Principles of Chemistry 1
Principles of Chemistry 1 CHEM 1211
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Popular in Chemistry
This 28 page Class Notes was uploaded by Ms. Ilene Kshlerin on Sunday October 11, 2015. The Class Notes belongs to CHEM 1211 at Columbus State University taught by Staff in Fall. Since its upload, it has received 33 views. For similar materials see /class/221222/chem-1211-columbus-state-university in Chemistry at Columbus State University.
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Formatted Indent First line 025quot Right 025quot Electronic Structure of Atoms Electronic Structure of Atoms 61 a b C d a b C a b Speed is distance traveled per unit time Measure the distance between the point where the stone is dropped and a second reference point possibly the shore line Using a stop watch measure the elapsed time between when the stone is dropped and when the rst wave reaches the second reference point Be sure to drop the stone exactly at the rst reference point Find the ratio of distance to time Measure the distance between two wave crests or troughs or any analogous points on two adjacent waves Better yet measure the distance between two crests or analogous points that are several waves apart and divide by the number of waves that separate them One way to make this distance measurement is to use a standard Take a photo of the event and include an object of known dimensions in the picture Measure the distances described above on the picture Obtain a conversion factor to actual distances by measuring the object on the picture and comparing the photo dimensions to the true dimensions Since speed is distance time and wavelength is distance we can calculate frequency by dividing speed by wavelength v cA We can measure frequency of the wave by dropping a cork in the water and counting the number of times per second it moves through a complete cycle of motion No visible radiation has wavelengths of 4 X 10 7 to 7 X 10 7 In much shorter than 1 cm Energy and wavelength are inversely proportional Photons of the longer 1 cm radiation have less energy than visible photons Radiation of 1 cm or 1 X 10 2 probably a microwave oven m is in the microwave region The appliance is The glowing stove burner is an example of black body radiation the observational basis for Plancks quantum theory The wavelengths emitted are related to temperature with cooler temperatures emitting longer wavelengths and hotter temperatures emitting shorter wavelengths At the hottest setting the burner emits orange visible light At the cooler low setting the burner emits longer wavelengths out of the visible region and the burner appears black If the burner had a super setting the emitted wavelengths would be shorter 131 Formatted Space After 04 line Line spacing single Formatted Font 12 pt Bold Formatted Right 025quot 1 i u um u Ch in E b c 65 a b H 1 6 66 a b C than those of orange light and the glow color would be more blue See Figure 64 for color variation with wavelength Increase The rainbow has shorter wavelength blue light on the inside and longer wavelength red light on the outside See Figure 64 Decrease Wavelength and frequency are inversely related Wavelength increases so frequency decreases going from the inside to the outside of the rainbow The light from the hydrogen discharge take is not a continuous spectIum so not all visible wavelengths will be in our hydrogen discharge rainbow Starting with the shortest wavelengths it will be violet followed by blueiviolet and blue green on the inside Then there will be a gap and nally a red band See the H spectIum in Figure 612 n 1 n 4 n 2 Wavelength and energy are inversely proportional the smaller the energy the longer the wavelength In order of increasing wavelength and decreasing energy iiin2ton4lt ivn3ton1 lt iin3ton2lt in1ton2 w2x will be positive atg zero at all values of x and have two maxima with larger magnitudes than the maximum in MX The greatest probability of nding the electron is at the two maxima in wz x at x n 2 and 37E 2 There is zero probability of nding the electron at x 7 This value is called a node Formatted Font Formatted Indent Left 0quot First line 0quot Formatted Font 12 pt Bold Formatted Font 12 pt Formatted Font color Black 1 4 u um b p dumbbell shape node at the nucleus c The lobes in the contour representation would extend farther along the y axis A larger principle quantum number 4p vs 3p implies a greater average distance from the nucleus for electrons occupying the orbital a In the leftmost box the two electrons cannot have the same spin The Pauli principle states that no two electrons can have the same set of quantum numbers Since the first three quantum numbers describe an orbital the fourth quantum number ms must have different values for two electrons in the same orbital their l spins must be opposite b Flip one of the arrows in the leftmost box so that one points up and the other down c Group 6A The drawing shows three boxes or orbitals at the same energy so it must represent p orbitals Since some of these p orbitals are partially filled they must be the valence orbitals of the element Elements with four valence electrons in their p orbitals belong to group 6A Radiant Energy 69 a Meters m b 1 seconds s 1 C 610 meters second m 871 or m s a Wavelength A and frequency v are inversely proportional the proportionality constant is the speed of light c v cA b Light in the 21L230 nm range is in the ultraviolet region of the spectrum These wavelengths are slightly shorter than the 400 nm shortiwavelength boundary of the visible region a True b False The frequency of radiation decreases as the wavelength increases c False Ultraviolet light has shorter wavelengths than visible light See Solution 610b d False Electromagnetic radiation and sound waves travel at different speeds a False Electromagnetic radiation passes through water SeeFigure B The fact that you can see objects through a glass of water should make this clear b True c False Infrared light has lower frequencies than visible light d False A foghorn blast is a form of sound waves which are not accompanied by oscillating electric and magnetic fields Formatted Font 12 pt Bold Formatted Font 12 pt 1 1 u um u Salve 614 AnalyzePlan Use the electromagnetic spectrum in Figure 64 to determine the wavelength of each type of radiation put them in order from shortest to longest wavelength Formatted Indent Left 0quot Hanging 059quot Formatted NLL 059quot Space After Formatted NLL Indent Left 059quot Space After 2 pt Formatted Ind Space After 0 p