Chem 209 Final Booklet Solutions

Transcription

1 Chem09 Final Booklet Chem 09 Final Booklet Solutions 1 of 38

2 Solutions to Equilibrium Practice Problems Chem09 Final Booklet Problem 3. Solution: PO 4 10 eq The expression for K 3 5 P O 4 eq eq PO 4 10 init 1 M In (a) Q 1 3, the reaction proceeds to the right. 5 5 P O 1 M 1 M 4 init init PO 4 10 init 5.0 M In (b) Q 3.7 3, the reaction proceeds to the left. 5 5 P O 8.0 M 0.70 M 4 init init of 38

3 Problem 4. Solution: Chem09 Final Booklet To determine the final concentrations, the first thing needed are the initial reactant concentrations and an expression for the reaction coefficient Q. [A (g) ] = and Q 1.00mol 0.50 L ABinit A B = 4.00 M, [AB (g) ] = [B (g) ] = (8.00).0 (4.00) (8.00) init init.00mol 0.50 L = 8.00 M, The direction of the reaction needs to be determined. To do this, we compare Q vs K. Since Q =.0 > K = 0.5, the reaction proceeds to the left. To determine what final concentrations will be from initial concentrations, a handy tool the Initial, Change, Equilibrium (ICE) table can be used. All compounds involved in the reaction are included in an ICE table as follows, with the species that will be consumed on the left side, and the species that will be produced on the right side. Since the reaction proceeds to the left, A and B will be formed and AB will be consumed. Concentration (M) AB (g) A (g) B (g) Initial 8.00 M 4.00 M 8.00 M Change -x +x +x Equilibrium 8.00 x x x As the reaction proceeds, x moles of A (g) and B (g) are formed as x moles of AB (g) are consumed. Make sure to consider stoichiometric coefficients appropriately. The equilibrium values are simply the sum of the initial + change concentrations. Substitute the equilibrium concentrations into K, and solve for x (remember that K is defined for the reaction in the way that it was initially described): K so, ABeq A B (8.00 x) 0.5 (4.00 x) (8.00 x) eq eq 3 of 38

4 (8.00 x) 0.5(4.00 x)(8.00 x) 64 3x 4x 0.5(3 1 x x ) 64 3x 4x 16 6x 0.5x 3.5x 38x 48 0 Chem09 Final Booklet To solve for x, the quadratic formula must be used, 38 ( 38) 4(3.5)(48) b b 4ac x 1.46 or 9.39, a (3.5) Since x 9.39 = > 8.00 (which would leave the equilibrium concentration of AB at equilibrium negative), then the x = 1.46 So, final concentrations are: [A] = = 5.46 M [B] = = 9.46 M [AB] = 8.00 (1.46) = 5.08 M To check, substitute these concentrations into the Equilibrium constant expression, ABeq A B (5.08) K 0.5, matches up. (5.46) (9.46) eq eq 4 of 38

5 Chem09 Final Booklet Problem 5. Answer: K c 4 [ NO] [ H ] [ NO ] Solution: Pure solids and pure liquids and solvents are not included in the expression. Products go over Reactants, and coefficients in the equation are written as superscripts. Problem 6. Solution: K c = [CO (g) ] Remember: Solids and liquids are not included in the expression. Problem 7. Solution: Equation is equal the double and reverse of equation1 therefore K p = K p1 = ( ) = 40.6 Problem 8. Answer: Equilibrium Constant = K 1 1/ = 1 Solution: The second equation if reversed (1/K) and halved (K1/). Combining these two gives 1/K1/. K 1 Problem 9. Answer: The reaction equation is: The relationship between K c and K p is: CO (g) + O (g) CO (g) K c = K p = K c (RT) n gas In this case there are 3 moles of gas in the reactants and moles of gas in the products, so n = -1 So solving for K p : K p = K c (RT) n gas = ( ) [(0.081 L atm mol K ) (98 K)] 1 = of 38

6 Problem 10. Solution: Chem09 Final Booklet N (g) + C H (g) HCN (g) i: c: +x +x -x e: x x 1.00-x Q=1>K so then rxn goes to the left (1.00 x) (1.00 x) K c (1.00 x) (1.00 x) K c 1 x K K c c Problem 11. Solution: COCl (g) CO (g) +Cl (g) initial: change: -x +x +x equil: 0.04-x x x x Kc = 0.04 x x + Kcx 0.04Kc = 0 Kc Kc (4)(0.04)( Kc) x of 38

