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CHAPTER 10 HW – SOLUTIONS
INTERMOLECULAR FORCES
1.) In each, identify the intermolecular force pointed to by an arrow.
IMF Dipole-dipole force London dispersion force (LDF) Hydrogen bond
2.) Elemental nitrogen (N2) can exist in the liquid state, and is used by dermatologists to freeze off warts and skin tags (at 1 atm its highest temperature is –196 ˚C). In the pure liquid phase, N2 molecules are held together through intermolecular forces. Identify the IMF pointed to by an arrow.
3.) Genetic information is housed in DNA, which has a double helical structure. The strands of DNA are held together through intermolecular forces (IMFs). In the portion of a DNA strand shown below, which connects the nucleic bases cytosine and guanine, identify the IMF pointed to by the arrow.
H Cl
H Cl HCC
HH
H
HH
HCC
HH
H
HH
HN
OH
NO
Cytosine
NCN
CC C
N
O
Guanine
NC N
CCC
NC
NO
N
HH
H
HH
HH
H
N NN N
LDF
LDF
Hydrogen bond
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4.) For each, sketch the intermolecular forces present in the pure liquid phase.
Draw a second identical molecule to the first, and show how they interact through the strongest possible intermolecular force. Use dashed lines to show the IMF, and identify it.
HBr NH3 HF
Drawing
IMF Dipole-dipole force Hydrogen bond Hydrogen bond
H2CO O2 CH3OH
Drawing
IMF Dipole-dipole force London dispersion force Hydrogen bond
5.) What is the strongest type of intermolecular force present in the liquid state of each of the following?
a. Xe London dispersion force (LDF) c. H2O Hydrogen bonds
b. CO Dipole-dipole force d. H2S Dipole-dipole force
BOILING POINT
6.) Phosphorus, arsenic (As), and antimony (Sb) are in the same family of the periodic table. Explain why the boiling point increases from PH3 (–88 ˚C) < AsH3 (–62 ˚C) < SbH3 (–17 ˚C).
The molar mass increases as shown: PH3 has a lower MM than SbH3 (easy to spot as P is on row 3 of the periodic table and Sb is on row 5).
As MM increases, the strength of London Dispersion forces (LDF) also increases. Therefore the boiling point is (low to high) PH3 < AsH3 < SbH3 as the intermolecular force strength increases in this series from weakest to strongest.
H Br
H Br
δ+ δ-
δ+ δ-H N
H
H
H N
H
H
δ+ δ- δ+
δ-δ+
δ+δ+
δ+H F
H F
δ+ δ-
δ+ δ-
HC
H
O
HC
H
O
δ-
δ-
δ+
δ+
O OO O
H C
H
H
OH
HC
H
H
OH
δ-δ+
δ+
δ-
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7.) Butane (C4H10) is a common fuel used in lighters (e.g. butane torches) and has a boiling point of –0.5 ˚C. Acetone (C3H6O) is used as nail polish remover and has a boiling point of 56.1 ˚C.
a. Is each material a liquid or gas at room temperature? Butane: gas Acetone: liquid
b. Use specific intermolecular forces to explain why acetone has a higher boiling point than butane.
Acetone has stronger intermolecular forces and so has a higher boiling point. Acetone can form dipole-dipole forces while butane forms London Dispersion Forces (LDFs). Dipole-dipole forces are stronger than LDFs.
8.) In each pair, decide which has the higher boiling point. Then use specific intermolecular forces to explain each answer.
a. Ar vs. HCl
Ar atoms interact with other Ar atoms through LDFs (weak), while HCl interacts with other HCl molecules through dipole-dipole forces (stronger). The substance with the stronger IMFs has a higher boiling point, so HCl has the ↑ bp. (FYI, actual bp: Ar: –186 ˚C, HCl: –85 ˚C).
b. C5H12 vs. C6H14
Both molecules interact in the liquid phase through LDF’s, but C6H14 has more atoms and a higher molar mass, which means it has stronger LDFs than C5H12. This means C6H14 has a higher boiling point (FYI, actual bp: C5H12: 36.2 ˚C, C6H14 : 69 ˚C).
