1. (5 points) State which of the following sets of quantum numbers would bepossible and which would not. Using one sentence, explain what is wrong withthe quantum numbers that are not possible.
a) n = 4, l = 4, ml = 4, ms =1/2
b) n =4, l = 3, ml = 2, ms = 1/2
c) n = 4, l =5, ml = 0, ms = -1/2
d) n = 4, l = -2, ml = 0, ms = -1/2
e) n = 4, l = 3, ml = 5, ms = -1/2
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2. (8 points) )Which of electron configuration represents a violation of auf bau,Hund’s rule, and/or the Pauli principle?Explain briefly your choices.
1s 2s 2pa ↑↓ ↑↓ ↑ ↑ ↓
b ↑↓ ↑↓ ↑↓ ↑
c ↑↓ ↑↓ ↓ ↓ ↓
d ↑ ↑↓ ↑↓ ↑↓
e ↑↓ ↑↓ ↑ ↑ ↑
Give the set of quantum numbers for each of the electrons in the p orbitals inorbital diagram (a)
3. (2 points) How many angular nodes does an 12f orbital have?
How many radial nodes does a 7p orbital have?
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4. Name the element with the following electron configuration:
[a] 1s22s22p63s23p2
[b] 1s22s22p63s23p63d104s24p64d105s25p5
[c] [Ar]4s13d5
[d] [Xe]4f146s2
5. (4 points) Using the periodic table write the expected ground state electronconfigurations fora) The third element in Group-4A
b) Element number 114 (yes even if it doesn't yet exist)
c) The elements with two unpaired 3d electrons.
d) The halogen with electrons in the 6p atomic orbitals.
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6. (8 points) Determine which of the following statements are true and explainwhy or why not based on shielding effects, quantum shielding, and/or Zeff.
a) Ionization energies increase down a group
b) The ionization energy of an cation is larger than that of the parent atom.
c) The sodium ion is smaller than the potassium ion
d) The fluoride anion is smaller than the fluorine atom
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7. Penetration is the process by which an outer electron moves through theregion occupied by the core electrons to spend part of its time closer to thenucleus. Penetration and the resulting effects on shielding cause an energylevel to spit into sublevels of differing energy. Use the penetration effect toexplain the difference in relative orbital energies of a 2s and a 2p electron inthe same atom.
8. The first ionization energy of the chlorine atom is 1251 kJ/mol. Which of thefollowing values would be a more likely ionization energy for the iodine atom:1000 kJ/mol or 1400 kJ/ mol. Explain your reasoning based on periodic trendsAND Zeff.
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9. Little is known about the properties of astatine, At because of its rarity andhigh radioactivity. Nevertheless, it is possible for us to make manypredictions about its properties. Do you expect the element to a gas, a liquidor a solid at room temperature? Explain. What is the formula for thecompound that forms when astatine reacts with sodium metal?
10. Answer the following questions clearly. Please explain your choice.:
a) Of the elements S, Se, Cl, which one has the largest radius?
b) Which is larger, the chlorine atom or the chlorine ion?
c) Which has the largest ionization potential, N, P, or As?
d) Which has the largest radius, O2–, N3–, F–?
e) Which one of the following ions would you expect to have the smallestradius? Cr6+, P3-, Cl–, K+, or Ti4+
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11. (a) Explain the following trend in Lattice energy based on ion size and thedefinition of lattice energy:
BeH2 3205 kJ/molMgH2 2791 kJ/molCaH2 2410 kJ/molSrH2 2250 kJ/molBaH2 2121 kJ/mol
(b) The lattice energy of ZnH2 is 2870 kJ/mol. Based on the data given inpart a, the radius of the zinc ion is expected to be closest to that of whichgroup 2A element? Use a complete sentence to explain your choice.
12. Using electronegativity values, predict which bond in the following groups willbe most polar. Show your work.
a) C—H, Si—H, P—H
b)Al—Br, Ga—Br, In—Br, Tl—Br
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13. Determine which of the structures listed below would be the most plausiblestructure based on formal charges.
