Baixe Resolução Atkins Principios de quimica e outras Exercícios em PDF para Química, somente na Docsity! CHAPTER 1 ATOMS: THE QUANTUM WORLD 1.2 radio waves < infrared radiation < visible light < ultraviolet radiation 1.4 (a) 8 1 142.997 92 10 m s ( )(7.1 10 s )λ− −× ⋅ = × 1 8 1 14 1 7 2.997 92 10 m s 7.1 10 s 4.2 10 m 420 nm λ − − − × ⋅ = × = × = (b) 8 1 182.997 92 10 m s ( )(2.0 10 s )λ− −× ⋅ = × 1 8 1 18 1 10 2.997 92 10 m s 2.0 10 s 1.5 10 m 150 pm λ − − − × ⋅ = × = × = 1.6 From Wien’s law: 3max 2.88 10 K m.λ −= × ⋅T 9 3 3 ( )(715 10 m) 2.88 10 K m 4.03 10 K − −× = × ≈ × T T ⋅ ⋅ 1.8 (a) 34 17 1 17 (6.626 08 10 J s)(1.2 10 s ) 8.0 10 J − − − = = × ⋅ × = × E hv (b) The energy per mole will be times the energy of one atom. 236.022 10× 23 1 34 17 1 7 4 (2.00 mol)(6.022 10 atoms mol ) (6.626 08 10 J s)(1.2 10 s ) 9.6 10 J or 9.6 10 kJ − − − = × × × ⋅ × = × × E 40 (c) 1 23 1 17 1 6 3 2.00 g Cu 63.54 g mol Cu (6.022 10 atoms mol )(8.0 10 J atom ) 1.5 10 J or 1.5 10 kJ − − − ⎛ ⎞ = ⎜ ⎟⋅⎝ ⎠ × × ⋅ × ⋅ = × × E − 1.10 1From and , .λ λ −= = =c v E hv E hc 34 8 1 9 19 1 23 1 19 1 5 1 1 (6.626 08 10 J s)(2.997 92 10 m s )(for one atom) 470 10 m 4.23 10 J atom (for 1.00 mol) (6.022 10 atoms mol )(4.23 10 J atom ) 2.55 10 J mol or 255 kJ mol E E − − − − − − − − − × ⋅ × ⋅ = × = × ⋅ = × ⋅ × ⋅ = × ⋅ ⋅ − 1.12 (a) false. UV photons have higher energy than infrared photons. (b) false. The kinetic energy of the electron is directly proportional to the energy (and hence frequency) of the radiation in excess of the amount of energy required to eject the electron from the metal surface. (c) true. 1.14 The wavelength of radiation needed will be the sum of the energy of the work function plus the kinetic energy of the ejected electron. 19 1 19 work function 2 kinetic 31 6 1 2 18 total work function kinetic 19 18 18 (4.37 ev)(1.6022 10 J eV ) 7.00 10 J 1 2 1 (9.10939 10 kg)(1.5 10 m s ) 2 1.02 10 J 7.00 10 J 1.02 10 J 1.72 10 J − − − − − − − − − = × ⋅ = × = = × × ⋅ = × = + = × + × = × E E mv E E E To obtain the wavelength of radiation we use the relationships between E, frequency, wavelength, and the speed of light: 41 1.28 (a) The highest energy photon is the one that corresponds to the ionization energy of the atom, the energy required to produce the condition in which the electron and nucleus are “infinitely” separated. This energy corresponds to the transition from the highest energy level for which n = 1 to the highest energy level for which n = .∞ 34 2 2 lower upper 15 1 2 2 18 1 1 (6.626 08 10 J s) 1 1(3.29 10 s ) 1 2.18 10 J − − − ⎛ ⎞ = ℜ − = × ⋅⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎛ ⎞× × −⎜ ⎟∞⎝ ⎠ = × E h n n (b) The wavelength is obtained from 1 1and , or , or .λ λ λ− −= = = =c v E hv E hc hcE 34 1 18 8 (6.626 08 10 J s)(2.997 92 10 m s ) 2.18 10 J 9.11 10 m 91.1 nm λ − 8 − − × ⋅ × ⋅ = × = × = − (c) ultraviolet 1.30 Because the line is in the visible part of the spectrum, it belongs to the Balmer series for which the ending n is 2. We can use the following equation to solve for the starting value of n: 8 1 14 1 9 15 1 2 2 1 2 14 1 15 1 2 2 2 2.99792 10 m s 6.91 10 s 434 10 m 1 1(3.29 10 s ) 1 16.91 10 s (3.29 10 s ) 2 λ − − − − − − × ⋅ = = = × × ⎛ ⎞ = × −⎜ ⎟ ⎝ ⎠ ⎛ ⎞ × = × −⎜ ⎟ ⎝ ⎠ cv v n n n 2 2 2 2 10.210 0.250 1 0.04 = − = n n 44 2 2 2 1 0.04 5 = = n n This transition is from the n = 5 to the n = 2 level. 