9.) In each pair, decide which has the higher heat of vaporization (DHvap). Then use specific intermolecular forces to explain each answer.
a. HF vs. HCl
HF can hydrogen bond in the liquid phase while HCl can only interact through dipole-dipole forces. Since hydrogen bonds are stronger than dip-dip forces, HF has a higher boiling point (FYI actual bp: HF: 20 ˚C, HCl: –85 ˚C). This means it will take more energy to boil the HF, so it has a higher heat of vaporization.
b. Cl2 vs. ICl
Cl2 (l) is held together by LDF, while ICl (l) has dipole-dipole forces. Dipole-dipole forces are stronger, so it takes more energy to break the IMFs between ICl, making it have a higher DHvap.
H C
H
H
C
H
H
C
H
H
C
H
H
H CC
C
O
H
H
H
HH H
butane acetone
Page 4
10.) SiH4 has a heat of vaporization (DHvap) of 12.10 kJ/mol. Should the DHvap of CH4 be higher or lower than 12.10 kJ/mol? Explain your answer.
SiH4 has a higher molar mass than CH4, so has stronger LDFs and should have a higher boiling point. With stronger IMFs, it should require more energy to vaporize the SiH4, meaning SiH4 will have a higher DHvap. The DHvap of CH4 should be lower than 12.10 kJ/mol.
11.) Water has a relatively high boiling point for its molecular size because it forms hydrogen bonds. Which is not a reason why water forms especially strong intermolecular forces? (Multiple choice.)
a. There is a large d+ on each H atom in water. b. The O–H bond is very polar. c. The H atom is very small, allowing for close contact of partial charges between water
molecules. d. The O–H covalent bonds are very strong.
LIQUID PROPERTIES
12.) Mercury atoms forms strong metallic bonds. Knowing this, explain why liquid mercury spilled from an old thermometer does not spread into a thin layer, but instead forms spherical droplets. Use a discussion of intermolecular forces to explain this preferred shape.
If the mercury atoms are bound by strong metallic bonds, each atom will want to make as many Hg-Hg connections as possible. Since there are fewer Hg-Hg connections on the surface (as compared to the interior), a substance with strong connections will try to minimize its surface area.
For a certain volume, a sphere has the smallest surface area of any shape, which is why liquid Hg prefers this shape.
13.) Some shampoos use stearyl alcohol (C18H38O) and cetyl alcohol (C16H34O) as thickeners (and for other purposes). Explain why these compounds cause a solution to become highly viscous.
H CH
HCH
HCH
HCH
HCH
HCH
HCH
HCH
HCH
HCH
HCH
HCH
HCH
HCH
HCH
HCH
HO H
cetyl alcohol
Cetyl alcohol is a very long molecule, which causes a solution to be viscous because the long strands become entangled in one another and make it difficult for individual molecules to slide past one another.
Page 5
14.) Multiple choice:
A water droplet falling and rebounding from a pool of water was captured by a high speed camera, and is shown below. The shape of the water droplet is best explained by ...
a. Water’s surface tension b. Capillary action c. Water’s viscosity d. Water’s low boiling point
Blood hemoglobin is routinely measured in a doctor’s office by a finger prick and withdrawal of a sample of blood into a small glass tube. The blood moves into the glass tube because of ...
a. Blood’s surface tension b. Capillary action c. Blood’s viscosity d. Blood pressure
(Follow-up) On the molecular level, blood moves into the glass tube because ...
a. The water molecules in blood are attracted to the glass through strong intermolecular forces
b. The water molecules in blood strongly hydrogen bond to each other c. A portion of the water molecules in blood have a high kinetic energy d. The water molecules vaporize into the tube
Corn syrup is made by treating corn starch with acid and enzymes to break down starch into glucose, which is why corn syrup is sometimes referred to as glucose syrup. Considering the structure of glucose (C6H12O6) shown below, corn syrup is a viscous liquid because the glucose molecules ...
