CHAPTER 7 ATOMIC STRUCTURE AND PERIODICITY. Questions

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1 CHAPTER ATOMIC STRUCTURE AND PERIODICITY Questions. The equations relating the ters are ν c, E hν, and E /. Fro the equations, wavelength and frequency are inversely related, photon energy and frequency are directly related, and photon energy and wavelength are inversely related. The unit of Joule (J) kg /s. This is why you ust change ass units to kg when using the debroglie equation. 0. Frequency is the nuber of waves (cycles) of electroagnetic radiation per second that pass a given point in space. Speed refers to the distance a wave travels per unit tie. All electroagnetic radiation (EMR) travels at the sae speed (c, the speed of light.99 0 ). However, each wavelength of EMR has its own unique frequency,. The photoelectric effect refers to the phenoenon in which electrons are eitted fro the surface of a etal when light strikes it. The light ust have a certain iniu frequency (energy) in order to reove electrons fro the surface of a etal. Light having a frequency below the iniu results in no electrons being eitted, whereas light at or higher than the iniu frequency does cause electrons to be eitted. For light having a frequency higher than the iniu frequency, the excess energy is transferred into kinetic energy for the eitted electron. Albert Einstein explained the photoelectric effect by applying quantu theory.. The eission of light by excited atos has been the key interconnection between the acroscopic world we can observe and easure, and what is happening on a icroscopic basis within an ato. Excited atos eit light (which we can observe and easure) because of changes in the icroscopic structure of the ato. By studying the eissions of atos, we can trace back to what happened inside the ato. Specifically, our current odel of the ato relates the energy of light eitted to electrons in the ato oving fro higher allowed energy states to lower allowed energy states. 3. Exaple.3 calculates the debroglie wavelength of a ball and of an electron. The ball has a 34 wavelength on the order of 0. This is incredibly short and, as far as the wave- particle duality is concerned, the wave properties of large objects are insignificant. The electron, with 0 its tiny ass, also has a short wavelength; on the order of 0. However, this wavelength is significant because it is on the sae order as the spacing between atos in a typical crystal. For very tiny objects like electrons, the wave properties are iportant. The wave properties ust be considered, along with the particle properties, when hypothesizing about the electron otion in an ato.

2 CHAPTER ATOMIC STRUCTURE AND PERIODICITY 3 4. The Bohr odel was an iportant step in the developent of the current quantu echanical odel of the ato. The idea that electrons can only occupy certain, allowed energy levels is illustrated nicely (and relatively easily). We talk about the Bohr odel to present the idea of quantized energy levels. 5. For the radial probability distribution, the space around the hydrogen nucleus is cut up into a series of thin spherical shells. When the total probability of finding the electron in each spherical shell is plotted versus the distance fro the nucleus, we get the radial probability distribution graph. The plot initially shows a steady increase with distance fro the nucleus, reaches a axiu, then shows a steady decrease. Even though it is likely to find an electron near the nucleus, the volue of the spherical shell close to the nucleus is tiny, resulting in a low radial probability. The axiu radial probability distribution occurs at a distance of n fro the nucleus; the electron is ost likely to be found in the volue of the shell centered at this distance fro the nucleus. The n distance is the exact radius of innerost (n ) orbit in the Bohr odel. 6. The widths of the various blocks in the periodic table are deterined by the nuber of electrons that can occupy the specific orbital(s). In the s block, we have one orbital (R 0, R 0) that can hold two electrons; the s block is two eleents wide. For the f block, there are degenerate f orbitals (R 3, R!3,!,!, 0,,, 3), so the f block is 4 eleents wide. The g block corresponds to R 4. The nuber of degenerate g orbitals is 9. This coes fro the 9 possible R values when R 4 ( R!4,!3,!,!, 0,,, 3, 4). With 9 orbitals, each orbital holding two electrons, the g block would be eleents wide. The h block has R 5, R!5,!4,!3,!,!, 0,,, 3, 4, 5. With degenerate h orbitals, the h block would be eleents wide.. If one ore electron is added to a half-filled subshell, electron-electron repulsions will increase because two electrons ust now occupy the sae atoic orbital. This ay slightly decrease the stability of the ato.. Size decreases fro left to right and increases going down the periodic table. Thus, going one eleent right and one eleent down would result in a siilar size for the two eleents diagonal to each other. The ionization energies will be siilar for the diagonal eleents since the periodic trends also oppose each other. Electron affinities are harder to predict, but atos with siilar sizes and ionization energies should also have siilar electron affinities. 9. The valence electrons are strongly attracted to the nucleus for eleents with large ionization energies. One would expect these species to readily accept another electron and have very exotheric electron affinities. The noble gases are an exception; they have a large IE but have an endotheric EA. Noble gases have a filled valence shell of electrons. The added electron in a noble gas ust go into a higher n value atoic orbital, having a significantly higher energy, and this is very unfavorable. 30. Electron-electron repulsions becoe ore iportant when we try to add electrons to an ato. Fro the standpoint of electron-electron repulsions, larger atos would have ore favorable (ore exotheric) electron affinities. Considering only electron-nucleus attractions, saller atos would be expected to have the ore favorable (ore exotheric) EAs. These trends are exactly the opposite of each other. Thus the overall variation in EA is not as great as ionization energy in which attractions to the nucleus doinate.

3 4 CHAPTER ATOMIC STRUCTURE AND PERIODICITY 3. For hydrogen and one-electron ions (hydrogen-like ions), all atoic orbitals with the sae n value have the sae energy. For polyatoic atos/ions, the energy of the atoic orbitals also depends on l. Because there are ore nondegenerate energy levels for polyatoic atos/ions as copared to hydrogen, there are any ore possible electronic transitions resulting in ore coplicated line spectra. 3. Each eleent has a characteristic spectru because each eleent has unique energy levels. Thus the presence of the characteristic spectral lines of an eleent confirs its presence in any particular saple. 33. Yes, the axiu nuber of unpaired electrons in any configuration corresponds to a iniu in electron-electron repulsions. 34. The electron is no longer part of that ato. The proton and electron are copletely separated. 35. Ionization energy applies to the reoval of the electron fro an ato in the gas phase. The work function applies to the reoval of an electron fro the solid eleent. M(g) M + (g) + e ionization energy; M(s) M + (s) + e work function 36. Li + ions are the sallest of the alkali etal cations and will be ost strongly attracted to the water olecules. Exercises Light and Matter 3. ν c s MHz Hz s ; c 39. ν s c v / s s 3.0 E hν J s s J/photon photon J photons ol J/ol 40. E hν J s n 9 0 n.0 0 J/photon.0 0 photon J photons ol J/ol