7 Problem 1. Solution: Kc [ CO ][ H ] [ CO][ H O] 5.10 Chem09 Final Booklet CO + H O CO +H I 0.1M C -x -x +x +x E 0.1-x 0.1-x 0.1+x 0.1+x (0.1 x) x x Kc 5.10 (0.1 x) x x x x x x x 1.x x 0.034M Therefore, at equilibrium [H] = x = M = 0.134M Problem 13. Solution: Only C is true An increase in volume will shift the equilibrium to the right thus causing an increase in the total moles of CO at equilibrium. Problem 14. Solution: a) shift to the left b) no effect c) shift to the right d) shift to the right Problem 15. Solution: The reaction will shift left forming Ni(CO) 4(g) to reach equilibrium Problem 16. Answer: C Solution: Q = [NO] [Cl ] [NOCl] = (1.) (0.56) (1.3) = Q=K, therefore we are already at equilibrium. 7 of 38

8 Chem03 Final Booklet Solutions to Acids and Bases Practice Problems Problem 17. Solution: Equal volumes of 0.1 M NaF and 0.1 M HF Adding an acid and the salt of its conjugate base can form a buffer, this is the case in (b). Problem 18. Solution: 1 and 3. A buffer normally consists of a weak acid and its conjugate base in roughly equal amounts. The ratio should be no greater than 0.1 to 10 of weak base to weak acid. In 1, the ratio of acid to base is 3:1. In 3 the ratio is 4:1. In, there is a greater amount of strong base than acid, so all of the acetic acid is consumed so it is not a buffer. 4 is not a buffer solution since all of the weak acid is consumed with strong base, 5 is not a buffer because sodium acetate is a weak base, so you have a weak base with a strong base in solution. Problem 19. Solution: B This is a buffer solution, therefore: [ A ] ph pka [ HA] eq eq log log( ) log 1.74 Problem 0. Solution: This is a buffer solution, therefore: [ A ] eq ph pka log [ HA ] eq [ CH COONa] = = 4.75 log [ CH3COONa ] Therefore log =0 and [ CH3COONa ] =1 [CH3COONa] = 0.10M # moles CH3COONa = 0.10 moles Problem 1. Solution: For this equilibrium [H+] = [In-] = 10-8 = 1 x 10-8 Ka = 1 x 10-6 = [ H ][ In [ HIn] ] = -8 (1x 10 ) [ HIn] [HIn] = 1 x 10-10, therefore [HIn]/[In- ] = 1 x 10-10/1 x 10-8 =0.01 = 1/100 8 of 38

9 Problem. Solution: ph = 8.91 Chem03 Final Booklet ph = pka + = = 8.91 CN log HCN 1.0 log.0 Problem 3. Solution: C This is a buffer system as there are equal amounts of conjugate acid and base, but the addition of H+ can change the ph slightly. The original ph of the buffer is approximately equal to pka. When H+ ions are added from the strong acid HCl, A- is converted into HA. Therefore the addition of 0.01 moles H+ produces 0.01 moles HA and consumes 0.01 moles A-. nha= = 0.51 mol na- = = 0.49 mol [H+]=Ka x nha/na- = (1.8 x 10-4) (0.51/0.49) = x 10-4 ph = -log (1.873 x 10-4) = 3.77 Problem 4. Solution: C There are more moles of carbonate buffer in solution c than any of the other solutions Problem 5. Solution: E The solution is a buffer with ph above 7. A buffer is resistant to both addition of strong acid and strong base and the concentration of the hydronium ion is not more than the hydroxide ion (ph >7). Problem 6. Solution: E HCN + H O CN + H 3 O + I 0.5 y 0 C E 0.5 y Ka = 6. x = [(1x10-7)( Y+1x10-7)]/ 0.5 Y=0.003M x 1L = 0.003moles Mass = g/molNaCN x moles = 0.15g 9 of 38