a. Have multiple oxygen atoms b. Minimize their surface area by staying
connected to one another c. Are strongly attracted to the plastic container
through hydrogen bonding d. Hydrogen bond to each other at multiple
sites, making it difficult for them to slide past one another
C
CC
C
CO C
O
H H
HH
HO
HO
H OH
OH
HH
H
Page 6
VAPOR PRESSURE
15.) Dichloromethane (CH2Cl2) is a solvent sometimes used to decaffeinate coffee beans. It is a liquid at room temperature, and has a vapor pressure of 330 mmHg at 20 ˚C. Should its vapor pressure be higher or lower than 330 mHg at 30 ˚C? Explain why.
When temperature increases, vapor pressure also increases (they are directly proportional). This is because at a higher temperature a greater percentage of the molecules have enough energy to break the intermolecular forces and move to the gas phase (to evaporate). Therefore, there are more moles of CH2Cl2 in the gas phase, and CH2Cl2 should have a vapor pressure (Pvap) greater than 330 mmHg at 30 ˚C.
16.) Which graph (A–D) correctly depicts the relationship between vapor pressure and temperature?
17.) Which has the lowest vapor pressure at 25 ˚C in each set?
a. Cl2, Br2, I2 I2 (↑MM, ↑LDF, less vapor) c.
b. H2O, CH4, H2 H2O (H bonds, ↑IMF) (Dip-dip H-bonds LDF )
18.) The following graph compares vapor pressure to temperature for 3 compounds: carbon disulfide, ethanol, and heptane. Based on the graph, which compound has the lower boiling point? Briefly explain your answer.
Temp (˚C) Vap
or p
ress
ure
(mm
Hg)
At any temperature (for example 30˚C), carbon disulfide has the higher vapor pressure of the three. This must be because it has weaker IMFs and a lower boiling point, producing more gas a certain temperature. As boiling point ↓ vapor pressure ↑. CS2 = lowest boiling point.
CC
C
O
H
H
H
HH H
CC
CH
H
H
HH H
H H
CC
OHH
HH
H H
Page 7
19.) The following temperature and vapor pressure data was collected for benzene (C6H6).
T (˚C) Vapor Pressure, Pvap, (mmHg)
1/T (1/K) ln Pvap
7.6 40. 0.003562 3.7
26.1 100. 0.003342 4.61
60.6 400. 0.002996 5.99
80.1 760. 0.002831 6.63
a. Using the data, what is the normal boiling point of benzene (in ˚C)? 80.1 ˚C (Pvap = 760 mmHg)
b. Calculate the inverse Kelvin temperature and natural log of the vapor pressure for the first data point (7.6 ˚C, 40. mmHg). Enter these values into the table above.
c. Using the plot of inverse Kelvin temperature and ln Pvap (including the slope and y-intercept data), calculate the heat of vaporization (DHvap) of benzene in kJ/mol.
ln$𝑃&'() = − ∆-./01
2345 + 𝑐 slope of line = –
∆-./01
DHvap = – slope (R) → DHvap = – (–4027 K) × 8.:3;<=>?@∙B
× 3C=3DDD=
= 33.49 kJ/mol
20.) Temperature and vapor pressure data are collected for alcohol (ethanol) and a graph is plotted with the natural log of the vapor pressure on the y-axis and inverse Kelvin temperature on the x-axis. The graph produces a straight line which has a slope of –5142 and y-intercept of 21.28. Use the graphical data to determine the heat of vaporization of alcohol in kJ/mol.
As previously, DHvap = – slope (R) DHvap = – (–5142 K) × 8.:3;<=>?@∙B
× 3C=3DDD=
= 42.75 kJ/mol
21.) Estimate the normal boiling point of ether and ethanol (in Kelvin) from the graph of vapor pressure versus temperature.