4 CHAPTER ATOMIC STRUCTURE AND PERIODICITY 5 4. The wavelength is the distance between consecutive wave peaks. Wave a shows 4 wavelengths, and wave b shows wavelengths. Wave a: Wave b: Wave a has the longer wavelength. Because frequency and photon energy are both inversely proportional to wavelength, wave b will have the higher frequency and larger photon energy since it has the shorter wavelength. c.99 0 ν.5 0 s E J s J Because both waves are exaples of electroagnetic radiation, both waves travel at the sae speed, c, the speed of light. Fro Figure. of the text, both of these waves represent infrared electroagnetic radiation. 4. Referencing Figure. of the text,. 0 0 electroagnetic radiation is X rays. c / s.99 6 ν 0. 0 s Fro the wavelength calculated above, 0. MHz electroagnetic radiation is FM radiowaves J s.99 0 / s E J The J/photon electroagnetic radiation is visible (green) light. The photon energy and frequency order will be the exact opposite of the wavelength ordering because E and ν are both inversely related to. Fro the previously calculated wavelengths, the order of photon energy and frequency is: FM radiowaves < visible (green) light < X rays longest shortest lowest ν highest ν sallest E largest E

5 6 CHAPTER ATOMIC STRUCTURE AND PERIODICITY 43. E photon n J s n.3 0 J J photon.3 0 J ato C photon atos C 44. E photon hν, E photon J s J photon J photons ol kj 000 J. 0 6 kj/ol E photon J s J photon J photons ol kj 000 J. 0 kj/ol X rays do have an energy greater than the carbon-carbon bond energy. Therefore, X rays could conceivably break carbon-carbon bonds in organic copounds and thereby disrupt the function of an organic olecule. Radiowaves, however, do not have sufficient energy to break carbon-carbon bonds and are therefore relatively harless. 45. The energy needed to reove a single electron is: 9. kj ol ol kj J E, E J s J n kj ol ol kj J to reove one electron E, E J s J n 4. Ionization energy energy to reove an electron. 0 E photon E photon hν and ν c. So ν c and E.

6 CHAPTER ATOMIC STRUCTURE AND PERIODICITY E photon J s n. 0 J kj ol ol.4 0 kj atos ato.4 0 ato J ionization energy per ato E, E J s J n No, it will take light having a wavelength of 34.4 n or less to ionize gold. A photon of light having a wavelength of 5 n is longer wavelength and thus lower energy than 34.4 n light. 49. a. 0.% of speed of light b. h, v J s kg kg Note: For units to coe out, the ass ust be in kg because J s h J s 3.4 v kg n 0 n. This nuber is so sall that it is insignificant. We cannot detect a wavelength this sall. The eaning of this nuber is that we do not have to worry about the wave properties of large objects. 50. a. b. h J s.3 v.65 0 kg ( ) h h J s, v v kg h, v h v J s ( ).6 0 kg This particle is probably a proton or a neutron. 5. h, v h v ; for.0 0 n.0 0 :

7 CHAPTER ATOMIC STRUCTURE AND PERIODICITY v J s kg For.0 n : v J s kg Hydrogen Ato: The Bohr Model 53. For the H ato (Z ): E n J/n ; for a spectral transition, E E f! E i : E. 0 J nf n i where n i and n f are the levels of the initial and final states, respectively. A positive value of E always corresponds to an absorption of light, and a negative value of E always corresponds to an eission of light. a. E!. 0 J 3!. 0 J 4 9 E!. 0 J (0.500! 0.)! J The photon of light ust have precisely this energy ( J). E E photon hν, E J s J n Fro Figure., this is visible electroagnetic radiation (red light). b. E. 0 J J E J s J n This is visible electroagnetic radiation (green-blue light). c. E -. 0 J! J J s J This is ultraviolet electroagnetic radiation. 0.6 n

8 CHAPTER ATOMIC STRUCTURE AND PERIODICITY 54. a. E. 0 J 3 4! J E J s J n Fro Figure., this is infrared electroagnetic radiation. b. E!. 0 J 4 5! J E J s J c. E!. 0 J 3 5! J n (infrared) E J s J. 6 0 n (infrared) a. 3 E a b b. 4 n c c. Energy levels are not to scale b a a c 3 b. 5 4 E c. 5 3 n Energy levels are not to scale. 5. The longest wavelength light eitted will correspond to the transition with the sallest energy change (sallest E). This is the transition fro n 6 to n 5.

9 0 CHAPTER ATOMIC STRUCTURE AND PERIODICITY E. 0 J J J s n E J The shortest wavelength eitted will correspond to the largest E; this is n 6 n. E. 0 J 6. 0 J J s n E. 0 J 5. There are 4 possible transitions for an electron in the n 5 level (5 4, 5 3, 5, and 5 ). If an electron initially drops to the n 4 level, three additional transitions can occur (4 3, 4, and 4 ). Siilarly, there are two ore transitions fro the n 3 level (3, 3 ) and one ore transition for the n level ( ). There are a total of 0 possible transitions for an electron in the n 5 level for a possible total of 0 different wavelength eissions. 59. E. 0 J nf n i. 0 J J E photon J s / s n E.09 0 J Because wavelength and energy are inversely related, visible light ( n) is not energetic enough to excite an electron in hydrogen fro n to n 5. E. 0 J J J s / s n E J Visible light with 40.4 n will excite an electron fro the n to the n 6 energy level. 60. a. False; it takes less energy to ionize an electron fro n 3 than fro the ground state. b. True

10 CHAPTER ATOMIC STRUCTURE AND PERIODICITY c. False; the energy difference between n 3 and n is saller than the energy difference between n 3 and n ; thus the wavelength is larger for the n 3 n electronic transition than for the n 3 n transition. E and are inversely proportional to each other (E /). d. True e. False; n is the first excited state, and n 3 is the second excited state. 6. Ionization fro n corresponds to the transition n i n f 4, where E 4 0. E E 4! E!E J E photon E J s J n To ionize fro n, E E! E!E J J s n J 6. E E 4! E n -E n. 0 J n E photon J s J E photon E.36 0 J. 0, n n 6.0, n E E photon hν J s s J E!4.5 0 J because we have an eission.!4.5 0 J E n E 5!. 0 J 5, n n 5 0.0, n 0.50, n 4, n The electronic transition is fro n 5 to n. 64. E E photon J s J

11 CHAPTER ATOMIC STRUCTURE AND PERIODICITY E -5.00! J because we have an eission. 0 J E E n. 4, 0.004, n n n 0 J Quantu Mechanics, Quantu Nubers, and Orbitals n 65. a. p v kg s 3 kg p x h, 4π x h J s π p (9. 0 kg / s) b. x h J s π p kg 0.00 / s c. The diaeter of an H ato is roughly.0 0 c. The uncertainty in position is uch larger than the size of the ato. d. The uncertainty is insignificant copared to the size of a baseball. 66. Units of E C t J s, the sae as the units of Planck's constant. kg kg Units of (v) C x kg s J s s s s 6. n,, 3,... ; l 0,,,... (n - ); l -l... -, -, 0,,,...+l 6. p: n, l is not possible; 3f: n 3, l 3 is not possible; d: n, l is not possible; In all three incorrect cases, n l. The axiu value l can have is n -, not n. 69. b. For l 3, l can range fro -3 to +3; thus +4 is not allowed. c. n cannot equal zero. d. l cannot be a negative nuber. 0. a. For n 3, l 3 is not possible. d. s cannot equal!. e. l cannot be a negative nuber. f. For l, l cannot equal. The quantu nubers in parts b and c are allowed.