1.32 (a) The easiest way to approach this problem is to plot the radial probability distribution, 2 23s( )P r r R= ⋅ , and, using the plot, determine the position of the nodes. Another approach is to take the derivative of this probability distribution with respect to r and set this derivative equal to zero to find the position of the extrema of the curve. Taking the latter approach, one finds that the minima are at 3s( is given in Table 1.2)R 9 (3 3) 9 (3 3) and 2 2 ⎛ ⎞ ⎛+ ⋅ − ⋅ ⎜ ⎟ ⎜⎜ ⎟ ⎜ ⎝ ⎠ ⎝ o oa a ⎞ ⎟⎟ ⎠ , or 7.10 and 1.90⋅ ⋅o oa a . (b) o 2 3 42 4d o o Z Z1/ 96 5 6 2 2 −⎛ ⎞⎛ ⎞ = −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ Zr ar rR Z e a a As in part (a) above, the easiest way to find the node is to plot the probability distribution function and find the position of the node by inspection. Taking the derivative of the probability distribution function and setting it equal to zero, one finds three extrema for the 4d distribution: ( )o o12 , 2 7 13 , a a+ ( oand 2 7 13 a− ) . The first of these three is the position of the node while the latter two are the positions of the maxima in the distribution. 1.34 The dxy orbital will have its lobes pointing between the x and y axes, while the orbital will have its lobes pointing along the x and y axes.2 2x y−d 45 x x 2 2x y d −xyd 1.36 The equation demonstrated in example 1.8 can be used: 0 02(0.65 ) / 2 3 0 0 2 3 0 e ( 0.65 , , ) 0.27 (0, , ) 1 ψ θ φ π ψ θ φ π − = = = ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ a a r a a a 1.38 The radial probability distribution may be found by integrating the full wavefunction, ψ(r,θ,φ), over all possible values of θ and φ. Since s- orbitals are spherically symmetric and are not a function of θ or φ, integration of any s-orbital over all θ and φ always gives the same result: 2 2 2 2 0 0 0 0 2 2 2 0 ( ) sin ( ) sin ( ) 2 ( ) (4 ) π π π π π ψ θ θ φ ψ θ θ φ ψ φ ψ π ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ ⎡ ⎤ = =⎢ ⎥ ⎣ ⎦ ∫ ∫ ∫ ∫ ∫ r d d r d d r d r The sinθ term in the equation above is needed to correct for the differential volume element in spherical polar coordinates. Likewise, to integrate over all possible values of r one must evaluate the integral: where again the r 2 2 0 (4 ) ( )π ψ ∞ ∫ r r dr 2 term corrects for the differential volume element in spherical polar coordinates. From this expression it is clear that the probability distribution is . 2 2(4 ) ( )π ψ r r 46 2 2 2 2 2 0 1 0 2 0 3 0 4 0 1 2 3 4 1 1 1 1 πε πε πε πε πε ⎛ ⎞ − − − − = − + + +⎜ ⎟ ⎝ ⎠ e e e e e r r r r r r r r The first four terms are the attractive terms between the nucleus and each electron, and the last six terms are the repulsive interactions between all the possible combinations of electrons taken in pairs. (b) The number of attractive terms is straightforward. There should be one term representing the attraction between the nucleus and each electron, so there should be a total of n terms representing attractions. The number of repulsive terms goes up with the number of electrons. Examining the progression, we see that n = 1 2 3 4 5 6 7 # of repulsive terms = 0 1 3 6 10 15 21 Hence, the addition of an electron adds one rab term for each electron already present; so the difference in the number of