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) 2.997 92 × 108 m s⋅ −^1 = ( λ)(7.1 × 1014 s −^1 )
8 1 14 1 7
2.997 92 10 m s 7.1 10 s 4.2 10 m 420 nm
− − −
= ×^ ⋅
×
= × =
(b) 2.997 92 × 108 m s⋅ −^1 = ( λ)(2.0 × 1018 s −^1 )
8 1 18 1 10
2.997 92 10 m s 2.0 10 s 1.5 10 m 150 pm
− − −
= ×^ ⋅
×
= × =
1.6 From Wien’s law: T λmax = 2.88 × 10 −^3 K m.⋅
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.
6.022 × 1023
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
(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 From c = v λ and E = hv E , = hc λ−^1.
34 8 1 9 19 1 23 1 19 1 5 1 1
(for one atom) (6.626 08^10 J s)(2.997 92^10 m s^ ) 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. work function^19 1 kinetic^2 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:
The frequency of the given line is:
8 -1 13 - 9
2.9979 10 m s (^) 5.073 10 s
ν λ 5910 10 − m
= = × = ×
×
c
This frequency is closest to the frequency resulting from the n = 9 to n = 6 transition:
2 2 15 2 2 13 2 1
(^1 1) 3.29 10 Hz 1 1 5.08 10 Hz
= ℜ ⎛^ − ⎞^ = × ⎛^ − ⎞= ×
⎜⎝ (^) n n ⎟⎠ ⎜⎝ ⎟⎠
1.24 (a) The Rydberg equation gives v when ℜ = 3.29 × 1015 s−^1 , from which
one can calculate λ from the relationship c = v λ.
2 2 12 8 1
2 2 12 8 1 15 1 8
and 2.997 92 10 m s 1 1
2.997 92 10 m s (3.29 10 s ) 1 1 1 25 9.49 10 m 94.9 nm
λ λ
λ λ
−
− − −
= ℜ ⎛^ − ⎞
= = × ⋅
= ℜ ⎛^ − ⎞
× ⋅ = × ⎛⎜^ − ⎞⎟
= × =
v (^) n n c v c (^) n n
(b) Lyman series (c) This absorption lies in the ultraviolet region.
1.26 Here we are searching for a transition of He+^ whose frequency matches that of the n = 2 to n = 1 transition of H. The frequency of the H transition is:
H (^2 )
= ℜ ⎛⎜^ − ⎞⎟^ = ⎛⎜ ℜ
A transition of the He+^ ion with the same frequency is the n = 2 to n = 4 transition:
ν He += (Z )^2 ℜ ⎛⎜⎝^212 − 412 ⎞⎟⎠^ = (2 )^2 ℜ ⎛⎜⎝^163 ⎞⎟⎠^ =⎛⎜⎝^34 ⎞ ⎟⎠ℜ
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 = ∞.
2 2 34 lower upper (^15 12 ) 18
(^1 1) (6.626 08 10 J s)
(3.29 10 s ) 1 1 1 2.18 10 J
−
− −
= ℜ ⎛^ − ⎞= × ⋅
× × ⎛⎜^ − ⎞⎟
= ×
E h (^) n n
(b) The wavelength is obtained from
c = v λ and E = hv , or E = hc λ −^1 , or λ= hcE −^1.
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
− 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 : (^8 114 ) 9 (^15 12 ) 1 2 (^14 1 15 12 ) 2
2.99792 10 m s (^) 6.91 10 s 434 10 m (3.29 10 s )^1
6.91 10 s (3.29 10 s ) 1 1 2
− − − −
− −
= = ×^ ⋅ = ×
×
= × ⎛^ − ⎞
× = × ⎛^ − ⎞
v^ c
v (^) n n
n
22
22
0.210 0.250^1
n
n
x x
d xy^ d^ x^2 −y^2
1.36 The equation demonstrated in example 1.8 can be used: 2(0.65 0 ) / 0 2 0 03 2 03
e ( 0.65 , , ) (^) 0. (0, , ) (^1)
− = (^) = = ⎛ ⎞ ⎜⎝ ⎟⎠
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 sin θ 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
∞ ∫ 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 (4 π ) ψ ( ) r^2 r^2.
