for BF 3 , BCl 3 , BBr 3 , BI 3 In order to determine the intermolecular forces you must
know the type of molecule. Is it covalent, polar-covalent or ionic For covalent molecules draw the Lewis structure and
determine if it is polar. The boron halides are all trigonal planar, non-polar molecules. So what is the intermolecular force
LDF. How will LDF differ for BF 3 , BCl 3 , BBr 3 and BI 3 As the size of an atom or molecule increases, the polarizability
increases and thus LDF increases. Now you can answer the question: because atom size increases down the group, F < Cl <
Br < I, the LDF forces increase down the group, BF 3 < BCl 3 < BBr 3 < BI 3 and the boiling points will increase down the group,
BF 3 < BCl 3 < BBr 3 < BI 3 . The boiling points are: BF 3 –101 o C, BCl 3 12 o C, BBr 3 91 o C, BI 3 210 o C. Quite a difference!
Review Problems
1. Circle the formula that best fits each of the following descriptions:
a) greatest electronegativity Al C Na N
b) longest bond length HCl HF HI HBr
c) largest ionic radius Mg 2+ F – Na + O 2–
d) least polar bond C – O C – N O – H
e) smallest atomic radius S Na Si Al
2. Draw the Lewis structure for SeF 2 and PO 3– 3 , predict the molecular geometry, and if the molecule is polar indicate the
direction of the net dipole.
3.
–
a) Draw the Lewis structures for each of the following compounds. Nitrogen is the central atom in both NO 2 and NO – 3 .
NO + –
NO 2
–
NO 3
b) Which one of the following has the longest NO bond length, NO + , NO – 2 , or NO – 3
c) Which one of the following has the shortest NO bond length, NO + , NO – 2 , or NO – 3
4. a) Using the molecular orbital model write the electronic configuration and bond order of N 2 and O 2 b) Are N 2 and O 2
paramagnetic or diamagnetic What does it mean when a molecule is paramagnetic How can we observe the
magnetism of a molecule experimentally c) Does N 2 or O 2 have a higher bond energy Why
5. What is the hybridization of the carbon atoms in CH 2 =CH and CH 2 2=C=CH 2 For each of these molecules are all the
atoms in the same plane
6. For each of the following molecules, indicate the number of σ-bonds, π-bonds, the steric number for each of the C and N
atoms and hybridization on each of the C and N atoms. What are the angles a, b, and c in each of the molecules
H
• O• •
H C C C H
••
H O N O
•
••
••
H
7. Which of the following bonds is least polar C – O Br – Br O – H H – F
8. Order Rb, F and O in increasing electronegativity. Order Na, Si, Ne and Cl in increasing atomic radius.
9. Indicate the number of σ-bonds and π–bonds in each of the following molecules. a) N 2 b) CO c) O 2
2- 2-
10. Draw the Lewis structure for CO 3 including resonance structures. Determine the hybridization on the carbon in CO 3 and
give the molecular geometry
11. What is the hybridization on each of the carbons in butane, CH 3 CH 2 CH 2 CH 3 . Are all the carbon and hydrogen atoms in the
same plane
12. How many electrons are involved in the π-bonding in benzene, C 6 H 6 What is the hybridization on the carbon atoms Are
all the carbon and hydrogen atoms in the same plane Why or why not
13. Which one of the following molecules, C 2 H 2 , C 2 H 4 , and C 2 H 6 , has the shortest bond distance between the carbon bonds
Why
14. Write the electronic configuration using the molecular orbital model of the following species and determine which has the
largest dissociation energy O 2 , O – 2 , O 2– 2+
2 , O 2
15. Write the electronic configuration and number of unpaired electrons for the following: O, Al 3+ , C, Se 2–
16. a) Draw the Lewis structure for SF 6 . How many electrons does S share Is this consistent with the octet rule
b) For SF 6 , valence shell expansion of S allows additional electrons to be accepted in which orbitals Is valence shell
expansion observed for second row elements
c) What is the hybridization of S in SF 6
17. Which elements most commonly form π-bonds
18. Which has the highest boiling point, BF 3 , BCl 3 , BBr 3 , BI 3 Why
19. What is the hybridization on carbon in O=C=O
20. Why does hydrogen have such low boiling and melting points
21. Highest boiling point. Explain. a) He, Ne, Ar, Kr, Xe b) NH 3 , PH 3 c) HF, HCl
22. Predict the boiling point of H 2 O if H-bonding between water molecules did not occur (use Fig 16.4 Pg. 732 text). Would
water be a liquid at room temperature without H-bonding
2– 2–
23. Why can't carbon form a compound such as CCl 6 while tin can form SnCl 6
24. Why can carbon form π–bonds with carbon, nitrogen and oxygen while other group 4 elements, Si, Ge, Sn and Pb, cannot
••
2

25.
