Chapter 6
The Structure of Atoms
Chapter 6
The Structure of Atoms
INSTRUCTOR’S NOTES
This chapter and the next one (Chapter 7) are aimed at giving students a firm foundation for understanding of the key
theories of chemistry. Therefore, we devote a total of about 9 lectures to these chapters, with much of that time
devoted to electron configurations—particularly of ions—and to periodic properties.
Challenges for teaching this material include:

Pre-existing misconceptions (orbits versus orbitals)

Non-intuitive aspects of quantum mechanics (wave nature of particles, quantized energy)

The concept of quantum numbers which may have no precedent for students

An image representing a probability (orbital) rather than a definitive physical entity
A careful study of the figures should help in their understanding of these ideas.
Students should be urged to master both the new concepts and the mathematical relationships among them. It may
be hard for them to appreciate the significance of the quantum model, but they will soon see consequences of this
world in periodic properties, bonding, polarity and later in spectroscopy.
New to this edition is the incorporation of a section on electron spin and magnetism into this chapter, material that
previously was included in the chapter on electron configurations and periodic trends. We have included resources
for demonstrating the paramagnetism of liquid oxygen below, and the demonstration can be repeated when covering
molecular orbital theory in Chapter 9. Students have experience with magnetism so the inclusion of this topic in the
chapter should provide them with something familiar to relate to the new information which they are learning.
Continuing the study of magnetism through the application of the well-known MRI technique should provide some
motivation to comprehend the theory behind it.
SUGGESTED DEMONSTRATIONS
1.
Line Spectra

One place where our students have some difficulty is in understanding the ideas of “line spectra.” This
subject is not so important when discussed in terms of the Bohr model as it is in understanding the general
idea of the absorption and emission of energy by atoms or molecules. To help students understand line
spectra, we give each of them a small diffraction grating. (Large sheets from a scientific supply house are
cut into small pieces, and each piece is mounted in cardboard and held with a staple.) Using these gratings,
students can view the spectra of light from discharge tubes in the classroom.

Hughes, E. Jr.; George, A. “Suitable Light Sources and Spectroscopes for Student Observation of Emission
121
Chapter 6
The Structure of Atoms
Spectra in Lecture Halls,” Journal of Chemical Education 1984, 61, 908.

Johnson, Kristin A.; Schreiner, Rodney. “A Dramatic Flame Test Demonstration,” Journal of Chemical
Education 2001, 78, 640.

Maynard, J.H. “Using Hydrogen Balloons To Display Metal Ion Spectra,” Journal of Chemical Education
2008, 85, 519.
2.
Properties of Waves

If you have access to a laser pointer, the Optical Transform Kit available from the Institute for Chemical
Education (http://ice.chem.wisc.edu) can be used to project diffraction patterns on the wall of a lecture
room.
3.
Atomic Orbitals

To illustrate the shapes of atomic orbitals, we use a set of wooden orbital models. These are currently
obtainable from Klinger Educational Products Corp. (http://www.klingereducational.com)
4.
Paramagnetism of Oxygen

Shakhashiri, B. Z. “Preparation and Properties of Liquid Oxygen,” Chemical Demonstrations: A Handbook
for Teachers of Chemistry; University of Wisconsin Press, 1985; Vol. 2, pp. 147-152.

Shimada, H.; Yasuoda, T.; Mitsuzawa, S. “Observation of the Paramagnetic Property of Oxygen by Simple
Method,” Journal of Chemical Education 1990, 67, 63.

122
See the ChemistryNow website for a demonstration of paramagnetic liquid oxygen.
Chapter 6
The Structure of Atoms
SOLUTIONS TO STUDY QUESTIONS
6.1
(a) microwaves
(b) red light
(c) infrared light
6.2
(a) red, orange, yellow
(b) blue
(c) blue
6.3
(a) green
6.4
6.5
=
(a)  =
c
2.998  108 m  s–1
=
= 261 m

1150  103 s–1
225 m ·
1 wavelength
= 0.863 wavelengths
261 m
(b)  =
c
2.998  108 m  s–1
=
= 3.06 m