Formatted Space After 12 pt Indent Left 0quot Hanging 0 pt Wavelength of Xirays lt ultraviolet lt green light lt red light lt infrared lt radio waves Check These types of radiation should read from left to right on Figure 64 Wavelength of a gamma rays lt d yellow visible light lt e red visible light lt b 931 MHz FM radio waves lt c 680 kHz or 0680 MHz AM radio waves ent Left 0quot Hanging 059quot t AnalyzePlan These questions involve relationships between wavelength frequency and the speed of light Manipulate the equation v c A to obtain the desired quantities paying attention to units Salve 8 a vc7t 2998 X10 m X 1 gtlt 1 314 X 1011551 quotma 9 pace 39 p 955 pm 1 X 10 6 m 2998 X 108 m 1 s 1cv X 7545 X 10 7 111 545mm b s 550 X 10 Formatted Space After 0 pt Formatted Indent First line 0quot Don tadjust space between Latin and Asian text c The radiation in b is in the visible region and is visible to humans The N1 X 10 3 In radiation in a is in the microwave region near the infrared edge and is not visible 150 X 104m 1 2998 108 d 500 ps X s X S 1X105ps Check Confirm that powers of 10 make sense and units are correct 2998 108 1 A a vc7t X 1 gtlt 7710300 X 1017 s 1 100A 1 X 10 m 2998 X 108 m 15 Ac v X 7394 X 10 3 m b s 76 X 1010 c The 1 X 10 9 In radiation in a is Xirays and can be observed by an Xiray detector Radiation b is microwave 1 X 10 15 s X 2998 X 108 m 1 fs 5 d 255 fs X 764 X 10 5 m764 pm AnalyzePlan v c A change nrn gt rn 2998 X 108 m X 1 1 Salve vc9t 7gtlt779688gtlt101 1 s 1 15 436nm 1gtlt10 m Formatted Font 12 pt Bold Formatted Font 12 pt Formatted Font color Black 1 4 u um The color is blue Check 3000 gtlt 105500 X 10 9 6 X 101 1 s l units are correct 2998 X 108 m X 1 X 779613 X 1014 s 1 ls 489nm 1gtlt10 m 618 v cQt The laser emits visible light the color is green to blueigreen Quantized Energy and Photons 619 a Quantization means that energy can only be absorbed or emitted in speci c amounts or multiples of these amounts This minimum amount of energy is called a quantum and is equal to a constant times the frequency of the radiation absorbed or emitted E hv b In everyday activities we deal with macroscopic objects such as our bodies or our cars which gain and lose total amounts of energy much larger than a single quantum hv The gain or loss of the relatively minuscule quantum of energy is unnoticed 620 Plancks original hypothesis was that energy could only be gained or lost in discreet amounts quanta with a certain minimum size The size of the minimum energy change is related to the frequency of the radiation absorbed or emitted AE hv and energy changes occur only in multiples o hv Einstein postulated that light itself is quantized that the minimum energy of a photon a quantum of light is directly proportional to its frequency E hv If a photon that strikes a metal surface has less than the threshold energy no electron is emitted from the surface If the photon has energy equal to or greater than the threshold energy an electron is emitted and any excess energy becomes the kinetic energy of the electron 621 AnalyzePlan These questions deal with the relationships between energy wavelength Formatted Indent Left 0quot Hanging 059quot and frequency Use the relationships E hv hc A to calculate the desired quantities Pay attention to units Salve 2998 108 a EhvhcA6626 gtlt 103415 gtlt X 1 X 17mg 1 s 438nm 1 X 10 m 454x10 19 J 12 b Ehv6626 gtlt 103415 X 447 X 1041 5 2998 108 c AhcE6626 gtlt 103415 gtlt X 692 X 10 8 m 1 287 X 10 J 692 nm This radiation is in the ultraviolet region Check Units are correct and powers of 10 are reasonable Formatted Indent Left 059quot Formatted Font 12 pt Bold Formatted Font 12 pt 1 I u um u 3 Formatted Indent Left 0quot Hanging 098quot a EhcA 6626 X 10 34 Is X W X f x 1quot Space After Opt mm X In 184 X 1043 Formatted Space After 2 pt c The relationship v E h requires energy in photon Change kIInol to I photon and divide by h 6 M 6699 X 1046 S b Ehv6626 X 1034 gtlt 247 k 1 X 103 1 mol Formatted Indent Left 098quot Right 013quot 7 X 2 t 619 X 1013 s 1 1 7 X X S Afte Inol k 6022 X 103 photons 6626 X 10 34 Is pace 39 ms AnalyzePlan Use E hc 7 pay close attention to units Salve 2998 108 1 a EhCA6626X103IJSXX 1 Xii 1 s 33 pm 1 X 10 In 60 X 10 20 J 2998 108 1 Formatted Indent Left 098quot Space After 0 pt EhcA6626x10 34SXXX 1 0154nIn 1 X 10 In 129 X 1045 Check 66 X 333 X 1034 X 103104 m 6 X 1020 66 X 3015 X 1034 X 10310 s 120 X 10mm 12 X 1015 The results are reasonable We expect the longer wavelength 33 pm radiation to have the lower energy Formatted Space After 2 pt Don t adjust space between Latin and Asian text b The 33 pin photon is in the infrared and the 0154 nIn 154 X 10 10 In photon is in the Xiray region the Xiray photon has the greater energy Formatted Space After 3 pt E hv 3 AM6626 gtlt 103415 gtlt 10101 669 X 1048 S 6 FM 6626 X 133415 gtlt 9831A651 X 1045 S The FM photon has about 100 times more energy than the AM photon AnalyzePlan Use E hcA to calculate photon Avogadro s number to calculate IInol photonI the result from part a to calculate photons in 100 In Pay attention to units Formatted Font 12 pt Bold Formatted Font 12 pt Formatted Font color Black 111m 11 m a E hCA 6626 x 10 341s X 2998 x 10 m mzz X 1019 PM 325 X 10 9 In s 611 X 10 19 photon 61122 10 19 6022 1023 h t b 1 gtlt wan X 105 Jrnol 368 kJrnol 1 photon 1 rnol 3 c 1ph0t0r119 X 100 rn X 1 X 10 164 X 1015 photons orma e pace r p 6112 X 10 J 1rn Check Powers of 10 orders of magnitude and units are correct 1 rnol 941 X 103 I X 6022 X 1023 photons 1563 X 10 18 156 X 10718Jphoton rnol O2 127 X 10 7 rn127nrn AihCE 7 6626 X 103415 X 2998x108 rn T 1563 X 1018 According to Figure 6A this is ultraviolet radiation AnalyzePlan E hcA gives photon Use this result with Is given to calculate photons s A Formatted Font Italic Formatted NLLIndent Left 0quot Space After 0 pt Formatted Space After 2 pt a The N1 X 10 6 In radiation is infrared but very near the visible edge 20126 X 10 19 6626 10 34 2998 108 b Ephoimmck X I s x f m S 987 X 10 9 In 201 X 10 19 Iphoton 052 1 h t I X Lori 81 X 1016 photonss 32 5 20126 X 10 Check 7 X 31000 X 1073 1 gtlt 1081079 m 21 X 1020 m 21 X 10719 Iphoton space between Latin and Asian text 05302gtlt 110 19 0008 X 1019 8 X 1016 photonss Units are correct powers of 10 are reasonable a The radiation is rnicrowave 6626 x 10 341s 2998 x 108 m E hcA gtlt 55957 X 10 23 b PM 355 X 10 3 In 1 560x103923 Iphoton Formatted Space After 12 pt Formatted Font 12 pt Bold Formatted Font 12 pt 1 1 u um u 55957 X 10 23 I X 32 X 108 photons X 60s 60 min 1 s X 6446 X 10 11 1 photon 1 min 