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Unformatted text preview: Chapter 15 : Quantum Mechanics and Atomic
Structure 35.2 The wavelength of this chemical wave is 1.2 cm; its frequency is 1/42 s“. The speed of propagation is the precinct of these two values, 0029 cm 3‘1. 15.4 Substitute in the equation 5 m )w. Assume that the gamma rays are propagating through a.
vacuum so c x 2.9979 x 3.03 m sol. The wavelength is then r: 2.9979X108ms‘1 _12
**;~W~105“0 m This equals 0.0106 A. 15.6 {21) Substituting in c a ALI gives the frequency of 488 nm light to be 6.14% x 10H S”. (b) We divide twice the distance from the earth to the moon by the'speecl of light. The answer
is 2.5 s. 15.8 The wavelength of the ultrasonic wave is its speed of propagation divided by its frequency. It
comes out to 0.030 m, or,3.0 cm. The resolution in the sonic image is never better than the
wavelength of sound used.’ If lower frequency (V = 8000 s“) is used, then the wavelength is
longer (18.7 cm) and all detail in the image of the fetus is lost. 15.10 The red light has lower frequency and hence lower energy than the green light. If green light
l ejects no electrons from the copper surface, then red light also ejects no electrons. 15.12 The maximum wavelength of light is the wavelength that supplies photons just energetic
enough to overcome the...work function of the surface: l A __ c __ if _ (6.626 x1043} s)(2.9979 X 108 In 3—1) L _ z #7
l y E 4.41 x 1019 J 4'50 X18 m 15.14 The wavelength of the radiation must be short enough to make the photon energetic enough
to eject an electron from the surface of the tungsten. The work function is 7.29 x 10‘19 J, which is supplied by photons of waveEength 27 2 nm or shorter (computed using A = fie/E}. if
theaejeeteéeelsetson~hasrasveloeitywof~2200ﬂx~170§~rnssipitsvkineticgenergyeis 1.82% 1.0—1.8 ery application of the formula I{.E. 2 1/2 ms2 with m = 9.109 x 10‘31 kg and v as given. The Chapter 1 The Nature and Conceptual Basis of Chemistry 143 energy required from the photon is the work function plus this kinetic energy; the sum equals
2.55 x 10"“18 J. The corresponding A is 77.9 nm. 15.16 A wavelength of 454 nm is in the biue region of the spectrum. 15.18 The energy lost by the potassium atom is carried away by a photon.“ The waveiength of the
photon is he (6.626 x 10‘34J s)(2.9979 x 103 In 3””)  ..._: : ”D? m
_. AB 49 x10'19J 4.1x10 m 410 nm A This waveiength is in the violet region of the spectrum. 15.20 (a) The energy carried by a. photon is the product of Planck’s constant and its frequency
E = by. In this case E = 3.8 ><10~19 J.
(b) The power of the laser is 10 W, which is 10 J 3"”. Hence: 10: x ( Iphoton ) 22.6 X1019 photon T 3.82 x 10"“:9 J s 15.22 The three emission ﬁnes connect each possibie pair of levels (see diagram). 11.9
10.1 E/ev 0
The corresponding waveiengths of emitted light; are given by he 12.3982 x 10”? m A : XE : erlVl (recall Example 15.3). Substituting Vthr = 10.1, 21.9, and 1.8 V (the last of these is the voitage
difference between the two excited states) gives wavelengths of A £1.23 X104 m, 1.04 x10'7 m, 6.9 ><10”7 m which can be written as 123, E04, and 690 nm. 15.24 (a) The energy change to remove. an electron from a groundstate atom is 1.602 x 10—19 J AE 313.6 eV( 16V )e 2.179 x10“13 J The wavelength of a. photon supplying this amount of energy is A " lag _ (6.626 x 10—34 Jr‘s)(2.99?9 >< 108 marl) = 9.12 X10"8 : .
