Big Idea 2: “Bonding and Phases” AP LEVEL ASSESSMENT
DUE: Mon 3/2
1. D Ca is a metal, and F is a nonmetal. Their EN differ sufficiently for them to
create an ionic bond, which is stronger than all other bond types except for
network covalent bonding. Which NH3 does not exhibit.
2. B Interstitial alloys form when atoms of greatly different sizes combine. The
aluminum atoms would have a chance to fit in between the comparatively
larger lead atoms.
KEY
In this molecule, sulfur forms dsp3 hydride orbitals, which have a trigonal bipyramid
structure. Because SF4 has one unshared electron pair, the molecule take
the “seesaw” or “folded square” shape.
Choices (A) and (B), CH4 and NH4+, each have 8 valence electrons distributed in
the same Lewis dot structure and shape, shown below for NH4+.
3. D When drawing Lewis diagrams for the four choices, the first three are
completely non-polar. Only OF2, with a bent shape, would exhibit any
dipoles.
In these molecules, the central atom forms sp3 hybrid orbitals, which have a
tetrahedral structure. There are no unshared electron pairs on the central
atom, so the molecules are tetrahedral.
Choice (D) AlCl4- has 32 val e− distributed in the same Lewis dot structure and
shape, shown below:
4. A Only atoms with at least three energy levels (n = 3 and above) have empty dorbitals that additional electrons can fit into, thus expanding their octet.
5. B Moles of H2 = 2.00 g ÷ 2.00 g • mol-1
Moles of Ne = 10.00 g ÷ 20.0 g • mol-1
Total moles = 1.50 mol.
XNe = mol Ne ÷ total mol
= 1.00 mol
= 0.500 mol
= 0.500 mol ÷ 1.50 mol
= 0.33
7. D Both gases only have dispersion forces. The more electrons a gas has, the
more polarizable it is and the stronger the intermolecular forces are.
8. A The gas molecules have the same amount of KE due to their temperature
being the same. Via KE = ½ mv2, if KE is the same, then the molecule with
less mass must correspondingly have a higher velocity.
9. B First, C2H6, is not an ionic substance and thus has no lattice energy. Next, LiF
is composed of ions with charges + and -, and will not be as strong as the
two compounds which have ions with charges 2+ and -. Finally, MgCl2 is
smaller than CaBr2, meaning it will have a higher lattice energy, as
(according to Coulomb’s law) atomic radius is inversely proportional with
bond energy.
10. C Ozone exhibits two resonance structures, shown below. Taking either bond
position yields a bond order of (1 + 2) ÷ 2 = 1.5.
In these molecules, the central atom form sp3 hybrid orbitals, which have a
tetrahedral structure. There are no unshared electron pairs on the central
atoms, so the molecules are tetrahedral.
13. C Resonance is used to describe a situation that lies between single and double
bonds, so the bond length would also be expected to be between that of
single and double bonds The best answer here is 140 pm, or choice (C).
14. B The Lewis dot structures for the
answer choices are shown right.
Only PH3 has a single unshared
electron pair on its central atom.
15. A H2 experiences only dispersion forces
and has the lowest boil point.
N2 also experiences only dispersion
forces, but is larger than H2 and has
more electrons, so it has stronger
interactions with other molecules.
NH3 is polar and undergoes hydrogen
bonding, so it has the strongest
intermolecular interactions and the
highest boiling point.
11. B A liquid with a low boiling point must be held together by weak IMF’s, of which
dispersion forces are the weakest kind.
12. C SF4 has 34 val e− distributed in the Lewis dot structure and shape below:
16. B The bond that holds HCl together is a
covalent bond with a large polarity.
The bond that holds NO together is
also polar covalent, but its polarity is
very small because N and O are so
close together on the periodic table.
Cl2 is nonpolar because they share
electrons equally, and SO3 is nonpolar because it is symmetrical (trigonal
planar). Nonpolar molecules have dipole moments of zero.
17. D From Dalton’s law, the partial pressure of a gas depends on the number of
moles of the gas that are present. The total number of gas present is
1.5 + 3.0 + 0.5 = 5.0 mol
If there are 3 moles of N, then 3/5 of the pressure must be due to nitrogen.