1 hr 64 X 10 11 Ihr 629 AnalyzePlan Use E hv and v cA Calculate the desired characteristics of the photons Compare Emm and E 120 to calculate maximum kinetic energy of the emitted electron Solve a Ehv6626 x 10 Js x 109 x 10 5 squot 72 x 10quotquot J b A 2998 X 108 m 1 s 2 75 107 275 C V X 7 X m Ilm ls 109 X 1015 2998 X 108 m 1 111111 E h 16626 10 34 7 7 C 120 C X l S X X 120nm X 1 X 1079 m 1656 X 10481136 X 1018 J The excess energy of the 120 nm photon is converted into the kinetic energy of the emitted electron Ek E120 r Emm 1655 X 1019 722 X 1019 93 X 10 19Ielectron Check E 120 must be greater than Emm in order for the photon to impart kinetic energy to the emitted electron Our calculations are consistent with this requirement Formatted Indent Left 059quot Space After 12 pt Don t adjust space between Latin and Asian text 441 X 10 19 I 6626 X 103415 Formatted Indent Left 0quot Hanging 098quot 630 a v Eh 66556 X 1014 666 X 1014 s 1 734 8 b AhcE 63921X 11001911 X 239998 X 10 m 450 X 10 7 m450nm X S C Em hCA 6626 x 10 9 I s X 2998 x 10 m 4525 X 106 453 X 106 439 X 10 m s Formatted Indent Left 098quot Space After 9 pt Don t adjust space between Latin and Asian text EK E439 Emm 4525 gtlt 1051917441 X 10519 0115 X 1019 11 X 10520 d One electron is emitted per photon Calculate the number of 439 nm photons in 100 11 The excess energy in each photon will become the kinetic energy of the electron it cannot be pooled to emit additional electrons 1 10 6 1 h t 1 X7 X p 0 on e 221 X 1012 electrons 100 X X 7 I p 4525 X 10 19 1photon Formatted Font 12 pt Bold Formatted Font 12 pt Bohi s Model Matter Waves Electronic Structure of Atoms U 141 11 m 631 When applied to atoms the notion of quantized energies means that only certain energies can be gained or lost only certain values of AE are allowed The allowed values of AE are represented by the lines in the emission spectra of excited atoms Formatted In dent Left 0quot Hanging 059quot Space After 9 pt 632 a According to Bohr theory when hydrogen emits radiant energy electrons are moving from a higher allowed energy state to a lower one Since only certain energy states are allowed only certain energy changes can occur These allowed energy changes correspond A hcAE to the wavelengths of the lines in the emission spectrum of hydrogen When a hydrogen atom changes from the ground state to an excited state the Formatted Space After 9pt single electron moves further away from the nucleus so the atom llexpands b 633 AnalyzePlan An isolated electron is assigned an energy of zero the closer the electron comes to the nucleus the more negative its energy Thus as an electron moves closer to the nucleus the energy of the electron decreases and the excess energy is emitted Conversely as an electron moves further from the nucleus the energy of the electron increases and energy must be absorbed Salve a As the principle quantum number decreases the electron moves toward the nucleus and energy is emitted b An increase in the radius of the orbit means the electron moves away from the nucleus energy is absorbed c An isolated electron is assigned an energy of zero As the electron moves to the Formatted Space After 9pt Dontadjusr n 3 state closer to the H nucleus its energy becomes more D 39 Space between Lam and As39an text and energy is emitted 634 a Absorbed b Emitted c Absorbed Formatted Ind Space After 9 p nt Left 0quot Hanging 098quot Egrmatted NL1 Indent Left 0quot First line 635 AnalyzePlan Equation 65 E 7218 X 10 1Bl1n2 Salve E2 7218 X 10 13122 7545 X 1049 E6 7218 X 103913162 760556 X m20 70606 X 1019 AE E 6 4 E2 0606 X 1019 4 4545 X 1019 4844 X 10719 I 484 X 10719 A hcAE 6626 10 34 2998 108 X Jsx410x10 7nn410nm S 4844 X 10 9 J The visible range is 4007700 nrrL so this line is visible the observed color is violet Formatted Font 12 pt Bold Formatted Font 12 pt Check We expect E 6 to be a more positive or less negative than E 2 and it is AE is 1quot1 1 394quot i Aiwm 11 r 11 um u positive which indicates emission The orders of magnitude make sense and units are correct 636 a AE7218 X 10 18 J 27112 218 X 10 18 1117116 72044 X 10 18 t 1 7204 X 10 18 I 2044 10718 v E X 7M3084 X 1015 308 X 1015 s 1 6626 X 10 15 8 Acvw gtlt 1755972 X 10 8 m 15 3084 X 10 I Since the sign of AB is negative radiation is emitted Formatted Indent Left 098quot Space After 12 pt b AE 7218 X 1018 114e 125 74578 X 1019 7458 X 1019 4578 X 10719 6909 X 1014 691 X 1014 51 A 2998 X 108 ms V 6626 X 103415 6909 X 10145 A 434 X 10 7 m Visible radiation is emitted Formatted Indent Left 098quot Space After 12 pt Don t adjust space between Latin and Asian text c AE 7218 X 1018 I 136 7 19 1817 X 1019 182 X 1019 1817 10 19 2998 108 v X 2742 X 10 274 X 10 squot A X fs 6626 X 10 Is 2742 X 10 s l A 109 X 10 6 m Radiation is absorbed 9 Formatted Indent Left 098quot Space After 12 pt Don t adjust space between Latin and Asian text 637 a Only lines with n 2 represent AE values and wavelengths that lie in the visible portion of the spectrum Lines with n 1 have larger AE values and shorter wavelengths that lie in the ultraviolet Lines with n gt 2 have smaller AE values and lie in the lower energy longer Wavelength regions of the electromagnetic spectIum b AnalyzePlan Use Equation 67 to calculate AE then A hcAE Salve ni 3nf 2 AE 7218 X 10718 7 7218 X 10718 114719 f 6626 X 10 34 15 X 2998 X 108 ms 211 E C 7218 X 10718 114719 656 X 10 7 m This is the red line at 656 nm 6626 X 10 34 15 X 2998 X 108 ms ni4nf2 7thcE 718 7218 X 10 147116 486 X 10 7 m This is the blue line at 486 nm Formatted Font 12 pt Bold Formatted Font 12 pt Formatted Font color Black 111m 11 m 6626 X 10 34 15 X 2998 X 108 rns 18 434 X 10 7 rn 7218 X 10 1147125 n 5nf2 AhcE This is the violet line at 434 nrn Check The calculated wavelengths correspond well to three lines in the H emission spectrum in Figure 612 so the results are sensible 638 a Transitions with n 1 have larger AE values and shorter wavelengths than those with n 2 These transitions will lie in the ultraviolet region 6626 X 104415 gtlt 2998 X 108 ms ni2n 1 AhcE 121x10 7m b 7218 X 10 18 111714 6626 10 34 2998 108 n 3nf 1 AhcELO3 X 10 7 In 7218 X 10 111719 34 8 Formatted Indent Left 098quot Ri ht o13quot n 4 nf 1 A hcE W 0972 X 10 7 m Space After 12 pt 9 7218 X 10 18 1117116 Formatted In 1 X 10 9 m 8 dent Left 0quot Hanging 093quot 639 a 938 mm X 17 938 X 10 m this line is in the ultraviolet region space After 0 pt nm b AnalyzePlan Only lines with nf 1 have a large enough AE to lie in the ultraviolet region see Solutions 637 and 638 Solve Equation 67 for n recalling that AB is negative