Slide
Upcoming SlideShare
Loading in...5
×
 

Slide

on

  • 3,478 views

 

Statistics

Views

Total Views
3,478
Views on SlideShare
3,475
Embed Views
3

Actions

Likes
0
Downloads
20
Comments
0

1 Embed 3

http://www.slideshare.net 3

Accessibility

Upload Details

Uploaded via as Microsoft PowerPoint

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

    Slide Slide Presentation Transcript

    • Chapter 1 Chemistry and Measurement
    • Example of The law of conservation of mass: Aluminum powder burns in ? oxygen to produce a substance called aluminum oxide. A sample of 2.00 grams of aluminum is burned in oxygen and produces 3.78 grams of aluminum oxide. How many grams of oxygen were used in this reaction? aluminum + oxygen = aluminum oxide 2.00 g + oxygen = 3.78 g oxygen = 1.78 g Copyright © Houghton Mifflin Company. All rights reserved. 1|2
    • Potassium is a soft, silvery-colored metal that melts at 64 C. It reacts ? vigorously with water, with oxygen, and with chlorine. Identify all of the physical properties and chemical properties given in this description. Physical Property Chemical Property Soft Reacts with water Silvery-colored Reacts with oxygen Melting point (64 C) Reacts with chlorine Copyright © Houghton Mifflin Company. All rights reserved. 1|3
    • Copyright © Houghton Mifflin Company. All rights reserved. 1|4
    • Perform the following calculation and round your answer to the correct ? number of significant figures: 92.35(0.456 0.421) Calculator answer: 3.23225000 The answer should be rounded to two significant figures (92.35 X 0.035): 3.2 Copyright © Houghton Mifflin Company. All rights reserved. 1|5
    • In winter, the average low temperature in interior Alaska is ? -30. F (two significant figures). What is this temperature in degrees Celsius and in kelvins? 5ο C tC tF 32ο F ο 9 F ο ο 5ο C tC 30. F 32 F ο 9 F 5ο C tC 62ο F ο 9 F tC 34.4444444ο C tC 34ο C Copyright © Houghton Mifflin Company. All rights reserved. 1|6
    • 1K tK tC 273.15 K 1ο C 1K ο tK 34 C ο 273.15 K 1 C tK 34 K 273.15 K tK 239.15 K tK 239 K Copyright © Houghton Mifflin Company. All rights reserved. 1|7
    • Oil of wintergreen is a colorless liquid ? used as a flavoring. A 28.1 g sample of oil of wintergreen has a volume of 23.7 mL. What is the density of oil of wintergreen? m 28.1g m d V 23.7 m L V 28.1 g d 23.7 mL g d 1.18565491 mL g d 1.19 mL Copyright © Houghton Mifflin Company. All rights reserved. 1|8
    • The dimensions of Noah’s ark were reported as 3.0 102 cubits by 5.0 ? 101 cubits. Express this size in units of feet and meters. (1 cubit = 1.5 ft) 1 cubit 1.5 f t 3 f t 1 yd 1 yd 0.9144 m (exact) 1.5 f t 2 1 1.5 ft 3.0 10 cubits 5.0 10 cubits 1 cubit 1 cubit 4.5000000 102 ft 1 7.5000000 10 ft 4.5 102 ft by 7.5 101ft 75 ft Copyright © Houghton Mifflin Company. All rights reserved. 1|9
    • 1 cubit 1.5 ft 3 ft 1 yd 1 yd 0.9144 m (exact) 2 1 4.5 10 ft by 7.5 10 ft 75 ft 1 yd 0.9144 m 1 yd 0.9144 m 4.5 10 ft 2 75 ft 3 ft 1 yd 3 ft 1 yd 1.37160000 102 m 22.8600000 m 1.4 102 m by 23 m Copyright © Houghton Mifflin Company. All rights reserved. 1 | 10
    • Chapter 2 Atoms, Molecules, and Ions Copyright © Houghton Mifflin Company. All rights reserved. 2 | 11
    • Write the nuclide symbol for the atom that has 19 protons and 20 neutrons. Atomic number: Z = 19 The element is potassium, K. Mass number: A = 19 + 20 = 39 The nuclide symbol is 39 19 K Copyright © Houghton Mifflin Company. All rights reserved. 2 | 12
    • An element has four naturally occurring isotopes. The mass and percentage of each isotope are as follows: Percentage Abundance Mass (amu) 1.48 203.973 23.6 205.9745 22.6 206.9759 52.3 207.9766 What is the atomic weight and name of the element? Copyright © Houghton Mifflin Company. All rights reserved. 2 | 13
    • To find the portion of the atomic weight due to each isotope, multiply the fraction by the mass of the isotope. The atomic weight is the sum of these products. Fractional Mass (amu) Mass From Abundance Isotope 0.0148 203.973 3.01880040 0.236 205.9745 48.6099820 0.226 206.9759 46.7765534 0.523 207.9766 108.771762 Total = 207.177098 The atomic weight is 207 amu; the element is lead. Copyright © Houghton Mifflin Company. All rights reserved. 2 | 14
    • What is formula of the ionic compound of Mg2+ and N3-? The common multiple of the charges is 6, so we need three Mg2+ and two N3-. The resulting formula is Mg3N2 Copyright © Houghton Mifflin Company. All rights reserved. 2 | 15
    • Common Monatomic Ions of the Main-Group Elements Period IA IIA IIIA IVA VA VIA VIIA 1 H- 2 Li+ Be2+ N3- O2- F- 3 Na+ Mg2+ Al3+ S2- Cl- 4 K+ Ca2+ Ga3+ Se2- Br- 5 Rb+ Sr2+ In3+ Sn2+ Te2- I- 6 Cs+ Ba2+ Tl3+, Pb2+ Bi3+ Tl+ Copyright © Houghton Mifflin Company. All rights reserved. 2 | 16
    • What are the names of the following ionic compounds? – BaO – Cr2(SO4)3 BaO is barium oxide. Cr2(SO4)3 is chromium(III) sulfate or chromic sulfate. Copyright © Houghton Mifflin Company. All rights reserved. 2 | 17
    • What are the chemical formulas for the following ionic compounds? – potassium carbonate – manganese(II) sulfate The ions K+ and CO32- form K2CO3 The ions Mn2+ and SO42- form MnSO4 Copyright © Houghton Mifflin Company. All rights reserved. 2 | 18
    • What are the names of the following compounds? – OF2 – S4N4 – BCl3 OF2 is oxygen difluoride S4N4 is tetrasulfur tetranitride BCl3 is boron trichloride Copyright © Houghton Mifflin Company. All rights reserved. 2 | 19
    • What are the formulas for the following binary molecular compounds? – carbon disulfide – nitrogen tribromide – dinitrogen tetrafluoride The formula for carbon disulfide is CS2. The formula for nitrogen tribromide is NBr3. The formula for dinitrogen tetrafluoride is N2F4. Copyright © Houghton Mifflin Company. All rights reserved. 2 | 20
    • A compound whose common name is green vitriol has the chemical formula FeSO4·7H2O. What is the chemical name of this compound? FeSO4·7H2O is iron(II) sulfate heptahydrate. Copyright © Houghton Mifflin Company. All rights reserved. 2 | 21
