CHEMISTRY 103 – Practice Problems #3 Chapters 6.5 – 9.3

http://www.chem.wisc.edu/areas/clc (Resource page) Prepared by Dr. Tony Jacob

Suggestions on preparing for a chemistry exam:

1. Organize your materials (quizzes, notes, etc.). 2. Usually, a good method to prepare for a chem exam is by doing lots of problems. Re-reading a section of a chapter is fine, but

re-reading entire chapters takes up large amounts of time that generally is better spent doing problems. 3. Old exams posted by your instructor should be completely worked though. Old exams give you a sense of how long the exam

will be, the difficulty of the problems, the variability of the problems, and the style of your instructor! Quizzes written by your instructor are also a good source of problems to work through.

CHAPTER 6 1. I. Which of the following sets of quantum numbers (n, l, ml, ms) is a valid set of quantum numbers? a. (2, 3, 2, 1/2) b. (4, 4, 2, 1/2) c. (4, 3, 4, 1/2) d. (4, 3, 2, 11/2) e. (7, 3, 2, 1/2) II. What is a possible set of quantum numbers (n, l, ml, ms) for the last electron added to In in the ground state? a. (5, 2, 2, 1/2) b. (5, 1, 2, 1/2) c. (5, 1, -1, -1/2) d. (4, 1, 1, 1/2) e. (3, 0, 0, 1/2) 2. What subshell is described by n = 5 and l = 2? a. 2h b. 5h c. 5p d. 5s e. 5d 3. What orbital is described by n = 3, l = 1, ml = 1? 4. I. What is the maximum number of electrons that can have each of these quantum numbers (not a multiple choice question). a. n = 4 b. n = 5, ms = +1/2 c. n = 3, l = 2 d. n = 2, l = 1 e. n = 4, l = 3, ms = 1/2 II. What is the maximum number of orbitals that can have each of these quantum numbers (not a multiple choice question). a. n = 3 b. n = 4, l = 2 c. n = 5, l = 1 d. n = 5, l = 5 e. n = 3, l = 2, ml = 1 5. How many spherical nodes would be in the subshell occupied by the last electron in a Cd atom? a. 1 b. 2 c. 3 d. 4 e. 5 6. How many spherical nodes and surface nodes does each subshell have? a. 4s b. 3d c. 4p 7. Which subshell penetrates closest to the nucleus? a. 3p b. 2p c. 3d d. 2s e. 4d 8. Draw the electron probability graph, Ψ2 versus r, for each subshell. On each graph, label the axes, note on the graph where the spherical nodes are, and note where the surface nodes are and the number of surface nodes that exist at that point. a. 2s b. 2p c. 5p c. 3d d. 4d 9. Identify the subshell and draw all the orbital(s) for the subshell: a. n = 3, l = 1 b. n = 2, l = 0 (ice cream time)

10. Which statement below is correct? a. The 2s subshell will penetrate closer to the nucleus than the 2p subshell. b. Keeping electrons unpaired as long as possible and maximizing spin comes from the Pauli Exclusion Principle. c. The spin quantum number, ms, can take on values of ±1. d. The subshell that fills after 4s is 4p. e. The probability of finding an electron in the 1s subshell is zero at a point far away from the nucleus. 11. In a multi-electron system, which of the following subshells has the lowest energy? a. 3s b. 3p c. 2p d. 3d e. 2s 12. The electronic configuration for As+1 is: a. 1s22s22p63s23p64s23d104p3 b. 1s22s22p63s23p64s23d104p4 c. 1s22s22p63s23p64s24p2 d. 1s22s22p63s23p64s23d104p2 e. 1s22s22p63s23p64s24d104p2 13. a. What is the electron configuration of Se in the ground state? Do not use shorthand notation. b. What is the electron configuration of Ni in the ground state? Use a noble gas shorthand. c. What are the valence electrons for N+ in the ground state? 14. a. What is the electron configuration for Rh in the ground state? Do not use a noble gas shorthand. b. How many unpaired electrons does Rh have in the ground state? c. How many unpaired electrons does Rh+3 have in the ground state? 15. a. What neutral atom has the electron configuration in the ground state: 1s22s22p63s23p64s23d6? b. What -2 ion has the electron configuration in the ground state: 1s22s22p63s23p64s23d104p4? c. What +2 ion has the electron configuration in the ground state: 1s22s22p63s23p63d5? 16. Answer the 4 questions below. If no element satisfies the criteria, write “no element”. a. List all neutral second period element that have 2 unpaired electrons. b. List all neutral transition metal elements with 3d e- only (no 4d or 5d e-) that have 4 unpaired electrons. c. List all neutral elements with a valence shell 3s23px (x > 0) that will have the fewest number of unpaired electrons. d. Which transition metal(s) with 3d e- only (no 4d or 5d e-) and a charge of +1 will have zero unpaired electrons? 17. Which atom or ion in the ground state has the following electron configuration, 1s22s22p63s23p63d5? a. Mn b. V c. Ti+2 d. Mn+2 e. Co-2 18. Which of the following atoms/ions in the ground state have more than 2 unpaired electrons? a. O- only b. N only c. Zr+2 only d. Se+ only e. more than one of the above 19. Using an orbital box diagram, draw the electron configuration for Se. 20. What is the smallest atomic number for a neutral atom that has a filled d-subshell in the ground state? a. 48 b. 30 c. 2 d. 29 e. 18 21. Which atom or ion has the most paramagnetic in the ground state? a. Ni+2 b. V+2 c. Ca d. Sb+5 e. Ne (nap time)