AE 2.179 x 1048 J m 91 2 “m 144 Instructor’s Manual T0 ACCOMPANY Principles of Modern Chemistry, 5th Edition (13) Soive the equation «gmoz : AE for v and substitute ms 2(2.179><19”13J) 6 _1
m1/ 2 We . 1
” m 9.199x10~31kg 21“ 9 ms Converting to miles per hour gives 1ft lmi 36005
— 6 —1 W m . 6 . —1
v—2.19>< 10 rue (0.3048 )(5280ft)( 1h ) 489x10 mih (c) For thermal excitation, kg?" m AE AE ._ 2.179X 10—18.}
kn __ 1.381x 10‘23 J K—1 Tm 21.58x105K 15.26 According to the Bohr model, the radius of a oneelectron atom or ion is “2 2 ﬂ _.
r = 73610 = goes x 10 1‘ m) Substitution of Z :2 2 for helium and n x 5 gives 7‘ = 6.61 x 30”“) In. The energy of any state
of a. one~electron atom or ion is given by E a —————(2.18 x 10*18 J) In the case of a He+ ion in the n 2 5 state, this energy equals M349 x 18“"19 J. Removing
the electron means changing the energy of the atom to E m 0. The change in energy of one
atom is this ﬁnal vaiue minus the initial value, or +3.49 x. 19”“) J. For a mole of atoms the
energy change is Avogadro’s number times larger or 210 id. The energy of the He+ ion in
the n :2 3 state is ~9.69 X 10‘19 J. The change in energy of the ion in the 5 we 3 transition
is the n x 3 energy (the ﬁnai energy) minus the n z 5 energy (the initiai energy). Hence
AE : w620 X 10—19 J. The transition gives off energy, as shown by its negative AE. The
frequency of the photon that carries this energy away is 9.36 x 10“ s”, and the aneIength
is 320 11m. 15.28 in the spectrum of Be3+ (a Z n 4 ion) the series of iines analogous to the Lyman series’ of atomic hydrogen has wavelengths equal to 1/16 of the wavelengths of the Lyman series; the
frequencies in this series are 16 times the frequencies in the Lyman series. A similar scaling
occurs with the series analogous to the Balmer series. These conclusions follow from the
dependence of the energy of the states of hydrogemlike ions on Zz. Thus, for the frequencies
in the Lymamlike series V:(ie)xa29x1015(i— : )3”
12 n
ﬁnai The v’s for the first three nﬁnafs are 3.95 X 1016, 4.68 x .1016, and 4.94 X 1016 5—1. The
corresponding waveiengths are 7.59, 6.41, and 6.07 nm, in the x—ray region. For the frequencies in the Rainier«like series V;(16)x3.29><30i5(1§~ 21 )s'1
2 nﬁnai The 12’s for the first three ”ﬁna]:S are 7.31 x 1015, 9.87 x 1015, and 1.11 x 1016 5"”. The
corresponding wavelengths are 41.0, 30.4, and 27.1 nm, in the ultraviolet region. Chapter 1 The Nature and Conceptual Basis of Chemistry 145 15.30 (a) In the ground state of the standing wave, there is one half~wavelength along the bond, L.
Hence A1 m 2.0 A. In the ﬁrst excited state, n z 2; there are two halfmwavelengths along the
bond, so A2 2 L 21.0 A. (b) The number of nodes is one less than the quantum number describing the standing wave.
Hence, there is 1 node. 15.32 (a) We know the mass of the electron so we can calculate the o of the electrons from their
kinetic energy, from the relationship K.E. m 1/2 mevz. It is 6.614x 106 In 3—1. The wavelength is 0.110 nm (using A = 11/32 = h/mev).
(b) The helium atom moving at 353 In 5"“1 gives A 2 0.282 mm.
(c) The krypton atom moving at 299 m s~1 has A : 0.0159 nm. 15.34 (a) We use the Heisenberg uncertainty principle:
. AmA(mv) _>_ h/élvr with m := 9109 x 10‘31 kg and v = 3.0 x 108 m :6“. We ﬁnd An: R 1.93 x 10'13 m. This
means that if We know nothing about the speed an electron, we can know its location to at best £0002 A. (1)} Because the helium atom is much more massive its A23 is much smaller. By a similar
computation it is 2.65 x 10“” In. ' 15.36 __ ,.hc # (6.626 X 10““ 3 #29979 X 1‘38 m 5"!) — 49
as __ A _ 809 x104 m H. 2.433 x10 J
as = ”2 [(2)2 + (1)2 +(1}z_(1)2_(1)2m(1)2] = hg [33
8 me)? 8 meL'z
_34 2
L = (6.626 x10 J s) {3] z 853 x1040 :11 a 8.53 A 1‘ 6(9109 x 1031 kg)(2.483 x 10“”19 1) 15.38 (a) This combination is not allowed because m5 is never equal to zero. If mS were changed to
i ii / 2, this combination would be allowad. (b) This combination is allowed. It speciﬁes a 23~electron.