• Show the calculations for formal charges• Explain your choice in one or two sentences based on formal charge,elecronegativity, size, etc.
C N O N C O C O N
14. Write the resonance structures (using the best formal charge model) forClO2–, chlorite. Show all formal charges.
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15. Draw the lewis structure for SeCl4.
16. The space shuttle orbiter utilizes the oxidation of methyl hydrazine bydinitrogen tetroxide. Use bond energies to estimate the ∆H for the reactionbelow.
5N2O4(l) + 4 N2H3CH3(l) → 12 H2O(g) + 9 N2(g) + 4CO2(g)
N NOO
O ON N
HH
H C H3dinitrogen tetroxide methyl hydrazine
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17. Write the best (formal charges, octet rule etc) Lewis structure for SO2Cl2.What is the molecular and electron pair geometry? What type of hybridizedorbitals are needed for this molecule?
Will it have the same molecular and electron pair geometry as OCl2? Justifyyour answer by comparing Lewis structures. Is this a polar molecule? why?
c) Show the complete hybridization process for the oxygen in oxygendichloride. Clearly label what the electrons are used for (bonding, lone pair) inthe hybridization scheme. You do not have to show chlorine in the scheme.
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18. The structures below show the stick and ball drawing of three possible shapesof an AF4 molecule. A is the general symbol for the central atom. Theterminal atoms are fluorine. The lone pairs are not shown.a) Identify the molecular geometry for each of the shapes.b) For each shape, give VSEPR (electron pair or electron domain) geometry onwhich the molecular geometry is based.c) For each shape, how many lone pairs are there on atom A?d) For each shape, what are approximate bond angles?
A B C
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19. When two or more compounds have the same composition but a differentarrangement of atoms (or geometries), we call them isomers. Isomers arecomposed of the same collection of atoms, but they differ in one or morechemical or physical property. There are several types of isomers found inchemistry. In a geometric isomer, the bonds are the same, but the orientationor arrangement of the bonds is different. PF2Cl3 has three geometricisomers. One of these isomers has a dipole of zero; two do not.
(a) Draw the geometric isomers of PF2Cl3 using VSEPR. Clearly show thethree dimensionality of each molecule. (Please label your drawings)(b) State which of the three molecules in part (a) has no dipole.(c) Explain BRIEFLY why your choice has no dipole.
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20. (5 points) Using the valence orbital diagram (box diagram), show how theatomic orbitals of the central atom in IF4— leads to the appropriate hybridorbitals. Fill the hybrids with the correct number of electrons please.
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21. The azide, N3– molecule is linear with two N—N bonds of equal length, 1.16Å.
a) Draw the best Lewis structure for the azide ion, showing formal charges.
b) Using VBT, what hybridization scheme would you expect at each nitrogen inthe azide molecule to have? (Describe theVSEPR geometry, show hybridizedorbitals containing electrons, link these orbitals to make bonds and sketch themolecule showing the sigma bonds and π bonds for full points.)
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22. What factor accounts for each of the following differences in bond length?
a) I2 has a longer bond than F2.
b) N—F is shorter than N—Br.
c) N—N is longer than N≡N
23. Consider the molecules H2 and He2 and the ion H2—.(a)Sketch the molecular orbitals of these two molecules.(b)How many electrons are there in each species?c)What is the bond order for each species?d)Now consider the ions HeH— and He+. Which one is more stable?
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24. A chloride of antimony is a solid at room temperature, but vaporizes whenheated. When 2.359 g of this compound are vaporized in an evaluated 1.00 Lchamber at 581K, the pressure in the bulb is 376 mmHg. What is the molecularweight of this chloride of antimony, assuming the vapor is an ideal gas?
25. Gaseous iodine pentafluoride, IF5, can be prepared by the reaction of solidiodine and gaseous fluorine: I2(s) + 5 F2(g) → 2 IF5(g).A 5.00 -L flask is charged with 10.0 g of F2(g) and 10.0 g of I2(s) . The reactionproceeds until one of the reagents is completely consumed. After the reactionis complete, the temperature in the flask is 125 °C. What is the partialpressure of the IF5 in the flask? What is the mole fraction of IF5 in theflask?