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11 Chem03 Final Booklet Problem 7. Solution: E A buffer normally consists of a weak acid and its conjugate base in roughly equal amounts. The ratio should be no greater than 0.1 to 10 of weak base to weak acid. In 1, the ratio of acid to base is 3:1. In 3 the ratio is 4:1. In, there is a greater amount of strong base than acid. Problem 8. Solution: C or E In order to determine which solutions are able to act as buffer solutions, determine what ions will be found in the solution and whether those ions are acidic, basic or spectator ions. If acidic and/or basic ions are found, calculate the amount of ions that are present. For solution A) all ions present (H+, NO3-, and Na+) are spectator ions. No buffer abilities possible. For solution B) all ions present (Na+, OH- and Cl-) are spectator ions. No buffer abilities possible. For solution C) This is a 1:1 ratio of conjugate acid to its conjugate base. This IS a buffer. For solution D) the HCl completely dissociates to form H+ and Cl- ions. NH3 can form an equilibrium where NH3 + H+ NH4+. However, HCl and its dissociated ions are present in 0.01mol amounts the same as the amount of NH3. Therefore, all the NH3 is used up in reacting with the HCl, so no buffering abilities are possible. For solution E) the NaOH completely dissociates to form Na+ and OH- ions. The 0.004mol of OH- will react with the 0.01mol of HF to form 0.004mol of F- and have 0.006mol left of HF since the OH- is the limiting reagent. This means we have HA and A- both present in a ratio of 3:. This is a buffer. Problem 9. Solution: C B + HO BH+ + OH- Calculate from the given ph, the concentration of OH- ions that dissociate form from the reaction of the base with water. poh = 14 ph = 14 (8.88) = 5.1 [OH-] = 10-pOH = = x 10-6M Assume 1 L of solution, therefore the [B] = 0.40mol/L and [BH+] = 0.50mol/L. K = [BH +][OH ] [B] (0.50)( x 10 6) = = 4.74 x of 38

12 Problem 30. Solution: D Chem03 Final Booklet Because equal volumes of the acid and weak base are being mixed, all concentrations (M) can be treated as moles (mol). HNO3 is a strong acid so it completely dissociates HNO3 H+ + NO NH3 will react with the H+ released by the HNO3 to form NH4+. Initially there is 0.3M or 0.3moles of NH3. Upon addition of 0.1moles H+ (from the HNO3), 0.1mol of NH3 will react to form 0.1mol of NH4+, leaving 0.mol NH3 unreacted. Therefore, NH3 + H+ NH4+ Using the following equation, solve for [H+]. [H+] = Ka x [HA] [A-] Kw = Ka x Kb so that Ka = Kw = 1.0 x = 5.56 x Kb 1.8 x 10-5 [H+] = 5.56 x x [0.1] =.78 x 10-10M [0.] ph = -log[h+] = -log[.78 x 10-10] = of 38

13 Problem 31. Solution: B Chem03 Final Booklet Write out the equations that are occurring in the solution described above. Because both solutes are being added to 1L of water, all concentrations (M) can be treated as moles (mol). CH5COONa is a soluble salt so it completely dissociates CH5COONa CH5COO- + Na HCl is a strong acid so it completely dissociates HCl H+ + Cl CH5COO- will react will all the H+ to form CH5COOH. Initially there is 0.1 mole of CH5COO-. When 0.01mol of H+ is added, the CH5COO- reacts leaving 0.09mol. There is also 0.1mol of CH5COOH to start, but when the CH5COO- reacts with the H+, it forms 0.01mol more CH5COOH so that the total amount of CH5COOH is 0.11mol. CH5COOH CH5COO- + H+ Before HCl is added After HCl is added Calculate [H+] using the following equation. (n = moles) [H+] = Ka x nha = 1.41 x 10-5 x (0.11) = 1.7 x 10-5M na- (0.09) Calculate ph from the [H+]. ph = -log[h+] = -log[1.7 x 10-5] = of 38

