Chemical Bonding Solutions

Pavel Sedach Chem201 Final Booklet Chemical Bonding Solutions Problem 1. Answer: B Formal Charge = Valence electrons lone pair electrons bonds FC F = 7 6 1 = 0 FC P = 5 2 3 = 0 FC S = 6 4 2 = 0 Problem 2. Answer: X is Carbon (C). FC X = x 0 4 = 0, Therefore x = 4 and that is the number of valence electrons in the element. Giving us carbon. Problem 3. Answer: C There are 4 lone pairs of electrons. 1 of 38

Pavel Sedach Chem201 Final Booklet Problem 4. Answer: D Resonance is defined as the movement of electrons through π (pi) bonds. π bonds are any bonds beyond a single bond. For instance a double bond has 1 π bond and a triple bond has 2 π bonds. If there are only single bonds, there are no π bonds, therefore there is no resonance. a) N 2 b) S 3 2 c) 3 d) C 4 Methane is the only compound with only single bonds. Moving electrons in this structure results in broken bonds this is another useful way to think of structures that can and can t have resonance. e) 2 You may have thought oxygen only has one structure but, because it has a double bond, there is a minor structure that can occur. Problem 5. Answer: D 2 of 38

Pavel Sedach Chem201 Final Booklet 3 of 38

VSEPR Solutions Problem 6. Answer: C Solution: A triple bond corresponds to 2 bonds and 1 bond. The two single bonds consist of one bond each. Therefore there is a total of three bonds and two bonds. Problem 7. Answer: a) N(1): sp3 N(2): sp2 (1): sp2 (2): sp3 b) i) 109 I ii) 120 iii) 6 lone pairs Solution: N(1) has 3 bonding pairs and 1 lone pairs of electrons sp3 hybridized N(2) has 3 bonding pairs of electrons sp2 hybridized (1) has 1 bonding pair and 2 lone pairs of electrons sp2 hybridized (2) has 1 bonding pair and 3 lone pairs of electrons sp3 hybridized i) N(1) is sp3 hybridized; tetrahedral structure; 109 bond angle ii) N(2) is sp2 hybridized; trigonal planar; 120 bond angle iii) By the Lewis Structure, there are 6 lone pairs of electrons. Problem 8. Answer: D Solution: From the Lewis diagram of the molecule we can see that the shape of the molecule is T- Shaped. 4 of 38

Problem 9. Which of the following statements is/are correct for the formate ion C 2? Answer: C Solution: the oxidation number of the C atom is +2, there are only 2 plausible contributing structures, and C 2 = 18 e- = AX3 = trigonal planar Problem 10. a) Answer: b) Answer: Sigma bonds = 4 Pi bonds = 1 c) Answer: A single bond contains one σ bond, whereas a double bond consists of one σ and one п bond. σ bonds are covalent bonds between electron pairs in the area between two atoms. п bonds are a covalent bond between parallel p orbitals and the electron pairs shared are below & above the line joining the 2 atoms. C N 2p orbitals sp 2 orbitals п bond σ bond d) Answer: There are 3 electron pairs around the central C atom, and thus it is sp2 hybridized. Thus the bond angle is 120. 5 of 38

Problem 11. Predict the geometric shape of Cl2- ion. Answer: Bent Solution: Total valence electrons: 7 + 2(6) + 1 = 20 electrons Center Cl has 2 bonding pairs and 2 lone pairs of electrons sp 3 hybridized & tetrahedral formation. Since there are 2 lone pairs of electrons, the structure is bent-shaped. Problem 12. Draw the Lewis structure for the peroxymonosulfate ion, ---S3-, and estimate the -- bond angle. Answer: < 109 Solution: -1 The center oxygen in -- has 4 pairs of electrons (2 lone pairs of electron and 2 bonding pairs of electrons); as such, this part of the molecule has a tetrahedral structure. The bond angle is less than 109.5 due to the 2 lone pairs of electrons that bend the structure more than bonding pairs of electrons. The tetrahedral formation means the electrons are distributed in sp3 orbitals. 6 of 38