i (c) This combination is allowed. It speciﬁes a ”idelectron
l (d) This combination is not; allowed. The quantum number I is never negative. 15.40 (a) 301' (la) 73 (c) 517. 15.42 {a} A 3d—orbital has 0 radial and 2 angular nodes; (b) a 7gworbitai has 2 radial and 4 angular
nodes; [(3) a Sporbital has 3 radial and 1 angular node.  15.44 (a) 3/2
R{3p) ‘“ i“ (31:) (60 — 0’?) exp (ME) A node occurs when R(3p} = 0. The function equals zero when 63 w» (72 = 0 which means it
equals zero when 0‘ ﬂ 6. Therefore” the node is at r a 660 2 6(0.526 >< 16“J m) = 3.17 x 1010 m = 3.}? A 146 Instructor’s Manual To ACCOMPANY Principles of Modern Chemistry, 5th Edition (13) 2 2 3/2 2 cr
R(3s) _. m (as) (27 — 180+ 20“ ) exp ("El Nodes occur at the roots of 27 — 180 + 202 m 0
M
r = W a 1.902, 7.098 ﬁxa
GD r1 2 1.902(s.529 x 10‘1“ m)m 1.9113 r2 : 7.098(0529 X10"10 m) m 3.75 A
15.46 (a) P: [Nel3s23p3 (b) Tc: [Kr]4d5532 (c) Ho: [Xerlle2 15.48 Li” {132252) is diamagnetic; B+ (1322.92) is diamagnetic;
F" (1322522396) is diamagnetic; A134“ (132322106) is diamagnetic.
All of the preceding have an even number of electrons and all electrons paired.
S“ (13223221363333295) is paramagnetic.
Ari" has the same ground—state conﬁguration as S“ and is therefore also paramagnetic.
Br+ ([Ar]3d1°4334p‘1) is paramagnetic; Te“ ([Krl4d105325p5) is paramagnetic.
A species with an odd number of electrons must be paramagnetic; species with an even number
of electrons may be diamagnetic or paramagnetic. 15.50 (a) The species is an atom and is therefore electrically neutral. It has 76 electrons. The
element with Z :2 76 is osmium. (b) The ion is F‘. (c) The ion is Ag5"‘“. 15.52 (a) This element; would follow the completion of the n = 7 row in the periodic table and
would have Z 211%.. (b) We note that 13'? exceeds :19 by 18. This diﬁ'erence would be neatly explained by the
intermediate filling; of nine 5g orbitais by 18 electrons before the closing of the seventh row of
the periodic table. 15.54 The three noblegas atoms would have Z = 3 (corresponding to “153”), Z W 16 (“1532332199
”), and Z = 27 (“1.932532p93533p9 ”). 15.56 (a) Sin (b) Ca (c) I" (d) Ge (e) Rb. 15.58 (a) The two are isoelectronic, and 82‘ has a lesser nuclear charge; it is larger. (b) The Tl+
is larger. Loss of two electrons to give Tl3+ reduces electron—electron repulsions and so allows
contraction. (c) The Ce8+ ion is larger, considering the lanthanide contraction. (d) The I‘
ion is larger; its outer electrons are in the n r 5 sheli. 15.60 For both K and Ga, the outer electrons are in the 4.9 shell, outside of the stable Ar core. K
has the lower ionization energy because of its lower nuclear charge. The second 43 electron in
Ca only partially'lshields the ﬁrst. K" is isoelectronic with Ar; Ca+ is isoeiectronic with K.