14 Chem03 Final Booklet Solutions to Electrochemistry Practice Problems Problem 33. Solution: List the oxidation and reduction steps: Reduction: Oxidation: Br (g) + e Br (aq) + Ag (aq) + e Ag (s) E cathode = V E anode = V E cell = E cathode - E anode = V V = 0.81 V E V ln n cell K eq or E 0.059V log n cell K eq log K log K eq eq K eq necell V (0.81V ) V Problem 34. Solution: The half-cell reactions are as follows (note: these are not at standard state or E cell = 0 V): Reduction: Oxidation: Pb + (aq) + e Pb (s) E cathode =? Pb + (aq) + e Pb (s) E anode =? E cell E cathode= E anode = E = cathode E - anode V log = V 0.100M 0.059V 1.15 log [ Pb V 1 E cell V = log [ Pb ] ] [Pb + (aq) ] = Solubility PbSO 4 Pb (aq) SO 4 Initial Some 0 0 Change -s +s +s Equilibrium Some-s S S K sp = [Pb + (aq) ][SO 4 ] K sp = ( ) = of 38

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16 Problem 35. Answers: Cr = +3, O = - Ca = +, C = +4, H = +1, O = - Fe = +3, C = +4, O = - Chem03 Final Booklet Problem 36. a) Answer: 0.15 V Pb + + e Pb E o = 0.13 V Co Co + + e E o = +0.8 V o E cell = 0.8 V 0.13V = 0.15 V (b) Answer: B Reduction occurs at the cathode; therefore, the lead electrode is the cathode. Problem 37. Answer: MnO 4 (aq) > Zn (s) > I (aq) > I (aq) > Zn + (aq) > MnO (aq) An oxidizing agent gets reduced therefore MnO 4 (aq) is the strongest oxidizing agent as it has the largest E o. Problem 38. Solution: Br (aq) will be reduced and I (aq) Br (aq) + e Br (aq) I (aq) E = 1.09 V I (aq) + e E = 0.54 V will be oxidized. 16 of 38

17 Problem 39. What is the value of E cell? Chem03 Final Booklet Answer: 0.55 V Br (aq) + e Br (aq) I (aq) Br (aq) + I (aq) Then put the Ecell into the Nerst: E = 1.09 V I (aq) + e E = 0.54 V Br (aq) + I (aq) E cell = 0. 55V E cell = 0.55 ( ) (log ( [0.] [0.1] [0.][10] ) = 0.66V Problem 40. Answer: D>B>A>C Solution: Reducing agent undergoes oxidation (most easily oxidized = strongest reducing agent) From A + + B A + B + B can oxidize to B+, B is a stronger reducing agent than A. From A + + C no reaction C cannot oxidize to C+ A is a stronger reducing agent than C. From B + + D B + D + D can oxidize to D + D is a stronger reducing agent than B. The decreasing order of reactivity (most easily oxidized to least easily oxidized) is: D>B>A>C, where D is the strongest reducing agent. 17 of 38

18 Problem 41. Answer: B + Given: Hg (aq) + Br (aq) To Find: anode reaction Hg Br (s) ANODE allows oxidation to take place, which is the loss of electrons. - Answers A, C and E are incorrect as they are reduction reactions. - Answer D is incorrect as the charges on both sides are NOT balanced. Hg (l) acts as an intermediate ANODE Hg (l) + Br (aq) + CATHODE: Hg (aq) Chem03 Final Booklet Hg Br (s) + e + e Hg (l) Problem 4. Answer: Ag + only Cu Cu + + e E = 0.34V So addition of this to the cathode reaction should be a positive number for the reaction to proceed. The only one that this works for is Ag + Problem 43. Answer: 0.5 V 3 (I (s) + 6 H O (l) HIO 3(aq) + 10 H e ) E = 1.0V 5 (ClO 3(aq) 3 I (s) + 5 ClO 3(aq) + 6 H e Cl aq + 3 H O (l) ) E = +1.45V + 3 H O (l) 6 HIO 3(aq) + 5 Cl (aq) E = 0.5V Problem 44. Answer: 0.6 V In the reaction above, Sn + (aq) is being oxidized to Sn 4+ (aq) (it lost e ) and Fe 3+ (aq) is being reduced to Fe + (aq) (gaining 1 e ). E cell = E reduction + E oxidation. Since all E values are given as E reduction values, the E red value for Fe in the reaction is V. The E value for Sn in the reaction must be reversed because Sn is undergoing oxidation, therefore E ox = 0.15V. E cell = E reduction + E oxidation = (+0.77V) + ( 0.15V) = 0.6V 18 of 38