Problem 13. a) Applying your knowledge of VSEPR, indicate the geometry of the three atoms indicated in acetic acid. C 3 b) What is the =C- bond angle? c) Indicate the orbital hybridization of the three atoms with the arrows. Answer: a) 3C- : 4 bonding pairs of electrons, thus it is tetrahedral 3 bonding pairs of electrons; thus it is trigonal planar C 3 4 bonding pairs of electrons; thus it is tetrahedral. b) The geometry around the central C carbon is trigonal planar structure. Thus, the bond angle is 120. C 3 c) 3C - C - Tetrahedral structure = sp3hybridization 3C - C - Trigonal planar structure = sp2 planar 3C - C - Tetrahedral structure = sp3 hybridization 7 of 38

Problem 14. a) Cl2CS (thiophosgene) (Carbon is central atom, Cl are equivalent). Lewis: *All formal charges are zero. Total valence electrons = 24 VSEPR AX3 Electronic geometry: trigonal planar Molecular geometry: trigonal planar b) PF6 Lewis Structure *Formal Charge on F = 0, Formal Charge on P = -1. Total valence electrons = 48 AX6 Electronic geometry: octahedral Molecular geometry: octahedral 8 of 38

Problem 15. a) N3 a) Total # of electrons = 1 + 5 + 3(6) = 24 (2) (1) (3) Formal charge: n (1) 6 [4 + ½ (4)] = 0 n (2) 6 [6 + ½ (2)] = -1 n (3) 6 [4 + ½ (4)] = 0 n N 5 [0 + ½ (8)] = +1 n 1 [0 + ½ (2)] = 0 Molecular drawing: b) 3 Structure around central : 4 electron pairs = bent Total # of valence electrons: 3(6) = 18 Lewis structure: Structure around central N: 3 electron pairs = trigonal planar (1) (2) (3) Formal charge on: (1) 6 [6 + ½ (2)] = -1 (2) 6 [2 + ½ (6)] = + 1 (3) 6 [4 + ½ (4)] = 0 Molecular structure around central = bent 9 of 38

Prep101 Chem201 Final Booklet Practice Problems Problem 1. Solution: Answer: B Non-zero zero zero with zero with non-zero Problem 2. Answer: B resonance resonance Solution: The most ionic bond will have the most electronegative species and the least electronegative species. Electronegativity increases as you move across and up the periodic table. F is the most electronegative and Li, here, is the least E.N. Problem 3. Answers: N 3 Br: dipole induced dipole, London dispersion ClBr ClBr: dipole dipole, London dispersion C 2 C: induced dipole dipole, London dispersion b) All 3 molecules are non-polar, so only London dispersion forces London dispersion forces increase in magnitude as MW increase As IM forces increase, boiling point increases Br 2 has highest MW, so has strongest IM forces and highest boiling point bp(n 2 ) < bp( 2 ) < bp(br 2 ) c) Compound Type of intermolecular forces C 4 Non-polar; London dispersion forces C 3 Polar; London dispersion forces & -bonding CCl 4 Non-polar; London dispersion forces that are greater than those in C 4 or C 3 The greater the intermolecular forces present, the lower the vapor pressure. (lowest vapor pressure) CCl4 < C3 < C4 (highest vapor pressure) 10 of 38