12 CHAPTER ATOMIC STRUCTURE AND PERIODICITY 3. ψ gives the probability of finding the electron at that point.. The diagras of the orbitals in the text give only 90% probabilities of where the electron ay reside. We can never be 00% certain of the location of the electrons due to Heisenburg s uncertainty principle. Polyelectronic Atos 3. 5p: three orbitals 3 d z : one orbital 4d: five orbitals n 5: l 0 ( orbital), l (3 orbitals), l (5 orbitals), l 3 ( orbitals), l 4 (9 orbitals); total for n 5 is 5 orbitals. n 4: l 0 (), l (3), l (5), l 3 (); total for n 4 is 6 orbitals. 4. p, 0 electrons (l when n ); 6 d, electrons (specifies one atoic orbital); 4f, 4 x y electrons ( orbitals have 4f designation); p y, electrons (specifies one atoic orbital); s, electrons (specifies one atoic orbital); n 3, electrons (3s, 3p, and 3d orbitals are possible; there are one 3s orbital, three 3p orbitals, and five 3d orbitals). 5. a. n 4: l can be 0,,, or 3. Thus we have s ( e ), p (6 e ), d (0 e ), and f (4 e ) orbitals present. Total nuber of electrons to fill these orbitals is 3. b. n 5, l +: For n 5, l 0,,, 3, 4. For l,, 3, 4, all can have l +. Four distinct orbitals, thus electrons. c. n 5, s +/: For n 5, l 0,,, 3, 4. Nuber of orbitals, 3, 5,, 9 for each value of l, respectively. There are 5 orbitals with n 5. They can hold 50 electrons, and 5 of these electrons can have s +/. d. n 3, l : These quantu nubers define a set of 3d orbitals. There are 5 degenerate 3d orbitals that can hold a total of 0 electrons. e. n, l : These define a set of p orbitals. There are 3 degenerate p orbitals that can hold a total of 6 electrons. 6. a. It is ipossible to have n 0. Thus no electrons can have this set of quantu nubers. b. The four quantu nubers copletely specify a single electron in a p orbital. c. n 3, s +/: 3s, 3p, and 3d orbitals all have n 3. These nine orbitals can each hold one electron with s +/; 9 electrons can have these quantu nubers d. n, l : this cobination is not possible (l for n ). Zero electrons in an ato can have these quantu nubers. e. n, l 0, l 0: these define a s orbital that can hold electrons.

13 4 CHAPTER ATOMIC STRUCTURE AND PERIODICITY. a. Na: s s p 6 3s ; Na has unpaired electron. or s s p 3s 3s b. Co: s s p 6 3s 3p 6 4s 3d ; Co has 3 unpaired electrons. s s p 3s 3p or 4s 3d 3d c. Kr: s s p 6 3s 3p 6 4s 3d 0 4p 6 ; Kr has 0 unpaired electrons. s s p 3s 3p 4s 3d 4p. The two exceptions are Cr and Cu. Cr: s s p 6 3s 3p 6 4s 3p 5 ; Cr has 6 unpaired electrons. s s p 3s 3p or or 4s 4s 3d 3d Cu: s s p 6 3s 3p 6 4s 3d 0 ; Cu has unpaired electron. s s p 3s 3p or 4s 4s 3d

14 CHAPTER ATOMIC STRUCTURE AND PERIODICITY 5 9. Si: s s p 6 3s 3p or [Ne]3s 3p ; Ga: s s p 6 3s 3p 6 4s 3d 0 4p or [Ar]4s 3d 0 4p As: [Ar]4s 3d 0 4p 3 ; Ge: [Ar]4s 3d 0 4p ; Al: [Ne]3s 3p ; Cd: [Kr]5s 4d 0 S: [Ne]3s 3p 4 ; Se: [Ar]4s 3d 0 4p 4 0. Cu: [Ar]4s 3d 9 (using periodic table), [Ar]4s 3d 0 (actual) O: s s p 4 ; La: [Xe]6s 5d ; Y: [Kr]5s 4d ; Ba: [Xe]6s Tl: [Xe]6s 4f 4 5d 0 6p ; Bi: [Xe]6s 4f 4 5d 0 6p 3. The following are coplete electron configurations. Noble gas shorthand notation could also be used. Sc: s s p 6 3s 3p 6 4s 3d ; Fe: s s p 6 3s 3p 6 4s 3d 6 P: s s p 6 3s 3p 3 ; Cs: s s p 6 3s 3p 6 4s 3d 0 4p 6 5s 4d 0 5p 6 6s Eu: s s p 6 3s 3p 6 4s 3d 0 4p 6 5s 4d 0 5p 6 6s 4f 6 5d * Pt: s s p 6 3s 3p 6 4s 3d 0 4p 6 5s 4d 0 5p 6 6s 4f 4 5d * Xe: s s p 6 3s 3p 6 4s 3d 0 4p 6 5s 4d 0 5p 6 ; Br: s s p 6 3s 3p 6 4s 3d 0 4p 5 *Note: These electron configurations were predicted using only the periodic table. The actual electron configurations are: Eu: [Xe]6s 4f and Pt: [Xe]6s 4f 4 5d 9. Cl: ls s p 6 3s 3p 5 or [Ne]3s 3p 5 Sb: [Kr]5s 4d 0 5p 3 Sr: s s p 6 3s 3p 6 4s 3d 0 4p 6 5s or [Kr]5s W: [Xe]6s 4f 4 5d 4 Pb: [Xe]6s 4f 4 5d 0 6p Cf: [Rn]s 5f 0 * *Note: Predicting electron configurations for lanthanide and actinide eleents is difficult since they have 0,, or electrons in d orbitals. This is the actual Cf electron configuration. 3. a. Both In and I have one unpaired 5p electron, but only the nonetal I would be expected to for a covalent copound with the nonetal F. One would predict an ionic copound to for between the etal In and the nonetal F. I: [Kr]5s 4d 0 5p 5 5p b. Fro the periodic table, this will be eleent 0. Eleent 0: [Rn]s 5f 4 6d 0 p 6 s c. Rn: [Xe]6s 4f 4 5d 0 6p 6 ; note that the next discovered noble gas will also have 4f electrons (as well as 5f electrons).