98.1  106 s–1
225 m ·
1 wavelength
= 73.6 wavelengths
3.06 m
5.0  102 nm ·
E=
hc

=
410 nm ·
c
=

10–9 m
= 5.0  10–7 m
1 nm
(6.626  10–34 J  s)(2.998  108 m  s –1 )
= 4.0  10–19 J/photon
5.0  10–7 m
4.0  10–19 J/photon ·
6.6
2.998  108 m  s–1
= 5.04  1014 s–1
5.95  10–7 m
10–9 m
= 5.95  10–7 m
1 nm
(b) 595 nm ·
6.02  1023 photons
= 2.4  105 J/mol photons
1 mol
10–9 m
= 4.1  10–7 m
1 nm
=
c

=
2.998  108 m  s–1
= 7.3  1014 s–1
4.1  10–7 m
E = h = (6.626  10–34 J·s)(7.3  1014 s–1) = 4.8  10–19 J/photon
4.8  10–19 J/photon ·
6.02  1023 photons
= 2.9  105 J/mol photons
1 mol
According to the text, red light has an energy of 1.75  105 J/mol
2.9  105 J
= 1.7
1.75  105 J
Violet light is 1.7 times more energetic than red light.
123
Chapter 6
6.7
The Structure of Atoms
396.15 nm ·
=
c

10–9 m
= 3.9615  10–7 m
1 nm
2.99792  108 m  s–1
= 7.5676  1014 s–1
3.9615  10–7 m
=
E = h = (6.62607  10–34 J·s)(7.5676  1014 s–1) = 5.0144  10–19 J/photon
5.0144  10–19 J/photon ·
6.8
6.02214  1023 photons
= 3.02  105 J/mol photons
1.00 mol
285.2 nm is in the ultraviolet region, 383.8 nm is just at the edge of the visible region, and 518.4 nm is in
the visible region. The most energetic line has the shortest wavelength, 285.2 nm.
285.2 nm ·
E=
6.9
hc

=
10–9 m
= 2.852  10–7 m
1 nm
(6.6261  10–34 J  s)(2.9979  108 m  s–1 ) 6.0221  1023 photons
·
= 4.194  105 J/mol
2.852  10–7 m
1 mol
(d) FM radiowaves
(c) microwaves
(a) yellow light
(b) X-rays
—increasing energy per photon
6.10
(b) radiowaves
(a) microwaves
(d) red light
(e) ultraviolet radiation
(c) -rays
—increasing energy per photon
6.11
2.0  102 kJ/mol ·
=
6.12
hc
(6.626  10–34 J  s)(2.998  108 m  s –1 )
=
= 6.0  10–7 m (visible region)
E
3.3  10–19 J
10–9 m
= 5.4  10–7 m
1 nm
540 nm ·
E=
1 mol
103 J
·
= 3.3  10–19 J/photon
6.02  1023 photons 1 kJ
hc

=
(6.626  10–34 J  s)(2.998  108 m  s –1 )
= 3.7  10–19 J/photon
5.4  10–7 m
This radiation does not have enough energy to activate the switch. This is also true for radiation with
wavelengths greater than 540 nm.
6.13
(a) The most energetic line has the shortest wavelength, 253.652 nm.
(b) 253.652 nm ·
=
10–9 m
= 2.53652  10–7 m
1 nm
c
2.997925  108 m  s–1
=
= 1.18190  1015 s–1

2.53652  10–7 m
E = h = (6.626069  10–34 J·s)(1.18190  1015 s–1) = 7.83135  10–19 J/photon
(c) The 404.656 nm line is violet, while the 435.833 nm line is blue.
124
Chapter 6
6.14
The Structure of Atoms
(a) The infrared region
(b) None of the lines mentioned are in the spectrum shown in Figure 6.7. None of the lines listed are in
the visible region.
(c) The most energetic line has the shortest wavelength, 837.761 nm.
(d) 865.438 nm ·
=
c

=
10–9 m
= 8.65438  10–7 m
1 nm
2.997925  108 m  s–1
= 3.46406  1014 s–1
8.65438  10–7 m
E = h = (6.626069  10–34 J·s)(3.46406  1014 s–1) = 2.29531  10–19 J/photon
6.15
Violet; ninitial = 6 and nfinal = 2
6.16
 = 121.6 nm (1.216  10–7 m)
=
c