Normal b.p. ether: ~310 K
(Where Pvap = 1 atm, or 760 mmHg)
Normal b.p. ethanol: ~350 K
m = –4027 b = 18.04
Page 8
22.) The vapor pressure of ammonia (NH3) is 2.00 atm at –18.7 ˚C and 20.00 atm at 50.1 ˚C. Calculate the heat of vaporization (in kJ/mol) of NH3.
ln 2EFEG5 =
∆-./01
2 34G− 3
4F5 → DHvap =
1∙@H2IFIG5
2 FJGKFJF5
P1 = 2.00 atm T1 = –18.7 ˚C + 273.15 = 254.5 K
P2 = 20.00 atm T2 = 50.1 ˚C + 273.15 = 323.3 K
DHvap = 8.:3;< L
MNO∙P∙@H2G.QQ/RMGQ.QQ/RM5
2 FSGS.SPK
FGTU.TP5
= K3V.3 L
MNO∙P
2D.DD:DV:FPKD.DD:VWVFP5
= K3V.3 L
MNO∙P
2KD.DDD8:XFP5 = 22800
=>?@
× 3C=3DDD=
= 22.8 kJ/mol
23.) The city of Denver, Colorado is nicknamed the Mile High City, as its official elevation is 5280 ft (or 1 mile) above sea level. Water boils at 95.0 ˚C in Denver. Knowing that water has a normal boiling point of 100.0 ˚C and a DHvap of 40.7 kJ/mol, what is the atmospheric pressure in Denver (in torr)?
𝑙𝑛 [𝑃3𝑃W\ =
𝛥𝐻&'(𝑅
[1𝑇W−1𝑇3\
Approach 1: normal bp = T1 and P1 ; use antilog
It shouldn’t matter which (T,P) pair is T1P1, they just need to be paired.
Normal boiling point P1 = 760 torr T1 = 100.0 ˚C + 273.15 = 373.2 K
P2 = ? T2 = 95.0 ˚C + 273.15 = 368.2 K
EFEG
= 𝑒∆c./0
d 2 1𝑇2−1𝑇15 =
P2 = EF
𝑒∆𝐻𝑣𝑎𝑝𝑅 i 1𝑇2
− 1𝑇1j =
kXDl?mm
n2UQ.opLFMNO 52
FQQQLFpL 52 MNO∙P
q.SFUTL52F
Srq.GPsF
SoS.GP5
= kXDl?mm
n$UqtTP)2Q.QQGoFrFPsQ.QQGrqQ
FP5
= kXDl?mm
n$UqtTP)2Q.QQQQSrFP5
= 640 torr
Page 9
Approach 2: change which is T1 and P1 for possibly easier algebra
Normal boiling point P2 = 760 torr T2 = 100.0 ˚C + 273.15 = 373.2 K
P1 = ? T1 = 95.0 ˚C + 273.15 = 368.2 K
EFEG
= 𝑒∆c./0
d 2 1𝑇2−1𝑇15 = P2 × 𝑒
∆c./0d 2 1
𝑇2−
1𝑇15
P1 = 760 torr × 𝑒2UQ.opLFMNO 52
FQQQLFpL 52
MNO∙Pq.SFUTL52
FSoS.GPK
FSrq.GP5
= 760 torr × 𝑒$;8V<B)2D.DDWX8DFP–D.DDWk3X
FP5 = 760 torr × 𝑒$;8V<B)2KD.DDDD:X
FP5
= 760 torr × 0.83 = 640 torr
Approach 3: normal bp = T1 and P1 ; use log rules for possibly easier algebra
Normal boiling point P1 = 760 torr T1 = 100.0 ˚C + 273.15 = 373.2 K
P2 = ? T2 = 95.0 ˚C + 273.15 = 368.2 K
𝑙𝑛 2EFEG5 = v-./0
12 34G− 3
4F5 → ln P1 – ln P2 =
v-./01
2 34G− 3
4F5
ln P2 = ln P1 – v-./01
2 34G− 3
4F5
ln P2 = ln (760 torr) – 2;D.kC=3>?@
5 23DDD=3C=
5 2 >?@∙B8.:3;<=
5 2 3:X8.WB
− 3:k:.WB
5
ln P2 = ln (760 torr) – $4895𝐾) 20.002716 3B− 0.002680 3
B5
ln P2 = ln (760 torr) – $4895𝐾) 20.000036 3B5
ln P2 = 6.457 P2 = e6.457 = 640 torr
Page 10
24.) Carbon disulfide (CS2) has a similar Lewis structure to carbon dioxide (CO2), but has different properties. CS2 is a liquid at room temperature, while CO2 is a gas, and CS2 is highly flammable while CO2 is not.
