1. (5 points) State which of the following sets of quantum numbers would be possible and which would not. Using one sentence, explain what is wrong with the 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

Page 1 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 2p a ↑↓ ↑↓ ↑ ↑ ↓

b ↑↓ ↑↓ ↑↓ ↑

c ↑↓ ↑↓ ↓ ↓ ↓

d ↑ ↑↓ ↑↓ ↑↓

e ↑↓ ↑↓ ↑ ↑ ↑

Give the set of quantum numbers for each of the electrons in the p orbitals in orbital diagram (a)

3. (2 points) How many angular nodes does an 12f orbital have?

How many radial nodes does a 7p orbital have?

Page 2 4. Name the element with the following electron configuration:

[a] 1s22s22p63s23p2

[b] 1s22s22p63s23p63d 104s24p64d 105s25p5

[c] [Ar]4s13d 5

[d] [Xe]4f146s2

5. (4 points) Using the periodic table write the expected ground state electron configurations for a) 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.

Page 3 6. (8 points) Determine which of the following statements are true and explain why 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

Page 4 7. Penetration is the process by which an outer electron moves through the region occupied by the core electrons to spend part of its time closer to the nucleus. Penetration and the resulting effects on shielding cause an energy level to spit into sublevels of differing energy. Use the penetration effect to explain the difference in relative orbital energies of a 2s and a 2p electron in the same atom.

8. The first ionization energy of the chlorine atom is 1251 kJ/mol. Which of the following 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 trends AND Zeff.

Page 5 9. Little is known about the properties of astatine, At because of its rarity and high radioactivity. Nevertheless, it is possible for us to make many predictions about its properties. Do you expect the element to a gas, a liquid or a solid at room temperature? Explain. What is the formula for the compound 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 smallest radius? Cr6+, P3-, Cl–, K+, or Ti4+

Page 6 11. (a) Explain the following trend in Lattice energy based on ion size and the definition of lattice energy:

BeH2 3205 kJ/mol

MgH2 2791 kJ/mol

CaH2 2410 kJ/mol

SrH2 2250 kJ/mol

BaH2 2121 kJ/mol

(b) The lattice energy of ZnH2 is 2870 kJ/mol. Based on the data given in part a, the radius of the zinc ion is expected to be closest to that of which group 2A element? Use a complete sentence to explain your choice.

12. Using electronegativity values, predict which bond in the following groups will be most polar. Show your work.

a) C—H, Si—H, P—H

b)Al—Br, Ga—Br, In—Br, Tl—Br

Page 7 13. Determine which of the structures listed below would be the most plausible structure 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) for – ClO2 , chlorite. Show all formal charges.

Page 8 15. Draw the lewis structure for SeCl4.

16. The space shuttle orbiter utilizes the oxidation of methyl hydrazine by dinitrogen tetroxide. Use bond energies to estimate the H for the reaction below.

5N2O4(l) + 4 N2H3CH3(l) → 12 H2O(g) + 9 N2(g) + 4CO2(g)

O O H H N N N N O O H CH3 dinitrogen tetroxide methyl hydrazine

Page 9 17. Write the best (formal charges, octet rule etc) Lewis structure for SO2Cl2. What is the molecular and electron pair geometry? What type of hybridized orbitals are needed for this molecule?

Will it have the same molecular and electron pair geometry as OCl2? Justify your answer by comparing Lewis structures. Is this a polar molecule? why?

c) Show the complete hybridization process for the oxygen in oxygen dichloride. Clearly label what the electrons are used for (bonding, lone pair) in the hybridization scheme. You do not have to show chlorine in the scheme.

Page 10 18. The structures below show the stick and ball drawing of three possible shapes of an AF4 molecule. A is the general symbol for the central atom. The terminal 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 on which 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

Page 11 19. When two or more compounds have the same composition but a different arrangement of atoms (or geometries), we call them isomers. Isomers are composed of the same collection of atoms, but they differ in one or more chemical or physical property. There are several types of isomers found in chemistry. In a geometric isomer, the bonds are the same, but the orientation

or arrangement of the bonds is different. PF2Cl3 has three geometric isomers. One of these isomers has a dipole of zero; two do not.