=
2.9979  108 m  s–1
= 2.465  1015 s–1
1.216  10–7 m
ninitial = 2 and nfinal = 1
6.17
6.18
6.19
(a) From n = 5 to n = 4, 3, 2, or 1
= 4 lines
From n = 4 to n = 3, 2, or 1
= 3 lines
From n = 3 to n = 2 or 1
= 2 lines
From n = 2 to n = 1
= 1 line
Total
= 10 lines possible
(b) Highest frequency (highest energy)
n = 5 to n = 1
(c) Longest wavelength (lowest energy)
n = 5 to n = 4
(a) From n = 4 to n = 3, 2, or 1
= 3 lines
From n = 3 to n = 2 or 1
= 2 lines
From n = 2 to n = 1
= 1 line
Total
= 6 lines possible
(b) Lowest energy
n = 4 to n = 3
(c) Shortest wavelength (highest energy)
n = 4 to n = 1
(a) n = 3 to n = 2
(b) n = 4 to n = 1
The energy levels are progressively closer at higher levels, so the energy difference from n = 4 to n = 1
is greater than from n = 5 to n = 2.
6.20
(a) n = 2 to n = 4 and (d) n = 3 to n = 5
125
Chapter 6
6.21
The Structure of Atoms
 1
1
1 
1
E = –Rhc  2 – 2
 = –1312 kJ/mol  2 – 2  = –1166 kJ/mol
n
n
1
3


initial 
 final
1166 kJ/mol ·
6.22
=
E
1.936  10–18 J
=
= 2.922  1015 s–1
h
6.6261  10–34 J  s
=
c
2.9979  108 m  s–1
=
= 1.026  10–7 m (ultraviolet region)

2.922  1015 s–1
 1
1 
1 
1
E = –Rhc  2 – 2
 = –1312 kJ/mol  2 – 2  = –63.78 kJ/mol
n
n
3
4


initial 
 final
63.78 kJ/mol ·
6.23
1 mol
103 J
·
= 1.936  10–18 J/photon
6.0221  1023 photons
1 kJ
1 mol
103 J
·
= 1.059  10–19 J/photon
6.0221  1023 photons
1 kJ
=
E
1.059  10–19 J
=
= 1.598  1014 s–1
h
6.6261  10–34 J  s
=
c
2.9979  108 m  s–1
=
= 1.876  10–6 m (infrared region)

1.598  1014 s–1
1m
2.5  108 cm
· 2
= 2.5  106 m/s
10 cm
1s
=
6.626  10—34 J  s
h
=
= 2.9  10–10 m
(9.109  10–31 kg)(2.5  106 m  s–1 )
mv
6.24
=
6.626  10–34 J  s
h
=
= 5.6  10–12 m
(9.11  10–31 kg)(1.3  108 m  s–1 )
mv
6.25
1.0  102 g ·
1 kg
= 0.10 kg
103 g
5.6  10–3 nm ·
v=
6.26
1m
= 5.6  10–12 m
109 nm
6.626  10–34 J  s
h
=
= 1.2  10–21 m·s–1
(0.10 kg)(5.6  10–12 m)
m
6.626  10–34 J  s
h
=
= 1.41  10–33 m
(1.50  10–3 kg)(313 m  s–1 )
mv
(a)  can be 0, 1, 2, 3
(b) m can be 0, ±1, ±2
126
6.626  10–34 J  s
h
=
= 2.2  10–34 m
(0.10 kg)(30 m  s –1 )
mv
103 m 1 hour
1 km
7.00  102 mile
·
·
·
= 313 m·s–1
0.6214 mile 1 km
3600 s
1 hour
=
6.27
=
Chapter 6
The Structure of Atoms
(c) n = 4,  = 0, m = 0
(d) n = 4,  = 3, m = 0, ±1, ±2, ±3
6.28
(a) The orbital type is d. It is a 4d orbital.
(b) When n = 5,  = 0, 1, 2, 3, and 4
=0
1 s orbital
=1
3 p orbitals
=2
5 d orbitals
=3
7 f orbitals
=4
9 g orbitals
There are a total of 52 or 25 orbitals in the n = 5 electron shell.
(c) In an f subshell there are 7 orbitals. m = 0, ±1, ±2, and ±3
6.29
6.30
n

