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**Unformatted text preview: **Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website CHAPTER 7 Quantum Theory and the Electronic
Structure of Atoms
INTRODUCTION
IN CHAPTER 7.1 FROM CLASSICAL PHYSICS TO QUANTUM
THEORY 2 WE LEARNED THAT AN ATOM IS MADE OF A NUCLEUS CONTAINING PROTONS AND NEUTRONS WITH ELECTRONS MOVING ABOUT
AT SOME DISTANCE FROM IT. THIS 7.2 THE PHOTOELECTRIC EFFECT DESCRIPTION, ALTHOUGH ESSENTIALLY 7.3 BOHR’S THEORY OF THE HYDROGEN
ATOM CORRECT, DOES NOT PROVIDE A DETAILED EXPLANATION OF ATOMIC
PROPERTIES. IT TURNS OUT THAT INDIVIDUAL ATOMS DO NOT BEHAVE LIKE ANYTHING WE KNOW IN THE MACROSCOPIC WORLD. 7.4 THE DUAL NATURE OF THE ELECTRON SCIENTISTS’ 7.5 QUANTUM MECHANICS RECOGNITION OF THIS FACT LED TO THE DEVELOPMENT OF A NEW BRANCH 7.6 QUANTUM NUMBERS
OF PHYSICS CALLED QUANTUM THEORY TO EXPLAIN THE BEHAVIOR OF 7.7 ATOMIC ORBITALS
THESE SUBMICROSCOPIC PARTICLES. QUANTUM 7.8 ELECTRON CONFIGURATION
THEORY ALLOWS US TO PREDICT AND UNDERSTAND THE CRITICAL ROLE THAT ELECTRONS PLAY IN CHEMISTRY. IN 7.9 THE BUILDING-UP PRINCIPLE ONE SENSE, STUDYING ATOMS AMOUNTS TO ASKING THE FOLLOWING QUESTIONS: 1. HOW MANY ELECTRONS ARE PRESENT IN A PARTICULAR
2. WHAT ENERGIES DO INDIVIDUAL ELECTRONS POSSESS?
3. WHERE IN THE ATOM CAN ELECTRONS BE FOUND? THE ATOM? ANSWERS TO THESE QUESTIONS HAVE A DIRECT RELATIONSHIP TO THE BEHAVIOR OF ALL SUBSTANCES IN CHEMICAL REACTIONS, AND THE
STORY OF THE SEARCH FOR ANSWERS PROVIDES A FASCINATING BACKDROP FOR OUR DISCUSSION. 243 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 244 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS 7.1 FROM CLASSICAL PHYSICS TO QUANTUM THEORY Early attempts by nineteenth-century physicists to understand atoms and molecules met
with only limited success. By assuming that molecules behave like rebounding balls,
physicists were able to predict and explain some macroscopic phenomena, such as the
pressure exerted by a gas. However, this model did not account for the stability of molecules; that is, it could not explain the forces that hold atoms together. It took a long
time to realize — and an even longer time to accept — that the properties of atoms and
molecules are not governed by the same physical laws as larger objects.
The new era in physics started in 1900 with a young German physicist named
Max Planck.† While analyzing the data on radiation emitted by solids heated to various temperatures, Planck discovered that atoms and molecules emit energy only in certain discrete quantities, or quanta. Physicists had always assumed that energy is continuous and that any amount of energy could be released in a radiation process. Planck’s
quantum theory turned physics upside down. Indeed, the flurry of research that ensued
altered our concept of nature forever.
PROPERTIES OF WAVES To understand Planck’s quantum theory, we must know something about the nature of
waves. A wave can be thought of as a vibrating disturbance by which energy is transmitted. The fundamental properties of a wave are illustrated by a familiar type — water waves. Figure 7.1 shows a seagull floating on the ocean. Water waves are generated by pressure differences in various regions on the surface of water. If we carefully
observe the motion of a water wave as it affects the motion of the seagull, we find that
it is periodic in character; that is, the wave form repeats itself at regular intervals.
Waves are characterized by their length and height and by the number of waves
that pass through a certain point in one second (Figure 7.2). Wavelength (lambda)
is the distance between identical points on successive waves. The frequency (nu) is
the number of waves that pass through a particular point in one second. In Figure 7.1,
the frequency corresponds to the number of times per second that the seagull moves
through a complete cycle of upward and downward motion. Amplitude is the vertical
distance from the midline of a wave to the peak or trough.