(3/5) (700 mmHg) = 420 mmHg
18. C From Dalton’s law, the partial pressure of a gas depends on the number of
moles of the gas that are present. If the mixture has twice as many moles of
He as Ne, then the mixture must be 1/3 Ne. So 1/3 of the pressure must be
due to Ne.
(1/3) (1.2 atm) = 0.4 atm
19. D From Dalton’s law, the partial pressure of N2 and H2O vapor must add up to
the total pressure in the container. The partial pressure of water vapor in a
closed container will be equal to the vapor pressure of H2O, so the partial
pressure of N2 is:
781 mmHg – 23 mmHg = 758 mmHg
20. A 11.2 L
1 mol
22.4 L
MW = gram ÷ mol
=
0.500 mol gas
=
22.0 g ÷ 0.500 mol = 44.0 g • mol-1
This is the molecular mass of CO2
21. D Density is measured in grams per liter. One mole of He gas has a mass of 4
grams and occupies a volume of 22.4 L at STP, so the density of He gas at
STP is:
4 _
22.4 g/L
22. D There are initially 3.0 mol of gas in the container. If they react completely, 2.0
moles of gas are produced. 2.0 mol of gas will exert exactly 2/3 as much
pressure as three moles of gas.
23. C From the ideal gas laws, for a gas sample at constant pressure and amount
V1 = V2
T1
T2
Solving for V2 we get
V2 = V1 T2
T1
So V1 is multiples by the factor
T2
T1
Remember to convert Celsius to Kelvin
127˚C + 273 = 400 K = 4
27˚C + 273
300 K
3
24. A We can find the number of moles of gas from PV = nRT. But remember to
convert 26˚C to 299 K. Then solve for n:
n = P V = (2.0) (6.0)
RT
(0.8) (299) mol
Now remember:
MW = grams =
mol
(10 g)
_
= (10 ) (0.8) (299)
(2.0) (6.0)
g/ mol
(2.0) (6.0) g/mol
(0.8) (299)
25. B First find the number of moles.
Moles = (molarity) (volume) = (5.0 M) (4.0 L) = 2.9 mol
MW = grams =
mol
240
2.0
_
= 120 g • mol-1
26. A Find out how many moles of Na+ were added.
Moles = (molarity) (volume) = (2 M) (0.5 L) = 1 mol
Because we get 2 mol of Na+ ions for every mol of Na2SO4 added, we need to
add only 0.5 mol of Na2SO4
27. C Hydrogen bonds are the strongest types of IMF’s when dealing with molecules
of similar size. Ethylene glycol has twice as many H bonds as ethanol
(acetone has none), and so it would have the highest boiling point.
28. B Vapor pressure arises from molecules breaking free from the INF’s holding
them together. Acetone, which has no H bonding, thus it has the weakest
IMF’s of the three and thus would have the highest vapor pressure.
29. D Water is very polar, and using the concept of “like dissolves like”, any
substance with polar molecules would be soluble in water. As all three
molecules are polar, all three liquids would be soluble in water.
30. (SFR) Use the principles of bonding and molecular structure to explain the following statements.
(a) The boiling point of argon is -186˚C, whereas the boiling point of neon is -246˚C
-Atoms of NG in the liquid phase are held together by Dispersion forces only
-Atoms with more e- are more easily polarized = Larger dispersion forces
-Ar has 18 e- to Ne 10 e-Ar has a higher boiling point
(b) Solid sodium melts at 98˚C, but solid potassium melts at 64˚C.
-Metals = metallic bonds
-K has a higher AR and as a result a less Zeff
-K’s increased AR = greater bond length and Coulomb’s law = weaker attractions = lower melting point
-K’s decreased Zeff = weaker attractions = lower melting point
(c) More energy is required to break up a CaO(s) crystal into ions than to break up a KF(s) crystal into ions.
-Both ionic bonds = crystal lattices
-Ca2+ O2- = greater charge differential than K+ and F- = stronger attraction according to Coulomb’s law = More energy required to break up
(d) Molten KF conducts electricity, but solid KF does not.
-Both ionic bonds composed of K+ and F- ions.
-Liquid (molten) state, ions are free to move and conduct (transport) electricity
-Solid ions are fixed in crystal lattice w/e− localized around them = no free charge to move and conduct electricity
31. (SFR) The carbonate ion CO32- is formed when carbon dioxide, CO2, reacts with slightly basic cold water.