22. Which ion/molecule is isoelectronic with PH3?

a. O3 b. N3- c. CH4 d. F2 e. none of the above

23. Which compound would be expected to possess the most similar chemical properties to CaCl2? a. SF2 b. NaCl c. MnCl2 d. MgBr2 e. CO2 24. Which atom or ion is the most paramagnetic in the ground state? a. Si b. Mn+2 c. N- d. Fe+2 e. P 25. Of the chemical species, P-, Ar, Co+3, S, Ni+2, how many have more than 2 unpaired electrons? a. 0 have more than 2 unpaired electrons b. 1 have more than 2 unpaired electrons c. 2 have more than 2 unpaired electrons d. 3 have more than 2 unpaired electrons e. 4 have more than 2 unpaired electrons CHAPTER 7 26. Place the following atoms in order of decreasing atomic radii: Cl, Se, Ge, Sn a. Se > Ge > Sn > Cl b. Sn > Ge > Se > Cl c. Cl > Se > Ge > Sn d. Cl > Ge > Se >Sn e. None of the above are correct. 27. Which material would be expected to be the least dense? a. sodium(s) b. aluminum(s) c. magnesium(s) d. need additional info 28. Which atom would most make sense to have the following ionization energies: IE1 = 578kJ/mol IE2 = 1817kJ/mol IE3 = 2745kJ/mol IE4 = 11,577kJ/mol a. Li b. Mg c. C d. Al e. none of the above 29. Rank the atoms from lowest to highest second ionization energy? Ca, Rb, K a. Ca < Rb < K b. Rb < Ca < K c. Rb < K < Ca d. K < Rb < Ca e. Ca < K < Rb 30. Which atom/ion will require the most energy to remove an electron? a. Na b. P- c. F+ d. Cl+ e. Si 31. Place the following atoms, N, O, S, in order of increasing first electron affinity (more exothermic/more negative values on the right). a. N < O < S b. N < S < O c. S < O < N d. O < S < N e. S < N < O 32. Which of the following reactions corresponds to the third ionization energy reaction? a. X-2(g) + e- → X-3(g) b. X-3(g) → X(g) + 3e- c. X(g) → X+3(g) + 3e- d. X+(g) → X+2(g) + e- e. X+2(g) → X+3(g) + e- 33. Sort these atoms/ions from smallest to largest in size. O-2, S-2, Se-2, F-, Ne (smallest) < < < < (largest) (take a walk)

34. Which process is the most exothermic? a. Fe(g) → Fe(g)+ + e- b. Fe+2(g) → Fe(g)+3 + e- c. Boiling 100g H2O(l) at 100˚C to H2O(g) at 100˚C. d. S-(g) + e- → S(g)-2 e. O(g) + e- → O(g)- 35. Which atom has the lowest third ionization energy? a. O b. F c. Mg d. P e. all have same IE3 36. Which reaction represents the first electron affinity for fluorine? a. F2(g) + e- → F2(g)- b. F(g) → F(g)+ + e- c. F2(g) + 2e- → 2F(g)- d. F(g)- + e- → F(g)-2 e. F(g) + e- → F(g)- 37. Determine the energy for each reaction below using the data given on the right. I. F(g) + Cl-(g) → F(g)- + Cl(g) IE1 for F = 1681kJ/mol