Prep101 Problem 4. Answer: D Chem201 Final Booklet There is more interaction between the atoms of carbon tetrachloride due to stronger London dispersion forces, which increase the boiling point with respect to methane. Chlorine is more polarizable than hydrogen because it has MANY more electrons that are also further from the nucleus, resulting in stronger London forces. Problem 5. Answer: B Solution: F is more likely to form hydrogen bonds with water because of the large dipole present in the F molecule. This allows the hydrogen bonds to form more readily. Problem 6. Solution: B Methanol experiences hydrogen bonding due to the polar nature of the molecule. Problem 7. Solution: D The lowest boiling point will occur in non-polar compounds. The only non-polar compound above is N 2. Problem 8. Solution: C Increasing the temperature increases the kinetic energy of the molecules in the liquid causing the molecules to move more quickly and increase the vapor pressure. Problem 9. Solution: A) 2 The boiling point depends on the type of intermolecular forces present. -bonding > dipole-dipole > London dispersion forces. C 4 has the lowest boiling point as there are only London dispersion forces present. S F 11 of 38 N -bonding -bonding -bonding -bonding London forces 2 strong dipoles 2 weak dipoles 1 very strong 3 weak dipoles dipole The order of intermolecular bonding strength & hence boiling temp is: (highest BP & bond strength) 2 > F > 2 S > N 3 > C 4 (lowest BP & bond strength)

Prep101 Chem201 Final Booklet Problem 10. Solution: A The higher the intermolecular forces, the higher the boiling point of the solution will be. 12 of 38

Prep101 Chem201 Final Booklet Practice Problems Problem 11. Answers: (i) C 3 C 3 (iv) 3 C C C 2 C Cl (v) (ii) C C 3 3 2 3 C C C C C C C 3 3 C (iii) C K Problem 12. Answers: (i) butanal (ii) Cyclopentane (iii) 3-chlorobutanoic acid (iv) butylamine (v) 2-ethyl-1-pentene (vi) (cis)-2-hexene (vii) methoxyethane (viii) benzoic acid (ix) 2-pentyne (x) 1,3-dichorobenzene 13 of 38

Prep101 Problem 13. Answers: A) B) C) N 2 Chem201 Final Booklet Problem 14. Answer: D D has the formula C 7 14. All the above structures have 7 carbons; therefore, you can just count the atoms. Problem 15. Answer: 9 Solution: heptane 2-methylhexane 3-methylhexane 2,4-dimethylpentane 2,3-dimethylpentane 3-ethylpentane 3,3-dimethylpentane 2,2,3-trimethylbutane 2,2-dimethylpentane 14 of 38

Prep101 Problem 16. Answer: C Solution: Chem201 Final Booklet A chiral carbon is one that has 4 different groups attached to it via σ bonds. A carbon containing double bonds does not count as asymmetric carbons since there are only 3 σ bonds. A) None B) None C) D) None Br N 2 N 2 C N 2 15 of 38

Prep101 Problem 17. Answer: Alkene, alkyne, ketone, alcohol Notes: Alkene (double bond), alkyne (triple bond), ketone (contains a carbonyl), alcohol (contains a hydroxyl) Chem201 Final Booklet NTE: ydroxyl (-) and carbonyl (C=) ARE NT functional groups. They are pieces of functional groups. For instance, an alcohol (R-) contains a hydroxyl. A ketone (R 2 C = ) contains a carbonyl. A carboxylic acid (R-C) contains BT a hydroxyl and carbonyl. Problem 18. Answer: B An ester group is one that contains RCR ( ) which is not present Problem 19. Answer: C 3 C 2 CC 2 C 2 C 3 * 3 C Cl *The middle two structures are achiral C C C C C C Cl 3 C C C * N 3 Problem 20. Consider the following molecules A, B, C, D and E. Answers: i. A and D or B and D ii. A and B iii. NNE Problem 21. Answers: (i) A and F (ii) C and E (iii) B, C, D and E are not optically active, thus, no rotation of plane polarized light. A and F are optically active but form racemic mixture. Thus, no rotation of plane polarized light. 16 of 38

Prep101 Problem 22. Answer: C 3 Chem201 Final Booklet C 3 Problem 23. Answer: 17 of 38

Prep101 Problem 24. Answer: Chem201 Final Booklet n reason nylon is so strong is that it is a high density polymer meaning it has long non-branching chains of high molecular weight. In the absence of the special amid functional group these chains could pack together tightly in a uniform structure and be fairly strong. owever, nylon s amide functional groups enable it to form hydrogen bonds between the nitrogen hydrogen of one strand and the carbonyl oxygen of an adjacent strand. These hydrogen bonds allow for tight packing and greatly increase its strength: N N N N N N 18 of 38