m
4
1
–1
4
1
0
4
1
1
n

m
5
2
–2
5
2
–1
5
2
0
5
2
1
5
2
2
6.31
When n = 4, there are four subshells, 4s, 4p, 4d, and 4f
6.32
When n = 5, there are five subshells, 5s, 5p, 5d, 5f, and 5g
6.33
(a) When n = 2, the maximum value of  is 1
(b) When = 0, m can only have a value of 0
(c) When = 0, m can only have a value of 0
6.34
(b) and (c) are valid sets of quantum numbers
(a) incorrect; when n = 3, the maximum value of  is 2
(d) incorrect; when = 3, m can only have values of 0, ±1, ±2, or ±3
6.35
(a) none; when  = 0, m can only have a value of 0
(b) 3 orbitals
(c) 11 orbitals
(d) 1 orbital
6.36
(a) 7 orbitals
(b) 25 orbitals
127
Chapter 6
The Structure of Atoms
(c) None;  cannot have a value equal to n
(d) 1 orbital
6.37
6.38
(a) ms can only have values of ±1/2
ms = +1/2
(b) when  = 1, m can only have values of 0, ±1
m = 0
(c) the maximum value of  is (n – 1)
=2
(a) the maximum value of  is (n – 1)
=1
(b) ms can only have values of
±1/2
ms = +1/2
(c) when  = 1, m can only have values of 0, ±1
m = 0
6.39
2d and 3f cannot exist. They do not follow the  = 0, 1, 2, …, n – 1 rule
6.40
2f and 1p are incorrect designations. They do not follow the  = 0, 1, 2, …, n – 1 rule
6.41
(a) 2p
(b) 3d
(c) 4f
128
n
2
2
2
3
3
3
3
3
4
4
4
4
4
4
4

1
1
1
2
2
2
2
2
3
3
3
3
3
3
3
m
–1
0
1
–2
–1
0
1
2
–3
–2
–1
0
1
2
3
Chapter 6
The Structure of Atoms
6.42
(a) 5f
(b) 4d
(c) 2s
n
5
5
5
5
5
5
5
4
4
4
4
4
2

3
3
3
3
3
3
3
2
2
2
2
2
0
m
–3
–2
–1
0
1
2
3
–2
–1
0
1
2
0
6.43
(d) 4d
6.44
(d) s orbital
6.45
The value of  indicates the number of nodal surfaces through the nucleus
(a) 2s:  = 0, zero nodal surfaces
(b) 5d:  = 2, two nodal surfaces
(c) 5f:  = 3, three nodal surfaces
6.46
The value of  indicates the number of nodal surfaces
(a) 4f:  = 3, three nodal surfaces
(b) 2p:  = 1, one nodal surface
(c) 6s:  = 0, zero nodal surfaces
6.47
(a) correct
(b) Incorrect; the intensity of a light beam is independent of frequency and is related to the number of
photons of light with a certain energy.
(c) correct
6.48
The Lyman series is found in the ultraviolet region and the Balmer series is in the visible region.
6.49
s
0 nodal surface
p
1 nodal surface
d
2 nodal surfaces
f
3 nodal surfaces
6.50
orbital
maximum number in a given shell
s
1
p
3
d
5
129
Chapter 6
The Structure of Atoms
f
6.51
6.52
7
 value
orbital type
3
f
0
s
1
p
2
d
s orbital
px orbital
x
y
6.53
6.54
orbital type
number of nodal surfaces
s
1
0
p
3
1
d
5
2
f
7
3
 1
1 
1
1
E = –Rhc  2 – 2
 = –1312 kJ/mol  2 – 2  = –16.04 kJ/mol
ninitial 
6 
5
 nfinal
16.04 kJ/mol ·
6.55
number of orbitals in a given subshell
1 mol
103 J
·
= 2.663  10–20 J/photon
6.0221  1023 photons
1 kJ
=
E
2.663  10–20 J
=
= 4.019  1013 s–1
h
6.6261  10–34 J  s
=
c
2.9979  108 m  s–1
=
= 7.460  10–6 m