19 Problem 45. Answer: A only Chem03 Final Booklet Calculate the E cell for all three reactions. In statement I), E cell = -1.86V (Cd + (aq) is being reduced and Cl (g) is being oxidized). In statement II), E cell = -0.9V (Sn + (aq) is being oxidized and reduced into Sn 4+ (aq) and Sn respectively). In statement II) E cell = 0.91V (Sn is being oxidized and Fe3+ is being reduced). E cell values that are positive means that the redox reactions will occur spontaneously, while a negative E cell value means the redox reaction is not spontaneous. Problem 46. Answer: I, II, and III Solution: Statement I) is true becausecu + has a more positive E reduction value than Cr 3+, which means that Cu + is a better oxidizing agent than Cr 3+ (remember that an oxidizing agent oxidizes other substances and becomes reduced in the process). Problem 47. Answer: In the reaction given above, Ag (aq) o overall E cell for the reaction is calculated as is being reduced to Ag while Ni is being oxidized to Ni + (aq). The o E cell = E reduction + E oxidation The reduction half of the reaction is given as V. In order to calculate the E value for the oxidation half, rearrange the E o cell equation to solve for E oxidation. o E oxidation = E cell E reduction = (+1.06V) (+0.80V) = +0.6V. The E oxidation value is the opposite sign from the E reduction value, so in order to solve for the E reduction n value as asked, the sign on the E oxidation value must be reversed. Therefore, E reduction for Ni + (aq) = 0.6V 19 of 38

20 Chem03 Final Booklet Problem 48. Answer: n = 4 H + O H O can be broken down into its half-cell reactions: H 4 H e O + 4 H e H O Since there is an exchange of four electrons, n = 4 in the Nernst Equation. Problem 49. Answer: 0.1 V Solution: E = E o 0.057V n lnq = 0.15V 0.057V [ Co ln [ Pb ] = 0.1V ] 0 of 38

21 Chem03 Final Booklet Problem 50. (a) When current is allowed to flow, which species is oxidized? Solution: MnO H e Mn H O E o = 1.51 V Cr H O Cr O H e E o = 1.33 V Oxidation is a loss of electrons. Cr 3+ is being oxidized to Cr O 7 (b) When current is allowed to flow, which species is reduced? Solution: MnO 4 is being reduced to Mn + (c) What is the value of E cell? Solution: MnO H e Mn H O E o = 1.51 V Cr H O Cr O H e E o = 1.33 V MnO 4 + Cr H O Cr O H + + Mn + E o = 0.18V (d) What is the oxidation state of Cr in Cr O 7? Solution: oxid. # = [7(-) + ]/ = +6 1 of 38

22 Chem03 Final Booklet Solutions to Kinetics Practice Problems Problem 51. Solution: v k[ I ] [ S O ] v k I S O m n m n [ ] [ 8 ] m n k(0.080) (0.040) m n k(0.040) (0.040) v v 1 3 m m m n k(0.080) (0.040) m n k(0.080) (0.00) n n 1 rate k[ I ] [ S O ] k rate [ I ][ S O ] k M s (0.080) (0.040) Problem 5. Solution: Rate = k[h O] x [CH 3 Cl] y x Rate1 k(0.0100) x Rate k(0.000) x =.5, therefore x = and the reaction is second-order in H O x Rate1 k(0.0100) x Rate k(0.000) ( y 3 ) = 3.69 ( 3 )y =, therefore y = 1 and the reaction is first-order in CH 3 3Cl Rate = k[h O] [CH 3 Cl] 1 of 38

23 Problem 53. Solution Chem03 Final Booklet [ A] 0 ln[ A] t ln[ A] kt, ln kt 1/ [ A ] t1/ [ A] 0 for t1/ [ A] t 1/ [ A] 0 [ A] 0 ln ln ln kt1/ [ A] t [ A] 1/ 0 ln t1/ k 0 1/ 1/ Problem 54. Solution: If 0% decomposes, then 80% of the sample remains. ln[ A] ln[ A] kt rearrange 0 ln 0.8[A] o k(50s) [A] o k sec 1 t 1/ ln k 155 sec Problem 55. Solution: (a) v v k[ CH NNCH ] n n k[ CH3NNCH 3] n ( ) n (.0510 ) n n 1 Therefore, the reaction is the first order. Rate = k [CH 3 NNCH 3(g) ], therefore k = (b) ln[a] t = ln[a] 0 kt M / s = 5.46 x 10-5 s-1 M ln[ ] = ln[ ] ( ) t -5.7 = ( )t ( )t =.30 t = s = 11.7 hours (c) [A]10% = = M Rate = k[a] = ( s-1)( M) = M/s 3 of 38