(a) Draw the geometric isomers of PF2Cl3 using VSEPR. Clearly show the three 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.

Page 12 20. (5 points) Using the valence orbital diagram (box diagram), show how the — atomic orbitals of the central atom in IF4 leads to the appropriate hybrid orbitals. Fill the hybrids with the correct number of electrons please.

Page 13 – 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 in the azide molecule to have? (Describe theVSEPR geometry, show hybridized orbitals containing electrons, link these orbitals to make bonds and sketch the molecule showing the sigma bonds and bonds for full points.)

Page 14 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?

Page 15 24. A chloride of antimony is a solid at room temperature, but vaporizes when heated. When 2.359 g of this compound are vaporized in an evaluated 1.00 L chamber at 581K, the pressure in the bulb is 376 mmHg. What is the molecular weight of this chloride of antimony, assuming the vapor is an ideal gas?

25. Gaseous iodine pentafluoride, IF5, can be prepared by the reaction of solid

iodine 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 reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is 125 °C. What is the partial

pressure of the IF5 in the flask? What is the mole fraction of IF5 in the flask?

Page 16 26. Assume that a single cylinder of an automobile engine has a volume of 600.0 cm 3. If the cylinder is full of air at 80°C and 0.980 atm, how many moles of

O2 are present? The mole percent of oxygen in air is 20.95 %. How many

grams of Octane (C8H18) could be combusted by this quantity of O2, assuming

complete combustion with the formation of CO2 and H2O?

27. Two gases in adjoining vessels were brought into contact by opening a stopcock between them. The one vessel measured 0.250 L and contained NO at 800.0 torr and 220.00K. The other measure 0.100L and contained oxygen at 600.0

torr and 220.00K. The reaction to form N2O4(s) exhausts the limiting reagent

complete. Neglecting the vapor pressure of N2O4(s) (which is minimal) what is the pressure and composition of the gas remaining at 220.00K after completion

of the reaction? What is the weight of N2O4(s) formed?

Page 17 28. (4 points) A gas whose you wish to know effuses through an opening at a rate one fourth as fast as that of helium gas. What is the molar mass of the unknown gas?

29. Given the following information about CO2, CS2, and CSe2, How do the

strengths of the intermolecular forces vary from CO2 to CS2 to CSe2?

• diselenide (CSe2) is a liquid at room temperature. The normal boiling point is 125°C, and the is -45.5°C. • is also a liquid at room temperature with a normal boiling point of 46.5°C and a melting point of -116.5°C. • is a gas at room temperature. It does not have a normal boiling point, but rather sublimes at -78°C.

Page 18 30. In each of the following groups of substances, pick the one that has the given property. Use your knowledge of the different types of intermolecular forces (ion-dipole, dipole-dipole, hydrogen bonding, instantaneous dipole (London forces, dispersion forces) to help you make your choice. Justify your answer with one sentence by describing the type of intermolecular force and how it effects that property.

a) Greatest viscosity: CH3CH2CH2CH3, CH3CH2OH, or HOCH2CH2OH EXAMPLE: Viscosity is based on shape, molar mass, and strong intermolecular 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 most electrons and can form the most hydrogen bonds. Ethylene glycol has the highest 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

Page 19 31. The vapor pressure curves for compound A and B are drawn below. Both A and B have diopoles. a) Which compound has the stronger dipole? b) Which compound is more volatile? Please explain your answers.

Vapor pressure curves for A and B 700 Compound A 600 Compound B 500

400 pressure 300

200 vapor

100 -20 0 20 40 60 80 100

32. Mt Kilimanjaro in Tanzania is the tallest peak in Africa (19,340 ft). If the barometric pressure at the top of the mountain is 350 torr, at what temperature will water boil there? H vap = 40.67 kJ/mol at 100°C

Page 20 33. I wanted stewed prunes for my breakfast in winter. Thinking ahead, I picked some 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. The prunes plumped up. I ate them with a small amount of cream. YUMMY!

Explain what happened to the fruit in terms of osmosis.