a. Using specific intermolecular forces, explain why CS2 is a liquid at room temperature while CO2 is a gas.
If CS2 is a liquid while CO2 is a gas, CS2 must have a higher boiling point than CO2. This is because it has a higher molar mass, which means it has stronger London Dispersion Forces (LDFs), which then requires a higher temperature to break.
b. A compound’s flammability is partially dependent on its vapor pressure, as vapors are what burn, not liquids. If insufficient vapors are present, a compound will not ignite. The minimum vapor pressure of CS2 that a match can ignite is 11.3 mmHg (this is the “flash point”). At what temperature (in ˚C) does this occur? CS2 has a DHvap of 28.7 kJ/mol and a normal boiling point of 46.5 ˚C.
Normal boiling point P1 = 760 mmHg T1 = 46.5 ˚C + 273.15 = 319.7 K
P2 = 11.3 mmHg T2 = ?
𝑙𝑛 2EFEG5 = v-./0
12 34G− 3
4F5 → 2 3
4G5 = �𝑙𝑛 2EF
EG5� � 1
v-./0� + 2 3
4F5
34G
= ln 2kXD>>-�33.:>>-�
5 × 8.:3;<=>?@∙B
× 3C=3DDD=
× 3>[email protected]=
+ 3
:3V.kB
34G
= 0.00122 + 0.003128 = 0.00435 T2 = 3
D.DD;:< = 230. K
230. K – 273.15 = –43 ˚C
25.) A pressure cooker (kitchen appliance) is a pot with a tight seal, allowing pressure to build. Most have a setting of 15 psi, which means the pressure is 15 psi (~1 atm) above atmospheric pressure (therefore a total pressure of ~2 atm at sea-level). At sea-level, water boils at ~119 ˚C inside a pressure cooker, and therefore food cooks more quickly. Explain why water boils at a higher temperature at 2 atm compared to 1 atm.
Water boils when its vapor pressure matches the atmospheric pressure (or in the pressure cooker, whatever pressure it’s surrounded by). At 2 atm, water has to have a vapor pressure of 2 atm (or 1520 mmHg) to start boiling. A higher temperature is needed to produce this higher vapor pressure.
Page 11
MULTIPLE PHASE CHANGES
26.) A beaker of water is set atop a hotplate and brought to a boil. Although energy is continually supplied via the hotplate, explain why the temperature remains at 100 ˚C while water remains in the beaker.
Energy is used during the boiling process to break intermolecular forces and convert the liquid to gas. There is no residual energy for the water molecules to absorb in order to increase their average motion (KEavg); therefore the temperature does not increase.
27.) Calculate the heat energy (q) associated with each transformation.
a. 150.0 g of ice at 0.0 ˚C is melted into water at 0.0 ˚C. q=?
Melt ice: q = DHfusn = X.DWC=3>?@
× 3>[email protected]�
× 150.0 g = 50.1 kJ
MMH2O = 2(1.01) + 16.00 = 18.02 g/mol
b. 150.0 g of water at 0.0 ˚C is frozen into ice at 0.0 ˚C.
Same math, reverse process. q = –50.1 kJ
c. 150.0 g of water at 75.0 ˚C is completely boiled (turned to 100.0 ˚C steam).