15 6 CHAPTER ATOMIC STRUCTURE AND PERIODICITY d. This is chroiu, which is an exception to the predicted filling order. Cr has 6 unpaired electrons, and the next ost is 5 unpaired electrons for Mn. Cr: [Ar]4s 3d 5 4s 3d 4. a. As: s s p 6 3s 3p 6 4s 3d 0 4p 3 b. Eleent 6 will be below Po in the periodic table: [Rn]s 5f 4 6d 0 p 4 c. Ta: [Xe]6s 4f 4 5d 3 or Ir: [Xe]6s 4f 4 5d d. At: [Xe]6s 4f 4 5d 0 6p 5. Note that eleent (when it is discovered) will also have electrons in the 6p atoic orbitals (as well as electrons in the p atoic orbitals). 5. a. The lightest halogen is fluorine: s s p 5 b. K: s s p 6 3s 3p 6 4s c. In: [Kr]5s 4d 0 5p d. C: s s p ; Si: s s p 6 3s 3p 6. a. This ato has 0 electrons. Ne b. S c. The predicted ground state configuration is [Kr]5s 4d 9. Fro the periodic table, the eleent is Ag. Note: [Kr]5s 4d 0 is the actual ground state electron configuration for Ag. d. Bi: [Xe]6s 4f 4 5d 0 6p 3 ; the three unpaired electrons are in the 6p orbitals.. Hg: s s p 6 3s 3p 6 4s 3d 0 4p 6 5s 4d 0 5p 6 6s 4f 4 5d 0 a. Fro the electron configuration for Hg, we have 3s, 3p 6, and 3d 0 electrons; total electrons with n 3. b. 3d 0, 4d 0, 5d 0 ; 30 electrons are in the d atoic orbitals. c. p 6, 3p 6, 4p 6, 5p 6 ; each set of np orbitals contain one p z atoic orbital. Because we have 4 sets of np orbitals and two electrons can occupy the p z orbital, there are 4() electrons in p z atoic orbitals. d. All the electrons are paired in Hg, so one-half of the electrons are spin up ( s +/) and the other half are spin down ( s!/). 40 electrons have spin up.. Eleent 5, Uup, is in Group 5A under Bi (bisuth): Uup: s s p 6 3s 3p 6 4s 3d 0 4p 6 5s 4d 0 5p 6 6s 4f 4 5d 0 6p 6 s 5f 4 6d 0 p 3 a. 5s, 5p 6, 5d 0, and 5f 4 ; 3 electrons have n 5 as one of their quantu nubers b. l 3 are f orbitals. 4f 4 and 5f 4 are the f orbitals used. They are all filled, so electrons have l 3.

16 CHAPTER ATOMIC STRUCTURE AND PERIODICITY c. p, d, and f orbitals all have one of the degenerate orbitals with l. There are 6 orbitals with l for the various p orbitals used; there are 4 orbitals with l for the various d orbitals used; and there are orbitals with l for the various f orbitals used. We have a total of orbitals with l. Eleven of these orbitals are filled with electrons, and the p orbitals are only half-filled. The nuber of electrons with l is ( e ) + ( e ) 3 electrons. d. The first electrons are all paired; one-half of these electrons (56 e ) will have s!/. The 3 electrons in the p orbitals singly occupy each of the three degenerate p orbitals; the three electrons are spin parallel, so the p electrons either have s +/ or s!/. Therefore, either 56 electrons have s!/ or 59 electrons have s!/. 9. B: s s p n l l s s 0 0 +/ s 0 0!/ s 0 0 +/ s 0 0!/ p*! +/ *This is only one of several possibilities for the p electron. The p electron in B could have l!, 0 or + and s +/ or!/ for a total of six possibilities. N: s s p 3 n l l s s 0 0 +/ s 0 0!/ s 0 0 +/ s 0 0!/ p - +/ p 0 +/ (Or all p electrons could have s!/.) p + +/ 90. Ti : [Ar]4s 3d n l l s 4s / 4s 4 0 0!/ 3d 3! +/ Only one of 0 possible cobinations of l and s for the first d electron. For the ground state, the second d electron should be in a different orbital with spin parallel; 4 possibilities. 3d 3! +/

17 CHAPTER ATOMIC STRUCTURE AND PERIODICITY 9. Group A: valence electron; ns ; Li: [He]s ; s is the valence electron configuration for Li. Group A: valence electrons; ns ; Ra: [Rn]s ; s is the valence electron configuration for Ra. Group 3A: 3 valence electrons; ns np ; Ga: [Ar]4s 3d 0 4p ; 4s 4p is the valence electron configuration for Ga. Note that valence electrons for the representative eleents of Groups A-A are considered those electrons in the highest n value, which for Ga is n 4. We do not include the 3d electrons as valence electrons because they are not in n 4 level. Group 4A: 4 valence electrons; ns np ; Si: [Ne]3s 3p ; 3s 3p is the valence electron configuration for Si. Group 5A: 5 valence electrons; ns np 3 ; Sb: [Kr]5s 4d 0 5p 3 ; 5s 5p 3 is the valence electron configuration for Sb. Group 6A: 6 valence electrons; ns np 4 ; Po: [Xe]6s 4f 4 5d 0 6p 4 ; 6s 6p 4 is the valence electron configuration for Po. Group A: valence electrons; ns np 5 ; : [Rn]s 5f 4 6d 0 p 5 ; s p 5 is the valence electron configuration for. Group A: valence electrons; ns np 6 ; Ne: [He]s p 6 ; s p 6 is the valence electron configuration for Ne. 9. a. valence electrons; 4s b. 6 valence electrons; s p 4 c. valence electrons; s p 5 d. 3 valence electrons; 5s 5p e. valence electrons; 3s 3p 6 f. 5 valence electrons; 6s 6p O: s s p x p y ( ); there are no unpaired electrons in this oxygen ato. This configuration would be an excited state, and in going to the ore stable ground state ( ), energy would be released. 94. The nuber of unpaired electrons is in parentheses. a. excited state of boron () b. ground state of neon (0) B ground state: s s p () Ne ground state: s s p 6 (0) c. exited state of fluorine (3) d. excited state of iron (6) F ground state: s s p 5 () Fe ground state: [Ar]4s 3d 6 (4) p 3d 95. None of the s block eleents have unpaired electrons. In the p block, the eleents with either ns np or ns np 4 valence electron configurations have unpaired electrons. For eleents -36, these are eleents C, Si, and Ge (with ns np ) and eleents O, S, and Se (with ns np 4 ). For the d block, the eleents with configurations nd or nd have two unpaired electrons. For eleents -36, these are Ti (3d ) and Ni (3d ). A total of eleents fro the first 36 eleents have two unpaired electrons in the ground state.