Page 21 Answer Key for Test “practice exam 3 w final”, 4/30/13 No. in No. on Q-Bank Test Correct Answer 6.5 13 1 a) n ≠ l or zero n>l b) n ≠ l n is always greater than l c) 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 atom in 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 and stays relatively constant down a family. Fluorine is an n=2 atom while Cesium is an n=5 atom. Fluorine is a small, non-metal atom with a high Zeff while cesium is very large and metallic. When cesium adds an electron, the 5s sub shell is filled, a stabilizing effect. When an electron is added to fluorine, the n=2 shell is filled, not just a sub-shell-this is more stabilizing, the electron is closer to the nucleus (n values!), a stabilizing effect. FALSE-Ionization energy is 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 remove electrons. Therefore, the ionization energy of the anion is lower than the ionization energy of the parent because it has a lower Zeff.TRUE-lithium is an atom in n=2 shell, while rubidium is an ion in n=5 shell. The electron in the 5s orbital is further from the nucleus (does not penetrate as well) as the 2s electron in lithium atom. When removing an electron to make the ion, both particles are smaller than the parent atom, but lithium ions have a smaller cloud than rubidium because overall, there are fewer electrons. Fewer electrons mean a smaller ion. TRUE-for similar reasons as answer C. The chloride ion has electrons added to orbitals that are further from the nucleus than fluorine. This means they do not penetrate the nucleus, as effectively, so the 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 7 the 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 electron density 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 Page 1 Answer Key for Test “practice exam 3 w final”, 4/30/13 No. in No. on Q-Bank Test 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 is period 5 compared to chlorine in period 3. The quantum shielding going down a family increases faster than the Z (the actual nuclear charge). The Zeff is therefore decreasing as the n quantum value increases. Because the Zeff is small, the attraction for the outer most electron is weaker in the iodine atom than in the chlorine atom leading to a larger sized atom and a lower ionization potential. 7.8a 4 9 I expect astatine to be a solid. In group 7A, the phase of the elements change from a gas phase for fluorine, to the solid phase for iodine, iodine is much heavier than fluorine. since At is the next element in the series, it should be similar to iodine. The formula would be NaAt. 7.9 6 10 a) S and CL are in the same period, n=3, while Se is right below S in n=4. The radius is dependent on the Zeff and quantum effects. Since S is in n= 3 and Se is in n=4, Se is bigger than S due to quantum effects, (size increases with increasing n. Cl is in the same period as S, but to the left of S. This means that for about the same amount of shielding, Cl has a larger Zeff than S, and therefore 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 more electrons and more shielding than the chlorine atom. As shielding increases, Zeff decreases so the attraction between the outermost electron and the nucleus also decreases. The chloride ion must be larger.

c)Ionization energy increases across a period and decreases down a period as a general trend-again due to quantum shielding and Zeff. N, P, and As are congers, or members of the same group. The electron is placed in quantum shells that are further from the nucleus, and are easier to remove. N should have the larger ionization energy because it's outermost electron is closest to the nucleus and therefore hardest to remove.

d) In an isoelectron series (to which these ions belong), as the negative charge on the ion increases, the size increases of the ion increases. As more electrons are added, the shielding increases. Nitride would have the largest radius.

e) In this series, both cations and anions are presented. Cations with larger positive charges tend to be smaller than the ions and atoms in the same Page 2 Answer Key for Test “practice exam 3 w final”, 4/30/13 No. in No. on Q-Bank Test Correct Answer isoelectronic series. Removing electrons decreases shielding so the outermost inner-core electrons feel a stronger nuclear charge. There are less electrons to compete for penetration. The increased Zeff results in a decrease of the radius.

8.2 4 11 Lattice energy is the energy needed to separate an ion pair in the solid phase to make ions in the gas phase. Lattice energy is dependent on two factors: the added radii of the ion pair and the charge. Lattice energy is inversely proportional to the distance (d) between the nuclie and proportional to the product 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 this series increases as n, the principal quantum number, increases. This means that Be is the smallest ion while Ba is the largest. The anion radius remains constant. The charge for both the cation and the anion are constant. The data shows the lattice energy decreases down group 2A for this series of hydrides because the size of the cation is increaseing, increasing the distance between the two radii.