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Periodic Properties Practice Problems Problem 16. Answer: B Solution: will have the highest 1s orbital energy because it has the least positive charge on the nucleus. Problem 17. Answer: a) V 5+ < Ti 4+ < Sr 2+ < Br - The electron configurations are... # of protons Sr 2+ : [Kr] 38 Br - : [Kr] 35 V 5+ : [Ar] 23 Ti 4+ : [Ar] 22 Sr 2+ and Br - has the same number of electrons; however, Zeff is greater for Sr 2+ due to a greater # of protons, resulting in a greater + charge that pulls the electrons closer to the nucleus. Thus, Sr 2+ is smaller than Br -. V 5+ and Ti 4+ has the same number of electrons; however, Zeff is greater for V 5+ due to a greater # of protons. Thus V 5+ is smaller than Ti 4+. Sr 2+ and Br - are larger than V 5+ and Ti 4+ because there are more electrons held in a larger subshell. (smallest) V 5+ < Ti 4+ < Sr 2+ < Br - (largest) b) Cl > Br > I Problem 18. Solutions a) Electron affinity becomes more negative from left to right because Cl has higher Zeff than S. b) For k, valence electron is in 4s orbital while valence electron in Li is in 2s orbital. 4s orbital is larger than 2s. c) Ionization of Xe removes 5p electron while ionization of Kr removes 4p electron. Ionization energy 1 2 n d) Zeff is larger in than in B. 20 of 38

Problem 19. a) Mg, Ionization energy increase as you move up a group and it also increases as you move right across a period. b) Mg, Radius increase as you move left along a period and as you move down a group. c) Ca<Be<P<Cl< Electronegativity increases as you up a group and as you move right across a period. F is the most electronegative element and Cs is the least electronegative element. Problem 20. Answer: a) [Ar] or [Ne]3s 2 3p 6 b) S 2- c) S 2- d) Ca 2+ e) S 2- < Ar < Ca 2+ Solution: a) [Ar] or [Ne]3s 2 3p 6 b) Ar, Ca 2+ and S 2- all have the same number of electrons; however, Ar has 18 protons, Ca 2+ has 20, and S 2- has 16. Thus S 2- has the least Zeff since it has the smallest charge pulling on the electrons. c) The species with the least favorable electron affinity is S 2- because it has the smallest Zeff (a smaller positive charge to pull electrons towards the nucleus). d) Ca 2+ has the greatest Zeff because it has the greatest number of protons pulling on the 18 electrons, resulting in the electrons being closer to the nucleus. Ionization energy is the energy required to remove an electron. It is most difficult to remove an electron from Ca 2+, as this will involve the removal of a core electron instead of a valence electron, and Ca 2+ has the greatest Zeff. It is easier to remove an electron from S 2- versus Ar because S 2- has a smaller Zeff. S 2- < Ar < Ca 2+ 21 of 38

Problem 21. has a noble gas electron configuration? has the smallest ionization energy? has an atomic number Z = 13? has a half-filled sub-shell with l = 2? Kr Cs Al Cr is a hydrogen-like species? e + has only one 4s electron? Cr have two unpaired electrons?, C, V 3+ has only two d electrons with n = 3? V 3+ is diamagnetic? has the largest radius? has only one electron with l = 1? has the largest number of unpaired electrons? are transition metal species? Kr Cs Al Cr V 3+, Cr 22 of 38