18 CHAPTER ATOMIC STRUCTURE AND PERIODICITY The s block eleents with ns for a valence electron configuration have one unpaired electron. These are eleents H, Li, Na, and K for the first 36 eleents. The p block eleents with ns np or ns np 5 valence electron configurations have one unpaired electron. These are eleents B, Al, and Ga (ns np ) and eleents F, Cl, and Br (ns np 5 ) for the first 36 eleents. In the d block, Sc ([Ar]4s 3d ) and Cu ([Ar]4s 3d 0 ) each have one unpaired electron. A total of eleents fro the first 36 eleents have one unpaired electron in the ground state. 9. We get the nuber of unpaired electrons by exaining the incopletely filled subshells. The paraagnetic substances have unpaired electrons, and the ones with no unpaired electrons are not paraagnetic (they are called diaagnetic). Li: s s ; paraagnetic with unpaired electron. s N: s s p 3 ; paraagnetic with 3 unpaired electrons. p Ni: [Ar]4s 3d ; paraagnetic with unpaired electrons. 3d Te: [Kr]5s 4d 0 5p 4 ; paraagnetic with unpaired electrons. 5p Ba: [Xe]6s ; not paraagnetic because no unpaired electrons are present. 6s Hg: [Xe]6s 4f 4 5d 0 ; not paraagnetic because no unpaired electrons. 5d 9. We get the nuber of unpaired electrons by exaining the incopletely filled subshells. O: [He]s p 4 p 4 : two unpaired e O + : [He]s p 3 p 3 : three unpaired e O : [He]s p 5 p 5 : one unpaired e Os: [Xe]6s 4f 4 5d 6 5d 6 : four unpaired e Zr: [Kr]5s 4d 4d : two unpaired e S: [Ne]3s 3p 4 3p 4 : two unpaired e F: [He]s p 5 p 5 : one unpaired e Ar: [Ne]3s 3p 6 3p 6 zero unpaired e

19 30 CHAPTER ATOMIC STRUCTURE AND PERIODICITY The Periodic Table and Periodic Properties 99. Size (radius) decreases left to right across the periodic table, and size increases fro top to botto of the periodic table. a. S < Se < Te b. Br < Ni < K c. F < Si < Ba All follow the general radius trend. 00. a. Be < Na < Rb b. Ne < Se < Sr c. O < P < Fe All follow the general radius trend. 0. The ionization energy trend is the opposite of the radius trend; ionization energy (IE), in general, increases left to right across the periodic table and decreases fro top to botto of the periodic table. a. Te < Se < S b. K < Ni < Br c. Ba < Si < F All follow the general IE trend. 0. a. Rb < Na < Be b. Sr < Se < Ne c. Fe < P < O All follow the general IE trend. 03. a. He b. Cl c. Eleent 6 is the next oxygen faily eber to be discovered (under Po), eleent is the next alkali etal to be discovered (under Fr), and eleent 0 is the next alkaline earth etal to be discovered (under Ra). Fro the general radius trend, eleent 6 will be the sallest. d. Si e. Na + ; this ion has the fewest electrons as copared to the other sodiu species present. Na + has the sallest nuber of electron-electron repulsions, which akes it the sallest ion with the largest ionization energy. 04. a. Ba b. K c. O; in general, Group 6A eleents have a lower ionization energy than neighboring Group 5A eleents. This is an exception to the general ionization energy trend across the periodic table. d. S ; this ion has the ost electrons copared to the other sulfur species present. S has the largest nuber of electron-electron repulsions, which leads to S having the largest size and sallest ionization energy. e. Cs; this follows the general ionization energy trend.

20 CHAPTER ATOMIC STRUCTURE AND PERIODICITY a. Sg: [Rn]s 5f 4 6d 4 b. W c. Siilar to chroiu oxide copounds/ions, SgO 3, Sg O 3, SgO 4, and Sg O soe likely possibilities. are 06. a. Uus will have electrons. [Rn]s 5f 4 6d 0 p 5 b. It will be in the halogen faily and will be ost siilar to astatine (At). c. Uus should for charged anions like the other halogens. Like the other halogens, soe possibilities are NaUus, Mg(Uus), C(Uus) 4, and O(Uus) d. Assuing Uus is like the other halogens, soe possibilities are UusO -, UusO -, UusO 3 -, and UusO As: [Ar]4s 3d 0 4p 3 ; Se: [Ar]4s 3d 0 4p 4 ; the general ionization energy trend predicts that Se should have a higher ionization energy than As. Se is an exception to the general ionization energy trend. There are extra electron-electron repulsions in Se because two electrons are in the sae 4p orbital, resulting in a lower ionization energy for Se than predicted. 0. Expected order fro IE trend: Be < B < C < N < O B and O are exceptions to the general IE trend. The IE of O is lower because of the extra electron-electron repulsions present when two electrons are paired in the sae orbital. This akes it slightly easier to reove an electron fro O copared to N. B is an exception because of the saller penetrating ability of the p electron in B copared to the s electrons in Be. The saller penetrating ability akes it slightly easier to reove an electron fro B copared to Be. The correct IE ordering, taking into account the two exceptions, is B < Be < C < O < N. 09. a. More favorable EA: C and Br; the electron affinity trend is very erratic. Both N and Ar have positive EA values (unfavorable) due to their electron configurations (see text for detailed explanation). b. Higher IE: N and Ar (follows the IE trend) c. Larger size: C and Br (follows the radius trend) 0. a. More favorable EA: K and Cl; Mg has a positive EA value, and F has a ore positive EA value than expected fro its position relative to Cl. b. Higher IE: Mg and F c. Larger radius: K and Cl. Al(!44), Si(!0), P(!4), S(!00.4), Cl(!34.); based on the increasing nuclear charge, we would expect the electron affinity (EA) values to becoe ore exotheric as we go fro left to right in the period. Phosphorus is out of line. The reaction for the EA of P is: P(g) + e P (g) [Ne]3s 3p 3 [Ne]3s 3p 4