† Max Karl Ernst Ludwig Planck (1858 – 1947). German physicist. Planck received the Nobel Prize in Physics in 1918 for
his quantum theory. He also made significant contributions in thermodynamics and other areas of physics. FIGURE 7.1 Properties of water waves. The distance between
corresponding points on successive waves is called the wavelength, and the number of times
the seagull rises up per unit of
time is called the frequency. It is
assumed that the seagull makes
no effort to move, but simply
rides up and down as the wave
moves from left to right under it. Back Forward Main Menu Wavelength TOC Study Guide TOC Textbook Website MHHE Website 7.1 FROM CLASSICAL PHYSICS TO QUANTUM THEORY 245 Wavelength
Wavelength
Amplitude Amplitude Wavelength Direction of wave
propagation (a)
FIGURE 7.2 (a) Wavelength
and amplitude. (b) Two waves
having different wavelengths and
frequencies. The wavelength of
the top wave is three times that
of the lower wave, but its frequency is only one-third that of
the lower wave. Both waves have
the same amplitude. Amplitude (b) Another important property of waves is their speed, which depends on the type
of wave and the nature of the medium through which the wave is traveling (for example, air, water, or a vacuum). The speed (u) of a wave is the product of its wavelength and its frequency:
u (7.1) The inherent “sensibility” of Equation (7.1) becomes apparent if we analyze the physical
dimensions involved in the three terms. The wavelength ( ) expresses the length of a
wave, or distance/wave. The frequency ( ) indicates the number of these waves that
pass any reference point per unit of time, or waves/time. Thus the product of these
terms results in dimensions of distance/time, which is speed:
distance
time distance
wave waves
time Wavelength is usually expressed in units of meters, centimeters, or nanometers, and
frequency is measured in hertz (Hz), where
1 Hz 1 cycle/s The word “cycle” may be left out and the frequency expressed as, for example, 25/s
(read as “25 per second”).
EXAMPLE 7.1 Calculate the speed of a wave whose wavelength and frequency are 17.4 cm and
87.4 Hz, respectively.
Answer From Equation (7.1),
u 17.4 cm
17.4 cm
1.52 Similar problem: 7.8. 87.4 Hz
87.4/s 103 cm/s PRACTICE EXERCISE Calculate the frequency (in Hz) of a wave whose speed and wavelength are 713 m/s
and 1.14 m, respectively. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 246 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS FIGURE 7.3 The electric field
and magnetic field components
of an electromagnetic wave.
These two components have the
same wavelength, frequency, and
amplitude, but they vibrate in two
mutually perpendicular planes. z Electric field component
y x Magnetic field component ELECTROMAGNETIC RADIATION Sound waves and water waves are
not electromagnetic waves, but
X rays and radio waves are. A more accurate value for the speed
of light is given on the inside back
cover of the book. There are many kinds of waves, such as water waves, sound waves, and light waves.
In 1873 James Clerk Maxwell proposed that visible light consists of electromagnetic
waves. According to Maxwell’s theory, an electromagnetic wave has an electric field
component and a magnetic field component. These two components have the same
wavelength and frequency, and hence the same speed, but they travel in mutually perpendicular planes (Figure 7.3). The significance of Maxwell’s theory is that it provides
a mathematical description of the general behavior of light. In particular, his model accurately describes how energy in the form of radiation can be propagated through space
as vibrating electric and magnetic fields. Electromagnetic radiation is the emission
and transmission of energy in the form of electromagnetic waves.
Electromagnetic waves travel 3.00 108 meters per second (rounded off), or
186,000 miles per second in a vacuum. This speed does differ from one medium to another, but not enough to distort our calculations significantly. By convention, we use
the symbol c for the speed of electromagnetic waves, or as it is more commonly called,
the speed of light. The wavelength of electromagnetic waves is usually given in nanometers (nm).
EXAMPLE 7.2 The wavelength of the green light from a traffic signal is centered at 522 nm. What
is the frequency of this radiation?
