Chemistry by Steven Zumdahl EOC Solutions -- Chapter 12
Chemistry by Steven Zumdahl EOC Solutions -- Chapter 12 Chemistry 142
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CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY Light and Matter c 3.00 ▯ 10 m/s 10 ▯1 21. ν ▯ ▯ ▯2 = 3.0 × 10 s λ 1.0 ▯ 10 m ▯34 10 ▯ ▯23 h = E ν = 6.63 × 10 J s × 3.0 × 10 s = 2.0 × 10 J/photon ▯23 23 2.0 ▯ 10 J 6.02 ▯ 10 photons photon ▯ mol = 12 J/mol 22. The wavelength is the distance between consecutive wave peaks. Wave a shows4wave- lengths and wave b shows 8 wavelengths. ▯3 1.6 ▯ 10 m ▯4 Wave a: λ = = 4.0 × 10 m 4 ▯3 Wave b: λ = 1.6 ▯ 10 m = 2.0 × 10 m 8 Wave a has the longer wavelength. Because frequency and photon energy are both inversely proportional to wavelength, wave b will have the higher frequey and larger photon energy since it has the shorter wavelength. 8 ν ▯ c ▯ 3.00 ▯ 10 m/s = 1.5 × 10 s▯1 λ 2.0 ▯ 10 ▯4m ▯34 8 E ▯ hc ▯ 6.63 ▯ 10 J s ▯ 3.00 ▯ 10 m/s = 9.9 × 1022J λ 2.0 ▯ 10 ▯4m Because both waves are examples of electromagnetic radiation, both waves travel at the same velocity, c, the speed of light. From Figure 12.3 of the text, both of these waves represent infrared electromagnetic radiation. 484 CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 23. Referencing Figure 12.3 of the text, 2.12 × 100m electromagnetic radiation is X rays. c 2.9979 ▯ 10 m/s λ ▯ ▯ = 2.799 m ν 107.1 ▯ 10 s ▯1 From the wavelength calculated above, 107.1 MHz electromagnetic radiation is FM radio- waves. ▯34 8 hc 6.626 ▯ 10 J s ▯ 2.998 ▯ 10 m/s λ ▯ ▯ ▯19 E 3.97 ▯ 10 J ▯19 The 3.97 × 10 J/photon electromagnetic 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 tλ. From the previously calculated wavelengths, the order of photon energy and frequency is: s y a r X < t hg i l ) n e e r g ( e l b i s i v < s e v awo i d a r F t strghesst λ t s ehg i h lowest ν smallest E largest E ▯34 8 hc 6.626 ▯ 10 J s ▯ 2.998 ▯ 10 m/s 24. E photon ▯ 1m ▯ 150.nm ▯ 1 ▯ 10 nm 5 1photon 1atom C 01 × 89 . 1 J × ▯ = 1.50 × 10 atoms C 1.32 ▯10 ▯18J photon c 3.00 ▯ 10 m/s 25. a. ▯ = ▯ = 5.0 ×10 ▯6 m ν 6.0 ▯ 10 s ▯1 .MRE d e r a r f n i s i s i h t , 3 . 21 e r ug i F mo r F . b h = E . c ▯ = 6.63 × 10 ▯34J s × 6.0 × 103s▯1 = 4.0 × 4.0 ▯ 10▯20J 6.022 ▯ 10 23photons ▯ = 2.4 × 10 J/mol photon mol d. Frequency and photon energy are directly related (E = hv). Because 5.4 × 10 has a lower frequency than 6.0 × 10 13s▯1 EMR, the 5.4 × 10 energetic photons. 486 26. Ephoton 2.0 E photon 2.0 X rays do have an energy greater than the carbon-carbon bond energy. Therefore, X rays could conceivably break carbon-carbon bonds in organic compounds and thereby disrupt the function of an organic molecule. Radiowaves, however, do not have sufficient energy to break carbon-carbon bonds and are therefore relatively harmless. 27. The energy needed to remove a single electron is: E 890.1kJ 28. mol = E No, it will take light having a wavelength of 134.4 nm or less to ionize gold. A photon of light having a wavelength of 225 nm is longer wavelength and thus lower energy than 134.4 nm light. 29. The energy to remove a single electron is: : s i t h g i l m n - 4 5 2 f o y g r e n E n a c i f i n g i s n i s i t i t a h t l l a 488 . b 34. = r oF 35. . ) a C ( m u i c l a c s i t n e m e l e e h Hydrogen Atom: The Bohr Model 36. 