21 3 CHAPTER ATOMIC STRUCTURE AND PERIODICITY The additional electron in P - will have to go into an orbital that already has one electron. There will be greater repulsions between the paired electrons in P -, causing the EA of P to be less favorable than predicted based solely on attractions to the nucleus.. Electron affinity refers to the energy associated with the process of adding an electron to soething. Be, N, and Ne all have endotheric (unfavorable) electron affinity values. In order to add an electron to Be, N, or Ne, energy ust be added. Another way of saying this is that Be, N, and Ne becoe less stable (have a higher energy) when an electron is added to each. To rationalize why those three atos have endotheric (unfavorable) electron affinity values, let s see what happens to the electron configuration as an electron is added. Be(g) + e Be (g) N(g) + e N (g) [He]s [He]s p [He]s p 3 [He]s p 4 Ne(g) + e Ne (g) [He]s p 6 [He]s p 6 3s In each case soething energetically unfavorable occurs when an electron is added. For Be, the added electron ust go into a higher-energy p atoic orbital because the s orbital is full. In N, the added electron ust pair up with another electron in one of the p atoic orbitals; this adds electron-electron repulsions. In Ne, the added electron ust be added to a uch higher 3s atoic orbital because the n orbitals are full. 3. The electron affinity trend is very erratic. In general, EA decreases down the periodic table, and the trend across the table is too erratic to be of uch use. a. Se < S; S is ost exotheric. b. I < Br < F < Cl; Cl is ost exotheric. (F is an exception). 4. a. N < O < F, F is ost exotheric. b. Al < P < Si; Si is ost exotheric. 5. Electron-electron repulsions are uch greater in O than in S because the electron goes into a saller p orbital versus the larger 3p orbital in sulfur. This results in a ore favorable (ore exotheric) EA for sulfur. 6. O; the electron-electron repulsions will be uch ore severe for O + e O than for O + e O.. a. Se 3+ (g) Se 4+ (g) + e b. S (g) + e S (g) c. Fe 3+ (g) + e Fe + (g) d. Mg(g) Mg + (g) + e. a. The electron affinity of Mg + is H for Mg + (g) + e Mg + (g); this is just the reverse of the second ionization energy for Mg. EA(Mg + )!IE (Mg)!445 kj/ol (Table.5). b. IE of Cl is H for Cl (g) Cl(g) + e ; IE(Cl )!EA(Cl) 34. kj/ol (Table.) c. Cl + (g) + e Cl(g) H!IE (Cl)!55 kj/ol EA(Cl + )

22 CHAPTER ATOMIC STRUCTURE AND PERIODICITY 33 d. Mg (g) Mg(g) + e H!EA(Mg)!30 kj/ol IE(Mg ) Alkali Metals. It should be potassiu peroxide (K O ) because K + ions are stable in ionic copounds. K + ions are not stable; the second ionization energy of K is very large copared to the first. 0. a. Li 3 N; lithiu nitride b. NaBr; sodiu broide c. K S; potassiu sulfide c. ν s E hν J s s J c. For 59.0 n: ν s E hν J s s J For 59.6 n: ν c/ s ; E hν J The energies in kj/ol are: J 0 J kj 000 J kj 000 J kj/ol ol kj/ol ol 3. Yes; the ionization energy general trend is to decrease down a group, and the atoic radius trend is to increase down a group. The data in Table. confir both of these general trends. 4. It should be eleent with the ground state electron configuration [Rn]s 5f 4 6d 0 p 6 s. 5. a. 6 Li(s) + N (g) Li 3 N(s) b. Rb(s) + S(s) Rb S(s) 6. a. Cs(s) + H O(l) CsOH(aq) + H (g) b. Na(s) + Cl (g) NaCl(s) Connecting to Biocheistry c. ν c, ν n 9 0 n s

23 34 CHAPTER ATOMIC STRUCTURE AND PERIODICITY c. 0 n: ν n 9 0 n. 0 5 s 30 n: ν n s The copounds in the sunscreen absorb ultraviolet B (UVB) electroagnetic radiation having a frequency fro s to. 0 5 s. c. a. ν s b. Fro Figure., this is infrared electroagnetic radiation. c. E hν J s s J/photon photon J photons ol J/ol d. Frequency and photon energy are directly related (E hv). Because s EMR has a lower frequency than s EMR, the s EMR will have less energetic photons. c 30. S-type cone receptors: ν s n n s S-type cone receptors detect n light. Fro Figure. in the text, this is violet to green light. M-type cone receptors: n s n s M-type cone receptors detect n light. Fro Figure. in the text, this is blue to orange light.

24 CHAPTER ATOMIC STRUCTURE AND PERIODICITY 35 L-type cone receptors: n s n s L-type cone receptors detect n light. This represents green to red light. 3. O: s s p 4 ; C: s s p ; H: s ; N: s s p 3 ; Ca: [Ar]4s ; P: [Ne]3s p 3 ; Mg: [Ne]3s ; K: [Ar]4s 3. Cr: [Ar]4s 3d 5, 6 unpaired electrons (Cr is an exception to the noral filling order); Mn: [Ar]4s 3d 5, 5 unpaired e ; Fe: [Ar]4s 3d 6, 4 unpaired e ; Co: [Ar]4s 3d, 3 unpaired e ; Ni: [Ar]4s 3d, unpaired e ; Cu: [Ar] 4s 3d 0, unpaired e (Cu is also an exception to the noral filling order); Zn: [Ar]4s 3d 0, 0 unpaired e. 33. Fro the radii trend, the sallest-size eleent (excluding hydrogen) would be the one in the ost upper right corner of the periodic table. This would be O. The largest-size eleent would be the one in the ost lower left of the periodic table. Thus K would be the largest. The ionization energy trend is the exact opposite of the radii trend. So K, with the largest size, would have the sallest ionization energy. Fro the general IE trend, O should have the largest ionization energy. However, there is an exception to the general IE trend between N and O. Due to this exception, N would have the largest ionization energy of the eleents exained. 34. No; lithiu etal is very reactive. It will react soewhat violently with water, aking it copletely unsuitable for huan consuption. Lithiu has a low first ionization energy, so it is ore likely that the lithiu prescribed will be in the for of a soluble lithiu salt (a soluble ionic copound with Li + as the cation). Additional Exercises 35. E 30 kj ol E, ol kj J E J s Energy to ake water boil s T.99 0 J n 4. J 50.0 g 5.0 C J C g E photon J s J J s 50. J 0.9 s; J photon J.0 0 photons