4.019  1013 s–1
(a) green light
(b) Shorter wavelength corresponds to higher energy. Green light has a wavelength of 500 nm and red
light has a wavelength of 680 nm.
(c) green light
6.56
375 nm ·
E=
130
hc

=
10–9 m
= 3.75  10–7 m
1 nm
1 kJ
(6.626  10–34 J  s)(2.998  108 m  s –1 ) 6.022  1023 photons
·
· 3 = 319 kJ/mol
–7
10 J
3.75  10 m
1.00 mol
Chapter 6
6.57
The Structure of Atoms
(a)  =
c
2.998  108 m  s–1
=
= 0.35 m

850  106 s–1
(b) E = h = (6.626  10–34 J·s)(850  106 s–1) ·
(c) E =
hc

=
6.02  1023 photons
= 0.34 J/mol
1.00 mol
(6.626  10–34 J  s)(2.998  108 m  s –1 ) 6.02  1023 photons
·
= 2.8  105 J/mol
4.2  10–7 m
1.00 mol
2.8  105 J/mol
= 840,000
0.34 J/mol
(d) Violet light is 840,000 times more energetic than the radiation sent from cell phones.
6.58
E=
hc

=
(6.626  10–34 J  s)(2.998  108 m  s –1 )
= 4.2  10–19 J/photon
4.7  10–7 m
2.50  10–14 J
= 5.9  104 photons
4.2  10–19 J/photon
6.59
 1
1 
1
 1
He: E = –Z2Rhc  2 – 2
 = –(22)(1312 kJ/mol)  2 – 2  = 5248 kJ/mol
ninitial 
1 

 nfinal
 1
1
1 
 1
H: E = –Z2Rhc  2 – 2
 = –(12)(1312 kJ/mol)  2 – 2  = 1312 kJ/mol
n
n

1


initial 
 final
6.60
(i) (b) n = 7 to n = 6
(ii) (a) n = 7 to n = 1
(iii) (a) n = 7 to n = 1
6.61
1s, 2s = 2p, 3s = 3p = 3d, 4s
6.62
(a) 3
(d) 5
(g) 25
(b) 3
(e) 5
(h) 1
(c) 1
(f) 7
1.173  106 eV ·
1 mol
9.6485  104 J/mol
·
= 1.879  10–13 J/photon
6.0221

1023 photons
1 eV
6.63
=
E
1.879  10–13 J
=
= 2.836  1020 s–1
h
6.6261  10–34 J  s
=
c
2.9979  108 m  s–1
=
= 1.057  10–12 m

2.836  1020 s–1
131
Chapter 6
6.64
The Structure of Atoms
qeye = (11 g)(4.0 J/g·K)(3.0 K) = 130 J
E=
hc

=
(6.626  10–34 J  s)(2.998  108 m  s–1 )
= 1.7  10–24 J/photon
0.12 m
130 J
= 8.0  1025 photons
1.7  10–24 J/photon
103 m
1s
·
= 260 seconds (4.3 minutes)
1 km
2.998  108 m
6.65
7.8  107 km ·
6.66
(a) The most energetic line has the shortest wavelength, 357.9 nm
(b) blue-indigo
6.67
(a) size and energy
(b) 
(c) more
(d) 7
(e) 1
(f) d, s, p
(g) 0, 1, 2, 3, 4
(h) 16
(i) paramagnetic
6.68
(a) size and energy; shape
(b) 0, 1, 2
(c) f
(d) 4; 2; –2
(e) letter
p
d
 value
1
2
nodal planes
1
2
(f) f
(g) 2d, 3f
(h) n = 2, = 1, m = 2, ms = +½ is not valid
n = 4, = 3, m = 0, ms = 0 is not valid
(i) (i) 3
(ii) 9
(iii) none
(iv) 1
(j) paramagnetic
6.69
(a) (b) is diamagnetic, (c) is paramagnetic, and (a) is ferromagnetic
(b) Solid (a) is most strongly attracted to a magnetic field; solid (b) is least strongly attracted.
132
Chapter 6
6.70
The Structure of Atoms
540 nm ·
10–9 m
= 5.4  10–7 m
1 nm
=
c