    • Calcium chloride hexahydrate is used to melt snow on roads. What is the chemical formula of the compound? The chemical formula for calcium chloride hexahydrate is CaCl2·6H2O. Copyright © Houghton Mifflin Company. All rights reserved. 2 | 22
    • Balance the following equations: NH3 + O2  NO + H2O C2H5OH + O2  CO2 + H2O 4NH3 + 5O2  4NO + 6H2O C2H5OH + 3O2  2CO2 + 3H2O Copyright © Houghton Mifflin Company. All rights reserved. 2 | 23
    • Chapter 3
    • Calculate the formula weight of the following compounds from their formulas. Report your answers to three significant figures. – calcium hydroxide, Ca(OH)2 – methylamine, CH3NH2 Copyright © Houghton Mifflin Company. All rights reserved. 3 | 25
    • Calculate the formula weight of the following compounds from their formulas. calcium hydroxide, Ca(OH)2 Ca(OH)2 methylamine, CH3NH2 1 Ca 1(40.08) = 40.08 amu 2O 2(16.00) = 32.00 amu 2H 2(1.008) = 2.016 amu Total 74.095 3 significant figures 74.1 amu CH3NH2 1C 1(12.01) = 12.01 amu 1N 1(14.01) = 14.01 amu 5H 5(1.008) = 5.040 amu Total 31.060 3 significant figures 31.1 amu Copyright © Houghton Mifflin Company. All rights reserved. 3 | 26
    • AFirst, find the molar mass of ,HNO3: 0.253 mol sample of nitric acid, HNO3 contains HNO3. How many grams is this? 1 H 1(1.008) = 1.008 1 N 1(14.01) = 14.01 3 O 3(16.00) = 48.00 63.018 (2 decimal places) 63.02 g/mol Copyright © Houghton Mifflin Company. All rights reserved. 3 | 27
    • Next, using the molar mass, find the mass of 0.253 mole: 63.02 g 0.253 mole x 1 mole = 15.94406 g 15.9 g (3 significan figures) t Copyright © Houghton Mifflin Company. All rights reserved. 3 | 28
    • The average daily requirement of the essential amino acid leucine, C6H14O2N, is 2.2 g for an adult. What is the average daily requirement of leucine in moles? First, find the molar mass of leucine: 6C 6(12.01) = 72.06 2O 2(16.00) = 32.00 1N 1(14.01) = 14.01 14 H 14(1.008) = 14.112 2 decimal places 132.182 132.18 g/mol Copyright © Houghton Mifflin Company. All rights reserved. 3 | 29
    • Next, find the number of moles in 2.2 g: 1 mole 2.2 g x 132.18g 2 1.6643x 10 mol 2 1.7 x 10 mol or 0.017 mol (2 significan figures) t Copyright © Houghton Mifflin Company. All rights reserved. 3 | 30
    • The daily requirement of chromium in the human diet is 1.0 10-6 g. How many atoms of chromium does this represent? First, find the molar mass of Cr: 1 Cr 1(51.996) = 51.996 g/mol Now, convert 1.0 x 10-6 grams to moles: 23 6 1 mol 6.02 x 10 atoms 1.0 x 10 gx x 51.996g 1 mol =1.157781368 x 1016 atoms 1.2 x 1016 atoms (2 significant figures) Copyright © Houghton Mifflin Company. All rights reserved. 3 | 31
    • Lead(II) chromate, PbCrO4, is used as a paint pigment (chrome yellow). What is the percentage composition of lead(II) chromate? First, find the molar mass of PbCrO4: 1 Pb 1(207.2) = 207.2 1 Cr 1(51.996) = 51.996 4 O 4(16.00) = 64.00 323.196 (1 decimal place) 323.2 g/mol Copyright © Houghton Mifflin Company. All rights reserved. 3 | 32
    • Now, convert each to percent composition: 207.2 g Pb : x 100% 64.11% 323.20g 51.996g Cr : x 100% 16.09% 323.20g 64.00 g O: x 100% 19.80% 323.20g Check: 64.11 + 16.09 + 19.80 = 100.00 % Copyright © Houghton Mifflin Company. All rights reserved. 3 | 33
    • Benzene has the empirical formula CH. Its molecular weight is 78.1 amu. What is its molecular formula? Empirical formula w eight (12.01 1.008) amu 13.02 amu 78.1 6 13.02 Molecular formula C6H6 Copyright © Houghton Mifflin Company. All rights reserved. 3 | 34
    • Hexamethylene is one of the materials used to produce a type of nylon. It is composed of 62.1% C, 13.8% H, and 24.1% N. Its molecular weight is 116 amu. What is its molecular formula? 1 mol C 5.171 62.1 g C X 5.171 mol C 3 12.01 g C 1.720 1 mol H 13.69 13.8 g H X 13.69 mol H 8 1.008 g H 1.720 1 mol N 1.720 24.1 g N X 1.720 mol N 1 14.01 g N 1.720 Empirical formula C3H8N Copyright © Houghton Mifflin Company. All rights reserved. 3 | 35
    • The empirical formula is C3H8N. Molecular formula = (Empirical formula)n Find the empirical formula weight: 3(12.01) + 8(1.008) + 1(14.01) = 58.104 amu 116 n 2 Molecular formula: C H N 58.10 6 16 2 Copyright © Houghton Mifflin Company. All rights reserved. 3 | 36
    • Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) How many grams of CO2 are produced when 20.0 g of propane is burned? Copyright © Houghton Mifflin Company. All rights reserved. 3 | 37
    • C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) Molar masses C3H8: 3(12.01) + 8(1.008) = 44.094 g CO2: 1(12.01) + 2(16.00) = 44.01 g 1 mol C3H8 3 mol CO 2 44.01 g CO 2 20.0 g C3H8 X X X 44.094 g C3H8 1 mol C3H8 1 mol CO 2 59.8856987 g CO 2 3 59.9 g CO2 (3 significant figures) Copyright © Houghton Mifflin Company. All rights reserved. 3 | 38
    • Urea, CH4N2O, is used as a nitrogen fertilizer. It is manufactured from ammonia and carbon dioxide at high pressure and high temperature: 2NH3 + CO2(g)  CH4N2O + H2O In a laboratory experiment, 10.0 g NH3 and 10.0 g CO2 were added to a reaction vessel. What is the maximum quantity (in grams) of urea that can be obtained? How many grams of the excess reactant are left at the end of the reactions? Copyright © Houghton Mifflin Company. All rights reserved. 3 | 39
    • 2NH3 + CO2(g)  CH4N2O + H2O Molar masses: NH3 1(14.01) + 3(1.008) = 17.02 g CO2 1(12.01) + 2(16.00) = 44.01 g CH4N2O 1(12.01) + 4(1.008) + 2(14.01) + 1(16.00) = 60.06 g 1 mol NH 3 1 mol CH 4N2 O 60.06 g CH 4N2 O 10.0 g NH 3 X X X 17.02 g NH 3 2 mol NH 3 1 mol CH 4N2O 17.6 g CH 4N2 O 1 mol CO 2 1 mol CH 4N2 O 60.06 g CH 4N2 O 10.0 g CO 2 X X X 44.01 g CO 2 1 mol CO 2 1 mol CH 4N2 O 13.6 g CH 4N2O CO2 is the limiting reactant. 13.6 g CH4N2O will be produced. Copyright © Houghton Mifflin Company. All rights reserved. 3 | 40