for emission Solve A nf n M218 X 10 J n 1 4 1 1t 112 M218 X 1048 112 M218 X 1048 1 712 12 k 1 1 C M218 X 10 18 I M218 X 1048 n 7 176626 X 10 34 15 X 2998 X 108 rns 938 X 10 8 111 X 218 X 10 18 I 712 1 6 n values must be integers n6n1 Formatted Indent Left 059quot Space After 12 pt Don t adjust space between Latin and Asian text Check From Solution 638 we know that n gt 4 for A 938 nrn The calculated result is close to 6 so the answer is reasonable Formatted In Space A ter 2 Formatted Font 12 pt Bold Formatted Font 12 pt dent Left 0quot Hanging 098quot pt 2626 X 10 6m this line is in the infrared 9 640 a 2626mm X m 1 nrn 1 1 u um u b Absorption lines with n 1 am in the ultraviolet and with n 2 are in the visible Thus n 2 3 but we do not know the exact value of n Calculate the longest wavelength with n 3 n 4 If this is less than 2626 nIn 11 gt 3 6626 X 103415 gtlt 2998 X 108 ms 7218 X 10 18 1116719 This wavelength is shorter than 2626 X 10 6 In so n gt 3 tIy n 4 and solve for n as in Solution 639 1 hcE 1875 X 10 6 m Formatted Indent Left 098quot Space After 2 pt 712 7 117 6626 X 10 34 15 X 2998 X 108 ms 12 7 1 hc n r 1218 X 1048 2626 X 10 5 m X 218 X 10 I n 6 n 4 2 AnalyzePlan A i11kgim Change mass to kg and velocity to Ins in each mv 5 case Solve a 501m gtlt 1000B X 3 X l 138914ms 1 hr 1 km 60 mm 60 s 7 1s2 85kg 138916quot 1kg 1000g b 100g gtlt 00100 kg 34 2 F tt dI d t Left o9squots Afte 2 t 1 5 00100 kg 250 In c We need to calculate the mass of a single Li atom in kg 694gLi X 1kg X 111101 111101 Li 1000g 6022 X 1023 Li atoms 1152 X 10 26 115 X 10 26 kg 77 6626 X 10 34 kgm2 s 2 X 1 26 X 155 23x10quot3m ls 1152gtlt10 kg 25gtlt10 m A hInv change mass to kg and velocity to Ins 1 k quot mass of muon2068 gtlt 91094 X 1048 g X 10mg 18838 X 1048 188 X 1048 kg mam dent M 0 59 Spaceme39 3 pt 8 7 6626 X 10 34 kgmZs X 1 X s 7 152 18838 X 10 28 kg 885x103 ms 1 397 X 10 10 m 397 A AnalyzePlan Use v h H17 change wavelength to meters and mass of neutron back inside cover to kg Sn De Formatted Font 12 pt Bold Formatted Font 12 pt Formatted Font color Black 1 4 u um 0955 X 10 10 m m16749 X 10 27 kg 710 A 0955 A X M 1A 734 z V6626gtlt10 Zkgm 5X 1 727 X 1 710 A14X103ms ls 16749 X 10 kg 0955 X 10 In Check 66161 gtlt 10 3410 2710 10 e 4 X 103 ms Formatted Indent Left 059quot Space After 8 pt Don39t adjust space between La in and Aslan text Formatted Indent Left 059quot Space After 3 pt 644 mE 91094 X 10 31 kg back inside cover of text 7 6626 X 10 34 kgmZs X 1 X 7 152 91094 X 10 31 kg 938 X 106m A 775 X 1071116 1A 775 X 10 11 m X in 0775 A 1 X 10 m Since atomic radii and interatomic distances are on the order of 175 A Section 23 the wavelength of this electron is comparable to the size of atoms Formatted Indent Left 059quot Space After 12 pt 645 AnalyzePlan Use Ax 2 h4n m Av paying attention to appropriate units Note that the uncertainty in speed of the particle Av is important rather than the speed itself Salve g 150 X 1075 kg Av 001ms 1 a m150mggtlt7gtlt7 1000 mg 1000g 6626 gtlt 103415 fzsm X 10 27 4 X 10 27 m 4150 X 10 kg 001 ms AXE b m 1673 X 10 24 g 1673 X 10 27 kg Av 001 X 104 ms gt7 6626 X 10 34 15 47r1673 X 10 27 kg 001 X 104 ms Check The more massive particle in a has a much smaller uncertainty in position Formatted Indent Left 059quot Space After 12 pt 646 Ax 2 h4139JInAv use masses in kg Av in m s 23 X 10quot m 6626 gtlt 103415 6 X 1078 m 49109 X 10 kg 001 X 10 ms 6 6626 X 10 34 15 3 X 10 11 m 41r1675 X 10 kg 001 X 10 ms c For particles moving with the same uncertainty in velocity the more massive neutron has a much smaller uncertainty in position than the lighter electron In our model of the atom we know where the massive particles in the nucleus are located but we cannot know the location of the electrons with any certainty if Formatted Font 12 pt Bold Formatted Font 12 pt i i u um u we know their speed Quantum Mechanics and Atomic Orbitals 647 a The uncertainty principle states that there is a limit to how precisely we can simultaneously know the position and momentum related to energy of an electron The Bohr model states that electrons move about the nucleus in precisely circular orbits of known radius each permitted orbit has an allow ed energy associated with it Thus according to the Bohr model we can know the exact distance of an electron from the nucleus and its energy This violates the uncertainty principle Formatted AL Indent First line 0quot Tab stops t 059quot b deBroglie stated that electrons demonstrate the properties of both particles and waves that each particle has a wave associated with it A wave function is the mathematical description of the matter wave of an electron c Although we cannot predict the exact location of an electron in an allowed energy state we can determine the likelihood or probability of finding an electron at a particular position or within a particular volume This statistical knowledge of electron location is called the probability density or electron density and is a function of V2 the square of the wave function 141 648 a The Bohr model states with 100 certainty that the electron in hydrogen can be found 053 A from the nucleus The quantum mechanical model taking the wave nature of the electron and the uncertainty principle into account is a statistical model that states the probability of finding the electron in certain regions around the nucleus While 053 A might be the radius with highest probability that probability would always be less than 100 b The equations of classical physics predict the instantaneous position direction of motion and speed of amacroscopic particle they do not take quantum theory or the wave nature of matter into account For macroscopic particles these are not significant but for microscopic particles like electrons they are crucial Schrodinger s equation takes these important theories into account to produce a statistical model of electron location given a specific energy c The square of the wave function has the physical signi cance of an amplitude o Formatted Space After 9 pt Don t adjust space 2 between Latin and Asian text probability The quantity u at a given point in space is the probability of locating the electron within a small volume element around that point at any given instant The total probability that is the sum of L412 over all the space around the nucleus must equal 1 Formatted Indent Left 0quot Hanging 098quot Space After 2 pt 649 a The possible values of l are n r 1 to 0 n 4V 1 3 2 1 0 Formatted Space After 9 pt Don t adjust space b The possible values of In are fl to 1 I 2 ml 72 71 0 1 2 between