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26. Assume that a single cylinder of an automobile engine has a volume of 600.0cm3. If the cylinder is full of air at 80°C and 0.980 atm, how many moles ofO2 are present? The mole percent of oxygen in air is 20.95 %. How manygrams of Octane (C8H18) could be combusted by this quantity of O2, assumingcomplete combustion with the formation of CO2 and H2O?
27. Two gases in adjoining vessels were brought into contact by opening a stopcockbetween them. The one vessel measured 0.250 L and contained NO at 800.0torr and 220.00K. The other measure 0.100L and contained oxygen at 600.0torr and 220.00K. The reaction to form N2O4(s) exhausts the limiting reagentcomplete. Neglecting the vapor pressure of N2O4(s) (which is minimal) what isthe pressure and composition of the gas remaining at 220.00K after completionof the reaction? What is the weight of N2O4(s) formed?
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28. (4 points) A gas whose molar mass you wish to know effuses through anopening at a rate one fourth as fast as that of helium gas. What is the molarmass of the unknown gas?
29. Given the following information about CO2, CS2, and CSe2, How do thestrengths of the intermolecular forces vary from CO2 to CS2 to CSe2?
• Carbon diselenide (CSe2) is a liquid at room temperature. The normalboiling point is 125°C, and the melting point is -45.5°C.• Carbon disulfide is also a liquid at room temperature with a normal boilingpoint of 46.5°C and a melting point of -116.5°C.• Carbon dioxide is a gas at room temperature. It does not have a normalboiling point, but rather sublimes at -78°C.
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30. In each of the following groups of substances, pick the one that has the givenproperty. Use your knowledge of the different types of intermolecular forces(ion-dipole, dipole-dipole, hydrogen bonding, instantaneous dipole (Londonforces, dispersion forces) to help you make your choice. Justify your answerwith one sentence by describing the type of intermolecular force and how iteffects that property.
a) Greatest viscosity: CH3CH2CH2CH3, CH3CH2OH, or HOCH2CH2OHEXAMPLE: Viscosity is based on shape, molar mass, and strongintermolecular forces. all of the molecules are linear and have london forces.Butane (MM = 58) has only london forces it is a gas at near room temperature.ethanol (MM =45) can form a hydrogen bond. it is a liquid at room temperature.ethylene glycol has the largest molar mass (MM=61), therefore the mostelectrons and can form the most hydrogen bonds. Ethylene glycol has thehighest viscosity.
b) Smallest (lowest cohesion) surface tension: H2O, CH3CN, or CH3OH
c) Smallest vapor pressure at 25°C: CO2, H2O, or SO2
d) Highest boiling point: Cl2, Br2, or I2
e) Strongest hydrogen bonding: NH3, PH3, or SbH3
f) Greatest heat of vaporization: H2O, H2S, or H2Se
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31. The vapor pressure curves for compound A and B are drawn below. Both A andB have diopoles.a) Which compound has the stronger dipole?b) Which compound is more volatile?Please explain your answers.
100806040200-20100
200
300
400
500
600
700Compound A
Compound B
Vapor pressure curves for A and B
vapo
r pr
essu
re
32. Mt Kilimanjaro in Tanzania is the tallest peak in Africa (19,340 ft). If thebarometric pressure at the top of the mountain is 350 torr, at whattemperature will water boil there? ∆H vap = 40.67 kJ/mol at 100°C
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33. I wanted stewed prunes for my breakfast in winter. Thinking ahead, I pickedsome ripe plums and dried them in the sun for several weeks during the summer.I noticed that the plums dried up to make prunes. I stored them in my pantry.When winter came, I boiled the prunes in water, sugar and lemon juice. Theprunes plumped up. I ate them with a small amount of cream. YUMMY!
Explain what happened to the fruit in terms of osmosis.