    • 2NH3 + CO2(g)  CH4N2O + H2O To find the excess NH3, we find how much NH3 reacted: 1 mol CO 2 2 mol NH 3 17.02 g NH 3 10.0 g CO 2 X X X 44.01 g CO 2 1 mol CO 2 1 mol NH 3 7.73460577g NH3 7.73 g NH3 reacted Now subtract the amount reacted from the starting amount: 10.0 g at start -7.73 g reacted 2.27 g remains 2.3 g NH3 is left unreacted. (1 decimal place) Copyright © Houghton Mifflin Company. All rights reserved. 3 | 41
    • 2NH3 + CO2(g)  CH4N2O + H2O When 10.0 g NH3 and 10.0 g CO2 are added to a reaction vessel, the limiting reactant is CO2. The theoretical yield is 13.6 g of urea. When this reaction was carried out, 9.3 g of urea was obtained. What is the percent yield? Theoretical yield = 13.6 g 9.3 g x 100% Actual yield = 9.3 g 13.6 g = 68% yield (2 significant figures) Copyright © Houghton Mifflin Company. All rights reserved. 3 | 42
    • Chapter 4 Chemical Reactions
    • Determine whether each of the following reactions occurs. If it does, write the molecular, ionic, and net ionic equations. KBr + MgSO4  NaOH + MgCl2  K3PO4 + CaCl2  Copyright © Houghton Mifflin Company. All rights reserved. 4 | 44
    • Classify the following as strong or weak acids or bases: a. KOH b. H2S c. CH3NH2 d. HClO4 Copyright © Houghton Mifflin Company. All rights reserved. 4 | 45
    • Write the molecular, ionic, and net ionic equations for the neutralization of sulfurous acid, H2SO3, by potassium hydroxide, KOH. Copyright © Houghton Mifflin Company. All rights reserved. 4 | 46
    • Write the molecular, ionic, and net ionic equations for the reaction of copper(II) carbonate with hydrochloric acid. Copyright © Houghton Mifflin Company. All rights reserved. 4 | 47
    • Potassium permanganate, KMnO4, is a purple-colored compound; potassium manganate, K2MnO4, is a green- colored compound. Obtain the oxidation numbers of the manganese in these compounds. Copyright © Houghton Mifflin Company. All rights reserved. 4 | 48
    • What is the oxidation number of Cr in dichromate, Cr2O72-? Copyright © Houghton Mifflin Company. All rights reserved. 4 | 49
    • Balance the following oxidation-reduction reaction: FeI3(aq) + Mg(s)  Fe(s) + MgI2(aq) Copyright © Houghton Mifflin Company. All rights reserved. 4 | 50
    • You place a 1.52 g of potassium dichromate, K2Cr2O7, into a 50.0 mL volumetric flask. You then add water to bring the solution up to the mark on the neck of the flask. What is the molarity of K2Cr2O7 in the solution? Molar mass of K2Cr2O7 is 294 g. 1 mol 1.52 g 294 g 0.103 M 50.0 x 10-3 L Copyright © Houghton Mifflin Company. All rights reserved. 4 | 51
    • A solution of sodium chloride used for intravenous transfusion (physiological saline solution) has a concentration of 0.154 M NaCl. How many moles of NaCl are contained in 500. mL of physiological saline? How many grams of NaCl are in the 500. mL of solution? mol M V M olarm ass NaCl 58.4 g 0.154 M 0.500 L 58.4 g 0.0770m ol 0.0770 mol NaCl 1 m ol 4.50 g NaCl Copyright © Houghton Mifflin Company. All rights reserved. 4 | 52
    • A saturated stock solution of NaCl is 6.00 M. How much of this stock solution is needed to prepare 1.00 L of physiological saline solution (0.154 M)? M iVi M fVf (0.154 M )(1.00 L) Vi M fVf 6.00 M Vi Mi Vi 0.0257L or 25.7 mL Copyright © Houghton Mifflin Company. All rights reserved. 4 | 53
    • A soluble silver compound was analyzed for the percentage of silver by adding sodium chloride solution to precipitate the silver ion as silver chloride. If 1.583 g of silver compound gave 1.788 g of silver chloride, what is the mass percent of silver in the compound? Copyright © Houghton Mifflin Company. All rights reserved. 4 | 54
    • Molar mass of silver chloride (AgCl) = 143.32 g 1 mol AgCl 1 mol Ag 107.9 g Ag 1.788 g AgCl x x x 143.32g AgCl 1 mol AgCl 1 mol Ag = 1.346 g Ag in the compound 1.346 g Ag X 100% 1.583 g silver compound = 85.03% Ag Copyright © Houghton Mifflin Company. All rights reserved. 4 | 55
    • Zinc sulfide reacts with hydrochloric acid to produce hydrogen sulfide gas: ZnS(s) + 2HCl(aq) ZnCl2(aq) + H2S(g) How many milliliters of 0.0512 M HCl are required to react with 0.392 g ZnS? Copyright © Houghton Mifflin Company. All rights reserved. 4 | 56
    • ZnS(s) + 2HCl(aq) ZnCl2(aq) + H2S(g) Molar mass of ZnS = 97.47 g 1 mol ZnS 2 mol HCl 1 L solution 0.392 g ZnS x x x 97.47 g ZnS 1 mol ZnS 0.0512mol HCl = 0.157 L = 157 mL HCl solution Copyright © Houghton Mifflin Company. All rights reserved. 4 | 57
    • A dilute solution of hydrogen peroxide is sold in drugstores as a mild antiseptic. A typical solution was analyzed for the percentage of hydrogen peroxide by titrating it with potassium permanganate: 5H2O2(aq) + 2KMnO4(aq) + 6H+(aq)  8H2O(l) + 5O2(g) + 2K+(aq) + 2Mn2+(aq) What is the mass percent of H2O2 in a solution if 57.5 g of solution required 38.9 mL of 0.534 M KMnO4 for its titration? Copyright © Houghton Mifflin Company. All rights reserved. 4 | 58
    • 5H2O2(aq) + 2KMnO4(aq) + 6H+(aq)  8H2O(l) + 5O2(g) + 2K+(aq) + 2Mn2+(aq) Molar mass of H2O2 = 34.01 g 0.534 mol KMnO4 3 5 mol H2 O 2 34.01g H2 O 2 38.9 x 10 L x x x 1L 2 mol KMnO4 1 mol H2 O 2 = 1.77 g H2O2 1.77 g H2O 2 X 100% 57.5 g solution = 3.07% H2O2 Copyright © Houghton Mifflin Company. All rights reserved. 4 | 59
    • Chapter 5
    • A volume of oxygen gas occupies 38.7 mL at 751 mmHg and 21 C. What is the volume if the pressure changes to 359 mmHg while the temperature remains constant? PiVi Vf Pf (38.7 mL)(751mmHg) Vf (359 mmHg) = 81.0 mL (3 significant figures) Copyright © Houghton Mifflin Company. All rights reserved. 5 | 61
    • You prepared carbon dioxide by adding HCl(aq) to marble chips, CaCO3. According to your calculations, you should obtain 79.4 mL of CO2 at 0 C and 760 mmHg. How many milliliters of gas would you obtain at 27 C? Copyright © Houghton Mifflin Company. All rights reserved. 5 | 62