Answer Rearranging Equation (7.1) and replacing u with c because we are dealing with electromagnetic waves, we get c Recalling that 1 nm 1 10 9 m (see Table 1.3) and that c
3.00 write 522 nm
5.75 3.00 108 m/s, we 8 10 m/s
1 10 9 m
1 nm 1014/s, or 5.75 1014 Hz 14 Similar problems: 7.7, 7.8. Back Forward Comment Thus, 5.75
10 waves pass a fixed point every second. This very high
frequency is in accordance with the very high speed of light. Main Menu TOC Study Guide TOC Textbook Website MHHE Website 7.1 FROM CLASSICAL PHYSICS TO QUANTUM THEORY 247 10–3 10–1 10 103 105 107 109 1011 1013 1020 1018 1016 1014 1012 1010 108 106 104 X rays Ultraviolet Wavelength (nm) Gamma
rays
Type of radiation Visible Frequency (Hz)
Infrared Microwave 3
6 Radio waves 2
5 1
4 9 X ray Sun lamps Heat
lamps Microwave ovens,
police radar,
satellite stations 8 7 # 0 * UHF TV,
cellular
telephones FM radio,
VHF TV AM
radio (a)
FIGURE 7.4 (a) Types of electromagnetic radiation. Gamma
rays have the shortest wavelength
and highest frequency; radio
waves have the longest wavelength and the lowest frequency.
Each type of radiation is spread
over a specific range of wavelengths (and frequencies).
(b) Visible light ranges from a
wavelength of 400 nm (violet) to
700 nm (red). (b) PRACTICE EXERCISE What is the wavelength (in meters) of an electromagnetic wave whose frequency is
3.64 107 Hz?
Figure 7.4 shows various types of electromagnetic radiation, which differ from
one another in wavelength and frequency. The long radio waves are emitted by large
antennas, such as those used by broadcasting stations. The shorter, visible light waves
are produced by the motions of electrons within atoms and molecules. The shortest
waves, which also have the highest frequency, are associated with (gamma) rays,
which result from changes within the nucleus of the atom (see Chapter 2). As we will
see shortly, the higher the frequency, the more energetic the radiation. Thus, ultraviolet radiation, X rays, and rays are high-energy radiation.
PLANCK’S QUANTUM THEORY When solids are heated, they emit electromagnetic radiation over a wide range of wavelengths. The dull red glow of an electric heater and the bright white light of a tungsten light bulb are examples of radiation from heated solids.
Measurements taken in the latter part of the nineteenth century showed that the
amount of radiation energy emitted by an object at a certain temperature depends on
its wavelength. Attempts to account for this dependence in terms of established wave Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 248 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS theory and thermodynamic laws were only partially successful. One theory explained
short-wavelength dependence but failed to account for the longer wavelengths. Another
theory accounted for the longer wavelengths but failed for short wavelengths. It seemed
that something fundamental was missing from the laws of classical physics.
Planck solved the problem with an assumption that departed drastically from accepted concepts. Classical physics assumed that atoms and molecules could emit (or
absorb) any arbitrary amount of radiant energy. Planck said that atoms and molecules
could emit (or absorb) energy only in discrete quantities, like small packages or bundles. Planck gave the name quantum to the smallest quantity of energy that can be
emitted (or absorbed) in the form of electromagnetic radiation. The energy E of a single quantum of energy is given by
E h (7.2) where h is called Planck’s constant and is the frequency of radiation. The value of
Planck’s constant is 6.63 10 34 J s.
According to quantum theory, energy is always emitted in multiples of h ; for example, h , 2 h , 3 h , . . . , but never, for example, 1.67 h or 4.98 h . At the time
Planck presented his theory, he could not explain why energies should be fixed or quantized in this manner. Starting with this hypothesis, however, he had no trouble correlating the experimental data for emission by solids over the entire range of wavelengths;
they all supported the quantum theory.
The idea that energy should be quantized or “bundled” may seem strange, but the
concept of quantization has many analogies. For example, an electric charge is also
quantized; there can be only whole-number multiples of e, the charge of one electron.
Matter itself is quantized, for the numbers of electrons, protons, and neutrons and the
numbers of atoms in a sample of matter must also be integers. Our money system is
based on a “quantum” of value called a penny. Even processes in living systems involve quantized phenomena. The eggs laid by hens are quantized, and a pregnant cat
gives birth to an integral number of kittens, not to one-half or three-quarters of a kitten.
7.2 THE PHOTOELECTRIC EFFECT In 1905, only five years after Planck presented his quantum theory, Albert Einstein†
used the theory to solve another mystery in physics, the photoelectric effect, a
phenomenon in which electrons are ejected from the surface of certain metals exposed
to light of at least a certain minimum frequency, called the threshold frequency (Figure
7.5). The number of electrons ejected was proportional to the intensity (or brightness)
of the light, but the energies of the ejected electrons were not. Below the threshold
frequency no electrons were ejected no matter how intense the light.