25 36 CHAPTER ATOMIC STRUCTURE AND PERIODICITY k 000 k s s (about 3 inutes) 3. E J s J Fro the spectru, c is greenish yellow light. 00 c c 39. E!R H nf ni!. 0 J J 6 E J s / s J 0 00 c c 5 Fro the spectru, c is violet light, so the n 6 to n visible spectru line is violet. 40. Exceptions: Cr, Cu, Nb, Mo, Tc, Ru, Rh, Pd, Ag, Pt, and Au; Tc, Ru, Rh, Pd, and Pt do not correspond to the supposed extra stability of half-filled and filled subshells. 4. a. True for H only. b. True for all atos. c. True for all atos. 4. n 5; l -4, -3, -, -, 0,,, 3, 4; electrons 43. When the p and d orbital functions are evaluated at various points in space, the results soeties have positive values and soeties have negative values. The ter phase is often associated with the + and! signs. For exaple, a sine wave has alternating positive and negative phases. This is analogous to the positive and negative values (phases) in the p and d orbitals. 44. He: s ; Ne: s s p 6 ; Ar: s s p 6 3s 3p 6 ; each peak in the diagra corresponds to a subshell with different values of n. Corresponding subshells are closer to the nucleus for heavier eleents because of the increased nuclear charge. 45. The general ionization energy trend says that ionization energy increases going left to right across the periodic table. However, one of the exceptions to this trend occurs between Groups A and 3A. Between these two groups, Group 3A eleents usually have a lower ionization energy than Group A eleents. Therefore, Al should have the lowest first ionization energy value, followed by Mg, with Si having the largest ionization energy. Looking at the values for the first ionization energy in the graph, the green plot is Al, the blue plot is Mg, and the red plot is Si. Mg (the blue plot) is the eleent with the huge jup between I and I 3. Mg has two valence electrons, so the third electron reoved is an inner core electron. Inner core electrons are

26 CHAPTER ATOMIC STRUCTURE AND PERIODICITY 3 always uch ore difficult to reove than valence electrons since they are closer to the nucleus, on average, than the valence electrons. 46. a. The 4+ ion contains 0 electrons. Thus the electrically neutral ato will contain 4 electrons. The atoic nuber is 4, which identifies it as chroiu. b. The ground state electron configuration of the ion ust be s s p 6 3s 3p 6 4s 0 3d ; there are 6 electrons in s orbitals. c. d. e. Fro the ass, this is the isotope 50 4 Cr. There are 6 neutrons in the nucleus. f. s s p 6 3s 3p 6 4s 3d 5 is the ground state electron configuration for Cr. Cr is an exception to the noral filling order. 4. Valence electrons are easier to reove than inner-core electrons. The large difference in energy between I and I 3 indicates that this eleent has two valence electrons. This eleent is ost likely an alkaline earth etal since alkaline earth etal eleents all have two valence electrons. 4. All oxygen faily eleents have ns np 4 valence electron configurations, so this nonetal is fro the oxygen faily. a valence electrons. b. O, S, Se, and Te are the nonetals fro the oxygen faily (Po is a etal). c. Because oxygen faily nonetals for charged ions in ionic copounds, K X would be the predicted forula, where X is the unknown nonetal. d. Fro the size trend, this eleent would have a saller radius than bariu. e. Fro the ionization energy trend, this eleent would have a saller ionization energy than fluorine. 49. a. Na(g) Na + (g) + e IE 495 kj Cl(g) + e Cl (g) EA!34. kj Na(g) + Cl(g) Na + (g) + Cl g) H 46 kj b. Mg(g) Mg + (g) + e IE 35 kj F(g) + e F - (g) EA!3. kj Mg(g) + F(g) Mg + (g) + F (g) H 40 kj c. Mg + (g) Mg + (g) + e IE 445 kj F(g) + e F (g) EA!3. kj Mg + (g) + F(g) Mg + (g) + F (g) H kj

27 3 CHAPTER ATOMIC STRUCTURE AND PERIODICITY d. Using parts b and c, we get: Mg(g) + F(g) Mg + (g) + F (g) H 40 kj Mg + (g) + F(g) Mg + (g) + F (g) H kj Mg(g) + F(g) Mg + (g) + F (g) H 54 kj 50. Applying the general trends in radii and ionization energy allows atching the various values to the eleents. Ar : s s p 6 3s 3p 6 Mg : s s p 6 3s :.5 MJ/ol : 0.9 D : 0.35 MJ/ol :.60 D K : s s p 6 3s 3p 6 4s : 0.4 MJ/ol :.35 D Size: Ar < Mg < K; IE: K < Mg < Ar Challenge Probles h 5., where ass and v velocity; v v rs 3RT h 3RT h 3RT For one ato, R.345 J K ol ol atos J K ato J s, 3 3(.3 0 )(33 K) kg g Molar ass ato 3 g ol atos 3.0 g/ol The ato is sulfur (S). 5. E photon J s J E.39 0 J; the general energy equation for one-electron ions is E n!. 0 J (Z )/n, where Z atoic nuber.

28 CHAPTER ATOMIC STRUCTURE AND PERIODICITY 39 E. 0 J (Z), Z 4 for Be 3+ n f n i E J. 5 n f, 0 (4) nf 5 n f , n f 4 This eission line corresponds to the n 5 n 4 electronic transition. 53. a. Because wavelength is inversely proportional to energy, the spectral line to the right of B (at a larger wavelength) represents the lowest possible energy transition; this is n 4 to n 3. The B line represents the next lowest energy transition, which is n 5 to n 3, and the A line corresponds to the n 6 to n 3 electronic transition. b. Because this spectru is for a one-electron ion, E n. 0 J (Z /n ). To deterine E and, in turn, the wavelength of spectral line A, we ust deterine Z, the atoic nuber of the one electron species. Use spectral line B data to deterine Z. E J Z 3 Z Z 9 5 E J s( / s) J Because an eission occurs, E J. E J. 0 J 6 Z, Z 9.00, Z 3; the ion is Li Solving for the wavelength of line A: E (3) J E J s( J / s) n 54. For hydrogen: E!. 0 - J 5! J For a siilar blue light eission, He + will need about the sae E value.

29 40 CHAPTER ATOMIC STRUCTURE AND PERIODICITY For He + : E n!. E! n 4 f 0 J (Z /n ), where Z : 0 J!. 0 - J 4 6 n f 4, n, n f.949 f The transition fro n 4 to n 3 for He + should eit a siilar colored blue light as the n 5 to n hydrogen transition; both these transitions correspond to very nearly the sae energy change. 55. For one-electron species, E n R Z /n. IE is for the n n transition. So: H 4 IE E E E R H Z / n R H Z 4. 0 ol 4 kj ol J kj. 0 J (Z ); solving: Z 6 Eleent 6 is carbon (X carbon), and the charge for a one-electron carbon ion is 5+ ( 5). The one-electron ion is C A node occurs when ψ 0. ψ when σ + σ 0. ± Solving using the quadratic forula: σ () 4 4()() ± 0 4 σ.0 or σ.90; because σ r/a o, the nodes occur at r (.0)a o and at r (.90)a o.0 0 0, where r is the distance fro the nucleus. 5. For r a o and θ 0 (Z for H): ψ p z 4(π ) / / () / e cos ; ψ.46 0 For r a o and θ 90, ψ p z 0 since cos 90 0; ψ 0; there is no probability of finding an electron in the p z orbital with θ 0. As expected, the xy plane, which corresponds to θ 0, is a node for the p z atoic orbital. 5. a. Each orbital could hold 3 electrons. b. The first period corresponds to n which can only have s orbitals. The s orbital could hold 3 electrons; hence the first period would have three eleents. The second period corresponds to n, which has s and p orbitals. These four orbitals can each hold three electrons. A total of eleents would be in the second period. c. 5 d.