37. a. For hydrogen (Z = 1), the energy levels in units of joules are given by the equation E = CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY r oF n = 1 and n = 2: 1 1 2 ▯ 2 = 1 ▯ 0.25 = 0.75 1 2 Because the differences between 1/n values for consecutive energy levels decrease as n s a r e h t e go t r e s o l c t e g s l e v e l yg r e n e e h t , s e s a e r c n i , n e go r dyh r o f no i t i s n a r t l a r t c e p s a r oF . b ▯18 ΔE = ▯2.178 × 10 where n ind n afe the levels of the initial and final states, respectively. A positive value of ΔE always corresponds to an absorption of light, and a negative value of ΔE always corresponds to an emission of light. In the diagram, the red line is for the n = 3 to n = 2 transition. ΔE = ▯2.178 × 10 ▯18 ▯18 ΔE = ▯2.178 × 10 J × (0.2500 ▯ 0.1111) = ▯3.025 × × 520 . 3 ( yg r e n e s i h t y l e s i c e r p e v a h t s um t hg i l f o no t ohp e hT | ΔE| = E photon hν = hc 6.6261 ▯ 10 ▯ = = |ΔE| , 3 . 21 e r ug i F mo r F ▯ = 656.7 nm is red light so the diagram is correct for the red line. e h t r o f s i e n i l n e e r g e h t ,ma r g a i d e h t n I ΔE = ▯2.178 × 10 ▯18 hc 6.6261 ▯ 10 ▯ ▯ = |ΔE| , 3 . 21 e r ug i F mo r F ▯ = 486.4 nm is green-blue light. The diagram is consistent with this line. In the diagram, the blue line is for the n = 5 to n = 2 transition. ▯18 ΔE = ▯2.178 × 10 . n o i t a i d a r c i t e n g a m o r t c e l e n e e w t e b e c n e r e f f i d y g r e n e s i n e g o r d y h n i e t a t s d n u o r g . l : s a n e v i g s i e l o m r e p y g r e n CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY ▯18 23 2.178 ▯ 10 J ▯ 6.0221 ▯ 10 atoms ▯ 1kJ ▯ 1311.6 kJ atom mol 1000 J mol e H . b : Z = 2; IE = 1311.6 kJ/mol × 2 = 5246 kJ/mol (Assume n = 1 for all.) i L . c 2+: Z = 3; IE = 1311.6 kJ/mol × 3 = 1.180 × 10 kJ/mol 5+ 2 4 C . d : Z = 6; IE = 1311.6 kJ/mol × 6 = 4.722 × 10 kJ/mol 25+ 2 5 e F . e : Z = 26; IE = 1311.6 kJ/mol × (26) = 8.866 × 10 kJ/mol ▯34 8 47. Ephoton hc ▯ 6.6261 ▯ 10 J s ▯ 2.9979 ▯ 10 m/s = 7.839 × 10 λ 253.4 ▯ 10 ▯9 m ▯19 ΔE = ▯7.839 × 10 J because we have an emission. The general energy equation for one-electron ions is E = ▯2.178 × n where Z = atomic number. ▯18 2▯ 1 1 ▯ 3+ ΔE = ▯2.178 × 10 J (Z) ▯ 2 ▯ 2 ▯ , Z = 4 for Be ▯ nf ni ▯ ▯19 ▯18 2▯ 1 1 ▯ ΔE = ▯7.839 × 10 J = ▯2.178 × 10 (4) ▯ 2 ▯ 2 ▯ ▯ nf 5 ▯ 7.839 ▯ 10▯19 1 1 1 ▯18 ▯ = 2 , 2 = 0.06249, n f 4 2.178 ▯10 ▯ 16 25 nf nf e h t o t s dnop s e r r o c e n i l no i s s ime s i hT n = 5 → n = 4 electronic transition. Wave Mechanics and Particle in a Box 48. Units of ΔE × Δt = J × s, the same as the units of Planck's constant. Linear momentum p is equal to mass times velocity, p = mv. 2 2 m kg m kg m f o s t i n U ΔpΔx = kg × × m = = 2 × s = J × s s s s ▯32 49. a. Δp = mΔv = 9.11 × 10 ▯31kg × 0.100 m/s = 9.11 ▯ 10 kg m s h h 6.626 ▯ 10 ▯34 J s ΔpΔx ≥ ,Δx ▯ ▯ ▯32 4π 4πΔp 4 ▯ 3.142 ▯(9.11 ▯ 10 kg m/s) y l l a i t i n i n o i t u b i r t s i d y t i l 3 / 1 n e e w t e b n o r t c e l e e h t g n i d n i 496 CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY Orbitals and Quantum Numbers 58. Quantum numbers give the allowed solutions to Schrödinger equation. Each solution is an allowed energy level called a wave function or an orbital. Each wave function solution is described by three quantum numbers, n, ℓ, and m . The physical significance of the quantum