–
What is the hybridization of P in PF 6
26. Indicate the type of crystalline solid formed for each of the following substances.
Graphite, S 8 (See text Pg. 872, Fig. 19.16 a) Na, H 2 O, SiO 2 , CH 3 NH 2 , NH 4 Cl, white phosphorous, black phosphorous (See
text, pg 866, Fig. 19.12).
27. Indicate the molecular geometry, polarity, and most important (strongest) intermolecular force for each of
the following substances. a) BF 3 b) SO 2 c) C 6 H 6 d) Cl 2 e) CH 3 OH f) HCl g) NH 3
h.) CCl 4
28. A substance does not conduct electricity in either the solid or the liquid state and is almost insoluble in water. It melts at
53 o C and boils at 174 o C. These properties are characteristic of which one of the following crystalline solids: a) ionic
b) metallic c) molecular d) covalent (atomic network)
29. In the lab this week you worked with benzophenone and cyclohexanone. Given the structures below, explain why
benzophenone, C 13 H 10 O, is a solid at room temperature and why cyclohexanone, C 6 H 10 O, is at liquid at room temperature.
Are all the carbon, hydrogen and oxygen atoms in the same plane in benzophenone Are all carbon, hydrogen and oxygen
atoms in the same plane in cyclohexanone
O
O
benzophenone
cyclohexanone
30. The enthalpy of vaporization of water is 44 kJ/mol. On top of one of the peaks in Rocky Mountain National Park the
pressure of the atmosphere is 0.75 atm. a) Determine the boiling point of water at this location. b) Will your food take
more or less time to cook at this elevation
31. Consider the phase diagram for CO 2 , with its triple point, T, as shown.
a) What phases are present at point B
b) Dry ice, CO 2 (s), is solid at P = 1 atm and T = – 100 o C.
Starting at this point on the phase diagram, draw a line
showing the phase transition(s) which occur as the
temperature is increased from – 100 o C to 100 o C
at constant pressure.
c) According to the phase diagram, which phase is more
dense, liquid or solid
Pressure
B•
•
T
Temperature
Solutions
1. a) N b) HI c) O 2– d) C–N e) S
2. Lewis structure molecular geometry polarity
PO 3
3–
SeF 2
3. a)
b) NO 3 – . c) NO +
4 a) N 2 : (σ 2s ) 2 (σ 2s
*) 2 (π 2p ) 4 (σ 2p ) 2 Bond order = 3. O 2 : (σ 2s ) 2 (σ 2s
*) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p
*) 2 Bond order = 2.
b) N 2 is diamagnetic because is has no unpaired electrons and O 2 is paramagnetic because it has unpaired electrons.
Paramagnetic molecules are attracted to a magnetic field. c) The bond order for N 2 is greater than for O 2 and therefore N 2
has a greater bond energy.
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5. For CH 2 =CH both carbons are 2 sp2 hybridized and all the atoms are in the same plane. For CH 2 =C=CH 2 the two end
carbons are sp 2 and the central carbon is sp hybridized. Not all atoms are in the same plane: the two CH 2 groups are in
perpendicular planes.
6. Molecule A: six σ-bonds, two π-bonds, the hybridization on the left carbon is sp 3 , the other two carbons are sp hybridized;
angle a = 109 o , angle b = 180 o , angle c = 180 o Molecule B: four σ-bonds, one π-bond, the hybridization on the nitrogen
is sp 2 ; angle a = 109 o , angle b = 120 o , angle c = 120 o
7. Br 2 is least polar because it is non-polar.
8. Electronegativity: Rb < O < F Atomic Radius: Ne < Cl < Si < Na Why
9. a) N 2 has a triple bond; one σ-bond and two π-bonds. b) CO has a triple bond; one σ-bond and two π-bonds. c) O 2 has
a double bond; one σ-bond and one π-bond.
10. See problem 62, Chapter 13.
11 The carbons in butane are all sp 3 hybridized. The carbon and hydrogen atoms are not in the same plane. See Fig 22.4, pg.
974. Can you see how all the dipoles in butane cancel and that butane is non-polar Most hydrocarbons are essentially
non-polar.
12. There are three π-bonds and therefore there are six π-electrons. The hybridization on the carbon atoms is sp 2 . All the
carbon and hydrogen atoms are in the same plane because the sp 2 hybridization results in planar geometry. For
resonance to occur the p-orbitals on neighboring carbon atoms must overlap which requires the molecule to be planar.
13. C 2 H 2 has the shortest C–C bond because it is a triple bond whereas C 2 H 4 has a double bond and C 2 H 6 has a single bond.