24 Chem03 Final Booklet 1 1 Problem 56. Solution: Because this reaction is second order kt if 1/[A] was plotted vs. t [ A] [ A] o then the y-intercept would be 1/[A]0 and the slope would be k 1/[A] slope = +k 1/[A] Problem 57. Answer t Problem 58. Solution: Since Hrxn > 0, then Hproducts > Hreactants 4 of 38

25 Thus, Ea rev = = 5 kj/mol (draw an energy diagram to see this better) Chem03 Final Booklet Problem 59. Solution: Let T1 = 96 C = 369 K, and T = 5 C = 98 K. Using the Arrhenius Equation ln ( k 1 k ) = E a R ( 1 T 1 T 1 ) x 10 s E 1 1 a ln 1. x 10 s JK mol 369 K 98 K and solve for Ea, Ea = 1.8 kj/mol Problem 60. Solution: from Arrhenius equation since vk, then ln k 1 E 1 R T T k a 1 ln k ln v E a 1 1 k 1 v 1 R T T 1 ln(3) E a E a J / mol 77.4 kj / mol Problem 61. Solution: Since we have an elementary process, then rate of the reaction is 1 1 This reaction is second order therefore kt [ A] [ A] 1 0.1M 1 0.M 5 M-1 = k(35. min) k(35. min) k = 1.4 x 10-1 M-1min-1 o rate k[a], 5 of 38

26 Chem03 Final Booklet Problem 6. The steps are elementary, and the first step is the rate determining step. So the overall rate law only depends on the first step: rate = k[o3][no] 6 of 38

27 Prep101 Chem01 Final Booklet Periodic Properties Practice Problems Problem 63. Answer: a) V 5+ < Ti 4+ < Sr + < Br - The electron configurations are... # of protons Sr + : [Kr] 38 Br - : [Kr] 35 V 5+ : [Ar] 3 Ti 4+ : [Ar] Sr + and Br - has the same number of electrons; however, Zeff is greater for Sr + due to a greater # of protons, resulting in a greater + charge that pulls the electrons closer to the nucleus. Thus, Sr + 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 + 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 + < Br - (largest) b) Cl > Br > I Problem 64. 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 s orbital. 4s orbital is larger than s. c) Ionization of Xe removes 5p electron while ionization of Kr removes 4p electron. Ionization energy 1 n d) Zeff is larger in O than in B. Problem 65. 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. 7 of 38

28 Prep101 Chem01 Final Booklet c) Ca<Be<P<Cl<O 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 66. Answer: a) [Ar] or [Ne]3s 3p 6 b) S - c) S - d) Ca + e) S - < Ar < Ca + Solution: a) [Ar] or [Ne]3s 3p 6 b) Ar, Ca + and S - all have the same number of electrons; however, Ar has 18 protons, Ca + has 0, and S - has 16. Thus S - 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 - because it has the smallest Zeff (a smaller positive charge to pull electrons towards the nucleus). d) Ca + 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 +, as this will involve the removal of a core electron instead of a valence electron, and Ca + has the greatest Zeff. It is easier to remove an electron from S - versus Ar because S - has a smaller Zeff. S - < Ar < Ca + 8 of 38

29 Prep101 Chem01 Final Booklet Problem 67. 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 =? Kr Cs Al Cr is a hydrogen-like species? He + has only one 4s electron? Cr have two unpaired electrons? O, 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 9 of 38

30 Prep101 Chem01 Final Booklet Problem 68. 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 for K > IE for Ca because the ionization process for Ca leads to the formation of the stable noble gas configuration (Ca+) 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 + (g) + e Ca + (g) Ca + (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 < Se 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. 30 of 38