    • Vi = 79.4 mL Vf = ? Pi = 760 mmHg Pf = 760 mmHg Ti = 0°C = 273 K Tf = 27°C = 300. K TfVi Vf Ti (300. K)(79.4 mL) Vf (273 K) = 87.3 mL (3 significant figures) Copyright © Houghton Mifflin Company. All rights reserved. 5 | 63
    • Divers working from a North Sea drilling platform experience pressure of 5.0 101 atm at a depth of 5.0 102 m. If a balloon is inflated to a volume of 5.0 L (the volume of the lung) at that depth at a water temperature of 4 C, what would the volume of the balloon be on the surface (1.0 atm pressure) at a temperature of 11 C? Copyright © Houghton Mifflin Company. All rights reserved. 5 | 64
    • Vi = 5.0 L Vf = ? Pi = 5.0 101 atm Pf = 1.0 atm Ti = 4°C = 277 K Tf = 11°C = 284. K Tf PiVi Vf Ti Pf (284 K)(5.0 x 101 atm)(5.0L) Vf (277 K)(1.0 atm) = 2.6 x 102 L (2 significant figures) Copyright © Houghton Mifflin Company. All rights reserved. 5 | 65
    • Ideal Gas Law The ideal gas law is given by the equation PV=nRT The molar gas constant, R, is the constant of proportionality that relates the molar volume of a gas to T/P. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 66
    • A 50.0 L cylinder of nitrogen, N2, has a pressure of 17.1 atm at 23 C. What is the mass of nitrogen in the cylinder? V = 50.0 L PV P = 17.1 atm n RT T = 23°C = 296 K (17.1atm)(50.0 L) n 35.20m ol L atm 0.08206 (296 K) mol K 28.01 g mass 35.20 mol X mass = 986 g mol (3 significant figures) Copyright © Houghton Mifflin Company. All rights reserved. 5 | 67
    • Mm = 16.04 g/mol Mm P d P = 3.50 atm What is the density of methaneRT (natural gas), gas T = 125°C = 398 K CH4, at 125 C and 3.50 atm? g (16.04 )(3.50 atm) g mol d 1.72 d L L atm 0.08206 (398 K) (3 significan figures) t mol K Copyright © Houghton Mifflin Company. All rights reserved. 5 | 68
    • A 500.0-mL flask containing a sample of octane (a component of gasoline) is placed in a boiling water bath in Denver, where the atmospheric pressure is 634 mmHg and water boils at 95.0 C. The mass of the vapor required to fill the flask is 1.57 g. What is the molar mass of octane? (Note: The empirical formula of octane is C4H9.) What is the molecular formula of octane? Copyright © Houghton Mifflin Company. All rights reserved. 5 | 69
    • d = 1.57 g/0.5000 L dRT = 3.140 g/L Mm P P = 634 mmHg = 0.8342 atm T = 95.0°C = 368.2 K g L atm 3.140 0.08206 368.2K L mol K Mm (0.8342atm) g M m 114 mol (3 significan figures) t Copyright © Houghton Mifflin Company. All rights reserved. 5 | 70
    • Molar mass = 114 g/mol Empirical formula: C4H9 Empirical formula molar mass = 57 g/mol g 114 n mol 2 g 57 mol Molecular formula: C8H18 Copyright © Houghton Mifflin Company. All rights reserved. 5 | 71
    • When a 2.0-L bottle of concentrated HCl was spilled, 1.2 kg of CaCO3 was required to neutralize the spill. What volume of CO2 was released by the neutralization at 735 mmHg and 20. C? Copyright © Houghton Mifflin Company. All rights reserved. 5 | 72
    • First, write the balanced chemical equation: CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g) Second, calculate the moles of CO2 produced: Molar mass of CaCO3 = 100.09 g/mol 1 mol CaCO 3 3 1 mol CO 2 1.2 x 10 g CaCO 3 X X 100.09 g CaCO 3 1 mol CaCO 3 Moles of CO2 produced = 12 mol Copyright © Houghton Mifflin Company. All rights reserved. 5 | 73
    • nRT n = 12 mol V P = 735 mmHg P = 0.967 atm T = 20°C = 293 K L atm 12 mol 0.08206 (293 K) mol K V (0.967 atm) = 3.0 102 L (2 significant figures) Copyright © Houghton Mifflin Company. All rights reserved. 5 | 74
    • A 100.0-mL sample of air exhaled from the lungs is analyzed and found to contain 0.0830 g N2, 0.0194 g O2, 0.00640 g CO2, and 0.00441 g water vapor at 35 C. What is the partial pressure of each component and the total pressure of the sample? Copyright © Houghton Mifflin Company. All rights reserved. 5 | 75
    • 1 mol N2 L atm 0.0830 g N2 0.08206 308 K 28.01 g N2 mol K PN 0.749 atm 2 1L 100.0 mL 10 3 mL 1 mol O 2 L atm 0.0194 g O 2 0.08206 308 K 32.00 g O 2 mol K PO 0.153 atm 2 1L 100.0 mL 10 3 mL 1 mol CO 2 L atm 0.00640 g CO 2 0.08206 308 K 44.01 g CO 2 mol K PCO 0.0368atm 2 1L 100.0 mL 10 3 mL 1 mol H2 O L atm 0.00441 g H2 O 0.08206 308 K 18.01 g H2 O mol K PH O 0.0619atm 2 1L 100.0 mL 10 3 mL Copyright © Houghton Mifflin Company. All rights reserved. 5 | 76
    • PN2 0.749atm PO2 0.153atm PCO 2 0.0368atm PH2O 0.0619atm P PN2 PO2 PCO 2 PH2O P = 1.00 atm Copyright © Houghton Mifflin Company. All rights reserved. 5 | 77
    • The partial pressure of air in the alveoli (the air sacs in the lungs) is as follows: nitrogen, 570.0 mmHg; oxygen, 103.0 mmHg; carbon dioxide, 40.0 mmHg; and water vapor, 47.0 mmHg. What is the mole fraction of each component of the alveolar air? P PN2 PO2 PCO 2 PH2O 570.0 mmHg 103.0 mmHg 40.0 mmHg 47.0 mmHg P = 760.0 mmHg Copyright © Houghton Mifflin Company. All rights reserved. 5 | 78
    • Mole fraction of N2 Mole fraction of O2 570.0 mmHg 103.0 mmHg 760.0 mmHg 760.0 mmHg Mole fraction of CO2 Mole fraction of H2O 40.0 mmHg 47.0 mmHg 760.0 mmHg 760.0 mmHg Mole fraction N2 = 0.7500 Mole fraction O2 = 0.1355 Mole fraction CO2 = 0.0526 Mole fraction H2O= 0.0618 Copyright © Houghton Mifflin Company. All rights reserved. 5 | 79
    • You prepare nitrogen gas by heating ammonium nitrite: NH4NO2(s)  N2(g) + 2H2O(l) If you collected the nitrogen over water at 23 C and 727 mmHg, how many liters of gas would you obtain from 5.68 g NH4NO2? Copyright © Houghton Mifflin Company. All rights reserved. 5 | 80
    • P = 727 mmHg Molar mass NH4NO2 Pvapor = 21.1 mmHg = 64.04 g/mol Pgas = 706 mmHg nRT T = 23°C = 296 K V P 1 mol NH4NO 2 1 mol N2 5.68 g NH4NO2 X X 64.04 g NH4NO 2 1 mol NH4NO 2 = 0.08869 mol N2 gas Copyright © Houghton Mifflin Company. All rights reserved. 5 | 81
    • P = 727 mmHg Pvapor = 21.1 mmHg Pgas = 706 mmHg nRT V T = 23°C = 296 K P n = 0.08869 mol L atm 0.0887 mol 0.08206 (296 K) mol K V 1 atm 706 mmHg x 760 mmHg = 2.32 L of N2 (3 significant figures) Copyright © Houghton Mifflin Company. All rights reserved. 5 | 82