Problem 22. Answer: a) IE 1 for K > IE 1 for Ca because Zeff increases from left to right across a period, so Ca has a higher Zeff and it is therefore harder to remove an electron. IE 2 for K > IE 2 for Ca because the ionization process for Ca leads to the formation of the stable noble gas configuration (Ca2+) while the ionization reaction for K requires the destruction of a stable noble gas configuration. The former will always be lower in energy. b) K + (g) K 2+ (g) + e Ca + (g) Ca 2+ (g) + e c) The larger the negative charge, the greater the repulsion between electrons, the larger the radius. These are all atoms that have the configuration of Kr, so only charge influences radius. Therefore, Rb+ < Br < Se2 The ionization energy decreases as you go down Group 1 because as n increases it is easier to remove the outer electron. Radius increases as you go down Group A because as n increases, the radius increases. 23 of 38

Problem 23. a)answer: Ca2+ < K+ < Cl < S2 b) Answer: B < Be < < N c) Answer: Sr < Mg < S < F Problem 24. Answer: a) When AB is put into aqueous solution, it will dissociate into A- and B+ ions rather than A+ and B-. A has a greater ionization energy, and thus is less able to form A+ than B forming B+. Further, electron affinity for A is larger, and thus the reaction A A- is more favorable than B B-. b) A has a greater electronegativity; A has a larger electron affinity value and thus releases more energy when it gains an electron, making the reaction more favorable than B gaining an electron. c) A will be more to the right on the periodic table than B, since it has a higher ionization energy and larger electron affinity. Because A is further to the right, it has a greater Zeff and thus will be smaller than B. d) Since nonmetals are on the right hand side of the periodic table, and thus have more favorable electron affinity due to a greater Zeff, element A is the nonmetal. 24 of 38

Quantum Numbers Practice Problems Problem 25. Answer: C Solution: This gives the probability of finding an electron in a region of space. Problem 26. Answer: 2- has 10 electrons therefore Ne is isoelectronic with it. The electron configuration is 1s 2 2s 2 2p 6 Problem 27. Answer: D Solution: The electron configuration for N is 1s 2 2s 2 2p 3. The 1s and 2s levels will have be full, each consisting of 2 electrons with opposite spins (which rules out diagram A). The 2p level will have 3 electrons, one in each orbital (which rules out diagram E) and C). The best diagram will show all three electrons in the 2p orbitals with the same spin (i.e. either all facing up or down). Therefore, diagram D) is the best representation of an N atom in its ground state. Problem 28. Answer: C Solution: Sodium, Na, should be 1s 2 2s 2 2p 6 3s 1 Problem 29. Answer: a) [Ar]3d 1. The electronic configuration of Ti is [Ar]4s 2 3d 2. Electrons are removed from the s subshell first when transition metals, such as Ti, changes to an ionic state. Thus, the electron configuration of Ti 3+ is [Ar]3d 1. b) n = 3, l = 2, ml = -2,-1,0,1,2, ms = + ½, ½ c) Radial probability r 25 of 38

Problem 30. Answer: D The electron configuration for As is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 3 The valence electrons for the element As all reside in the 4th level so n = 4. The valence electrons only exist in s or p sublevels so l = 0 or 1. In the s sublevel there is only one orbital while in the p sublevel, there are 3 orbitals, therefore the value of ml = 0 for the s sublevel and ml = -1, 0, +1 for the p sublevel. ms = +1/2 or 1/2 since the spin of the electron in any orbital can only be up or down. The only answer that falls in range of all the possible choices is D. Problem 31. Answer: C Solution: l = 0, 1, (n 1), therefore if n = 3, then l cannot equal 3. 26 of 38

Problem 32. Answer: E a) n is the principal quantum number and can be any whole integer number. l is the secondary quantum number = 0 to n-1. ml is the third quantum number = -l to +l. Set II) is invalid because ml cannot equal -2 if l is 1. Set III) is invalid because ml cannot equal 3 if l is 2. Set IV) is invalid because l cannot equal 1 if n is 1. Problem 33. Answer: D Solution: These quantum numbers correspond to 4p The atom could be Br. Problem 34. K [Ar]4s 1 V 3+ [Ar]3d 2 Mo [Kr]5s 1 4d 5 Ru [Kr]5s 2 4d 6 Y 3+ [Kr] b) Sn = Tin Problem 35. [Ne]3s 2 [Ne]3s 2 3p 1 [Ar]4s 1 3d 5 [Kr]5s 2 4d 10 5p 4 Magnesium (Mg) Aluminum (Al) Chromium (Cr) Tellurium (Te) 27 of 38