Lam and As39an text Formatted AL Formatted Font 12 pt Bold Formatted Font 12 pt Electronic Structure of Atoms U 141 11 m 650 a For n 3 there are 3 1 values 2 1 0 and 9 ml values I 2 ml r2 r1 0 1 2 11ml 7101l0ml 0 Formatted Indent Left 0quot Hanging 098quot Space After 9 pt Don39t adjust space between Latin and Asian text b Forn 5 there are 5 1 values 4 3 2 1 0 and 25 ml values I 4 ml 74 to 4 13ml r3to3l2ml r2to2l1ml 71to1l00 In general for each principle quantum number 11 there are n levalues and n2 mlevalues For each shell there are 11 kinds of orbitals and 112 total orbitals Formatted Indent Left 102quot Space After 9 pt Don39t adjust space between Latin and Asian text 651 a 3p 11 3 l 1 b 2s n 21 0 c 4inampl3 d 5d n 51 2 Formatted Indent Left 059quot Space After 9 pt 652 a 211 210 2171 b 5 2 2 5 2 1 5 2 0 5 2 71 5 2 72 Formatted Indent Left 0quot Hanging 098quot Space After 9 pt 653 Impossible a 1p only I 0 is possible for n 1 d 2d for n 21 1 or 0 but not 2 Formatted Indent Left 0quot Hanging 059quot Space After 9 pt 654 a Permissible 2p Forbidden for l 0 ml can onl e ual 0 V Cl c Permissible 4d d Forbidden for n 3 the largest I value is 2 k Formatted Indent Left 059quot Space After 12 pt 6 55 Z Z Z A 657 a The 1s and 2s orbitals of a hydrogen atom have the same overall spherical shape The 2s orbital has a larger radial extension and one node while the 1s orbital has continuous electron density Since the 2s orbital is larger there is greater probability of nding an electron further from the nucleus in the 2s orbital Formatted Font 12 pt Bold Formatted Font 12 pt Formatted Font color Black FA ma 0 1 I 39 lnr 39 u in um u b A single 22 orbital is directional in that its electron density is concentrated along one of the three Cartesian axes of the atom The X27 orbital has electron densitV along both the x7 and Viaxes while the D7 orbital has densitV oan along thexiaxis 1m 10 1M1 A H 1 1111 161quot A r HtAi 6 I J 6 1111 P 11 1 39T39k A 111 11 X iv A w 1 1 1111 A 11 1 k111i A w 1 1 J a J s6 J J a the weeds c The average distance of an electron from the nucleus in a 3s orbital is greater than for an electron in a 2s orbital In general for the same kind of orbital the larger the 11 value the greater the average distance of an electron from the nucleus of the atom Formatted Line spacing 15 lines Don39t adjust space between Latin and Asian text Formatted AL d 1s lt 2p lt 3d lt 4f lt 6s In the hydrogen atom orbitals with the same 11 value are degenerate and energy increases with increasing 11 value Thus the order of increasing energy is given above In an s orbital there are n r 1 nodes b The 2pX orbital has one node the yz plane passing through the nucleus of the atom The 3s orbital has two nodes 7 Formatted AL c Probability density w2r is the probability of finding an electron at a single point r The radial probability function Pr is the probability of nding an electron at any point that is distance r from the nucleus Figure 618 contains plots of Pr vs r for 1s 2s and 3s orbitals The most obvious features of these plots are the radii of maximum probability for the three orbitals and the number and location of nodes for the three orbitals By comparing plots for the three orbitals we see that as n increases the number of nodes increases and the radius of maximum probability orbital size increases d 2s 2p lt 3s lt 4d lt 5s In the hydrogen atom orbitals with the same 11 value are degenerate and energy increases with increasing 11 value ManyElectron Atoms and Electron Configurations 659 a In the hydrogen atom orbitals with the same principle quantum number n have the same energy they are degenerate b In a manyielectron atom for a given nivalue orbital energy increases with increasing livalue s lt p lt d lt f Formatted Font 12 pt Bold Formatted Font 12 pt Electronic Structure of Atoms U 141 11 m 660 a The electron with the greater average distance from the nucleus feels a smaller attraction for the nucleus and is higher in energy Thus the 3p is higher in energy b Because it has alarger 11 value a 3s electron has a greater average distance from the chlorine nucleus than a 2p electron The 3s electron experiences a smaller attraction for the nucleus and requires less energy to remove from the chlorine atom 661 a 12712 b Electrons with opposite spins are affected differently by a strong inhomogeneous magnetic eld An apparatus with a strong inhomogeneous magnetic eld similar to the diagram in Figure 626 can be used to distinguish electrons with opposite spins c The Pauli exclusion principle states that no two electrons can have the same four quantum numbers Two electrons in a 1s orbital have the same 11 l and ml values They must have different his values 662 a The Pauli exclusion principle states that no two electrons can have the same four quantum numbers Formatted AL Indent First line39 0quot Tab stops Not at 059quot b An alternate statement of the Pauli exclusion principle is that a single orbital can hold a maximum of two electrons Thus the Pauli principle limits the maximum number of electrons in a main shell and its subshells which determines when a new row of the periodic table begins 663 AnalyzePlan Each subshell has an levalue associated with it For a particular levalue permissible mlevalues are fl to 1 Each mlevalue represents an orbital which can hold two electrons Salve a 6 b 10 c 2 d 14 664 a 4 b 14 c 2 d 2 665 a Each box represents an orbital b c No The electron con guration of Be is 1s22s2 There are no electrons in subshells Electron spin is represented by the direction of the halfearrows Formatted Font 12 pt Bold Formatted Font 12 pt U 1 1 1t 1119 cu that have degenerate orbitals so Hund s rule is not used 666 a Valence electrons are those beyond the previous noblegas or core electron con guration b Unpaired electrons are electrons that occupy orbitals singly That is when there is only one electron in an orbital this electron is unpaired c A P atom has ve valence electrons 3s23p3 Three of them those in the FormattedSpace After 12ptDon39tadjust space between Latin and Asian text degenerate 3p orbitals are unpaired 667 a Cs lte6s1 b Ni Ar4s 23d 3 c Se Ar4s23d1 4p 1 d Cd Kr5s24d10 e Ac Rn7s26d1 Formatted Indent Left 059quot Hanging 039quot Space After 8 pt Don39t adjust space between fPb Xe6sz4f145d 106132 Latin and Asian text Formatted Indent Left 0quot First line 0quot 668 a Ga Ar4s23d1 4p1 1 unpaired electron Formatted AL Indent First line 0quot Space After 39 39 ac y14pt on tajust space between Latin and Asian text Tab stops Not at 059quot C 2 b a39 f 0 unpalmd Elec39 ms Formatted