IE2 for F = 3374kJ/mol IE3 for F = 6050kJ/mol EA1 for F = -328kJ/mol

IE1 for Cl = 1251kJ/mol IE2 for Cl = 2297kJ/mol IE3 for Cl = 3826kJ/mol EA1 for Cl = -349kJ/mol

II. O(g)- + Al(g)+2 → O(g)-2 + Al(g)+3 IE1 for O = 1314kJ/mol IE2 for O = 3388kJ/mol IE3 for O = 5300kJ/mol EA1 for O = -141kJ/mol EA2 for O = 844kJ/mol

IE1 for Al = 577kJ/mol IE2 for Al = 1817kJ/mol IE3 for Al = 2745kJ/mol EA1 for Al = -42kJ/mol

III. O(g)- → O(g)+ + 2e- IE1 for O = 1314kJ/mol IE2 for O = 3388kJ/mol IE3 for O = 5300kJ/mol EA1 for O = -141kJ/mol EA2 for O = 844kJ/mol

CHAPTER 8 38. Draw the Lewis dot structure for each molecule. Include all resonance structures. a. IBr2

- b. BBr3 c. PO3-3 d. Cl2 e. IF4

- f. KrF4+2 g. SO2 h. HNO3 i. IBr3 j. CH2Cl2

k. OH l. N2 m. BeF2 n. C2H2 o. COCl2 p. C6H6 39. Which molecule violates the octet rule? a. BF4

- b. SiCl4 c. AsI3 d. SF4 e. none 40. What is the formal charge on the S in SO3? a. -2 b. -1 c. 0 d. +1 e. +2 (have some pizza)

41. I. Consider the following list of compounds. Rank the molecules from smallest to largest N–O bond length. NO2

- NO3- NO+ NO-

a. NO+ < NO- < NO2- < NO3

- b. NO3- < NO2

- < NO- < NO+ c. NO3- < NO+ < NO2

- < NO- d. NO+ < NO- < NO2

- = NO3- e. NO3

- = NO2- < NO- < NO+

II. Using the same compounds in the question above, which compound will have the greatest N–O bond enthalpy? 42. Consider the following bond lengths: C—O single bond: 1.43 Å C—O double bond: 1.23 Å C—O triple bond: 1.09 Å In the carbonate ion, CO3

-2, the most reasonable average C—O bond length would be: a. 1.43 Å b. 1.36 Å c. 1.23 Å d. 1.15 Å e. 1.09 Å 43. Which pairs of atoms will form the most ionic compound? a. nitrogen and oxygen b. chlorine and fluorine c. oxygen and oxygen d. sodium and oxygen e. phosphorus and oxygen 44. Shown below are four possible Lewis dot structures for SO3

-2 without resonance structures drawn. Decide which structure is best based on formal charges. Explain.

S

O

O OS

O

O OS

O

O OS

O

O O

-2

I II III IV 45. Draw the 3 Lewis dot resonance structures for thiocyanate, SCN- (C is in the middle). Based on formal charges, which resonance structure would be the better structure? The electronegativity values (Χ) are: ΧS = 2.5, ΧC = 2.5, and ΧN = 3.0. 46. I. In the molecule shown, CF2O, choose the structure with the correct locations of the δ+ or δ- symbols.

a.

! -

!+

CF F

O! -

! - b.

CF F

O

! -

! -

! -!+

c.

!+

! -! -

!+CF F

O

d.

!+ !+C

F F

O!-

! -

II. Would the C–O bond be nonpolar covalent, polar covalent, or ionic? 47. Identify each of the following bonds as nonpolar covalent, polar covalent, or ionic. a. F–I _________________________ b. C–C _________________________ c. Si-Si _________________________ d. Ca–O _________________________ e. H–O _________________________ (watch some TV)

48. Calculate the change in enthalpy for the reaction using the bond energies (kJ/mol) listed below. C2H4 + H2O → CH3–CH2–OH C–C = 348; C=C = 614; C≡C = 839; C–H = 413; C–O = 358; C=O = 799; O–H = 463 49. What is the Cl–Cl bond dissociation energy given the bond energies (kJ/mol) listed below. H2CO(g) + 2Cl2(g) → Cl2CO(g) + 2HCl(g) ΔHrxn = -208kJ C–C = 348; C=C = 614; C≡C = 839; C–H = 413; C–O = 358; C=O = 799; C–Cl = 328; H–Cl = 431 a. 484kJ b. 242kJ c. 26.5kJ d. 53.0kJ e. -208kJ CHAPTER 9 50. Which of the following compounds have tetrahedral molecular geometry? a. PCl4- b. SeF4