31 Prep101 Chem01 Final Booklet Problem 69. a)answer: Ca+ < K+ < Cl < S b) Answer: B < Be < O < N c) Answer: Sr < Mg < S < F Problem 70. 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. 31 of 38

32 Chem09 Final Booklet Chemical Bonding Solutions Problem 71. Answer: B Formal Charge = Valence electrons lone pair electrons bonds FC F = = 0 FC P = 5 3 = 0 FC S = 6 4 = 0 Problem 7. 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 73. Answer: C There are 4 lone pairs of electrons. 3 of 38

33 VSEPR Solutions Chem09 Final Booklet Problem 74. Answer: C Solution: A triple bond corresponds to 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 75. Answer: a) N(1): sp3 N(): sp O(1): sp O(): sp3 b) i) 109 I ii) 10 iii) 6 lone pairs Solution: N(1) has 3 bonding pairs and 1 lone pairs of electrons sp3 hybridized N() has 3 bonding pairs of electrons sp hybridized O(1) has 1 bonding pair and lone pairs of electrons sp hybridized O() 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() is sp hybridized; trigonal planar; 10 bond angle iii) By the Lewis Structure, there are 6 lone pairs of electrons. Problem 76. Answer: D Solution: From the Lewis diagram of the molecule we can see that the shape of the molecule is T- Shaped. 33 of 38

34 Chem09 Final Booklet Problem 77. Which of the following statements is/are correct for the formate ion HCO? Answer: C Solution: the oxidation number of the C atom is +, there are only plausible contributing structures, and HCO = 18 e- = AX3 = trigonal planar Problem 78. 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 atoms. C N p orbitals sp orbitals п bond σ bond d) Answer: There are 3 electron pairs around the central C atom, and thus it is sp hybridized. Thus the bond angle is of 38

35 Problem 79. Predict the geometric shape of ClO- ion. Answer: Bent Solution: Chem09 Final Booklet Total valence electrons: 7 + (6) + 1 = 0 electrons Center Cl has bonding pairs and lone pairs of electrons sp 3 hybridized & tetrahedral formation. Since there are lone pairs of electrons, the structure is bent-shaped. Problem 80. Draw the Lewis structure for the peroxymonosulfate ion, H-O-O-SO3-, and estimate the H-O-O bond angle. Answer: < 109 Solution: -1 The center oxygen in H-O-O has 4 pairs of electrons ( lone pairs of electron and bonding pairs of electrons); as such, this part of the molecule has a tetrahedral structure. The bond angle is less than due to the 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. 35 of 38

36 Chem09 Final Booklet Problem 81. a) Applying your knowledge of VSEPR, indicate the geometry of the three atoms indicated in acetic acid. O C H 3 OH b) What is the O=C-O bond angle? c) Indicate the orbital hybridization of the three atoms with the arrows. Answer: a) H3C- : 4 bonding pairs of electrons, thus it is tetrahedral 3 bonding pairs of electrons; thus it is trigonal planar O C H 3 OH 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 10. O C H 3 OH c) H3C - C - OH Tetrahedral structure = sp3hybridization H3C - C - OH Trigonal planar structure = sp planar H3C - C - OH Tetrahedral structure = sp3 hybridization 36 of 38

37 Chem09 Final Booklet Problem 8. a) ClCS (thiophosgene) (Carbon is central atom, Cl are equivalent). Lewis: *All formal charges are zero. Total valence electrons = 4 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 37 of 38

38 Chem09 Final Booklet Problem 83. a) HNO3 a) Total # of electrons = (6) = 4 () (1) (3) Formal charge: On O(1) 6 [4 + ½ (4)] = 0 On O() 6 [6 + ½ ()] = -1 On O(3) 6 [4 + ½ (4)] = 0 On N 5 [0 + ½ (8)] = +1 On H 1 [0 + ½ ()] = 0 Molecular drawing: b) O3 Structure around central O: 4 electron pairs = bent Total # of valence electrons: 3(6) = 18 Lewis structure: Structure around central N: 3 electron pairs = trigonal planar (1) () (3) Formal charge on: O(1) 6 [6 + ½ ()] = -1 O() 6 [ + ½ (6)] = + 1 O(3) 6 [4 + ½ (4)] = 0 Molecular structure around central O = bent 38 of 38