1.40 (a) To find the maximum in the radial probability distribution function for the 3 d -orbital, one can take the derivative of the distribution with respect to r , set the result equal to zero and solve for r: (^2 ) (^2 32 2 63) where 4 1 81 30
o
r d a o
P r R C r e C (^) a
32 32 32 (^2 6 2 6 5 2 6 26 ) 3 3
o o o ar ar ar o o
dP (^) C r e r e C e r r dr a a ≡ The non-trivial solution to this equation is found when: 6 5 5
Dividing both sides by : (^2 6) 0 and, therefore, 9 3
o
o o
r (^) a r r r (^) a r a
The position of the maximum in the distribution for the 3 d -orbital occurs when r = 9 a o. (b) The radial wavefunction for the 4 s -orbital is: (^32) 4 2 2 3 3 4 o (^) o o o 2 42
2 2 2 3 2 4 3 5 o o o
1 3 1 1 w^1 4 8 192 4 The derivative of the radial probability distribution, , with respect to is: 3 13 19 2 16 96
−
−
= ⎛^ − + − ⎞^ = ⎛^ ⎞
o
o
ra s s ra
R C (^) a r (^) a r (^) a r e C a P r R r dP (^) C e r r r r dr a a a
here
4 6 o 5 7 6 8 o o
a^ r
a r^ a r The non-trivial solutions to this equation are found when the polynomial in brackets is equal to zero. The roots are most easily found using a computer and are: 0.732 a o, 1.872 a o, 4 a o, 6.611 a o, 10.65 a o, 15.52 a o, and 24.62 a o. The last of these roots, 24.62 a o, is the position of the maximum of the probability distribution for the 4 s -orbital.
2 2 2 2 2 0 1 0 2 0 3 0 4 0 1 2 3 4
− − − − = − ⎛^ + + + ⎞
e (^) r e (^) r e (^) r e (^) r e ⎜⎝ (^) 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 r ab term for each electron already present; so the difference in the number of repulsive terms increases by n − 1 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
→ ∑ −^ ) n
n
The repulsive terms will have the form 2 0 ab 0 ab
− e − e (^) = e r r
where r ab represents the distance between the two electrons a and b. The total repulsive term will thus be 2 2 2 2 2
e (^) r + e (^) r + e (^) r + e (^) r + e (^) r + e 2 r 2 0 12 13 14 23 24 34
e ⎜⎝ (^) r r r r r r ⎟⎠
This gives
2 0 1 2 3 4 2 0 12 13 14 23 24 34
( )^1 1 1
πε
πε
= ⎛^ − ⎞ ⎛^ + + + ⎞
+ ⎛^ + + + + + ⎞
V r e 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
→
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 2 s -electrons will be shielded by the electrons in the 1 s - orbital and will thus experience a lower Zeff. (b) false. Because the 2 p - orbitals do not penetrate to the nucleus as the 2 s -orbitals do, they will experience a lower Zeff. (c) false. The ability of the electrons in the 2 s - orbital to penetrate to the nucleus will make that orbital lower in energy than the 2 p. (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
1 s^2 2 s^2 2 p^2.
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.
4 s^24 p
1.64 (a) This configuration is not possible because the maximum value l can have is n − 1 ; because n = 2, l max (^) =1. (b) This configuration is possible.
explained by observing that, as the three p -orbitals up through Group 15 are filled, each electron goes into a separate orbital. The next electron (for Group 16) goes into an orbital already containing an electron, so electron- electron repulsions are higher. This increased repulsion makes it easier to remove the additional electron from the Group 16 elements.
1.80 While Na has a smaller effective nuclear charge than K, the outermost electron occupies a 3 s -orbital. In potassium atoms, the outermost electron occupies a 4 s -orbital and so on average is much further from the nucleus than the electron in the 3 s -orbital of sodium. Despite the larger effective nuclear charge, the electron in potassium is more easily removed because it is, on average, further from the nucleus.