repulsive terms increases by 1−n for each additional electron. This relation can be written as a summation to give the total number of repulsive terms: number of repulsive terms = 1 ( 1 → )−∑ n n The repulsive terms will have the form 2 0 ab 0 ab ( )( ) 4 4πε πε − − = e e e r r where rab represents the distance between the two electrons a and b. The total repulsive term will thus be 2 2 2 2 2 0 12 0 13 0 14 0 23 0 24 0 344 4 4 4 4 4πε πε πε πε πε πε + + + + + e e e e e e r r r r r 2 r 2 0 12 13 14 23 24 34 1 1 1 1 1 1 4πε ⎛ ⎞ + + + + +⎜ ⎟ ⎝ ⎠ e r r r r r r This gives 49 2 0 1 2 3 4 2 0 12 13 14 23 24 34 1 1 1 1( ) 1 1 1 1 1 1 4 πε πε ⎛ ⎞⎛ ⎞− = + + +⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎛ ⎞ + + + + + +⎜ ⎟ ⎝ ⎠ eV r r r r r e r r r r r r The total number of attractive and repulsive terms will thus be equal to 1 ( 1 → + −∑ n n n ) 2 The point of this exercise is to show that, with each added electron, we add an increasingly larger number of e-e repulsive terms. 1.60 (a) false. The 2s-electrons will be shielded by the electrons in the 1s- orbital and will thus experience a lower Zeff. (b) false. Because the 2p- orbitals do not penetrate to the nucleus as the 2s-orbitals do, they will experience a lower Zeff. (c) false. The ability of the electrons in the 2s- orbital to penetrate to the nucleus will make that orbital lower in energy than the 2p. (d) false. There are three p-orbitals, and the electron configuration for C will be There will be two electrons in the p-orbitals, but each will go into a separate orbital and, as per quantum mechanics and Hund’s rule, they will be in these orbitals with the spins parallel (i.e., the spin magnetic quantum numbers will have the same sign) for the ground-state atom. (e) false. Because the electrons are in the same orbital, they must have opposite spin quantum numbers, m 2 2 21 2 2 .s s p s, because the Pauli exclusion principle states that no two electrons in an atom can have the same four quantum numbers. 1.62 The atom with a valence-shell configuration is germanium, Ge. The ground-state configuration is given by (d); the other configurations represent excited states. 24 4s p 1.64 (a) This configuration is not possible because the maximum value l can have is ; because n = 2, 1−n max 1.=l (b) This configuration is possible. 50 (c) This configuration is not possible because, for l = 4, ml can only be an integer from 3 to 3, i.e., can only equal 0, 1, 2, or 3.− + ± ± ±lm 1.66 (a) sulfur 2 4[Ne]3 3s p (b) cesium 1[Xe]6s (c) polonium 14 10 2 4[Xe]4 5 6 6f d s p (d) palladium 10[Kr]4d (e) rhenium 14 5 2[Xe]4 5 6f d s (f) vanadium 3 2[Ar]3 4d s 1.68 (a) Ga; (b) Na; (c) Sr; (d) Eu 1.70 (a) 4s; (b) 3p; (c) 3p; (d) 4s 1.72 (a) 5; (b) 2; (c) 7; (d) 12 For (d), note that the filled 3d-orbitals become core electrons, which are not available. The chemistry of Zn is dominated by its +2 oxidation number; and, consequently, Zn, Cd, and Hg are often grouped with the p- block elements and referred to collectively as the post transition metals. 