=
2.998  108 m  s–1
= 5.6  1014 s–1
5.4  10–7 m
 6.022  1023 photons 
5
E = h = (6.626  10–34 J·s)(5.6  1014 s–1) 
 = 2.2  10 J/mol photons
1
mol
photons


6.71
The pickle glows because the materials in the pickle are being excited by the addition of the energy (electric
current). The pickle has been soaked in brine (NaCl), so the electrons in the sodium atom are excited and
release energy as they return to lower energy states, providing yellow light. The same kind of light is
visible in many street lamps.
6.72
278 nm ·
10–9 m
= 2.78  10–7 m
1 nm
=
c

=
2.998  108 m  s–1
= 1.08  1015 s–1
2.78  10–7 m
 6.022  1023 photons 
5
E = h = (6.626  10–34 J·s)(1.08  1015 s–1) 
 = 4.30  10 J/mol photons
1 mol photons


Ultraviolet radiation; aspirin is colorless (does not absorb visible light)
6.73
(a)  =
1
= 5  10–4 cm = 5 m
2000 cm –1
(b) The low energy end is the right end of the spectrum; the high energy end is the left end of the spectrum
(c) The O—H interaction requires more energy.
6.74
Bohr’s circular orbit model contradicts the laws of classical physics, and Bohr had to artificially introduce
the concept of quantization to explain how these electron orbits could be stable.
6.75
(c) Electrons moving from a given level to one of lower n results in the emission of energy, which is
observed as light.
6.76
The square of the wave function is the probability of finding the electron within a given region of space,
also known as the electron density. This region of space where an electron of a given energy is most
probably located is its orbital. The units for 4r2 are 1/distance.
6.77
The electron behaves simultaneously as a wave and a particle. The modern view of atomic structure is
based on the wave properties of the electron, and describes regions around an atom’s nucleus in which there
is the highest probability of finding a given electron.
6.78
(b), (e) – (j)
6.79
(a) and (b)
6.80
N = 1, L = 1, M = –1, 0, +1
3 orbitals
N = 2, L = 2, M = –1, 0, +1
3 orbitals
133
Chapter 6
The Structure of Atoms
N = 3, L = 3, M = –1, 0, +1
3 orbitals
A total of 9 orbitals in the first three electron shells
6.81
A photon with wavelength of 93.8 nm has enough energy to promote an electron from n = 1 to n = 6.
From n = 6 to n = 5, 4, 3, 2, or 1
= 5 lines
From n = 5 to n = 4, 3, 2, or 1
= 4 lines
From n = 4 to n = 3, 2, or 1
= 3 lines
From n = 3 to n = 2 or 1
= 2 lines
From n = 2 to n = 1
= 1 line
Total
= 15 lines possible
The wavelengths for emissions with nfinal = 1, 2, and 3 are shown in Figure 6.10.
n = 6 to n = 5:
 1
1 
1
1
E = –Rhc  2 – 2
 = –1312 kJ/mol  2 – 2  = –16.04 kJ/mol
n
n
5
6


initial 
 final
16.04 kJ/mol ·
=
1 mol
103 J
·
= 2.663  10–20 J/photon
6.0221  1023 photons
1 kJ
(6.6261  10–34 J  s)(2.9979  108 m  s –1 )
hc
=
= 7.460  10–6 m (7460. nm)
2.663  10–20 J/photon
E
n = 6 to n = 4:
 1
1 
1
 1
E = –Rhc  2 – 2
 = –1312 kJ/mol  2 – 2  = –45.56 kJ/mol
ninitial 
6 
4
 nfinal
45.56 kJ/mol ·
=
1 mol
103 J
·
= 7.565  10–20 J/photon
23
6.0221  10 photons
1 kJ
(6.6261  10–34 J  s)(2.9979  108 m  s –1 )
hc
=
= 2.626  10–6 m (2626 nm)
7.565  10–20 J/photon
E
n = 5 to n = 4:
 1
1 
1
 1
E = –Rhc  2 – 2
 = –1312 kJ/mol  2 – 2  = –29.52 kJ/mol
n
n
4
5