Indent Left 0quot Don39t adjust space between Latin and Asian text c V Ar4s23d3 3 unpaired electrons d I Kr5s24d1 5p5 1 unpaired electron e Y Kr5s 24d1 1 unpaired electron f Pt Xe6sl4 145d33 2 unpaired electrons g Lu Xe6s24d145d1 1 unpaired electron s Formatted Space After 12 pt Don39t adjust space between Latin and Asian text 669 Mt Rn7525f146c17 FormattedIndent Left 0quot Hanging 059quot Space After 12 pt Don39t adjust space between Latin and Asian text 670 a Rn7s25f146d1 7p6 b None Element 118 would belong to Group 8A the Noble Gases These elements have completely filled subshells and orbitals a Groupsmhemme 671 a Mg Formatted Indent Left 0quot Hanging 098quot c Cr Space After 12 pt Formatted Font 12 pt Bold Formatted Font 12 pt Electronic Structure of Atoms U 141 11 m 672 a 7A halogens c 3A d the feblock elements Sm and Pu Formatted Indent Left 059quot Hanging 039quot Space After 12 pt 673 a The fifth electron would ll the 2p subshell same nevalue as 25 before the 35 b The Ne core has lled 2s and 2p subshells Either the core is He or the outer electron con guration should be 35 23p3 c The 3p subshell would ll before the 3d because it has the lower levalue and the same nevalue 674 Count the total number of electrons to assign the element a N He2522p37 Formatted Indent Left 059quot Hanging 039quot Line spacing Double Don39t adjust space between Latin and Asian text b Se Ar4sz3d 1 04p 1 c Rh Kr15524d7 Additional Exercises 675 a AA 16gtlt10397II1 45356X1039B36gtlt10393m AB 16gtlt 10 7m280gtlt 10 Bm 2998X108m 1 15 1 vcA v gtlt 84gtlt10 s b A 15 356 X 10 8 In 8 Vgwx37x1o scl ls 80gtlt10 rn c A ultraviolet B ultraviolet Formatted Space After 12 pt 676 a Elements that emit in the visible Ba dark blue Ca dark blue K dark blue Na yellow orange The other wavelengths are in the ultraviolet b Au shortest wavelength highest energy Na longest wavelength lowest energy a C A CV 2998 X 10 rns 1nm 455 mn Ba X 659 X 10145 1 X 10 9 m Formatted Indent Left 0quot Hanging 059quot Space After 12 pt Don39t adjust space between Latin and Asian text Formatted Font 12 pt Bold Formatted Font 12 pt 677 All electromagnetic radiation travels at the same speed 2998 X 108 ms Change rnjles to meters and seconds to some appropriate unit of time 1 1 u um u 16093 km X 1000 m 1 s 746gtlt10Emigtlt 7 x Bx 11111 1km 2998 X 10 m 1 min Formatted Indent Left 059quot Space After 12 F 667 mm pt 5 7 2998 X 108 ms X 111m a v cA 79 937 X 10 s 1 32011111 1 X 10 m b BMWW X lt1 X W 7 7 320 X 10 7 m 1000 mole 374 kJmol Formatted Indent First line 458quot Space After 6 pt c UVB photons have shorter wavelength and higher energy d Formatted Space After 12 pt Yes The higher energy UVB photons would be more likely to cause sunburn E hcA gt I photon total energy power gtlt time photons total energy I photon 7 6626 X 103415 gtlt 2998 X 108 ms E 9 25468 X 10 19 255 X 10 19 Jphoton 780 X 10 m 010 10 3 60 010 mWX X 69 min X 04140041 15 1mm 1 h t 04140 X p n 1626 X 1018 16 X 1018 photons 25468 X 10 19 J Ephoton hcA 7 6626 X 103415 gtlt 2998 X 108 1115 23 FormattedIndent Left 059quotS ace After 12 263 kImol E 26 X 10 12 C 1e 1 h t X w x 4623 X 107 16 X 107 photonss 1s 1602 X 10 c 1e 34 8 E hCA76626gtlt10 Isgtlt 2998X10 mgtlt 111m 7 X 1623 X 107 photon photon 7 630 nm 1 s 1gtlt107gm s 51 X 10 12 Is Formatted In 200 X 105 J 1mol photons 7 X Space After 4 dent Left 0quot Hanging 098quot t mol 6022 X 1023 photons p a 3321 X 10 19 332 X 10 19 Jphoton 757 6626 X 103415 gtlt 2998 X 108 ms E 3321 1019 I 598 X 10 7 m X b 7 c 598 X 10 7 In 598 nm is well within the visible portion of the electromagnetic spectIurn and corresponds to yellow or yelloworange light Red light with wavelengths near or greater than 700 nm does not have suf cient energy to Formatted Font 12 pt Bold Formatted Font 12 pt Formatted Font color Black 111m 11 m a b c e a b initiate electron transfer and darken the film Formatted Indent Left 0quot Hanging 098quot Right 013quot Space After 3 pt 2998 X108m X 1 1nm gtlt 77944088 X 10 441 X 10 s 1 s 680nm 1 X 10 m V cA Calculate Iphoton using E hc A change to kImol 6626 10 34 2998 108 E We X Js x 49213 X 10 19 292 X 10 19 Jphoton p 680 X 10 In 719 23 Formatted Indent Left 098quot S ace After 29213 X 10 I X 6022 X 10 photons X 1k 17592176kIm01 m p photon mol 1000 I Nothing The incoming incident radiation does not transfer sufficient energy to an electron to overcome the attractive forces holding the electron in the metal Formatted AL For frequencies greater than v0 any extra energy not needed to remove the electron from the metal becomes the kinetic energy of the ejected electron The kinetic energy of the electron is directly proportional to this extra energy Let Ema be the total energy of an incident photon Emm be the minimum energy required to eject an electron and E k be the extra energy that becomes the kinetic energy of the ejected electron Ema Emm Ek Ek Ema r Emm hvr hvo Ek hv 7 v0 The slope ofthe line is the value of h Plancks constant Lines with n 1 lie in the ultraviolet see Solution 630 and with n 2 lie in the visible see Solution 629 Lines with n 3 will have smaller AE and longer wavelengths and lie in the infrared Use Equation 67 to calculate AE then A hcAE ni 4 nf 3 AE 7218 X 10 e 7218 X 10 1197116 13911 13911 6626 X 10 34 15 X 2998 X 108 rns 7 718 7 187 X 10 6 In 218 X 10 19 116 A hcE 6626 X 103415 gtlt 2998 X 108 rns 7 718 7 128 X 10 6 In 218 X 10 19 125 In 5nf37thcE 6626 X 103415 gtlt 2998 X 108 rns 7 718 7 109 X 10 6 In 218 X 10 19 136 In 6nf37thcE Formatted Indent Left 098quot Space After 12 pt Don t adjust space between Latin and ian text Formatted Font 12 pt Bold Formatted Font 12 pt These three wavelengths are all greater than 1 pm or 1 X 10 6 m They are in the infrared close to the visible edge 07 X 10 6 m 1 1 u um u a Gaseous atoms of various elements in the sun s atmosphere typically have ground state electron configurations When these atoms are exposed to radiation from the sun the electrons change from the ground state to one of several allowed excited states Atoms absorb the wavelengths of light which correspond to these allowed energy changes All other wavelengths of solar radiation pass through the atmosphere unchanged Thus the dark lines are the wavelengths that correspond to allowed energy changes in atoms of the solar atmosphere The continuous background is all other wavelengths of solar radiation b The scientist should record the absorption spectrum of pure neon or othe FormattedSpace After 12pt elements of interest The black lines should appear at the same wavelengths regardless of the source of