(b) Looking at the data, the LE(BeH2)> LE(ZnH2)> LE(MgH2). From this one can conclude that the radius of the zinc cation must be larger than Mg ion, but smaller than Be ion becuase LE is related to the size of the ion. I think the size 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 one does the Lewis structures, the N—O—C is not stable at all due to formal charge considerations. This leaves us with N—C—O and C—N—O skeletons. Structure B has two relatively stable resonance structures. Because the electron density is delocalized between the N, C, and O atoms, the triple bond between C and N is longer than it should be. This structure is more stable than that of structure A. Structure A has two resonance structures also, but the first structure is unstable due to formal charge considerations. The first structure has a formal charge of 2— on the carbon and a positive charge on the nitrogen. If we have a choice, we minimize negative charges on the most electropositive atoms and localize charge density on the more electronegative atoms. The bond length for the triple bond is much shorter here since it is unlikely that the first resonance structure will form. Therefore structure A is .

Page 3 Answer Key for Test “practice exam 3 w final”, 4/30/13 No. in No. on Q-Bank Test Correct Answer

C N O C N O N C O N C O 2

B A 8.6 2 14

O Cl O O Cl O

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 the numbers in the table. do the math -21,619 kJ 9.2 1 17 O

Cl S O

Cl

the electron pair and the molecular geometry is tetrahedral. oxygen dichloride has the same electron pair geometry, but is angular or bent. It is definetly a polar molecule due to the lone pairs on the oxgen. the hybridization of oxygen dichloried 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 in the molecule allows the individual dipoles to cancel.

Page 4 Answer Key for Test “practice exam 3 w final”, 4/30/13 No. in No. on Q-Bank Test Correct Answer

Br Br Cl Cl P Cl Cl P Br Br Cl Cl Br P Cl Br Cl 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 the lowest formal charges. the terminal N atoms would be sp2 and the central N would be sp.

N N N

9.7 1 22 Bond lengths are controlled by four factors: (1) The smaller the principal quantum number of the valence orbitals, the shorter the bond. (2) The higher the bond multiplicity, the shorter the bond. (3) The higher the Zeff, the shorter the bond. (4) The larger the electronegativity 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 the principal quantum number increases the size of the atom increases. The inter-nuclear distance is greater in I2 than in F2 because the iodine atom is bigger than the F atom. In this case, we are comparing the overlap of nitrogen with a halogen. The Electronegativities of nitrogen, bromine, and fluorine are 3.0, 2.8, and 4.0 respectively. The inter-nuclear distance between 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 (Br has more shielding than F, the electrons are in higher energy orbitals, the orbitals are bigger). The N—F bond should be shorter. The factor in this case is multiplicity. When more electron density is placed between two atoms, the columbic interactions increase. This shrinks 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 of product, 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”, 4/30/13 No. in No. on Q-Bank Test 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. This is reflected in increased boiling points because the stronger the interaction, the harder it is to separate the particles from the bulk system and put them in the gas phase. (the same is true of melting points). Compounds that are gases at room temperature tend to have low intermolecular forces, while solids and liquids tend to have high intermolecular forces. the molecules presented in this problem belong to the same family. all contain carbon, but the group 6A atom increases in mass down the column. so the molar masses are increasing from the dioxide to the diselenide. as molar mass increases, the londong forces increase so it makes sense that the boiling points should also increase. 11.3 1 30 Iodine would have the highest boiling point because it is a solid, with the strongest London forces of the three. It is the biggest, heaviest, with the most electrons. B) the molecular mases are in order: 18, 41, 32, water has the strongest interaction with H-bond, followed by methanol, and then methylcyainde. CH3CN cannot hydrogen bond; it would have the lowest interaction of the three. C) H2O would have the smallest vapor pressure since it is a liquid at 25°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 (the last one) forms more hydrogen bonds than ethanol. Butane forms no hydrogen bonds and is a gas 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. Drying the prunes removes water from the fruit. More water is inside the fruit then outside, so the system will try to equalize by essentially pushing out the water. 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 the cells to try and equalize the water pressure.

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