+2 c. BrI4+ d. XeF2

-2 e. all do 51. Which of the following statements is false? a. A molecule with individual polar bonds can be nonpolar. b. A molecule with a central atom with 2 atoms and 2 lone pairs of electrons around it is a bent molecule. c. Repulsion between core electrons pairs is used to determine the shape of the molecule. d. The angle between two H atoms in H2Se is expected to be less than 109.5˚. e. Ammonia, NH3, will absorb microwave radiation. 52. Consider the following covalent bonds. Which bond will be the shortest? a. N–F b. P–Cl c. N–Cl d. P–F e. none of the above 53. For each molecule below, draw the Lewis dot structure, draw the electron domain geometry, and draw the molecular geometry. Re-draw the molecular geometry diagram and draw in vectors representing the bond polarity, and draw the net vector representing the net dipole if the molecule is polar otherwise write “no net vector” if the molecule is nonpolar. a. H2O b. BF3 c. IF2

- d. NH3 54. For ICl4-, TeCl4, XeF2, and CO2 molecules select the answer below that is incorrect. Determine the molecular geometry and polarity of the molecule. If all the answers given are correct, select answer "e". a. ICl4-: geometry shape: square planar; nonpolar b. TeCl4: geometry shape: seesaw; polar c. XeF2: geometry shape: linear; nonpolar d. CO2: geometry shape: linear; nonpolar e. All the answers are correct. 55. Which molecule will have an angle of 120˚? a. NF3 b. SF6 c. CF4 d. SeF4 e. none 56. Which molecule is polar? a. PH4

+ b. GeCl4-2 c. SiCl4 d. XeF4 e. none (almost done)

57. In Lewis dot structures, which electron interactions repel the most? a. bonding pair–bonding pair b. bonding pair–lone pair c. lone pair–lone pair d. since these are all electrons they are equivalent 58. How many of these molecules have a tetrahedral electron domain geometry around the central atom? CCl4 SF4 SiCl4-2 SeI2 KrF2 a. 1 b. 2 c. 3 d. 4 e. 5 59. A central atom that has 2 lone pairs and 3 bonding pairs of electrons around it will have a molecular shape: a. linear b. trigonal pyramid c. trigonal planar d. T-shape e. trigonal bipyramid 60. Which molecule is polar? a. I3

- b. SF6 c. XeF6+2 d. CI4 e. no molecule is polar

(yea done) ANSWERS 1. I. e {(2, 3, 2, 1/2) and (4, 4, 2, 1/2): l value invalid; (4, 3, 4, 1/2): ml invalid; (4, 3, 2, 11/2): ms invalid} II. c {Last electron is 5p electron; n = 5 and l = 1 for 5p; if l = 1, ml = -1, 0, 1} 2. e {n = 5 → 5; l = 2 → d; 5d} 3. 3px or 3py or 3pz {n = 3 → 3; l = 1 → p; 3p; the p orbitals are px, py, and pz; ml = 1 can specify any one of these orbitals} 4. I. a. 32 {2n2 = 2(4)2 = 32 electrons} b. 25 {2n2 = 2(5)2 = 50 electrons; only half will have ms = 1/2 → 25 electrons} c. 10 {subshell = 3d → 10 electrons} d. 6 {subshell = 2p → 6 electrons} e. 7 {subshell = 4f → 14 electrons; only half will have ms = 1/2 → 7 electrons} II. a. 9 {n2 = 32 = 9 orbitals} b. 5 {subshell = 4d → 5 orbitals} c. 3 {subshell = 5p → 3 orbitals} d. 0 {when n = 5 → l ≠ 5 → 0 orbitals} e. 1 {subshell = 3d; of the five 3d orbitals only one orbital has a ml = 1} 5. a {spherical nodes = n – l – 1; last electron in Cd = 4d → n = 4, l = 2; spherical nodes = 4 – 2 – 1 = 1} 6. a. 0 surface nodes; 3 spherical nodes {4s → n = 4 and l = 0; planar nodes = l = 0; spherical nodes = n – l -1 = 4-0-1 = 3}

b. 2 surface nodes; 0 spherical nodes {3d → n = 3 and l = 2; planar nodes = l = 2; spherical nodes = n – l -1 = 3-2-1 = 0} c. 1 surface node; 2 spherical nodes {4p → n = 4 and l = 1; surface nodes = l = 1; spherical nodes = n – l -1 = 4-1-1 = 2}

7. d {penetration: s > p > d and lower n values penetrate more; lower energy subshells penetrate more}

8. a. b. c.

c. d.