    • What is the rms speed of carbon dioxide molecules in a container at 23 C? 3RT kg m 2 u rms s2 Mm 3 8.3145 296 K mol K Recall 2 u rms kg m kg J 0.04401 s2 mol 5 m2 2 m urms 1.68x10 2 urms 4.10 x10 s s Copyright © Houghton Mifflin Company. All rights reserved. 5 | 83
    • Both hydrogen and helium have been used as the buoyant gas in blimps. If a small leak were to occur, which gas would effuse more rapidly and by what factor? 1 Rate H2 2.016 4.002 Rate He 1 2.016 4.002 Hydrogen will diffuse more quickly by a factor of 1.4. Copyright © Houghton Mifflin Company. All rights reserved. 5 | 84
    • Chapter 6
    • A person weighing 75.0 kg (165 lbs) runs a course at 1.78 m/s (4.00 mph). What is the person’s kinetic energy? m = 75.0 kg EK = ½ mv2 V = 1.78 m/s 2 1 m EK (75.0 kg) 1.78 2 s kg m 2 E K 119 2 119 J s (3 significan figures) t Copyright © Houghton Mifflin Company. All rights reserved. 6 | 86
    • In an endothermic reaction: The reaction vessel cools. Heat is absorbed. Energy is added to the system. q is positive. In an exothermic reaction: The reaction vessel warms. Heat is evolved. Energy is subtracted from the system. q is negative. Copyright © Houghton Mifflin Company. All rights reserved. 6 | 87
    • Sulfur, S8, burns in air to produce sulfur dioxide. The reaction evolves 9.31 kJ of heat per gram of sulfur at constant pressure. Write the thermochemical equation for this reaction. S8(s) + 8O2(g)  8SO2(g) 9.31 kJ 256.52 g S 8 ΔH 1 g S8 1 mol S 8 ΔH 2.39 10 3 kJ S8(s) + 8O2(g)  8SO2(g); H = –2.39 103 kJ Copyright © Houghton Mifflin Company. All rights reserved. 6 | 88
    • You burn 15.0 g sulfur in air. How much heat evolves from this amount of sulfur? The thermochemical equation is S8(s) + 8O2(g)  8SO2(g); H = -2.39 x 103 kJ Molar mass of S8 = 256.52 g 1 mol S 8 2.39 x 103 kJ q 15.0 g S 8 256.5 g S 8 1 mol S 8 q = –1.40 102 kJ Copyright © Houghton Mifflin Company. All rights reserved. 6 | 89
    • 8.8 x 103 kJ. If all this energy were to be provided by the combustion of glucose, C6H12O6, how many grams of glucose would have to be consumed by the man and the woman per day? C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l); Ho = -2.82 x 103 kJ The daily energy requirement for a 20-year-old man weighing 67 kg is 1.3 x 104 kJ. For a 20-year- old woman weighing 58 kg, the daily requirement is Copyright © Houghton Mifflin Company. All rights reserved. 6 | 90
    • C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l); H = -2.82 x 103 kJ For a 20-year-old man weighing 67 kg: 1 mol glucose 180.2g glucose 4 mglucose 1.3x10 kJ 3 2.82x10 kJ 1 mol glucose = 830 g glucose required (2 significant figures) For a 20-year-old woman weighing 58 kg: 1 mol glucose 180.2 g glucose 3 mglucose 8.8x10 kJ 3 2.82x10 kJ 1 mol glucose = 560 g glucose required (2 significant figures) Copyright © Houghton Mifflin Company. All rights reserved. 6 | 91
    • A piece of zinc weighing 35.8 g was heated from 20.00 C to 28.00 C. How much heat was required? The specific heat of zinc is 0.388 J/(g C). m = 35.8 g s = 0.388 J/(g C) q=m·s· t t = 28.00 C – 20.00 C = 8.00 C 0.388 J q 35.8 g 8.00 C gC q = 111 J (3 significant figures) Copyright © Houghton Mifflin Company. All rights reserved. 6 | 92
    • CH3NO2(g) + 3/4O2(g)  CO2(g) organic O(l) + 1burns Nitromethane, CH3NO2, an + 3/2H2 solvent /2N2(g) You placeaccording of the following reaction: in oxygen 1.724 g to nitromethane in a calorimeter with oxygen and ignite it. The temperature of the calorimeter increases from 22.23 C to 28.81°C. The heat capacity of the calorimeter was determined to be 3.044 kJ/°C. Write the thermochemical equation for the reaction. Copyright © Houghton Mifflin Company. All rights reserved. 6 | 93
    • We first find the heat evolved for the 1.724 g of nitromethane, CH3NO2. q rxn C cal Δt 3.044 kJ q rxn 28.81 C 22.23 C 20.03kJ C Now, covert that to the heat evolved per mole by using the molar mass of nitromethane, 61.04 g. - 20.03kJ 61.04 g CH3NO 2 q rxn 1.724 g CH3NO 2 1 mol CH3NO 2 H = –709 kJ CH3NO2(l) + ¾O2(g)  CO2(g) + 3/2H2O(l) + ½N2(g); H = –709 kJ Copyright © Houghton Mifflin Company. All rights reserved. 6 | 94
    • What is the enthalpy of reaction, H, for the reaction of calcium metal with water? Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g) This reaction occurs very slowly, so it is impractical to measure H directly. However, the following facts are known: H+(aq) + OH-(aq)  H2O(l); H = –55.9 kJ Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g); H = –543.0 kJ Copyright © Houghton Mifflin Company. All rights reserved. 6 | 95
    • Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g) 2H2O(l)  2H+(aq) + 2OH-(aq); H = +111.8 kJ Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g); H = –543.0 kJ Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g); H = –431.2 kJ Copyright © Houghton Mifflin Company. All rights reserved. 6 | 96
    • What is the heat of vaporization of methanol, CH3OH, at 25 C and 1 atm? Use standard enthalpies of formation (Appendix C). We want H for the reaction:CH3OH(l)  CH3OH(g) ΔHreaction nΔH f nΔH f products reactants kJ For liquidmethanol: ΔH f 238.7 mol kJ For gaseous methanol: ΔH f 200.7 mol kJ kJ ΔH vap 1mol 200.7 1mol 238.7 = +38.0 kJ mol mol Copyright © Houghton Mifflin Company. All rights reserved. 6 | 97
    • Methyl alcohol, CH3OH, is toxic because liver enzymes oxidize it to formaldehyde, HCHO, which can coagulate protein. Calculate Ho for the following reaction: 2CH3OH(aq) + O2(g)  2HCHO(aq) + 2H2O(l) Standard enthalpies of formation, ΔH fo: CH3OH(aq): -245.9 kJ/mol HCHO(aq): -150.2 kJ/mol H2O(l): -285.8 kJ/mol Copyright © Houghton Mifflin Company. All rights reserved. 6 | 98
    • We want H for the reaction: 2CH3OH(aq) + O2(aq)  2HCHO(aq) + 2H2O(l) ΔHreaction nΔH f nΔH f products reactants kJ kJ ΔH reacton o 2 mol 150.2 2 mol - 285.8 mol mol kJ kJ 2 mol 245.9 1mol 0 mol mol o ΔHreaction 300.4kJ 571.6kJ 491.8kJ o ΔHreaction 872.0kJ 491.8kJ o ΔHreaction 380.2kJ Copyright © Houghton Mifflin Company. All rights reserved. 6 | 99