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Answer Key for Test “practice exam 3 w final”, 4/30/13No. in
Q-BankNo. onTest Correct Answer
6.5 13 1 a) n ≠ l or zero n>lb) n ≠ l n is always greater than lc) works for me. The quantum numbers describe a 7p orbital
6.5 17 2 ?6.5 22 3 ?6.8 18 4 ?6.9 7 5 ?7.2 1 6
FALSE-electron affinity is the energy released (or not) when an electron is added to an atomin the gas phase. The more stable the configuration (half filled and filled sub shells or n-shells, the more energy released.) Electron affinity increases across a period andstays relatively constant down a family. Fluorine is an n=2 atom while Cesiumis an n=5 atom. Fluorine is a small, non-metal atom with a high Zeff whilecesium is very large and metallic. When cesium adds an electron, the 5s subshell is filled, a stabilizing effect. When an electron is added to fluorine, then=2 shell is filled, not just a sub-shell-this is more stabilizing, the electron iscloser to the nucleus (n values!), a stabilizing effect. FALSE-Ionization energyis the energy needed to remove an electron from an outer most shell in an atom(in the gas phase). Anions and parents have the same number of protons.Adding electrons lowers Zeff. Lowering the Zeff makes it easier to removeelectrons. Therefore, the ionization energy of the anion is lower than theionization energy of the parent because it has a lower Zeff.TRUE-lithium is anatom in n=2 shell, while rubidium is an ion in n=5 shell. The electron in the 5sorbital is further from the nucleus (does not penetrate as well) as the 2selectron in lithium atom. When removing an electron to make the ion, bothparticles are smaller than the parent atom, but lithium ions have a smallercloud than rubidium because overall, there are fewer electrons. Fewerelectrons mean a smaller ion. TRUE-for similar reasons as answer C. Thechloride ion has electrons added to orbitals that are further from the nucleusthan fluorine. This means they do not penetrate the nucleus, as effectively, sothe ion is larger when the n value is larger for the same family. Fluorine is n=2 while chlorine is n=3. If the rCl>rF, then the same is true for the ions.
7.2 6 7the 2s orbital is better at penetrating to the nucleus b/v the angular momentum q.n. is 0 vs.the 2p orbital which has an l=1. Orbitals with low n values represent small regions of electrondensity with the electron density close to the nucleus. Low l values mean fewer nuclear nodes.The lower l values shield the nucleus from higher l values. Because the p orbitals have a node
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Answer Key for Test “practice exam 3 w final”, 4/30/13No. in
Q-BankNo. onTest Correct Answer
at the nucleus it is harder for the electron to penetrate to the nucleus effectively.7.5 2 8
Chlorine has a higher zeffective than iodine because it has less core electrons than iodine. Iodine isperiod 5 compared to chlorine in period 3. The quantum shielding going down a family increasesfaster than the Z (the actual nuclear charge). The Zeff is therefore decreasing as the n quantumvalue increases. Because the Zeff is small, the attraction for the outer most electron is weaker in theiodine atom than in the chlorine atom leading to a larger sized atom and a lower ionization potential.
7.8a 4 9I expect astatine to be a solid. In group 7A, the phase of the elements changefrom a gas phase for fluorine, to the solid phase for iodine, iodine is muchheavier than fluorine. since At is the next element in the series, it should besimilar to iodine. The formula would be NaAt.
7.9 6 10a) S and CL are in the same period, n=3, while Se is right below S in n=4. Theradius is dependent on the Zeff and quantum effects. Since S is in n= 3 andSe is in n=4, Se is bigger than S due to quantum effects, (size increases withincreasing n. Cl is in the same period as S, but to the left of S. This meansthat for about the same amount of shielding, Cl has a larger Zeff than S, andtherefore a smaller radius. Cl is the smallest atom and Se is the largest.
b) Both atom and ion have the same nuclear charge. The chloride ion has moreelectrons and more shielding than the chlorine atom. As shielding increases,Zeff decreases so the attraction between the outermost electron and thenucleus also decreases. The chloride ion must be larger.
c)Ionization energy increases across a period and decreases down a period as ageneral trend-again due to quantum shielding and Zeff. N, P, and As arecongers, or members of the same group. The electron is placed in quantumshells that are further from the nucleus, and are easier to remove. N shouldhave the larger ionization energy because it's outermost electron is closest tothe nucleus and therefore hardest to remove.