neon a He is hydrogenilike because it is a oneielectron particle An He atom has two electrons The Bohr model is based on the interaction of a single electron with the nucleus but does not accurately account for additional interactions when two or more electrons are present Formatted AL Indent First line 0quot Tab stops t 059quot b Divide each energy by the smallest value to nd the integer relationship H 218 X 10 13218 X 10 18 1 Z 1 He 872 X 10 13218 X 10 18 k Z 2 112136 gtlt 10 17218 X 10 18 9 Z 3 The groundistate energies are in the ratio of 149 which is also the ratio ZZ the square of the nuclear charge for each particle The ground state energy for hydrogenilike particles is E R H ZZ By de nition n 1 for the ground state of a oneielectron particle c Z 6 for C E 7218 X 1013 62 7785 X 1017 Formatted Space After 12pt Don39t adjust space between Latin and Asian tex 1 10quot A hmv v hmA A 0711A gtlt 711 X 10 11 111 me 91094 X 10 31 kg 6626 X 10 34 15 91094 X 10 31 kg X 711 X 10 11 m 7 Formatted Indent Left 059quot Space After 12 102 X 10 ms pt 1 kg 0 m2 s2 V X 1 Plum Change keV to I electron Calculate v from kinetic energy A hmv Salve 1000 eV X 96485 k X 1000 X 1 mol 1 eV mol 1 k 6022 X 1023 electrons 2980 X 10 15 298 X 10715electron 186 keV gtlt Ek mv2 2v2 2Ek mv 2Ekm Formatted Font 12 pt Bold Formatted Font 12 pt Formatted Font color Black 1 4 u um 2 X 2980 X 10 15 lltg1112s2 12 9 1094 1031 k j 8089 X 107 809 X 107ms X g 6626 gtlt 103415 1kgmZsz A hmv gtlt 91094 X 10 31 kg gtlt 8089 X 107 ms 1 899 X 10 12 m 899 pm 689 An orbit is an exactly circular path with specified radius for an electron in an allowed energy state Each allowed orbit is denoted by a principle quantum number n An orbital is a region of space where there is a probability of nding an electron in an allowed energy state Orbitals are threedimensional volumes with characterich size shape and orientation and are denoted by the three quantum numbers n size 1 shape and m1 orientation An orbital is a statistical prediction while an orbit is a known location The notion of a known orbit violates the uncerminty principle see Solution 647 690 a l b n and 1 CL m s d mz 691 a Probability density wr2 is the probability of finding an electron at a single point at distance r from the nucleus The radial probability function 47oz is the probability of nding an electron at any point on the sphere de ned by radius r Pr 47E M0 2 Formatted stops Not a AL Indent First Iin t 9quot b The term 4m explains the differences in plots of the two functions Plots of the probability density wr2 for s orbitals shown in Figure 621 each have their maximum value at r 0 with n r 1 smaller maxima at greater values of r The plots of radial probability Pr for the same s orbitals shown in Figure 618 have values of zero at r 0 and the size of the maxima increases Pr is the product of Lpr2 and 47oz At r 0 the value of Lpr2 is finite and large but the value of 4m2 is zero so the value of Pr is zero As r increases the values of Lpr2 vary as shown in Figure 621 but the values of 4m2 increase continuously leading to the increasing size of Pr maxima as r increases Formatted Font 12 pt Bold Formatted Font 12 pt 1 r u um u PAT I 1 What the noble gas elements have in common are completed ns and np subshells Since the Pauli principle limits the number of electrons per orbital to two this leads to the rst three magic numbers 21s2 101s22s22p6 and 181s22s22p63s23p6 In the fourth row n r 1 d orbitals begin to ll as their energy falls below that of the np orbitals This leads to the next two magic numbers 361s22s22p63s23p64s23d1 4p6 and 541s22s22p63s23p64s23d1 4p65s24d1 5p6 In the sixth row the energy of the 4f orbitals falls below that of the n r 1d and np subshells and it fills This explains the nal magic number 860 p 1 up 1 ud up 1 ad up u mmudl up a The p z orbital has a nodal plane where z 0 This is the xy plane b The dXy orbital has four lobes and two nodal planes the two planes where x 0 and y 0 These are the yz and xz planes c The dXLVZ has four lobes and two nodal planes the planes where x2 r y2 0 These are the planes that bisect the x and y axes and contain the z axis a In the absence of a magnetic eld electrons with opposite rns values have the same energy Because electrons with opposite spins will have oppositely oriented magnetic elds only the interaction of the magnetic elds of the electrons with an external magnetic eld will cause the energies of the electrons to be different and observable b According to Figure 627 the particle with its magnetic eld parallel to the external eld will have the lower energy The left electron has its magnetic field oriented parallel to the described magnetic orientation so it will be lower in energy Formatted Indent Left 0quot Hanging 059quot Space After 12 pt Don39t adjust space between Latin and Asian text Formatted Space After 12 pt on39t adjust space between Latin and Asian text Formatted AL Formatted Font 12 pt Bold Formatted Font 12 pt Formatted Font color Black 111m 11 m Formatted Space After 12 pt c Microwave photons used to excite unpaired electrons in the ESR experiment have higher energy than radio wave photons used to excite nuclei in NMR Formatted Indent Left 0quot Hanging 098quot Space After 3 pt Formatted Space After 3 pt Formatted Space After 12 pt a This is the frequency of microwaves that excite the nuclei from one spin state to the other b AEhv6626 gtlt 103415 X 298 X 1045 450 X 106 S c Since AE 0 in the absence of a magnetic field it is reasonable to assume that the stronger the external field the greater AE In fact AB is directly proportional to eld strength Because AB is relatively small see part b the two spin states are almost equally populated with a very slight excess in the lower energy state The stronger the magnetic field the larger AE the greater number of nuclei in the lower energy spin state With more nuclei in the lower energy state more are able to absorb the appropriate radio wave photons and reach the higher energy state This increases the intensity of the NMR signal which provides more information and more reliable information than a weak absorption signal If Ins had three allowed values instead of two each orbital would hold three electrons instead of two Assuming that the same orbitals are available that there is no change in the n l and m1 values the number of elements in each of the first four rows would be Formatted NL1 1 row 1 orbital X 3 3 elements 2 1 row 4 orbitals X 3 12 elements 3rd row 4 orbitals X 3 12 elements 4 row 9 orbitals X 3 27 elements The siblock would be 3 columns