    • Chapter 7 Quantum Theory of An Atome
    • What is the wavelength of blue light with a frequency of 6.4 1014/s? = 6.4 1014/s c= so c = 3.00 108 m/s = c/ 8 m 3.00 x 10 c s λ 14 1 6.4 x 10 s = 4.7 10-7 m Copyright © Houghton Mifflin Company. All rights reserved. 7 | 101
    • What is the frequency of light having a wavelength of 681 nm? = 681 nm = 6.81 10-7 m c= so c = 3.00 108 m/s = c/ 8 m 3.00 x 10 c s 6.81x 10 7 m v = 4.41 1014 /s Copyright © Houghton Mifflin Company. All rights reserved. 7 | 102
    • The range of frequencies and wavelengths of electromagnetic radiation is called the electromagnetic spectrum. Copyright © Houghton Mifflin Company. All rights reserved. 7 | 103
    • The blue–green line of the hydrogen atom spectrum has a wavelength of 486 nm. What is the energy of a photon of this light? = 486 nm = 4.86 10-7 m E = h and c = 3.00 108 m/s c= so h = 6.63 10-34 J · s E = hc/ 34 8 m 6.63x10 Js 3.00 x 10 hc s E λ 4.86 x 10 7 m E = 4.09 10-19 J Copyright © Houghton Mifflin Company. All rights reserved. 7 | 104
    • What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes a transition from n = 6 to n = 3? 1 1 ΔE RH ni = 6 nf 2 ni 2 nf = 3 hc hc RH = 2.179 10-18 J ΔE so λ λ ΔE 18 1 1 ΔE 2.179 x 10 J 2 = -1.816 x 10-19 J 3 62 34 m 8 6.626 x 10 J s 2.998 x 10 s λ 19 1.094 10-6 m - 1.816 x 10 J Copyright © Houghton Mifflin Company. All rights reserved. 7 | 105
    • Compare the wavelengths of (a) an electron traveling at a speed that is one- hundredth the speed of light and (b) a baseball of mass 0.145 kg having a speed of 26.8 m/s (60 mph). Copyright © Houghton Mifflin Company. All rights reserved. 7 | 106
    • Electron me = 9.11 10-31 kg; v = 3.00 106 m/s 34 6.63 x 10 J s λ 2.43 10-10 m 31 m 6 9.11x 10 kg 3.00 x 10 s Baseball m = 0.145 kg; v = 26.8 m/s 34 6.63 x 10J s λ 1.71 10-34 m m 0.145 kg 26.8 Comment: This is such an s exceedingly small wavelength that the wave properties of a baseball cannot be detected by any existing measuring device. Copyright © Houghton Mifflin Company. All rights reserved. 7 | 107
    • Which of the following are permissible sets of quantum numbers? n = 4, l = 4, ml = 0, ms = ½ n = 3, l = 2, ml = 1, ms = -½ n = 2, l = 0, ml = 0, ms = ³/² n = 5, l = 3, ml = -3, ms = ½ (a) Not permitted. When n = 4, the maximum value of l is 3. (b) Permitted. (c) Not permitted; ms can only be +½ or –½. (b) Permitted. Copyright © Houghton Mifflin Company. All rights reserved. 7 | 108
    • Chapter 8 Electron Configuration and Periodicity
    • The building-up principle (or aufbau principle) is a scheme used to reproduce the ground-state electron configurations by successively filling subshells with electrons in a specific order (the building-up order). This order generally corresponds to filling the orbitals from lowest to highest energy. Note that these energies are the total energy of the atom rather than the energy of the subshells alone. Copyright © Houghton Mifflin Company. All rights reserved. 8 | 110
    • 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p This results in the following order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p Copyright © Houghton Mifflin Company. All rights reserved. 8 | 111
    • Another way to learn the building-up order is to correlate each subshell with a position on the periodic table. The principal quantum number, n, correlates with the period number. Groups IA and IIA correspond to the s subshell; Groups IIIA through VIIIA correspond to the p subshell; the ―B‖ groups correspond to the d subshell; and the bottom two rows correspond to the f subshell. This is shown on the next slide. Copyright © Houghton Mifflin Company. All rights reserved. 8 | 112
    • Copyright © Houghton Mifflin Company. All rights reserved. 8 | 113
    • Write the complete electron configuration of the arsenic atom, As, using the building-up principle. For arsenic, As, Z = 33. 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 Copyright © Houghton Mifflin Company. All rights reserved. 8 | 114
    • What are the electron configurations for the valence electrons of arsenic and zinc? Arsenic is in period 4, Group VA. Its valence configuration is 4s24p3. Zinc, Z = 30, is a transition metal in the first transition series. Its noble-gas core is Ar, Z = 18. Its valence configuration is 4s23d10. Copyright © Houghton Mifflin Company. All rights reserved. 8 | 115
    • Write an orbital diagram for the ground state of the nickel atom. For nickel, Z = 28. 1s 2s 2p 3s 3p 4s 3d Copyright © Houghton Mifflin Company. All rights reserved. 8 | 116
    • Which of the following electron configurations or orbital diagrams are allowed and which are not allowed by the Pauli exclusion principle? If they are not allowed, explain why? a. 1s22s12p3 a. Allowed; excited. b. 1s22s12p8 b. p8 is not allowed. c. 1s22s22p63s23p63d8 c. Allowed. d. 1s22s22p63s23p63d11 d. d11 is not allowed. e. Not allowed; electrons in one e. orbital must have opposite spins. 1s 2s Copyright © Houghton Mifflin Company. All rights reserved. 8 | 117
    • A representation of atomic radii is shown below. Copyright © Houghton Mifflin Company. All rights reserved. 8 | 118
    • Refer to a periodic table and arrange the following elements in order of increasing atomic radius: Br, Se, Te. Te is larger than Se. Se is larger than Br. Br < Se < Te Copyright © Houghton Mifflin Company. All rights reserved. 8 | 119
    • The size of each sphere indicates the size of the ionization energy in the figure below. Copyright © Houghton Mifflin Company. All rights reserved. 8 | 120
    • Copyright © Houghton Mifflin Company. All rights reserved. 8 | 121
    • Refer to a periodic table and arrange the following elements in order of increasing ionization energy: As, Br, Sb. Sb is larger than As. As is larger than Br. Ionization energies: Sb < As < Br Copyright © Houghton Mifflin Company. All rights reserved. 8 | 122