9. a. Three 3p orbitals: b. One 2s orbital: 10. a {b. → that’s Hund’s rule; c → ±1/2; d → 3d fills after 4s; e → the probability is always greater than 0} 11. e {order of filling from lowest to highest E: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, etc. from the Periodic Table}

12. d 13. a. 1s22s22p63s23p64s23d104p4 b. [Ar]4s23d8 c. 2s22p2 14. a. 1s22s22p63s23p64s23d104p65s24d7 b. 3 c. 4 15. a. Fe b. Ge-2 c. Mn+2 16. a. C and O {C: 2s22p2 and O: 2s22p4} b. Fe {Fe: 4s23d6; Cr does not have 4 unpaired electrons: 4s13d5} c. Ar {Ar: 3s23p6 – all electrons paired; 0 unpaired electrons} d. Cu {Cu: 4s13d10 – 1 unpaired electron; Cu+: 3d10 – 0 unpaired electrons} 17. d {With charged transition metals, remove the s electrons first.} 18. e {N = 3; Se+ = 3; Zr+2 = 2; O- = 1}

19. 4s3d 4p3p3s2p2s1s 20. d {the first d-subshell is the 3d subshell; it is filled first by Cu because of an exception with Cu electron configuration

(Cu: [Ar]4s13d10); Cu has 29p+ and therefore an atomic number of 29 is the correct answer; answer “b” is “close” to the correct answer and yields Zn (Zn: [Ar]4s23d10) but the exception with Cu makes it the correct answer}

21. b {V: [Ar]4s23d3; V+2: [Ar]3d3 since s electrons are removed first; note: Sb+5 – first remove the three 5p electrons and then the ion will resemble a transition metal; next, since it resembles a transition metal, remove the two 5s electrons; Sb0: [Kr]5s24d105p3; Sb+5: [Kr]4d10}

22. d {The number of electrons in PH3 is 15 + 3(1) = 18; only F2 has that number: 2(9) = 18e-} 23 d {elements in the same family have similar chemical properties; Mg and Ca-same family; Cl and Br-same family} 24 b {Mn: [Ar]4s23d5; Mn+2: [Ar]3d5 since s electrons are removed 1st} 25. b {P-: 2; Ar: 0; Co+3: 4; S: 2; Ni+2: 2} 26. b {Radii increase as you go to the left and down on the Periodic Table} 27. a {D = m/V; Na would be the largest in size and has the smallest mass so would be the least dense} 28. d {Since the IE4 is so large as compared to the other IE, the removal of the 4th electron breaks a noble gas

configuration. Only Al breaks a noble gas configuration during IE4: Al+3 → Al+4 + e-} 29. a {Look at positions of the +1 ions, Ca+, Rb+, K+ since the IE2 involves removing the 2nd electron from the

+1 ion. IE2 trend is higher as you go up and to the right on the Periodic Table. K+ is in the Ar position.} 30. c {removing an electron from a cation is more difficult than removing an electron from an anion or a neutral atom so the possible

answers are F+ and Cl+; the smaller the cation the harder it is to remove an electron; since F+ is smaller than Cl+ (recall F is smaller than Cl) it is harder to remove the next electron from F+}

31. a {N has EA ≈ 0; EA increase (more negative) up a period but 3rd period > 2nd period} 32. e {the third ionization energy refers to the third electron being removed} 33. Ne < F- < O-2 < S-2 < Se-2 {Ne is farther to the right and neutral so it’s smallest; O is larger than F and by making it a -2

anion it will definitely be larger than a -1 anion so this yields: Ne < F- < O-2; between O-2, S-2 and Se-2 since O < S < Se when neutral, the same holds for anions with the same charge so this yields: O-2 < S-2 < Se-2; combining these trends yields: Ne < F- < O-2 < S-2 < Se-2}

34. e {“a” is IE1 and “b” is IE3 – ionization energies are endothermic; “c” is endothermic; “d” is EA2 and EA2 are always endothermic; “e” is EA1 and EA1 is either 0 (zero) or negative (exothermic)}

35. d {IE increase as you go to the right and up on the Periodic Table; also, since we are considering the 3rd IE, think of the atoms as +2 ions: O+2, F+2, Mg+2, and P+2; assign these locations on the PT, i.e., O+2 has 6 electrons so it is in the C position and so on; P+2 is the only atom on the 3rd row and therefore has a lower IE3}