numbers are: n: Gives the energy (it completely specifies the energy only for the H atom or ions with . s l a t i b r o e h t f o e z i s e v i t a l e r e h t dn a ) no r t c e l e e no ℓ: Gives the type (shape) of orbital. m ℓ Gives information about the direction in which the orbital is pointing. The specific rules for assigning values to the quantum numbers n, ℓ, and m Section 12.9. In Section 12.10, the spin quantum number m isdiscussed.Sincewecannot locate electrons, we cannot see if they are spinning. The spin is a convenient model. It refers to the ability of the two electrons that can occupyanyspecificorbitaltoproducetwo differently oriented magnetic moments. 59. The 2p orbitals differ from each other in the direction in which they point in space. The 2p and 3p orbitals differ from each other in their size, energy, and number of nodes. A nodal surface in an atomic orbital is a surface in which the probability of finding an electron is zero. 60. a. This general shape represents a p orbital (ℓ = 1) and because there is a node in each of the lobes, this figure represents a 3p orbital (n = 3, ℓ = 1) b. This is an s orbital (ℓ = 0). And because there is one node present, this is a 2s orbital (n = 2, ℓ = 0). c. This is the shape of a specific d oriented orbital (ℓ = 2). This orbital is designated as a ni tneserp era sedon lanoi t idda on esuaceB d. z2 l a i b r3doz2 (n = 3, ℓ = 2). 61. 1p: n = 1, ℓ = 1 is not possible; 3f: n =3, ℓ = 3is notpossible; 2d: possible; in all three incorrect cases, n = ℓ. The maximum value ℓ can have is n 62. b. For ℓ = 3, m can range from -3 to +3; thus +4 is not allowed. ℓ . c n cannot equal zero. The quantum numbers in part a are allowed. 63. a. For n = 3, ℓ = 3 is not possible. . d m sannot equal ▯1. . e ℓ cannot be a negative number. . d ewo l l a e r a c dn a b s t r a p n i s r e bmun mu t n a uq e hT p 7 ; ) n o i t a n g i s e d f 4 e v a h s 498 CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY Polyelectronic Atoms 71. He: 1s ; Ne: 1s 2s 2p ; 6 Ar: 1s 2s 2p 3s 3p ; each peak in the diagram corresponds to a subshell with different values of n. Corresponding subshells are closer to the nucleus for heavier elements because of the increased nuclear charge. 72. In polyelectronic atoms, the orbitals of a given principal quantum level are not degenerate. In polyelectronic atoms, the energy order of the n = 1, 2, and 3 orbitals are (not to scale): 3d 3p E 3s 2p 2s 1s In general, the lower the n value for an orbital, the closer on average the electron can be to the nucleus, and the lower the energy. Within a specific n value orbital (like 2s vs. 2p or 3s vs. 3p vs. 3d), it is generally true that ns< E npE < nd. nf To rationalize this order, we utilize the radial probabilitydistributions. Inthe2sand2p distribution, notice that the 2s orbital has a small hump of electron density very near the nucleus. This indicates that an electron in the 2s orbital can be very close to the nucleus some of the time. The 2s electron penetrates to the nucleus more than a 2p electron, and with this penetration comes a lower overall energy for the 2s orbital as compared to the 2p orbital. In the n = 3 radial probability distribution, the 3selectron has two humps of electron density very close to the nucleus, and the 3p orbital has one hump very close to the nucleus. The 3s orbital electron is most penetrating, with the 3p orbital electron the next most penetrating, followed by the least penetrating 3d orbital electron. The more penetrating the electron, the lower the overall energy. Hence the 3s orbital is lower energy than the 3p orbitals which is lower energy than the 3d orbitals. 