↑T water: q = swmDT = (4.184 =�˚�
)(150.00 g)(100.0 ˚C – 75.0 ˚C) = 15,700 J × 3C=3DDD=
= 15.7 kJ
Boil water: q = DHvapn = ;D.kC=3>?@
× 3>[email protected]�
× 150.0 g = 339 kJ
Total: 15.7 kJ + 339 kJ = 355 kJ
28.) Calculate the amount of energy (as heat) needed to convert 5.00 g of water at 20.0 ˚C into steam at 115.0 ˚C.
↑T water: q = swmDT = (4.184 =�˚�
)(5.00 g)(100.0 ˚C – 20.0 ˚C) = 1,670 J × 3C=3DDD=
= 1.67 kJ
Boil water: q = DHvapn = ;D.kC=3>?@
× 3>[email protected]�
× 5.00 g = 11.3 kJ
MMH2O = 2(1.01) + 16.00 = 18.02 g/mol
↑T steam: q = ssteammDT = (2.06 =�˚�
)(5.00 g)(115.0 ˚C – 100.0 ˚C)
= 155 J × 3C=3DDD=
= 0.155 kJ
Total: 1.67 kJ + 11.3 kJ + 0.155 kJ = 13.1 kJ
Page 12
29.) A sample of sodium metal (0.250 g) is dropped into a mixture of 50.0 g water and 50.0 g ice, both at 0 ˚C. The sodium reacts with the water through the following equation, releasing energy as shown by the change in enthalpy. When the reaction is complete, has the ice melted or not? Show calculations to support your answer.
2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g) DH˚ = –368 kJ
Energy released: 0.250 g × 3>[email protected]�
× K:X8C=W>?@�'
= –2.00 kJ (energy released)
Energy to melt ice: q = DHfusn = 50.0 g ice × 3>[email protected]�
× X.DWC=3>?@
= 16.7 kJ
MMH2O = 2(1.01) + 16.00 = 18.012 g/mol
Only 2.00 kJ of energy is released, but it takes 16.7 kJ of energy to completely melt the ice.
The ice won’t fully melt.
PHASE DIAGRAMS
30.) For the following phase diagram,
a. Identify the phase(s) present at each point.
A: Solid
B: Liquid
C: Gas
D: Solid and gas
E: Solid, liquid, and gas
F: Liquid and gas
G: Liquid and gas
H: Supercritical fluid
b. Where is the triple point? E
c. Where is the critical point? G
d. On the diagram, label the normal melting point and normal boiling point.
Temperature
Pres
sure
GAS
LIQUID SOLID
Normal melting point
Normal boiling point
Page 13
31.) The following is a phase diagram of “Compound X”. Although phase diagrams are often not to scale, assume this diagram is to scale.
a. If Compound X was at room temperature in a lab at Butte College, in what phase(s) would it exist? (T ~ 25 ˚C, P ~ 760 mmHg): solid
b. Which process would occur if its temperature was increased, without changing pressure?
A: It would boil. B: It would condense. C: It would melt. D: It would freeze. E: It would sublime. F: It would deposit.
c. On the diagram, mark the following locations with a dot and letter:
Mark “A” anywhere Compound X exists only as a liquid.
Mark “B” anywhere Compound X is freezing.
Mark “C” anywhere Compound X exists as a supercritical fluid.
32.) The following heating curve and phase diagram are of two different substances. Use each to estimate the following physical properties.
a. Melting point: 20 ˚C b. Boiling point: 120 ˚C
c. Normal melting point: 170 K d. Normal boiling point: 260 K
33.) Butane (C4H10) is a gas at room temperature, but can be converted to a liquid if...
a. It is heated to a very high temperature. b. A high enough pressure is applied. c. Intermolecular forces are broken. d. X-rays are shined on it.
GAS
LIQUID SOLID
sublimation
A B
C
mp bp
Page 14
CRYSTALLINE SOLIDS
34.) Polonium (Z = 84) is a highly reactive metal with no stable isotopes, and is used as a radioactive heat source for satellites and the Moon rovers. Polonium crystallizes with a simple cubic unit cell. Calculate the following:
a. The number of polonium atoms in one unit cell.