14. O 2 (σ 2s ) 2 (σ 2s
*) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p
*) 2 Bond order = 2
–
O 2 (σ 2s ) 2 (σ 2s
*) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p
*) 3 Bond order = 1.5
2–
O 2 (σ 2s ) 2 (σ 2s
*) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p
*) 4 Bond order = 1
2+
O 2 (σ 2s ) 2 (σ 2s
*) 2 (σ 2p ) 2 (π 2p ) 4 Bond order = 3 = largest => Largest dissociation energy
15. O (1s) 2 (2s) 2 (2p) 4 Two unpaired electrons Al 3+ [Ne] or (1s) 2 (2s) 2 (2p) 6 No unpaired electrons
C (1s) 2 (2s) 2 (2p) 2 Two unpaired electrons Se 2– [Kr] No unpaired electrons
16. a) Octahedral geometry, sulfur shares 12 electrons. No it is not consistent with the octet rule. b) Sulfur accepts
additional electrons in d-orbitals. Valence shell expansion is not observed for second row elements. Why not
c) d 2 sp 3 hybridization on S.
17. π-bonds are commonly observed for carbon, oxygen and nitrogen. It is less common for sulfur to and phosphorous to
make π-bonds. Sulfur makes π-bonds with oxygen in SO 2 , but S is too big to make strong π-bonds with itself. This is
why unlike O 2 (g), a stable gas at room temperature, S 2 (g) is not observed. Sulfur is a solid at room temperature, S 8 (s).
18. BI 3 has the highest boiling point. These molecules are all covalent non-polar and therefore the larger the molecule, the
more polarizable, the stronger the LDF. 19. sp
20. H 2 has a low molar mass, it has only two electrons => low polarizability => weak LDF => low b.p.
21. a) Xe is the largest in size, greatest LDF=> higher b.p b) NH 3 has a higher b.p. due to H-bonding c) HF has a higher
b.p. due to H-bonding
22. A straight line extrapolation, following the trend of other group members, H 2 Te, H 2 Se and H 2 S, gives a boiling point of
H 2 O of approximately – 123 o C. Without H-bonding water would be a gas at room temperature and pressure.
23. Carbon follows the octet rule, it is not able to expand its valence shell. Sn can expand its octet by putting electrons into
d-orbitals.
24. Carbon has a small enough atomic radius to allow for effective overlap between the p-orbitals. The other group 4
elements are bigger and p-orbitals do not overlap effectively.
25. d 2 sp 3
26. Graphite (covalent network), S 8 (molecular), Na (metallic), H 2 O (molecular), SiO 2 (covalent network) CH 3 NH 2
(molecular), NH 4 Cl (ionic), white phosphorous (P 4 (s) molecular), black phosphorous (covalent network).
27. a) BF trigonal planar, non-polar, LDF b) SO 3 2 bent, polar, dipole-dipole c) C 6 H 6 planar, non-polar, LDF d) Cl 2
linear, non-polar, LDF e) CH 3 OH tetrahedral, polar, H-bonding f) HCl linear, polar, dipole-dipole g)NH 3
trigonal pyramid, polar, H-bonding h) CCl 4 non-polar, LDF 28. Molecular
29. In order to answer this question, determine the hybridization on the carbon atoms. All the carbon atoms in
benzophenone are sp 2 hybridized (trigonal planar geometry) and therefore all the carbon, hydrogen and oxygen atoms
are in the same plane. The planar geormetry results in a maximum surface area of contact between the molecules and
gives rise to relatively large LDF forces. The CO bond is polar so there are some dipole-dipole forces as well but the
LDF forces are dominant in benzophenone. In cyclohexanone, the carbon bound to the oxygen is sp 2 hybridized, the
other 5 carbons are sp 3 hybridized. The molecule is therefor not planar. See Figure 22.6, Pg. 980. LDF forces are
weaker and therefor cyclohexanone has a lower melting point and is a liquid at room temperature.
30. T = 365 K = 92.6 o C. Watch units!!! 1 kJ = 1000 J and you must convert degrees Celcius to degrees Kelvin.
31. a) At point B, solid and liquid CO 2 are in equilibrium. b) The horizontal line drawn is below the triple point and
shows that CO 2 (s) goes directly to a gas at P = 1 atm. c) Solid CO 2 is more dense than liquid CO 2 . You can see this
from the slope of the solid-liquid line. Keeping T constant while increasing P, draw a straight vertical line, liquid CO 2
will go to the more dense phase, solid CO 2 . Water is unusual in this respect, liquid water is more dense than ice. See
Fig. 16, Pg. 780 text. That is why ice floats on water.
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