Thus the comparison of the ionization energies of K+ and Ca+ echoes that of Ar and K. Ca+
has an appreciably lower ionization energy. in comparing K with Cal" (ale, the first ionization
energy of K with the second of Ca), one is comparing isoeiectronic species. The one with the
higher nuclear charge ( Cai“) should have the higher ionization energy, and does. 15.62 The electron afﬁnity of Cl is higher because (31‘ ion has a closedwshell conﬁguration While 8“”
does not. The ionisation energy of Cl is higher because the effective nuclear charge is greater
in Cl than in S due to incomplete shielding of the additional positive charge on the Cl nucleus. . Emacs: 5,2
.3
l
'i
l
i
i Chapter 1 The Nature and Conceptual Basis of Chemistry 147 15.64 Calcium does have a. positive afﬁnity for an electron, but the quoted number is very small. We
convert it from M moi—1 to J per atom and then use E m hrs/A to compute the wavelength.
Using the most recent value for the EA of Ca(2.0 kJ mol“) gives 60 x 10”5 m; using the earlier value of 1.7 kJ mol”1 gives 7.0 x 18”“5 m. Infrared radiation is energetic enough to
remove the extra. electron from Ca‘. 15.66 The time it takes for any electromagnetic waves to arrive from Cygnus A is its distance divided
by the speed of light. This is 3 x 102’1 m divided by 3.00 x 108 n1 3‘1 or 1016 s, which equals
3 x 108 years. In other words, Cygnus A is 300 million lightwyears away. The frequency of the radio wave is c divided by its wave length. It is 3.0 x 107' e“. 15.68 34 8 1
M he .... 6.626 x10“ J s x 3.00 x 10 m 6"" __ *16
AE‘"’” _ T ‘ 0.29 x 10*9 m ‘ 9'9 x 10 J
6.626 x 1034 3 X 3.00 x 108 ~1
AEAM radio : ' W220 m m S 3 99 X 10w28 3 The Xrey photon is capable of inducing a. chemical reaction for which the required energy is
6.0 x 105 kJ mol"1. (This exceeds the dissociation energy of every Chemical substance) The
AM photon has an energy of only 6.0 x 10““4 J mol'“1 and is ineffective in inﬂuencing chemical reactions.
15.?0 (6.)
E W :55 ﬂ (6.626 X 16“?"1 J s)(2.998 X108 m a”)
M /\ W (315 x 10“9 m)
m 6.306 x 1049.}: 3.94 eV = 380 kJ moi1
(b) Ephoton = KE + barrier RE: Iii — 6.306 x 10—19 J (6.626 x 10““ J s)(2.998 x 10'8 m 8‘1) .49
: ._._____...____c_.m m 6.366 10 3
(200 x 104 m) X 2 9.932 x 10—19 —~ 6.306 x 10*19 J
x 3.626 >< 10“19 J L6 2.26 eV e 215 k.) Incl—1 (c) Compute the momentum of the electrons and then their wavelength p : me 2x V mgr)? m me(2£;'max)
= (9.109 x 1031 kg) 2(3.e26 x 10—19 kg m2s2)
= 8.125 x 10"25 kg m 3‘1 h __ 6.626 x 105.3“ kg m2 s‘3 rmmmj 1*10 =0.815
p 8.126x1025kgms—1 85)”) m am An 148 15.72 15.74 15.76 15713 15.80 Instructor’s Manual To ACCOMPANY Principles of Modern Chemistri, 5th Edition v __ nh r _ £0122}? # ranhezme __ 382
ﬂ 2nmer h. arrZIezirne 1) ~ Qnmeegnzha W 26051}:
2 x (1.60218 x 10*19 C)2
+ w W = _ e —1
“(He )‘ 2(8.8542 x 10—12 o2 J—1 m—1)(1)(e.626 x 10—34 J s) 4 38 X 10 m S
—19 2
MUM) : 92 x (1.60218 x10 C) : 2.01 X 108 m 84 mm The speed of light is 2.9979 x 108 In s“. Relativistic effects wiii be very important in U914“,
but less important in Bel". The 05+ ion is a hydrogenlike ion with Z = 6. All transitions in its spectrum have frequencies
that ﬁt the following formula with Z a 5 and integrai n:
i/ 2 E u —Zz(3.29 X 1015 5W1) (—2—L—~ — 21 >
A ninitiai ”ﬁnal Suppose that green light has a wavelength range running from 500 to 550 nm (Fig. 15.3).