The photoelectric effect could not be explained by the wave theory of light.
Einstein, however, made an extraordinary assumption. He suggested that a beam of
light is really a stream of particles. These particles of light are now called photons.
Using Planck’s quantum theory of radiation as a starting point, Einstein deduced that
† Albert Einstein (1879 – 1955). German-born American physicist. Regarded by many as one of the two greatest physicists
the world has known (the other is Isaac Newton). The three papers (on special relativity, Brownian motion, and the photoelectric effect) that he published in 1905 while employed as a technical assistant in the Swiss patent office in Berne have
profoundly influenced the development of physics. He received the Nobel Prize in Physics in 1921 for his explanation of
the photoelectric effect. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 7.2 This equation has the same form as
Equation (7.2) because, as we will
see shortly, electromagnetic radiation is emitted as well as absorbed
in the form of photons. Incident
light + 249 each photon must possess energy E, given by the equation
E h where is the frequency of light. Electrons are held in a metal by attractive forces,
and so removing them from the metal requires light of a sufficiently high frequency
(which corresponds to sufficiently high energy) to break them free. Shining a beam of
light onto a metal surface can be thought of as shooting a beam of particles — photons
— at the metal atoms. If the frequency of photons is such that h is exactly equal to
the energy that binds the electrons in the metal, then the light will have just enough
energy to knock the electrons loose. If we use light of a higher frequency, then not
only will the electrons be knocked loose, but they will also acquire some kinetic energy. This situation is summarized by the equation e– h
Metal THE PHOTOELECTRIC EFFECT KE BE (7.3) where KE is the kinetic energy of the ejected electron and BE is the binding energy
of the electron in the metal. Rewriting Equation (7.3) as
KE – h BE shows that the more energetic the photon (that is, the higher its frequency), the greater
the kinetic energy of the ejected electron.
Now consider two beams of light having the same frequency (which is greater
than the threshold frequency) but different intensities. The more intense beam of light
consists of a larger number of photons; consequently, it ejects more electrons from the
metal’s surface than the weaker beam of light. Thus the more intense the light, the
greater the number of electrons emitted by the target metal; the higher the frequency
of the light, the greater the kinetic energy of the emitted electrons.
EXAMPLE 7.3 Voltage
Meter
source
FIGURE 7.5 An apparatus for
studying the photoelectric effect.
Light of a certain frequency falls
on a clean metal surface. Ejected
electrons are attracted toward the
positive electrode. The flow of
electrons is registered by a detecting meter. Calculate the energy (in joules) of (a) a photon with a wavelength of 5.00 104 nm
(infrared region) and (b) a photon with a wavelength of 5.00 10 2 nm (X-ray region).
Answer (a) We use Equation (7.2):
E From Equation (7.1), c/ (as explained in Example 7.2). Therefore
E hc
(6.63 34 J s)(3.00 108 m/s)
1 10 9 m
104 nm)
1 nm 10 (5.00
3.98 Similar problem: 7.15. h 10 21 J This is the energy of a single photon with a 5.00 104 nm wavelength.
(b) Following the same procedure as in (a), we can show that the energy of the photon that has a wavelength of 5.00 10 2 nm is 3.98 10 15 J. Thus an “X-ray”
photon is 1 106, or a million times, more energetic than an “infrared” photon in
our case.
PRACTICE EXERCISE The energy of a photon is 5.87 Back Forward Main Menu TOC 10 Study Guide TOC 20 J. What is its wavelength (in nanometers)? Textbook Website MHHE Website 250 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS Einstein’s theory of light posed a dilemma for scientists. On the one hand, it explains the photoelectric effect satisfactorily. On the other hand, the particle theory of
light is not consistent with the known wave behavior of light. The only way to resolve
the dilemma is to accept the idea that light possesses both particlelike and wavelike
properties. Depending on the experiment, light behaves either as a wave or as a stream
of particles. This concept was totally alien to the way physicists had thought about
matter and radiation, and it took a long time for them to accept it. We will see in Section
7.4 that a dual nature (particles and waves) is not unique to light but is characteristic
of all matter, including electrons.
7.3 BOHR’S THEORY OF THE HYDROGEN ATOM Einstein’s work paved the way for the solution of yet another nineteenth-century “mystery” in physics: the emission spectra of atoms.