(a)
(i) Draw the Lewis electron dot structure for the carbonate ion. Include resonance forms if they apply.
(ii) Draw the Lewis electron dot structure for carbon dioxide
(b) Describe the hybridization of carbon in the carbonate ion
-Central atom = C
-Forms 3 sigma bonds with oxygen atoms
-No free electrons
-sp2
(c)
(i) Describe the relative lengths of the three C – O bonds in the carbonate ion.
-Same length
-Resonances means that the varying single and double bonds results in identical bond length somewhere in-between single and double bonds.
(ii) Compare the average length of the C – O bonds in the carbonate ion to the average length of the C – O bonds in carbon dioxide.
- C – O bonds in CO32- have resonance between single and double bonds
- C – O bonds in CO2 are both double bonds
- C – O bonds in CO32- will be between single and double bonds and will thus be longer than the pure double bonds of CO 2
32. (SFR)
The graph above shows the changes in pressure with changing temperature of gas samples of helium and argon confined in a closed 2-L vessel.
(a)What is the total pressure of the two gases in the container at a temperature of 200K
-From reading the graph.
Ptotal
=
PHe + PAr
Ptotal
=
(1 atm) + (1.5 atm) =
(b) How many moles of helium are contained in the vessel?
2.5 atm
-From reading the graph:
@ 200 K = 1 atm
n = P V = (1.0 atm) (2.0L)
RT
(0.0821) (2.0 x 102 K)
=
(c) How many molecules of helium are contained in the vessel?
0.12 mol He
6.022 x 1023 molecules
1 mol
0.12 mol He
=
7.2 x 10 22 molecules He
(d) Molecules of which gas will have a greater distribution of velocities at 200 K? Justify your answer.
-Both have same temp so have same Energy
-As the mass of He < Ar have higher average velocisties
-The higher the average, the more distribution for individual molecules
(e) If the volume of the container were reduced to 1 L at a constant temperature of 300 K, what would be the new pressure of the helium gas.
P1V1
T1
=
(1.5 atm)( 2.0 L )
P2V2
T2
=
Since T is constant
P2 (1.0 L)
P1V1
=
P2V2
P2 = 3.0 atm
33. (SFR) A student preforms an experiment in which a butane lighter is held underwater directly beneath a 100-mL graduated cylinder which has been filled with water. The
switch on the lighter is pressed, and butane gas is released into the graduated cylinder. The student’s data table for this lab is as follows:
(a) Given the vapor pressure of water at 19.0˚C is 16.5 mmHg, determine the partial pressure of the butane gas collected in atmospheres.
745 mmH – 16.5 mmHg
=
729 mmHg
1 atm
760 mmHg
=
0.959 atm
(b) Calculate the molar mass of butane gas from the experimental data given.
20.432 g – 20.296 g
MW = grams
mole
=
=
0.136 g CH4
0.136 g CH4
2.74 x 10-3 mol
n = P V = (0.959 atm) (0.06840 L)
RT
(0.0821) (292 K)
=
=
2.74 x 10-3 mol CH4
49.6 g/mol
(c) If the formula of butane is C4H10, determine the percent error for the student’s results.
MW C4H10 = 58.12 g/mol
Percent Error
=
│Actual – Experimental Value │
Actual Value
= │58.12 – 49.6 │
58.08
= 14.5% error
(d) The following are common potential error sources that occur during this lab. Explain whether or not each error could have been responsible for the error in the student’s
results
(i) The lighter was not sufficiently dried before massing it after the gas was released.
-Mass of butane calculated will be low
-Result in MW = g ÷ mol would be a lower MM
-Consistent with student’s % error
(ii) The gas in the lighter was not held underwater long enough to sufficiently cool it to the same temperature of the water and was actually at a higher temperature
than the water.
-Moles of butane calculated will be high
-Result in MW = g ÷ mol would be a lower MM
-Consistent with student’s % error
(iii) Not all the butane gas released was collected in the graduated cylinder.
-Volume of gas would be artificially low = mole calculation based on it would thus be low
-Result in MW = g ÷ mol would be a higher MM
-NOT consistent with student’s % error