Insert these wavelengths expressed in meters, Z 2 6, and o in In 5‘1 into the formula. The
units cancel out and ( 21 _ 21 ):0.00460 to 0.90506 nﬁnai ”initiai Now, systematically try combinations of integers that give the desired result. Note that Tlﬁna]
must be less than ninitgal. The ﬁrst combination that works is minimal 2 8 and nﬁna] : 7 for which
1 1 __ 0
3%, w 333 — 0.0 478
(a) The loss in intensity of the tone is due to destructive interference between the two loud— speakers superseding constructive interference as one loudspeaker is moved closer Destructive
interference occurs when one wavetrain lags the other by 1,2,3 halfwavelengths. The
halfwavelength 1/2A is thus 0.16 (or 0.080 or 00533. . ) m, and A is 0.32 m. We take this
ﬁrst answer because losses of intensity of the tone Would presumably have been reported at
movement distances less than 0.16 In ifthe wavelength were shorter. (b) The frequency is the speed of the sound divided by its waveiength. It is 1.1 x 103 s‘l. h 6.626 x 10““ J s 45 _ as 1 8 _1
AE_4WAtMWW5.3X1G j  AV—WHM—tlwAt—aSXEOS
The IOWest excited state is (500 — 120) kJ moi”L above the ground state.
A127 2 g;
A _ he (6.626 x 104% s)(2.998 x 108 In 31) AE 5(500 W 120) kJ mol“(1/6.022 x 1023 mol“i)(103 J/kJ) ”T 15.84 Chapter 1 The Nature and Conceptual Basis of Chemistry 149 15.32 (a) A —— J3» —— 1 J 5 w 0 34
_mvm0.145kgx20ms—1—' m
(l3)
Ap m 0.145 kg x 2 m 5‘1 z 0.29 kg m 3—1
h i J s 1J s
> ——— = W :
ApAm __ 41v 4n hence Am 2 47Tl029 kg m 5*1) 0.27 m
(c) 2
_‘ cult2 _ 8.554 x 10—13 C J"1 m”1 (1 J s) _9
“0 ‘ magma ‘ 71“ x (1 <3)2 x 0.091 kg "W 2'82 X10 m “” 28 A
ha n2 n2 n2
E 1. _.__ 56 .JL _£
My“ 8m lot)? "l" L2 + L2
h? 9 112 h? 3 In?
a ' I m : 3
E1“ 8mL2(1/4+1+ ) 32mL2 Em 8mL3(4/4+1+1) 53ml;2
h? 17 13.2 hi“) 21 h2
m . 1 I w ‘ m 2 a
E3” 8 mt? (9/4” + i ) 323ng 13112 El“ 8 mL2 (1/4 + 4 +1) 327111;2
h? 3 h” 122 29 h?
z E 2 E x z x f m :
E41: 221 212 8 ml} (6) 4m}? E321 E3” 8 ml;2 {9/4 + 4 +1) 32ml,2 E11; is the ground state; the others are excited states. Note the double degeneracy of two of
the excited states and the triple degeneracy of a third. 15.86 (a) The probability density of ﬁnding an electron in the vicinity of any point is equal to the square of the wave function of the electron evaluated at that point. Thus, the probability
density of ﬁnding the ls electron of the H atom at a distance r from the nucteus is $2 : (l/wa§)exp(w2rlae) The portion in the ﬁrst parentheses equals 2.15 x i03u m”3 and the exponential term is unitless.