8 corners (1/8) = 1 sphere, or 1 Po atom
b. The packing efficiency for the metal, to 3 significant figures.
Packing = 3�(�nmn(US�m
S)
(Wm)S × 100 =
US�mS
8mS × 100 = 52.4%
c. The percentage of space unoccupied by polonium atoms in the metal (to 3 SFs).
Unoccupied = 100% – 52.4% = 47.6%
d. The length of one edge of the unit cell (length ‘l’) in picometers (pm). The radius of a polonium atom is 169 pm (where 1 × 1012 pm = 1 m).
l = 2r l = 2 × 169 pm = 338 pm
35.) Nickel metal crystallizes with a face-centered cubic unit cell. The radius of a nickel atom is 124 pm. What is the length of one edge of the unit cell of nickel (length ‘l’) in pm?
l 2 + l 2 = (4r)2
2 l 2 = 16r2
l 2 = 8 r2
l = √8 r = √8 (124 pm) = 351 pm
length l
length l
Page 15
36.) Silver metal crystallizes with a face-centered cubic unit cell. The edge of the unit cell (length ‘l’) is 407 pm. What is the radius of a silver atom in picometers?
l 2 + l 2 = (4r)2
2 l 2 = 16r2
@G
8 = r2 r =
@√8
= ;Dk(>√8
= 144 pm
37.) Calculate the packing efficiency (to 3 SFs) of the atoms in a face-centered cubic unit cell.
#Atoms: 8 corners (1/8) + 6 faces (1/2) = 4 spheres
Unit Cell length: l 2 + l 2 = (4r)2 l = √8 r (see question 30)
Packing = ;�(�nmn�(US�m
S)
(√8m)S × 100 =
FrS �mS
8√8mS × 100 = 74.0%
38.) Sodium metal crystallizes with a body-centered cubic unit cell. Calculate the number of Na atoms in the unit cell, and the coordination number of the sodium.
8 corners (1/8) + 1 middle = 2 Na atoms
Coordination number = 8 (look at the middle one)
39.) Copper metal crystallizes with a face-centered cubic unit cell. What is the coordination number of the copper?
Coordination number = 12 (4 on the same level, 4 in front of it, 4 in back of it)
40.) Sodium bromide crystallizes in the same manner as NaCl. Describe the unit cell of NaBr, including the type of unit cell (simple cubic, face-centered cubic, or body-centered cubic) and location of the cations and anions. Sketch one face of the unit cell.
The Br– ions (anions) are larger than the sodium ions and form a face-centered cubic unit cell. The smaller Na+ ions (cations) fit inside the cracks of the unit cell, four on each layer.
= Br–
= Na+
Page 16
41.) Multiple choice:
The unit cell with the highest coordination number is:
a. Simple cubic b. Body-centered cubic c. Face-centered cubic
Pure gold is 24 K (24 karat), but gold-silver alloys are much more commonly encountered in jewelry. 18 K “yellow” gold contains 75% Au, 12.5% Ag, and 12.5 % Cu. As the gold, silver, and copper atoms replace each other in the crystal lattice, 18 K gold is considered a ...
a. Substitutional alloy b. Interstitial alloy c. Ionic solid d. Covalent molecule
Cesium chloride forms the following unit cell, with a Cs+ ion at the middle of the unit cell shown and Cl– ions on the corners. Which is NOT true about this unit cell?
a. The Cl– ions form a simple cubic unit cell b. The unit cell is body-centered cubic c. The coordination number of the Cs+ ion is 8 d. The Cl– ions do not touch each other
Which is NOT true about the “electron sea model” to describe metallic bonding?
a. Valence electrons are shared by all metal atoms b. Valence electrons are stationary around each metal nucleus c. The metal nuclei form cations which attract the sea of valence electrons d. The model explains how metals conduct electricity and heat.
Quartz (SiO2) is classified as a(n)…
a. Ionic solid b. Molecular solid c. Network solid d. Metallic solid
Which solid should have the highest melting point?
a. C6H12O6 b. H2O c. H2C2O4 d. KBr