wide the piblock 9 columns wide and the diblock 15 columns wide Formatted Indent Left 059quot Space After 12 pt a Se Ar4s23d1 4p l b Rh Kr5s24d7 c Si Ne3s23p2 d Hg lte6s24f1 5d10 e Hf Xe6s24f145d2 Formatted Indent Left 059quot Hangin 039quot Space After 12 pt Don39t adjust space between Latin and Asian text Formatted Space After 18 pt Don39t adjust space between Latin and Asian x The core would be the electron configuration of element 118 If no new subshell begins to ll the condensed electron configuration of element 126 would be similar to those of elements vertically above it on the periodic chart Pu and Sm The condensed configuration would be 1188s26f6 On the other hand the 5g subshell could begin to ll after 8s resulting in the condensed con guration 1188s25g6 Exceptions are also possible likely Formatted Font 12 pt Bold Formatted Font 12 pt 1 r u um u Integrative Exercises 6 99 6100 6101 We know the wavelength of microwave radiation the volume of coffee to be heated and the desired temperature change Assume the density and heat capacity of coffee are the same as pure water We need to calculate i the total energy required to heat the coffee and ii the energy of a single photon in order to nd iii the number of photons required i From Chapter 5 the heat capacity of liquid water is 4184 g C To find the mass of 200 mL of coffee at 23 C use the density of water given in Appendix B 0997 g 200 mL gtlt 1994 199 g coffee 1 mL L804 gtlt 1994g gtlt 60 C723 C 3087 X 104 J 31 k g 2998 108 177 10 24 ii EhcA6626 gtlt 104415 gtlt X X 1 1 5 0112 m 1photon 17 X 1028 photons 1 h r iii 3087 X 104 I X Log 1774 X 10 I The answer has 2 sig figs because the temperature change 43 C has 2 sig figs Formatted Normal Indent Left 098quot Space After 6pt Formatted NL1Indent Lett 0quot Space After 0 pt AHRn AH 0294H 094H 039 AH M 0 2475kJe1423kJ1052kJ 1052 k 11101 03 1 11101 03 X 100017 1747 X 10 19 I 6022 X 1023 molecules 1k 7 O3 molecule 734 a AEhC7 AE6626gtlt10 Insgtlt2 998gtlt10 ms AE 1747 X 10 I Radiation with this wavelength is in the infrared portion of the spectrum Clearly processes other than simple photodissociation cause 0 3 to absorb ultraviolet radiation Formatted Indent Left 059quot Space After 0 pt Formatted Indent Left 059quot Don39t adjust space between Latin and Asian text 1137 X 1075111 a The electron configuration of Zr is Kr5s24d2 and that of Hf is Xe6s24f 5d2 Although Hf has electrons in f orbitals as the rare earth elements do the 4f subshell in Hf is lled and the 5d electrons primarily determine the chemical properties of the element Thus Hf should be chemically similar to Zr rather than the rare earth elements Formatted Font 12 pt Bold Formatted Font 12 pt Formatted Font color Black 1 4 u um C d 6102 a b C ZrC14s 4Nal gt Zrs 4NaCls This is an oxidationireduction reaction Na is oxidized and Zr is reduced 2er 25 4C12g 3Cs gt 2er14 s CO 2g 2COg 1 1Z0 2 lZCl 2330 ZCl 554ng0Z gtlt amp amp gtlt ng ZrC14 1232 g ZrOZ 2 mol ZrOZ 1 mol ZrCl4 In ionic compounds of the type MCl 4 and MO 2 the metal ions have a 4 charge indicating that the neutral atoms have lost four electrons Zr Kr5s24d2 loses the four electrons beyond its Kr core configuration Hf Xe6s24f145d2 similarly loses its four 6s and 5d electrons but not electrons from the complete 4f subshell Each oxide ion 02 carries a 27 charge Each metal oxide is a neutral compound so the metal ion or ions must adopt a total positive charge equal to the total negative charge of the oxide ions in the compound The table below lists the electron configuration of the neutral metal atom the positive charge of each metal ion in the oxide and the corresponding electron con guration of the metal ion i K Ar 451 1 Ar Ca Ar 4s2 2 Ar 111 Sc Ar 4s23d1 3 Ar 1v Ti Ar 4s23d2 4 Ar v V Ar 4s23d3 5 Ar vi Cr Ar 4s13d5 6 Ar Each metal atom loses all valence electrons beyond the Ar core configuration In K2 0 Sc 2 O 3 and V20 5 where the metal ions have odd charges two metal ions are required to produce a neutral oxide Formatted Space After 2 pt i potassium oxide calcium oxide iii scandiumIII oxide iv titanium IV oxide v vanadium V oxide vi chromium V I oxide Roman numerals are required to specify the charges on the transition metal ions because more than one stable ion may exist Recall that AH 0 for elements in their standard states In these reactions Ms and H 2 g are elements in their standard states Formatted Font 12 pt Bold Formatted Font 12 pt i K20s H2g gt2Ks H20g 1 1 u um u AH AH H 209 2AH Ks7AH K20s7AH H 2g AH 7 724182kJ 20773652k17071214kJ 3505 H2g gt 355 H20g AH AH H20gAH Cas74H CaOs 74H H 2g AH 7724182KJ 0776551k1707 3933kJ m Ti02s 2H2g gt Tis 2H 20g AH 2AH H20gAH Tis7AH Tio2s72AH H2g 2724182 0 7 79387 7 20 4551 k iv V205s 5H2g gt2Vs 5H20g AH 54H H 20g 24H Vs7AH v2o5s754H H 2g 5724182 20 7 715506 7 50 3415 k AH becomes more negative moving from left to right across this row of the Formatted Space After Zpt Line spacing periodic chart Since Sc lies between Ca and Ti the median of the two AH EECIsyigtzi tDontadju Space between Lam values is approximately 7785 kImol However the trend is clearly not linear Dividing the AH values by the positive charge on the pertinent metal ion produces the values 7363 118 7235 and 7310 The value between Ca 318 and Ti 4 7235 is Sc 3 r277 Multiplying r277 by 3 a value of approximately 7 830 k results A reasonable range of values for AH of Sc 2 O 3 s is then 7785 to r 830 kImol Formatted Indent Left 0quot Hanging 098quot pt BohI s theory was based on the Rutherford nuclear model of the atom That is Space After 0 Bohr theory assumed a dense positive charge at the center of the atom and a diffuse negative charge electrons surrounding it BohI s theory then specified the nature of the diffuse negative charge The prevailing theory before the nuclear model was Thomson s plum pudding or watermelon model with discrete electrons scattered about a diffuse positive charge cloud BohI s theory could not have been based on the Thomson model of the atom DeBroglie s hypothesis is that electrons exhibit both particle and wave properties Thomson s conclusion that electrons have mass is a particle property while the nature of cathode rays is a wave property De Broglie s hypothesis actually rationalizes these two seemingly contradictory observations about the properties of electrons Formatted Indent Left 098quot Space After 9 pt 238U 92 p 146 n 92 e 235U 92 p 143 n 92 e Formatted Font 12 pt Bold Formatted Font 12 pt In keeping with the de nition isotopes only the number of neutrons is different
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