36. e {EA1: X(g) + e- → X(g)-}

37. I. 21kJ/mol {F(g) → F(g)-: EA1(F); Cl-(g) → Cl(g) : -EA1(Cl); ΔHrxn = EA1(F) + (-EA1(Cl)) = -328 + (-(-349)) = 21kJ/mol II. 3589kJ/mol {O(g)- → O(g)-2: EA2(O); Al(g)+2 → Al(g)+3: IE3(Al); ΔHrxn = EA2(O) + IE3(Al) = 844 + 2745 = 3589kJ/mol} III. 1455kJ/mol {O(g)- → O(g) + e-: -EA1(O); O(g) → O(g)+ + e-: IE1(O); ΔHrxn = -EA1(O) + IE1(O) = -(-141) + 1314 =

1455kJ/mol} 38. See below for Lewis dot structures. 39. d {S has 10 electrons around it} 40. e {Draw Lewis dot structure; has resonance but this does not change FC for the S atom; S: 6 - 4 = 2} 41. I. a. {Longest bond has smallest bond order; NO+ has N-O triple bond, NO- has N-O double bond, NO2

- has N-O

with BO = 1.5 and NO3- has a N-O BO = 1.33; NO3

- has smallest bond order → longest bond length}

II. NO+ {Greater BO → greater bond dissociation energy; NO+ has N-O triple bond/greatest bond order → greatest bond dissociation energy}

42. b {From Lewis dot structures, including the resonance structures, the C—O bond is between a single and double bond (it's 11/3 bonds). Therefore, the bond length should be between 1.43 (single bond) and 1.23 Å (double bond).}

43. d {Most ionic means greatest ΔEN which means further apart from one another on the PT.} 44. Structure IV can be dropped because of the high FC. Structure I can be dropped because it has fewer zeros

than Structures II or III. The difference between Structures II and III is a -1 FC on the O atom and a FC of 0 on the S (Structure II) and the reverse of that for Structure III. Since O is more electronegative than S the O atom should have the more negative FC. Therefore Structure II is the best structure.

S

O

O OS

O

O OS

O

O OS

O

O O

-2

I II III IV

(-1)

(-1)(-1)

(+1)

(0)

(-1) (-1) (-1)

(0) (0)

(0) (0) (0)

(0) (-1) (-2)

45. Lower formal charges (closer to zero) are better. Two resonance structures have two 0’s and one –1 for formal charges. These are

better structures than the structure with a –2 formal charge. To determine between these 2 resonance structures which is better, consider the electronegativity of S and N. Since N has a greater EN, it should have the more negative formal charge. Hence, the resonance structure with 2 double bonds is the best structure in which the –1 formal charge is on the N and not the S.

:: S C N. .. . -

(-2)(+1) (0)S C N: :. . . .

-

(0) (0) (-1)

:: S C N. .. . -

(0)(-1) (0) 46. I. b {compare each pair of atoms; the more EN atom in the pair gets a δ- while the less EN atom gets a δ+}

II. polar covalent 47. a. polar covalent b. nonpolar covalent c. nonpolar covalent d. ionic e. polar covalent 48. -42kJ {ΔHrxn = bonds broken – bonds formed = [1(C=C) + 4(C–H) + 2(O–H)] – [1(C–C) + 5(C–H) + 1(C–O) + 1(O–H)] =

[1(614) + 4(413) + 2(463)] – [1(348) + 5(413) + 1(358) + 1(463)] = -42kJ} 49. b {ΔHrxn = [2(C–H) + 1(C=O) + 2(Cl–Cl)] – [(2(C–Cl) + 1(C=O) + 2(H–Cl)];

-208 = [2(413) + 1(799) + 2(x)] – [(2(328) + 1(799) + 2(431)]; x = 242} 50. b {draw the correct LDS} 51. c {valence electrons are used} 52. a {smaller atoms will have a small bond length} 53. See below. 54. e {Draw Lewis dot structure → Determine VSEPR molecular shape → Determine polarity} 55. d {Draw Lewis dot structures. Use VSEPR to determine shapes and angles.} 56. b {GeCl4

-2 is a “seesaw” structure which is always polar; “a” and “c” – tetrahedral with same atoms around central atom → nonpolar; “d” – square planar with same atoms around central atom → nonpolar}

57. c 58. b {CCl4 and SeI2} 59. d {from VSEPR table} 60. e {all the molecules are nonpolar}

38.

53.