73. Valence electrons are the electrons in the outermost principal quantum level of an atom (those electrons in the highest n value orbitals). The electrons in the lower n value orbitals are all inner core or just core electrons. The key is that the outermost electrons are the valence electrons. When atoms interact with each other, it will be the outermost electrons that are involved in these interactions. In addition, how tightly the nucleus holds these outermost electrons determines atomic size, ionization energy, and other properties of atoms. Elements in the same group have similar valence electron configurations and, as a result, have similar chemical properties. 74. The widths of the various blocks in the periodic table are determined by the number of electrons that can occupy the specific orbital(s). In the s block, we have one orbital (ℓ = 0, mℓ . n o r t c e l e e l g n i s a y f i c e p s s 6 ] e X [ : u E : e r a s n o i t a r u g r o d n i s n o r t c e l e 2 r o p3 s 3 p2 CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 501 s 1 : uC 22s 2p 3s 3p 4s 3d ; Cu has 1 unpaired electron. 1s 2s 2p 3s 3p or 4s 4s 3d 2 2 6 2 4 82. a. The complete ground state electron for this neutral atom is 1s 2s 2p 3s 3p . This atom has 2 + 2 + 6 + 2 + 4 = 16 electrons. Because the atom is neutral, it also has 16 protons, making the atom sulfur, S. 2 1 4 b. Complete excited state electron configuration: 1s 2s 2p ; this neutral atom has 2 + 1 + 4 = 7 electrons, which means it has 7 protons, which identifies it as nitrogen, N. c. Complete ground state electron configuration: 1s 2s 2p 3s 3p 4s 3d 4p ; this 1▯0 5 charged ion has 35 electrons. Because the overall charge is 1 ▯, this ion has 34 protons which identifies it as selenium. The ion is Se . ▯ 83. a. 2 valence electrons; 4s 2 b. 6valenceelectrons; 2s 22p 4 c. 7 valence electrons; 7s 7p 2 5 .3l5;n 25p 1 2 6 2 3 e. 8 valence electrons; 3s 3p .fl6;n 6p 2 2 6 2 6 2 10 6 2 10 6 2 14 10 84. Hg: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d a. From the electron configuration for Hg, we have 3s 2, 3p , and 3d electrons; 18total electrons with n = 3. 10 10 10 d3 . b , 4d , 5d ; 30 electrons are in d atomic orbitals. 6 6 6 6 c. 2p , 3p , 4p , 5p ; each set of np orbitals contain one p atomiz orbital. Because we have 4 sets of filled np orbitals and two electrons can occupy the p orbitalz there are 4(2) = 8 electrons in p ztomic orbitals. d. All the electrons are paired in Hg, so one-half of the electrons are spin-up ( m = +1/2) snd the other half are spin-down (m = ▯1s2); 40 electrons have spin-up. 85. Element 115, Uup, is in Group 5A under Bi (bismuth): Uup: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p 6 2 14 10 3 2 6 10 14 a. 5s , 5p , 5d , and 5f ; 32 electrons haven = 5 as one of their quantum numbers 14 14 b. ℓ = 3 are f orbitals. 4f and 5f are the f orbitals used. They are all filled so 28 electrons have ℓ = 3. n i e b d l u o h s n o r t c e l e d d n o CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY electrons. For elements 1-36, these are Ti (3d ) and Ni (3d ). A total of 8 elements from the first 36 elements have two unpaired electrons in the ground state. 