30 CHAPTER ATOMIC STRUCTURE AND PERIODICITY a. st period: p, q, r 0, s ±/ ( eleents) nd period: 3rd period: p, q, r 0, s ±/ ( eleents) p 3, q, r 0, s ±/ ( eleents) p 3, q 3, r -, s ±/ ( eleents) p 3, q 3, r 0, s ±/ ( eleents) p 3, q 3, r +, s ±/ ( eleents) 4th period: p 4; q and r values are the sae as with p 3 ( total eleents) b. Eleents, 4,, and 0 all have filled shells and will be least reactive. c. Draw siilarities to the odern periodic table. XY could be X + Y, X + Y, or X 3+ Y 3. Possible ions for each are: X + could be eleents, 3, 5, or 3; Y could be or. X + could be 6 or 4; Y could be 0 or. X 3+ could be or 5; Y 3 could be 9 or. Note: X 4+ and Y 4 ions probably won t for. XY will be X + (Y ) ; See above for possible ions. X Y will be (X + ) Y See above for possible ions. XY 3 will be X 3+ (Y ) 3 ; See above for possible ions. X Y 3 will be (X 3+ ) (Y ) 3 ; See above for possible ions. d. p 4, q 3, r!, s ±/ ( electrons) p 4, q 3, r 0, s ±/ ( electrons) p 4, q 3, r +, s ±/ ( electrons) A total of 6 electrons can have p 4 and q 3.

31 4 CHAPTER ATOMIC STRUCTURE AND PERIODICITY e. p 3, q 0, r 0; this is not allowed; q ust be odd. Zero electrons can have these quantu nubers. f. p 6, q, r 0, s ±/ ( electrons) p 6, q 3, r!, 0, +; s ±/ (6 electrons) p 6, q 5, r!4, -, 0, +, +4; s ±/ (0 electrons) Eighteen electrons can have p The third IE refers to the following process: E + (g) E 3+ (g) + e H IE 3. The electron configurations for the + charged ions of Na to Ar are: Na + : s s p 5 Al + : [Ne]3s Mg + : s s p 6 Si + : [Ne]3s P + : [Ne]3s 3p S + : [Ne]3s 3p Cl + : [Ne]3s 3p 3 Ar + : [Ne]3s 3p 4 IE 3 for sodiu and agnesiu should be extreely large copared with the others because n electrons are uch ore difficult to reove than n 3 electrons. Between Na + and Mg +, one would expect to have the sae trend as seen with IE (F) versus IE (Ne); these neutral atos have identical electron configurations to Na + and Mg +. Therefore, the s s p 5 ion (Na + ) should have a lower ionization energy than the s s p 6 ion (Mg + ). The reaining + ions (Al + to Ar + ) should follow the sae trend as the neutral atos having the sae electron configurations. The general IE trend predicts an increase fro [Ne]3s to [Ne]3s 3p 4. The exceptions occur between [Ne]3s and [Ne]3s 3p and between [Ne]3s 3p 3 and [Ne]3s 3p 4. [Ne]3s 3p is out of order because of the sall penetrating ability of the 3p electron as copared with the 3s electrons. [Ne]3s 3p 4 is out of order because of the extra electron-electron repulsions present when two electrons are paired in the sae orbital. Therefore, the correct ordering for Al + to Ar + should be Al + < P + < Si + < S + < Ar + < Cl +, where P + and Ar + are out of line for the sae reasons that Al and S are out of line in the general ionization energy trend for neutral atos.

32 CHAPTER ATOMIC STRUCTURE AND PERIODICITY 43 IE Na + Mg + Al + Si + P + S + Cl + Ar + Note: The actual nubers in Table.5 support ost of this plot. No IE 3 is given for Na +, so you cannot check this. The only deviation fro our discussion is IE 3 for Ar + which is greater than IE 3 for Cl + instead of less than. 6. The ratios for Mg, Si, P, Cl, and Ar are about the sae. However, the ratios for Na, Al, and S are higher. For Na, the second IE is extreely high because the electron is taken fro n (the first electron is taken fro n 3). For Al, the first electron requires a bit less energy than expected by the trend due to the fact it is a 3p electron versus a 3s electron. For S, the first electron requires a bit less energy than expected by the trend due to electrons being paired in one of the p orbitals. 6. Size also decreases going across a period. Sc and Ti along with Y and Zr are adjacent eleents. There are 4 eleents (the lanthanides) between La and Hf, aking Hf considerably saller. 63. a. As we reove succeeding electrons, the electron being reoved is closer to the nucleus, and there are fewer electrons left repelling it. The reaining electrons are ore strongly attracted to the nucleus, and it takes ore energy to reove these electrons. b. Al: s s p 6 3s 3p ; for I 4, we begin reoving an electron with n. For I 3, we reove an electron with n 3 (the last valence electron). In going fro n 3 to n, there is a big jup in ionization energy because the n electrons are uch closer to the nucleus on average than the n 3 electrons. Since the n electrons are closer to the nucleus, they are held ore tightly and require a uch larger aount of energy to reove copared to the n 3 electrons. In general, valence electrons are uch easier to reove than inner-core electrons.

33 44 CHAPTER ATOMIC STRUCTURE AND PERIODICITY c. Al 4+ ; the electron affinity for Al 4+ is H for the reaction: Al 4+ (g) + e Al 3+ (g) H I 4,600 kj/ol d. The greater the nuber of electrons, the greater the size. Size trend: Al 4+ < Al 3+ < Al + < Al + < Al 64. None of the noble gases and no subatoic particles had been discovered when Mendeleev published his periodic table. Thus there was no eleent out of place in ters of reactivity. There was no reason to predict an entire faily of eleents. Mendeleev ordered his table by ass; he had no way of knowing there were gaps in atoic nubers (they hadn't been discovered yet). h kg /s kg/ato v ( ) ato 6 kg atos ol 000 g kg 40. g/ol The eleent is calciu, Ca. Integrated Probles 66. a. ν E h.5 0 J s J s c.99 0 /s n 5 ν.3 0 s b. E photon and are inversely related (E /). Any wavelength of electroagnetic radiation less than or equal to 65 n ( # 65) will have sufficient energy to eject an electron. So, yes, 59-n EMR will eject an electron. c. This is the electron configuration for copper, Cu, an exception to the expected filling order. 6. a. An ato of franciu has protons and electrons. Franciu is an alkali etal and fors stable + cations in ionic copounds. This cation would have 6 electrons. Therefore, the electron configurations will be: Fr: [Rn]s ; Fr + : [Rn] [Xe]6s 4f 4 5d 0 6p 6

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