d) In an isoelectron series (to which these ions belong), as the negative chargeon the ion increases, the size increases of the ion increases. As moreelectrons are added, the shielding increases. Nitride would have the largestradius.
e) In this series, both cations and anions are presented. Cations with largerpositive charges tend to be smaller than the ions and atoms in the same
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Answer Key for Test “practice exam 3 w final”, 4/30/13No. in
Q-BankNo. onTest Correct Answer
isoelectronic series. Removing electrons decreases shielding so the outermostinner-core electrons feel a stronger nuclear charge. There are less electronsto compete for penetration. The increased Zeff results in a decrease of theradius.
8.2 4 11Lattice energy is the energy needed to separate an ion pair in the solid phaseto make ions in the gas phase. Lattice energy is dependent on two factors: theadded radii of the ion pair and the charge. Lattice energy is inverselyproportional to the distance (d) between the nuclie and proportional to theproduct of the charge. These compounds are hydrides of the group 2A metals.The cation portion of this series are in the same group. The radius of thisseries increases as n, the principal quantum number, increases. This meansthat Be is the smallest ion while Ba is the largest. The anion radius remainsconstant. The charge for both the cation and the anion are constant. Thedata shows the lattice energy decreases down group 2A for this series ofhydrides because the size of the cation is increaseing, increasing the distancebetween the two radii.
(b) Looking at the data, the LE(BeH2)> LE(ZnH2)> LE(MgH2). From this onecan conclude that the radius of the zinc cation must be larger than Mg ion, butsmaller than Be ion becuase LE is related to the size of the ion. I think thesize is closer to the size of the Mg ion becuase the LE are closer.
8.4 4 12 Si–H has the highest electronegativity difference. Al—Br also.8.5 1 13
The three ways that the atoms can be ordered are: N—C—O, N—O—C, and C—N—O. If onedoes the Lewis structures, the N—O—C is not stable at all due to formal chargeconsiderations. This leaves us with N—C—O and C—N—O skeletons. Structure B has tworelatively stable resonance structures. Because the electron density is delocalized betweenthe N, C, and O atoms, the triple bond between C and N is longer than it should be. Thisstructure is more stable than that of structure A. Structure A has two resonancestructures also, but the first structure is unstable due to formal charge considerations. Thefirst structure has a formal charge of 2— on the carbon and a positive charge on thenitrogen. If we have a choice, we minimize negative charges on the most electropositive atomsand localize charge density on the more electronegative atoms. The bond length for the triplebond is much shorter here since it is unlikely that the first resonance structure will form.Therefore structure A is fulminate.
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Answer Key for Test “practice exam 3 w final”, 4/30/13No. in
Q-BankNo. onTest Correct Answer
C N O N C ON C O
2ONC
A B
8.6 2 14
Cl OO O OCl
8.7 8 15 ?8.8 4 16 ∆H = bonds broken - bonds formed. make a list of the number and
types of bonds. [10 N=O, 10N-O, 5N-N] + [4N-N, 12N-H, 4N-C,12C-H] — [24 O-H,9 N trple bond N, 8 C=O. look up thenumbers in the table. do the math -21,619 kJ
9.2 1 17
S
O
Cl
OCl
the electron pair and the molecular geometry is tetrahedral. oxygen dichloridehas the same electron pair geometry, but is angular or bent. It is definetly apolar molecule due to the lone pairs on the oxgen. the hybridization of oxygendichloried is ≠ ≠ ≠ ≠ sp3
9.2 8 18 square planar, tetrahedral, see saw; octahedral, tetrahedral,trigonal bipyramidal; 2,0,1
9.3 3 19
The first structure has no dipole because the arrangement in of the atoms inthe molecule allows the individual dipoles to cancel.
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Answer Key for Test “practice exam 3 w final”, 4/30/13No. in
Q-BankNo. onTest Correct Answer
BrP ClBr
ClCl
BrP BrCl
ClCl
ClP ClBr
Br
Cl
no dipole dipole dipole
9.5 8 20 ?9.6 2 21
The best structure would have two double bonds, becuase this is the resonance structure with thelowest formal charges. the terminal N atoms would be sp2 and the central N would be sp.