1 89. The s block elements with ns for a valence electron configuration have one unpaired electrons. These are elements H, Li, Na, and K for the first 36 elements. The p block elements with ns np or ns np valence electron configurations have 2 1 elements B, Al, and Ga (ns np ) and elements F, Cl, and Br (ns np ) for the first 36 elements. In the d block, Sc ([Ar]4s23d ) and Cu ([Ar]4s 3d ) each have one unpaired electron. A total of 12 elements from the first 36 elements have one unpaired electron in the ground state. 90. O: 1s 2s 2p 2x2 y2( ↑↓ ↑↓ ); there are no unpaired electrons in this oxygen atom. This configuration would be an excited state, and in going to the more stable ground state ( ↑↓ ↑ ↑ ), energy would be released. 91. We get the number of unpaired electrons by examining the incompletely filled subshells. 2 4 4 s 2 ] e H [ : O 2p 2p : ↑↓ ↑ O +: [He]2s 2p p2 : ↑ ↑ O -: [He]2s 2p 5 2p : ↑↓ ↑↓ ↑ s 6 ] e X [ : s O 24f 5d 6 5d : ↑↓ ↑ 2 2 2 s 5 ] r K [ : r Z 4d 4d : ↑ ↑ 2 4 4 s 3 ] e N [ : S 3p 3p : ↑↓ ↑ 2 5 5 s 2 ] e H [ : F 2p 2p : ↑↓ ↑↓ ↑ s 3 ] e N [ : r A 3p 6 3p : ↑↓ ↑↓ ↑↓ 92. a. Excited state of boron 2 2 1 s 1 : e t a t s d n u o r g B 2s 2p c. Excited state of fluorine s 1 : e t a t s d n u o r g F s4]rA[ :2sats dnuorg eF The Periodic Table and Periodic Properties + 93. Ionization energy: P(g) ▯ P (g) + e ; electron affinity: P(g) + e ▯ P (g) 94. Across a period, the positive charge from the nucleus increases as protons are added. The number of electrons also increase, but these outer electrons do not completely shield the increasing nuclear charge from each other. The general result is that the outer electrons are t a i c o s s a e g n a h c y g r e n e e h t e F < P < O . f r S < e S < F < i S < O < P < . dn e r t ) E I ( yg r e n e e b o t n e g o l a h t x e n e h t s i 7 1 n a . ) n o i t p e c x e n a s i F ( c i m r e h ) d n e r t E I e h t s w o l l o ) d n e r t s u i d a r e h t s w o l l o f ( e d i r d y h m u i d o s ; H a N . f e d i r d CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 509 123. Size decreases from left to right and increases going down the periodic table. Thus going one element right and one element down would result in a similar size for the two elements diagonal to each other. The ionization energies will be similar for the diagonal elements since the periodic trends also oppose each other. Electron affinities are harder to predict, but atoms with similar sizes and ionization energies should also have similar electron affinities. 124. Energy to make water boil = s × m × ΔT = 4.18J × 50.0 g × 75.0°C = 1.57 × 10 J. oC g hc 6.626 ▯ 10 ▯34J s ▯ 2.998 ▯ 10 m/s Ephoton ▯ = 2.04 × 10▯24J λ 9.75 ▯ 10 ▯2 m 4 1s 4 1photon 27 1.57 × 10 J × 750.J = 20.9 s; 1.57 × 10 J × 2.04 ▯ 10 ▯24 J = 7.70 × 10 photons 125. 60 × 10 km × 1000m ▯ 1s = 200 s (about 3 minutes) km 3.00 ▯ 10 m 310.kJ 1mol 126. E = ▯ = 5.15 × 10 ▯2kJ = 5.15 × 10▯19J mol 6.022 ▯ 10 23 hc 6.626 ▯ 10 ▯4 J s ▯ (2.998 ▯ 10 m/s) λ = ▯ = 3.86 × 10 m = 386 nm E 5.15 ▯ 10 ▯19J hc 6.626 ▯ 10 ▯34J s ▯ 2.998 ▯ 10 m/s 100cm 127. λ = ▯ = 5.53 × 10 ▯7m × E 3.59 ▯ 10 ▯19J m ▯5 = 5.53 × 10 cm ▯5 From the spectrum, λ = 5.53 × 10 cm is greenish yellow light. ▯ 1 1 ▯ ▯18 ▯ 1 1 ▯ ▯19 128. ΔE = ▯R H▯ 2 ▯ 2 ▯ = ▯2.178 × 10 J ▯ 2 ▯ 2 ▯ = ▯4.840 × 10 J ▯ nf n i ▯ ▯ 2 6 ▯ hc 6.6261 ▯ 10▯34 J s ▯ 2.9979 ▯ 10 m/s ▯7 100cm λ = = ▯19 = 4.104 × 10 m × ΔE 4.840 ▯ 10 J m ▯5 = 4.104 × 10 cm ▯5 From the spectrum, λ = 4.104 × 10 cm is violet light, so the n = 6 to n = 2 visible spectrum line is violet. 