1.74 (a) 2; (b) 3; (c) 1; (d) 0 1.76 (a) ; (b) (c) (d) 2 5ns np 2 4 ;ns np 3 ( 1)+nd n s2 ; 2 2ns np 1.78 (a) As one goes across a period, a proton and an electron are added to each new atom. The electrons, however, are not completely shielded from the nucleus by other electrons in the same subshell, so the set of electrons experience an overall greater nuclear charge. (b) The ionization energies of the Group 16 elements of O, S, and Se lie somewhat lower than those of the Group 15 elements that preceed them. This exception may be 51 34 -27 12 34 2 1 27 12 3 1 (6.626 08 10 J s) (1.674 93 10 kg)(200 10 m) (6.626 08 10 kg m s ) (1.67493 10 kg)(200 10 m) 2.0 10 m s − − − − − − − × ⋅ = × × × ⋅ ⋅ = × × = × ⋅ v 1.98 A ground-state oxygen atom has four electrons in the p-orbitals. This configuration means that as one goes across the periodic table in Period 2, oxygen is the first element encountered in which the p-electrons must be paired. This added electron-electron repulsion energy causes the ionization potential to be lower. 1.100 molar volume = molar mass /density 3(cm mol )−⋅ 1 1(g mol )−⋅ 3(g cm )−⋅ Element Molar vol. Element Molar vol. Li 13 Na 24 Be 4.87 Mg 14.0 B 4.62 Al 9.99 C 5.31 Si 12.1 N 16 P 17.0 O 14.0 S 15.5 F 17.1 Cl 17.5 Ne 16.7 Ar 24.1 54 0 5 10 15 20 25 30 0 5 10 15 20 Atomic Number M ol ar V ol um e (c m 3 , m ol -1 ) s p s p The molar volume roughly parallels atomic size (volume), which increases as the s-sublevel begins to fill and subsequently decreases as the p- sublevel fills (refer to the text discussion of periodic variation of atomic radii). In the above plot, this effect is most clearly seen in passing from Ne(10) to Na(11) and Mg(12), then to Al(13) and Si(14). Ne has a filled 2p-sublevel; the 3s-sublevel fills with Na and Mg; and the 3p-sublevel begins to fill with Al. 1.102 In general, as the principal quantum number increases, the energy spacing between orbitals becomes smaller. This trend indicates that it doesn’t take very much change in electronic structure to cause the normal orbital energy pattern to rearrange. 1.104 (a) 0 0 2sin sin 3sin sin 0 2 6 π π π π π π ⋅ ⋅⎛ ⎞ ⎛ ⎞⋅⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⋅ ⋅⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ∫ L L x x dx L L L x L x L L = 55 (b) Below is a plot of the first two wavefunctions describing the one- dimensional particle-in-a-box and the product of these two wavefunctions. Notice that the area above zero in the product exactly cancels the area below zero, making the integral of the product zero. This happens whenever a wavefunction that is unaltered by a reflection through the center of the box (wavefunctions with odd n) is multiplied by a wavefunction that changes sign everywhere when reflected through the center of the box (wavefunctions with even n). ψn = 1 (ψn = 1)(ψn = 2) ψn = 2 1.106 (a) 1 2 ∆ ∆ ≥p x where 34 346.626 08 10 J s 1.054 57 10 J s 2 2π π − −× ⋅= = = × ⋅ h 34 351 (1.054 57 10 J s) 5.272 85 10 J s 2 − −∆ ∆ ≥ × ⋅ = × ⋅p x The minimum uncertainty occurs at the point where this relationship is an equality (i.e., using = rather than ).≥ The uncertainty in position will be taken as the 200 nm corresponding to the length of the box: ( ) 35 35 2 1 28 1 9 28 1 5.272 85 10 J s 5.272 85 10 kg m s 2.64 10 kg m s 200 10 m 2.64 10 kg m s − − − − − − − − ∆ ∆ = × ⋅ × ⋅ ⋅ ∆ = = × ⋅ ⋅ × = ∆ = ∆ = × ⋅ ⋅ p x p mv m v Because the mass of an electron is the uncertainty in velocity will be given by 28 319.109 39 10 g or 9.109 39 10 kg,−× × − 0 1 x 0 10 L 0 L x 56

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