1.82 (a) silicon (118 pm) > sulfur (104 pm) > chlorine (99 pm); (b) titanium (147 pm) > chromium (129 pm) > cobalt (125 pm); (c) mercury (155 pm) > cadmium (152 pm) > zinc (137 pm); (d) bismuth (182 pm) > antimony (141 pm) > phosphorus (110 pm)
1.84 (a) Ba 2 +^ ; (b) As^3 −^ ; (c) Sn 2 +
1.86 (a) Al; (b) Sb; (c) Si
1.88 From Appendix 2D, the radii (in pm) are Ge 122 Sb 141 Ge 2 +^90 Sb^3 +^89 The diagonal relationship between elements can often be attributed to the fact that the most common oxidation states for these elements give rise to ions of similar size, which consequently often show similar reaction chemistry.
1.90 (a) Ga and Si and (c) As and Sn. Note: (b) Be and Al exhibit a diagonal relationship. Because diagonal relationships often exist as a result of similarities in ionic radii, they can exist across the s and p blocks.
1.92 (c) hafnium and (d) niobium
1.94 (a) metal; (b) nonmetal; (c) metalloid; (d) metalloid; (e) nonmetal; (f) metalloid
1.96 (a) λ = 0.20 nm; E = hv or E = hc λ −^1 34 8 1 9 16
(6.626 08 10 J s)(2.997 92 10 m s ) 0.20 10 m 9.9 10 J
− − − −
= ×^ ⋅^ ×^ ⋅
×
= ×
E
This radiation is in the x-ray region of the electromagnetic spectrum. For comparison, the K α radiation from Cu is 0.154 439 0 nm and that from Mo is 0.0709 nm. X-rays produced from these two metals are those most commonly employed for determining structures of molecules in single crystals.
(b) From the de Broglie relationship p = h λ −^1 , we can write h λ −^1 = mv ,
or v = h m −^1 λ−^1 .For an electron, m e = 9.109 39 × 10 −^28 g. (Convert units to kg and m.) 34 31 12 34 2 - 31 12 6 1
(6.626 08 10 J s) (9.109 39 10 kg)(200 10 m) (6.626 08 10 kg m s ) (9.109 39 10 kg)(200 10 m) 3.6 10 m s
− − − − − −
= ×^ ⋅
× ×
= ×^ ⋅^ ⋅
× ×
= × ⋅ -
v
(c) Solve similarly to (b). For a neutron, m n = 1.674 93 × 10 −^24 g. (Convert units to kg and m.)
0
5
10
15
20
25
30
0 5 10 15 20 Atomic Numbe r
Molar Volume (cm
3 , mol
-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 2 p- sublevel; the 3 s- sublevel fills with Na and Mg; and the 3 p- 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
sin sin^2
sin sin 3 0 2 6
π π
π π π π
⎛ ⋅^ ⎞ ⎛⋅ ⋅ ⎞
= ⎛^ ⋅^ ⎞^ − ⎛^ ⋅ ⎞
∫
L
L
x x (^) dx L L L x L x L L =
(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 ).
ψ n = 1 (^) ( ψ n = 1)( ψ n = 2)
ψ n = 2
1.106 (a) (^) ∆ ∆ p x ≥ 12 =
where 6.626 08 10 34 J s (^34)
2 π 2 π 1.054 57^10 J s
× − ⋅ −
= = h = = × ⋅ (^1) (1.054 57 10 34 J s) 5.272 85 10 35 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 2 128 ) 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
9.109 39 × 10 −^28 g or 9.109 39 × 10 −^31 kg,
0 1 x
0 L 00 L^1
x x^ x
0 2 2 2 (^20) 2 2 2 2 2 2 2 2
sin sin^3
cos 2 2 sin^2 8 cos 4 4 sin^4 32
8 1 0 32 1 0 8 1 0 32 1 0 0
= ⎛^ ⎛^ ⎞+ ⎞−
= + − + − ⎡^ + − + ⎤
∫
L
L
x (^) x x L L L x x x L L L L x x x L L L L L L L
Because the integral is zero, one would not expect to observe a transition between the n = 1 and n = 3 states. (b) Again, evaluating the integral:
0 2 2 0
sin sin^2
(^1) cos sin 1 cos 3 3 sin 2 18
= ⎛^ ⎛^ ⎛^ ⎞^ + ⎛^ ⎞^ ⎞^ − ⎛ ⎛^ ⎞^ + ⎛^ ⎞ ⎞
∫
L
L
x (^) x x L L L x x x x x x L L L L L L
Given the L^2 term, we see that the integral, and therefore I , will increase as the length of the box increases.