    • Copyright © Houghton Mifflin Company. All rights reserved. 8 | 123
    • The electron affinity is > 0, so the element must be in Group IIA or VIIIA. The dramatic difference in ionization energies is at the third ionization. The element is in Group IIA. Copyright © Houghton Mifflin Company. All rights reserved. 8 | 124
    • For R2O5 oxides, R must be in Group VA. R is a metalloid, so R could be As or Sb. The oxide is acidic, so R is arsenic, As. Copyright © Houghton Mifflin Company. All rights reserved. 8 | 125
    • Chapter 9 Ionic and Covalent Bonding
    • Table 9.1 illustrates the Lewis electron-dot symbols for second- and third-period atoms. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 127
    • For chlorine, Cl, Z = 17, so the Cl- ion has 18 electrons. The electron configuration for Cl- is Give the electron configuration and the Lewis symbol for the chloride ion, Cl-. 1s2 2s2 2p6 3s2 3p6 - The Lewis symbol for Cl- is [ Cl ] Copyright © Houghton Mifflin Company. All rights reserved. 9 | 128
    • Manganese, Z = 25, has 25 electrons;. Its electron configuration is Give the electron configurations of Mn and Mn2+. 1s2 2s2 2p6 3s2 3p6 3d5 4s2 Mn2+ has 23 electrons. When ionized, Mn loses the 4s electrons first; the electron configuration for Mn2+ is 1s2 2s2 2p6 3s2 3p6 3d5 Copyright © Houghton Mifflin Company. All rights reserved. 9 | 129
    • Using the periodic table only, arrange the following ions in order of increasing ionic radius: Br-, Se2-, Sr2+. These ions are isoelectronic, so their size decreases with increasing atomic number: Sr2+ < Br- < Se2- Copyright © Houghton Mifflin Company. All rights reserved. 9 | 130
    • Using electronegativities, arrange the following bonds in order by increasing polarity: C—N Na—F O—H For C—N, the difference is 3.0 (N) – 2.5 (C) = 0.5. For Na—F, the difference is 4.0 (F) – 0.9 (Na) = 3.1. For O—H, the difference is 3.5 (O) – 2.1 (H) = 1.4. Bond polarities: C—N < O—H < Na—F Copyright © Houghton Mifflin Company. All rights reserved. 9 | 131
    • Writing Lewis Electron-Dot Formulas Calculate the number of valence electrons. Write the skeleton structure of the molecule or ion. The central atom is the one with less electronegativity. Distribute electrons to the atoms surrounding the central atom or atoms to satisfy the octet rule. Distribute the remaining electrons as pairs to the central atom or atoms. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 132
    • Write the electron dot formulas for the following: a. OF2 b. NF3 c. NH2OH, hydroxylamine Copyright © Houghton Mifflin Company. All rights reserved. 9 | 133
    • a. Count the valence electrons in OF2: O 1(6) F 2(7) F O F 20 valence electrons O is the central atom (it is less electronegative). Now, we distribute the remaining 16 electrons, beginning with the outer atoms. The last four electrons go on O. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 134
    • b. Count the valence electrons in NF3: N 1(5) F N F F 3(7) 26 valence electrons F N is the central atom (it is less electronegative). Now, we distribute the remaining 20 electrons, beginning with the outer atoms. The last two electrons go on N. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 135
    • c. Count the electrons in NH2OH: N 1(5) H N O H H 3(1) O 1(6) H 14 valence electrons N is the central atom (it is less electronegative than O). Now, we distribute the remaining six electrons, beginning with the outer atoms. The last two electrons go on N. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 136
    • Write electron-dot formulas for the following: a. CO2 b. HCN Copyright © Houghton Mifflin Company. All rights reserved. 9 | 137
    • a. Count the electrons in CO2: C 1(4) O 2(6) C O O 16 valence electrons C is the central atom. Now, we distribute the remaining 12 electrons, beginning with the outer atoms. Carbon does not have an octet, so two of the lone pairs shift to become a bonding pair, forming double bonds. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 138
    • b. Count the electrons in HCN: H 1(1) C 1(4) N 1(5) H C N 10 valence electrons. C is the central atom. The remaining electrons go on N. Carbon does not have an octet, so two of the lone pairs shift to become a bonding pair, forming a triple bond. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 139
    • Phosphorus pentachloride exists in solid state as the ionic compound [PCl4]+[PCl6]-; it exists in the gas phase as the PCl5 molecule. Write the Lewis formula of the PCl4+ ion. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 140
    • Count the valence electrons in PCl4+: P 1(5) Cl + Cl 4(7) -1 Cl P Cl 32 Cl P is the central atom. The remaining 24 nonbonding electrons are placed on Cl atoms. Add square brackets with the charge around the ion. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 141
    • Give the Lewis formula of the IF5 molecule. Count the valence electrons in IF5: I 1(7) F F 5(7) F F 42 valence electrons I F F I is the central atom. Thirty-two electrons remain; they first complete F octets. The remaining electrons go on I. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 142
    • Compare the formal charges for the following electron-dot formulas of CO2. O C O O C O Formal charge = group number – (number of bond pairs) – (number of nonbonding electrons) For the left structure: For the right structure: C: 4–4–0=0 C: 4–4–0=0 O: 6–2–4=0 O: 6 – 1 – 6 = –1 O: 6 – 3 – 2 = +1 The left structure is better. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 143
    • Chapter 10
    • The diagrams below illustrate molecular geometry and the impact of lone pairs on it for linear and trigonal planar electron-pair arrangements. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 145
    • Molecular geometries with a tetrahedral electron- pair arrangement are illustrated below. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 146