initial 
 final
29.52 kJ/mol ·
=
6.82
1 mol
103 J
·
= 4.902  10–20 J/photon
6.0221  1023 photons
1 kJ
(6.6261  10–34 J  s)(2.9979  108 m  s –1 )
hc
=
= 4.052  10–6 m (4052 nm)
4.902  10–20 J/photon
E
According to de Broglie’s equation,  =
h
, any moving particle has an associated wavelength. However,
m
a heavy particle such as a golf ball has an incredibly small wavelength that cannot be measured with any
134
Chapter 6
The Structure of Atoms
instrument now available.
6.83
(a) Group VII B, period 5
(b) n = 5,  = 0, m = 0
(c) 0.141  106 eV ·
1 mol
9.6485  104 J/mol
·
= 2.26  10–14 J/photon
6.0221  1023 photons
1 eV
=
E
2.26  10–14 J
=
= 3.41  1019 s–1
h
6.626  10–34 J  s
=
c
2.998  108 m  s–1
=
= 8.79  10–12 m

3.41  1019 s–1
(d) (i) HTcO4(aq) + NaOH(aq)  NaTcO4(aq) + H2O()
(ii) 4.5  10–3 g ·
4.5  10–3 g ·
1 mol Tc 1 mol NaTcO4
185 g
·
·
= 8.5  10–3 g NaTcO4
1 mol Tc
98 g
1 mol NaTcO4
1 mol Tc 1 mol HTcO4
40.0 g
1 mol NaOH
·
·
·
= 1.8  10–3 g NaOH
1 mol Tc
98 g
1 mol HTcO4 1 mol NaOH
(e) 1.5 mol NaTcO4 ·
1.5 mol NaTcO4 ·
2
6.84
(a) probability =
4r (e
185 g
1 mol
·
= 2.8  10–4 g NaTcO4
6
10  mol 1 mol NaTcO4
1
1 mol
·
= 1.5  10–4 mol/L
6
10  mol 0.0100 L
-r/a o
ao
2
) (d)
3
For r = ao = 52.9 pm and (d) = 1.0 pm,
-1 2
Probability =
4(e ) (1.0 pm)
= 0.010
52.9 pm
(b) For r = 0.50ao,
2
Probability =
4(0.5) (e
0.5 2
) (1.0 pm)
= 0.0070
52.9 pm
135
Chapter 6
The Structure of Atoms
For r=4.0ao,
2
Probability =
-4
2
4(4.0) (e ) (1.0 pm)
= 0.00041
52.9 pm
The probabilities are in accord with the surface density plot shown in Figure 6.12b.
SOLUTION TO APPLYING CHEMICAL PRINCIPLES: CHEMSITRY OF THE SUN
1.
2.
 =
=
c

c

=
2.998  108 m/s
= 5.102  1014 s -1
-9
587.6  10 m
=
2.998  108 m/s
= 5.271  10-7 m/s = 527.1 nm
14 -1
5.688  10 s
(6.6261  10-34 J  s)(2.9979  108 m/s)
E=
=
= 3.3726  10-19 J (589.00 nm light)
-9

589.00  10 m
hc
3.
E=
hc

=
(6.6261  10-34 J  s)(2.9979  108 m/s)
= 3.3692  10-19 J (589.59 nm light)
-9
589.59  10 m
Energy difference = 3.3726 × 10-19 J – 3.3692 × 10-19 J = 3.4 × 10-22 J
4.
E=
hc

=
(6.626  10-34 J  s)(2.998  108 m/s)
= 3.381  10-19 J/photon
587.6  10-9 m
(3.381 x 10-19 J/photon)(6.022 x 1023 photons/mol) = 2.036 x 105 J/mol = 203.6 kJ/mol
5. Balmer series: G, F, C
136