N N N
9.7 1 22Bond lengths are controlled by four factors: (1) The smaller the principal quantum number ofthe valence orbitals, the shorter the bond. (2) The higher the bond multiplicity, the shorterthe bond. (3) The higher the Zeff, the shorter the bond. (4) The larger theelectronegativity difference of the bonded atoms, the shorter the bond.These elements are in different rows of the periodic table. Iodine is n=5 and F is n=2. As theprincipal quantum number increases the size of the atom increases. The inter-nucleardistance is greater in I2 than in F2 because the iodine atom is bigger than the F atom. Inthis case, we are comparing the overlap of nitrogen with a halogen. The Electronegativities ofnitrogen, bromine, and fluorine are 3.0, 2.8, and 4.0 respectively. The inter-nuclear distancebetween NF vs. NBr should be shorter because the electronegativity difference is greater,the fluorine is smaller than bromine, the Zeff of fluorine is greater than that of bromine (Brhas more shielding than F, the electrons are in higher energy orbitals, the orbitals arebigger). The N—F bond should be shorter. The factor in this case is multiplicity. Whenmore electron density is placed between two atoms, the columbic interactions increase. Thisshrinks the inter-nuclear distance between the two atoms.
9.7 5 23 ?10.5 18 24 ?10.6 1 25
calculate the moles of iodine and fluorine = 0.263 mol F2 and 0.0394 mol I2. find LR via moles ofproduct, 0.0788 mol IF5 made. (Iodine LR) find amout of F2 remaining 0.0662 mol F2 remain.PV=nRT, toal P = (0.149 mol X R T)/V = 0.945 atm, 0.516 atm, 0.544 mole fraction
10.6 18 26 ?Page 5
Answer Key for Test “practice exam 3 w final”, 4/30/13No. in
Q-BankNo. onTest Correct Answer
10.6a 1 27 the remaining gas is nitrogen monoxide. the pressure of the remaining gas is0.300-atm. the mass of dinitrogen tetroxide formed is 0.402g.
10.8 7 28 ?11.2 10 29
Intermolecular strengths depend on the interactions between molecules. Thisis reflected in increased boiling points because the stronger the interaction,the harder it is to separate the particles from the bulk system and put themin the gas phase. (the same is true of melting points). Compounds that aregases at room temperature tend to have low intermolecular forces, while solidsand liquids tend to have high intermolecular forces. the molecules presented inthis problem belong to the same family. all contain carbon, but the group 6Aatom increases in mass down the column. so the molar masses are increasingfrom the dioxide to the diselenide. as molar mass increases, the londongforces increase so it makes sense that the boiling points should also increase.
11.3 1 30Iodine would have the highest boiling point because it is a solid, with the strongest Londonforces of the three. It is the biggest, heaviest, with the most electrons. B) the molecularmases are in order: 18, 41, 32, water has the strongest interaction with H-bond, followed bymethanol, and then methylcyainde. CH3CN cannot hydrogen bond; it would have the lowestinteraction of the three. C) H2O would have the smallest vapor pressure since it is a liquid at25°C and the others are gases. Neither of the other two compounds can hydrogen bond:carbon dioxide is a non polar gas, while sulfur dioxide is a polar gas. D) ethylene glycol (thelast one) forms more hydrogen bonds than ethanol. Butane forms no hydrogen bonds and is agas at room temperature. E) ammonia is the only one of the three that forms hydrogen bonds.F) water has the greatest heat of vaporization. The other two are gases at standard state.
11.5 12 31 ?13.3 6 32 ?13.5 1 33
Osmosis is the passage of water through a semi-perimable membrane. Dryingthe prunes removes water from the fruit. More water is inside the fruit thenoutside, so the system will try to equalize by essentially pushing out thewater. To re-hydrate the fruit, the water is added back by osmosis again.This time the fruit is low in water, so water rushes back in through into thecells to try and equalize the water pressure.
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