    • Molecular geometries for the trigonal bipyramidal electron-pair arrangement are shown on the next slide. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 147
    • Copyright © Houghton Mifflin Company. All rights reserved. 10 | 148
    • Molecular geometries for the octahedral electron- pair arrangement are shown below. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 149
    • Use the VSEPR model to predict the geometries of the following molecules: a. AsF3 b. PH4+ c. BCl3 a. Trigonal pyramidal. b. Tetrahedral. c. Trigonal planar. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 150
    • Using the VSEPR model, predict the geometry of the following species: a. ICl3 b. ICl4- a. T-shaped. b. Square planar. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 151
    • Copyright © Houghton Mifflin Company. All rights reserved. 10 | 152
    • Which of the following molecules would be expected to have a zero dipole moment? a. GeF4 b. SF2 c. XeF2 d. AsF3 a. GeF4 tetrahedral molecular geometry zero dipole moment b. SF2 bent molecular geometry nonzero dipole moment c. XeF2 linear molecular geometry zero dipole moment d. AsF3 trigonal pyramidal molecular geometry nonzero dipole moment Copyright © Houghton Mifflin Company. All rights reserved. 10 | 153
    • Hybrid orbitals have definite directional characteristics, as described in Table 10.2. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 154
    • F N N F F F Use valence bond theory to describe the bonding about an N atom in N2H4. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 155
    • The sp3 hybridized N atom is 1s sp3 Consider one N in N2F4: the two N—F bonds are formed by the overlap of a half-filled sp3 orbital with a half-filled 2p orbital on F. The N—N bond forms from the overlap of a half-filled sp3 orbital on each. The lone pair occupies one sp3 orbital. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 156
    • Use valence bond theory to describe the bonding in the ClF2- ion. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 157
    • The sp3d hybridized orbital diagram for the Cl- ion is sp3d 3d Two Cl—F bonds are formed from the overlap of two half-filled sp3d orbitals with half-filled 2p orbitals on the F atom. These use the axial positions of the trigonal bipyramid. Three lone pairs occupy three sp3d orbitals. These are in the equatorial position of the trigonal bipyramid. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 158
    • Describe the bonding about the C atom in formaldehyde, CH2O, using valence bond theory. O C H H Copyright © Houghton Mifflin Company. All rights reserved. 10 | 159
    • After hybridization, the orbital diagram for C is 1s sp2 2p The C—H bonds are formed from the overlap of two C sp2 hybrid orbitals with the 1s orbital on the H atoms. The C—O bond is formed from the overlap of one sp2 hybrid orbital and one O half-filled p orbital. The C—O bond is formed from the sideways overlap of the C 2p orbital and an O 2p orbital. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 160
    • Give the orbital diagram and electron configuration of the F2 molecule. Is the molecular substance diamagnetic or paramagnetic? What is the order of the bond in F2? Copyright © Houghton Mifflin Company. All rights reserved. 10 | 161
    • F2 has 18 electrons. The KK shell holds 4 electrons so 14 remain. * 2s 2s 2p 2p 2p 2p The molecular electron configuration is KK( 2s)2( 2s)2( 2p)4( 2p)2 ( *2p)4 The bond order is ½(8 - 6) = 1. The molecule is diamagnetic. For F2 and O2, 2p is lower in energy than 2p. This order would not affect the determination of bond order and magnetic properties for these molecules. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 162
    • A number of compounds of the nitrosonium ion, NO+, are known, including nitrosonium hydrogen sulfate, (NO+)(HSO4-). Use the molecular orbitals similar to those of homonuclear diatomic molecules and obtain the orbital diagram, electron configuration, bond order, and magnetic characteristics of the NO+ ion. Note: The stability of the ion results from the loss of an antibonding electron from NO. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 163
    • NO+ has 14 electrons. The KK shell holds 4 electrons, leaving 10 electrons for bonding. * 2s 2s 2p 2p 2p 2p The molecular electron configuration is KK( 2s)2( 2s)2( 2p)4( 2p)2 The bond order is ½(8 - 2) = 3. The ion has a diamagnetic molecular orbital electron configuration. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 164
    • Copyright © Houghton Mifflin Company. All rights reserved. 10 | 165
    • The valence-shell electron-pair repulsion (VSEPR) model predicts the shapes of molecules and ions by assuming that the valence-shell electron pairs are arranged about each atom so that electron pairs are kept as far away from one another as possible, thereby minimizing electron pair repulsions. The diagram on the next slide illustrates this. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 166
    • Two electron pairs are 180 apart ( a linear arrangement). Three electron pairs are 120° apart in one plane (a trigonal planar arrangement). Four electron pairs are 109.5° apart in three dimensions (a tetrahedral arrangement). Copyright © Houghton Mifflin Company. All rights reserved. 10 | 167
    • Five electron pairs are arranged with three pairs in a plane 120° apart and two pairs at 90°to the plane and 180° to each other (a trigonal bipyramidal arrangement). Six electron pairs are 90° apart (an octahedral arrangement). This is illustrated on the next slide. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 168
    • Copyright © Houghton Mifflin Company. All rights reserved. 10 | 169
    • These arrangements are illustrated below with balloons and models of molecules for each. Copyright © Houghton Mifflin Company. All rights reserved. 10 | 170