Problem 36. Answers a) 7, m l = l 0 + l where l = 3 as we are dealing with the f shell and m l = 3, 2, 1,0,1,2,3 b) 6, [Ar] 4s 3d c) 50, l = 0,1,2, to a maximum of n 1 i.e. 4. Therefore 5s = 2e (l = 0), 5p = 6e (l = 1), 5d = 10e (l = 2), 5f = 14e (l = 3), 5g = 18e (l = 4): total = 50. Problem 37. Answer a) Many solutions. Some examples are: anion = S 2- and cation K + K+ is [Ar] and S 2- is [Ar] b) Fe=26, for Fe: [Ar] 4s 2 3d 6, for Fe 2+ : [Ar] 3d 6 c) It is the ground state for copper. 28 of 38

Problem 39. Answer: B Solution: C has 2 unpaired electrons N has 3 unpaired electrons has 2 unpaired electrons F has 1 unpaired electrons Ne has 0 unpaired electrons Therefore, N will more strongly pulled into an in-homogeneous magnetic field. Problem 40. Atomic Number Configuration Species State Paramagnetic r Diamagnetic 1 1s 2 Ground Diamagnetic 7 1s 2 2s 2 2p 3 N Ground Paramagnetic 17 1s 2 2s 2 2p 6 3s 2 3p 6 Cl Ground Diamagnetic 11 1s 2 2s 2 2p 6 3p 1 Na Excited Paramagnetic 22 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 Ti 2+ Excited Diamagnetic 13 1s 2 2s 2 2p 6 Al 3+ Ground Diamagnetic Note: Ti 2+ is in an excited state because it would normally lose its 4s electrons first before its 3d electrons. The given configuration is correct for the neutral atom but NT the cation. Problem 41. Species Si Cr V 3- P Zn 2+ Number of unpaired electrons 2 4 4 3 2 29 of 38

Problem 42. i) 3p ii) 3d iii) 3s iv) 3d y 2 2 x y xy Answer: i) y ii) y x + + x iii) y iv) y + x x + Problem 43. rbital Number of nodal planes Number of additional degenerate orbitals 3d x 2 y 2 2 4 2py 1 2 30 of 38

Problem 44. Z Cross Section X Name of rbital 3dz 2 3dxz 2pz orbital Value of n 3 3 Likely 2 (nodes will show up if 3 or higher) Value of l 2 2 1 Total Number of Nodal Planes & Surfaces 2 nodal surfaces 2 nodal surfaces 2 nodal surfaces (one plane, one sphere) Problem 45. rbital l All values of ml Atom 1s 0 0 2p 1-1, 0, 1 B or F 3d 2-2, -1, 0, 1, 2 Sc 3f X X X 4p 1-1, 0, 1 Ga or Br 6s 0 0 Cs 31 of 38

Electromagnetic Radiation Practice Problems Problem 46. Answer: D Solution: Statements (a), (b), and (c) are all true. Problem 47. What is the energy of one mole of photons with a wavelength of 285 nm? Answer: A Solution: hc Js m E 34 8 (6.626 10 )(3.0 10 / s) 9 (285 10 m) 6.98 10 This is the energy per photon; we want the molar energy 19 6.98 10 19 J 6.022 1023 photons = 4.20 10 5 J photon mol J Problem 48. Answer: D Solution: Energy per photon = h = hc/ -34-1 8-1 (6.626 10 J s )(2.998 10 m s ) -9 257 10 m -19 7.73 10 J Problem 49. a) Answer: (iii) Solution: There are six possible transitions- b) Answer: (iii) Solution: Going from n=4 to n=1 will have the largest E. 32 of 38