129. When the p and d orbital functions are evaluated at various points in space, the results sometimes have positive values and sometimes have negative values. The term phase is often associated with the + and ▯ signs.Forexample,asinewavehasalternatingpositiveand negative phases. This is analogous to the positive and negative values (phases) in the p and d orbitals. e + ) g ( l C _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ) g ( l C + ) g ( aN ) g ( gM _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ) g ( F + ) g ( gM gM . c e + ) g ( F _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ : t e g ew , c dn a b s t r a p mo r F . d ) g ( F + ) g ( gM gM _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ) g ( F 2 + ) g ( gM . s no r t c e l e e c n e l a v 6 = 4 + 2 . a s 1 : e b t s u m n o i e h t f o n o i . e z i s e h t r e t a e r g e h t o t s d n o p s e r r o c d o i r e p t s r i f 8 2 . d . t hg i l mn 005 - 004 t c e t e d s r o t p e c e r : s r o t p e c e r e n o c . t hg i l mn 036 - 054 t c e t e d : s r o t p e c e r e n o c mn 007 - 005 t c e t e d s r o t p e c e r E , n o i n o r t c e l e - e n o a r o f s , s : A e n i l f o h s i n o i n o b r a c n o r t c e l e - e n o o t l a n o i t r o p o r p s i y t i l 0 0 1 × 5 1 . 2 ( = p = y t i l i b a b o r 0 1 × 0 . 1 s i t a h t n o r t c e l e n × e t a u l a v e 0 1 × 7 4 x o b r a h t i w d e l l i f e b l l i w s l e v e l ) s t n ) s t n e m e l e l a t o t 8 ( 3 = p h t . 3 = e s e h t e v a h n a c s n o r t c e l e o r 520 E I 3 for sodium and magnesium should be extremely large compared with the others because n = 2 electrons are much more difficult to remove than n =3electrons.BetweenNa 2+ Mg , one would expect to have the same trend as seen with IE (F) versus IE (Ne); these neutral atoms have identical electron configurations to Na 1s 2s 2p ion (Na ) should have a lower ionization energy than the 1s 2s 2p ion (Mg ). l A ( s n o i + 2 g n i n i a m e r e h T having 1he same electron configurations. The general IE trend predicts an increase from [Ne]3s to[Ne]3s [Ne]3s 3p and [Ne]3s 3p . [Ne]3s 3p is out of order because of the small penetrating ability of the 3p electron as compared with the 3s electrons. [Ne]3s 3p is out of order because of the extra electron-electron repulsions present when two electrons are paired in the same orbital. Therefore, the correct ordering for Al where P 2+andAr general ionization energy trend for neutral atoms. IE Note: The actual numbers in Table 12.6 support most of this plot. No IE you cannot check this. The only deviation from our discussion is IE for Ar than IE3for Cl Marathon Problem 153. a. Let λ = wavelength corresponding to the energy ?, and the ground state, n = 1. Use the information in pa difference ΔE 10 7). ΔE 1 ▯ n , . e . i ; X t n e m e l e f o l l e h s e 522 r o F . E , n o i n o r t c e l e - e n o a r o F . e H r o F s 5 ] r K [ : Y
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