At r equals 0 the exsonential term equals 1 and 1&2 is 2.15 x £030 m”. The probability of ﬁnding the electron exactly at a mathematical point is new because a point
has no volume to accommodate the electron. The small sphere centered at the nucleus has
however a volume of 1 ping (1.0 X 10—36 m3). Over the very short distance between the center and surface of this sphere $2 stays nearly constant at 2.15 >< 1030 311—3. It follows that the
probability of ﬁnding the electron within the small sphere is about: 33 .—.~ 2.15 x 103” mm3 x 1.00 ><10‘36 m3 = 2.15 x 106 (b) At a distance on or 52.9 pm (0.529 x 10”“) m) from the nucleus, the value of the functions
1,92 is less than it is at the nucleus. The exponential part of the function drops rapidly as 1
increases: 11:2(at an) m (2.15 x1030 m3)e“2'"/“° e 2.91 >< 1039 m3 Assume that 1/22 is constant throughout the 1 pm3 volume which the problem speciﬁes. The
chance of ﬁnding the electron at 52.9 pm in a. ﬁxed direction is p :— 2.91x 1029 {II'3 x is x 10""35 m3 m 2.91x 3.0—7 150 Instructor’s Manual T0 ACCOMPANY Principles of Modern Chemistry, 5th Edition 3 (c) A spherical shell cf thickness 1 pm and radius 52.9 pm has a volume:
Vsheu : JIM2m = @4529 pm)2 x 1 pm m 3.52 x 1032 m3 This substantially larger volume naturally has a greater probability of holding the electron
than does the tiny E pm3 volume element considered in part b: p : 2.91x 1029 nor‘3 x 3.52 x 10~32 m3 = 0.0102
15.88 (a) 3 1/2
Y(px)=(&w) sinﬂcesgo 7T 3 1/2
Y(py}m(g;) sinﬂsingo 3 1/2
Y{pz)=(E) c059 The radial part of the wave function need not be taken into account because it does not affect
the angular symmetry. 3/2023.) + Y2(py) + Yam} 2 (21—337?) ( i112 0:203ch + sin2 gain2 {,0 + 0052 8) 3 : (E?) (sin2 3(ccs2 (p + einL2 g9) + cos2 9) m (23:) Hence, the N atom is spherically symmetric because $2 is independent of angle. (b) The following species are spherically symmetric: F”, Na, Se”, Cu, M0, Sb, Au. 15.98 a
Z 71A)
5.7 1.87
58 1.82
59 1.82
69 1.81
61 , 1.81
62 1.80
63 (Eu) 2.08 (—
64 1.79
55 1.76
56 1.75
67 1.74
68 1.73
69 1.72 "(9 (Yb) 1.94 <—
11 1:22 Chapter 1 The Nature and Conceptual Basis of Chemistry 151 (a) The trend in atomic radius with Z for the rare earth elements is generally downward (the
so—called lanthanide contraction}. It is caused by the increasing nuclear charge that attracts the electrons in the 4 f sub—shell toward the nucleus. (b) The elements 63(Eu) and 70(Yb) are the exceptions to the trend. They contain, respec—
tively, a halfﬁlled and a ﬁlled 4f subshell. 15.92 The third ionization energy of Li is the AE for the process L12+(g) we Li3+(g) +e'. The ejec—
tion of a ls—electron from a lithium atom on the other hand is represented Li(g) we Li+ * (g)+e' .
where the star indicates that the Li+ ion is in an excited state. In both cases a ls—electron is
removed. The IE3 is larger because the ls—electron is removed against the full attraction of
the Z 2 3 nucleus whereas the removal of the ls—electron in the photoelectron spectroscopy
experiment is from a Z a 3 nucleus but assisted by repulsions from the other two electrons. 35.913: The ﬁrst ionization energy of K is 419 is] mol‘1 __ £3 __ (6.626 x 18—34.] s)(2.9979 x 108 m 3‘1)6.022 x 1023 mol"1
_ A; _. 650 X 10‘9 m
= 1.84 x 105 J moi”1 m 184 R3 morl AE1 4191a moi—1 : 184M moi—1 + A1172 AE; : 235 kJ mol1 he (6.626 x 10—34 J s)(3.00 X 10"3 In 5‘1)(6.022 x 1033 mol”) m = . = 509 x 10"“9
)‘2 AEQ 2.35 >< 105 J moi—1 m
This is the maximum wavelength that the second photon may have.
15.96 (a) Ground—state Al: 152232211638231).
(b)
—34 s ~1
AEmhﬁzﬁﬁwle Js}(2.998x16 ms ) :5'03x10_193 A 395 X 10—9 m
(c) The energy level diagram for Al...
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 Winter '07
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 Chemistry

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