Problem 50. Answer: B Solution: 1Watt = 1J/sec, therefore, Energy for all photons = 72 J / sec 5sec 360J Energy for one photon = 360J 3.63 10 20 9.91 10 photons 19 J 34 8 hc hc (6.626 10 )(3 10 ) E 547nm 19, therefore E 3.63 10 Problem 51. Answer: B Solution: E = h For one atom of helium: 2370kJ 1mol 1st Ionization energy = = 3.94 10-21kJ 23 mol 6.022 10 atoms 18 E 3.94 10 J 5.95 10 34 h 6.626 10 Js 15 Problem 52. a) Answer: b) Answer: E = hc λ = hν, c λ = ν c ν = λ = 3.00 108 ms 1 6.91 10 14 s 1 = 4.342 10 7 m = 434.2 nm E = hν = 6.262 10 34 Js 6.91 10 14 s 1 = 4.579 10 19 J photon 4.579 10 19 J 6.022 1023 photons 275.6 kj = photon mol mol 33 of 38

Molecular rbital Theory Solutions Problem 53. (a) Answer: (iii) Solution: Bond rder = 2 has 13 valence electrons Bonding electrons Antibonding electrons 2 2 2s 2 s 2 2 2p 2py 2pz Bond order = 2 2 2py (8 5) 2 1 2pz = 3 ½ = 1.5 (b) Answer: 2 is paramagnetic (meaning there are unpaired electrons); thus it is attracted towards a magnetic field. Problem 54. Answer: A Solution: The larger the bond order, the stronger the bond. B.. for Li 2 = 4 2 = 1 2 B.. for Be 2 = 6 2 = 0 2 B.. for 2 = 2 1 = 0.5 2 + B.. for e 2 = 2 1 = 0.5 2 B.. for e 2 = 2 2 = 0 2 34 of 38

Problem 55. For the homo-diatomic B 2 provide the following: a) A full and labelled molecular orbital diagram including bond order b) A representation of the highest energy orbital that is occupied by at least one electron c) Determine whether B 2 would have a longer or shorter bond than B 2. Explain your answer. Answers: a) Bond order = 2 0 2 = 1 b) It is an 2 orbital p c) B 2 has one more bonding electron, so the bond order of B 2 would be greater than for B 2. Therefore, B 2 will have a shorter bond than B 2. 35 of 38

Problem 56. Complete the following table: Ion Paramagnetic or Diamagnetic Bond rder Bond Length: Longer or Shorter than that of 2 2 + Para 2½ S N or CF A neutral diatomic molecule with the same M electron configuration using any combination of the atoms C, N,, F 2 2 + Dia 3 S N 2, C 2 Para 1½ L F 36 of 38

Problem 57. a) Answer: 2 C 2 M configuration: 2 1s 2 1s 2 2s 2 2 2 2 2s 2pz 2py 2pz diamagnetic B.. = 3 b) Answer: 2 2 M configuration: 1s2 *1s2 2s2 *2s2 2pz2 2px2 2py2 *2px2 *2py2 diamagnetic B.. = 1 Problem 58. Benzene (C 6 6 ) Answer: it is in resonance Problem 59. Answer: D Solution: Levels b and a are anti-bonding orbitals and levels d and c are bonding orbitals. no. of e in bonding Ms no. of e in antibondin g Ms Bond order =, therefore adding an electron to 2 level b, an anti-bonding orbital, decreases the bond order. Level c is a 2p orbital, d are orbitals therefore D is true. 37 of 38

Problem 60. The Which model can account for the bond order of ½ for 2 +? Which model can predict the detailed shapes of molecules using simple electrostatic arguments? Which model can rationalize the fact that liquid oxygen sticks to the poles of a strong magnet? Which model invokes the use of hybrid orbitals? Which model can most easily calculate the ionization energy of U + 91? Which model involves the concept of anti-bonding orbitals? Which model uses resonance to describe bonds that are not localized between two atoms? Lewis Structure X VSEPR X Valence Bond Theory X Molecular rbital Theory X X X X Bohr Theory X 38 of 38