EMISSION SPECTRA Ever since the seventeenth century, when Newton showed that sunlight is composed
of various color components that can be recombined to produce white light, chemists
and physicists have studied the characteristics of emission spectra, that is, either continuous or line spectra of radiation emitted by substances. The emission spectrum of
a substance can be seen by energizing a sample of material either with thermal energy
or with some other form of energy (such as a high-voltage electrical discharge if the
substance is gaseous). A “red-hot” or “white-hot” iron bar freshly removed from a hightemperature source produces a characteristic glow. This visible glow is the portion of
its emission spectrum that is sensed by eye. The warmth of the same iron bar represents another portion of its emission spectrum — the infrared region. A feature common to the emission spectra of the sun and of a heated solid is that both are continuous; that is, all wavelengths of visible light are represented in the spectra (see the visible
region in Figure 7.4).
The emission spectra of atoms in the gas phase, on the other hand, do not show
a continuous spread of wavelengths from red to violet; rather, the atoms produce bright
lines in different parts of the visible spectrum. These line spectra are the light emission only at specific wavelengths. Figure 7.6 is a schematic diagram of a discharge tube
that is used to study emission spectra, and Figure 7.7 shows the color emitted by hydrogen atoms in a discharge tube.
Every element has a unique emission spectrum. The characteristic lines in atomic
spectra can be used in chemical analysis to identify unknown atoms, much as fingerprints are used to identify people. When the lines of the emission spectrum of a known
element exactly match the lines of the emission spectrum of an unknown sample, the
identity of the sample is established. Although the utility of this procedure was recognized some time ago in chemical analysis, the origin of these lines was unknown until early in this century.
EMISSION SPECTRUM OF THE HYDROGEN ATOM In 1913, not too long after Planck’s and Einstein’s discoveries, a theoretical explana- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 7.3 FIGURE 7.6 (a) An experimental arrangement for studying the
emission spectra of atoms and
molecules. The gas under study is
in a discharge tube containing
two electrodes. As electrons flow
from the negative electrode to the
positive electrode, they collide
with the gas. This collision
process eventually leads to the
emission of light by the atoms (or
molecules). The emitted light is
separated into its components by
a prism. Each component color is
focused at a definite position, according to its wavelength, and
forms a colored image of the slit
on the photographic plate. The
colored images are called spectral lines. (b) The line emission
spectrum of hydrogen atoms. BOHR’S THEORY OF THE HYDROGEN ATOM 251 Photographic plate Slit
High
voltage
Line
spectrum Prism Discharge tube
Light separated into
various components
(a) 400 nm 500 600 700 (b) tion of the emission spectrum of the hydrogen atom was presented by the Danish physicist Niels Bohr.† Bohr ’s treatment is very complex and is no longer considered to be
correct in all its details. Thus, we will concentrate only on his important assumptions
and final results, which do account for the special lines.
When Bohr first tackled this problem, physicists already knew that the atom contains electrons and protons. They thought of an atom as an entity in which electrons
whirled around the nucleus in circular orbits at high velocities. This was an appealing
model because it resembled the motions of the planets around the sun. In the hydrogen atom, it was believed that the electrostatic attraction between the positive “solar”
proton and the negative “planetary” electron pulls the electron inward and that this
force is balanced exactly by the acceleration due to the circular motion of the electron.
Bohr ’s model of the atom included the idea of electrons moving in circular orbits, but he imposed a rather severe restriction: The single electron in the hydrogen
atom could be located only in certain orbits. Since each orbit has a particular energy
associated with it, the energies associated with electron motion in the permitted orbits
must be fixed in value, or quantized. Bohr attributed the emission of radiation by an
energized hydrogen atom to the electron dropping from a higher-energy orbit to a lower
one and giving up a quantum of energy (a photon) in the form of light (Figure 7.8).
Using arguments based on electrostatic interaction and Newton’s laws of motion, Bohr
showed that the energies that the electron in the hydrogen atom can possess are given
by
En
FIGURE 7.7 Color emitted by
hydrogen atoms in a discharge
tube. The color observed results
from the combination of the colors emitted in the visible spectrum. Back Forward Main Menu RH 1
n2 where RH, the Rydberg‡ constant, has the value 2.18 (7.4) 10 18 J. The number n is an † Niels Henrik David Bohr (1885 – 1962). Danish physicist. One of the founders of modern physics, he received the Nobel
Prize in Physics in 1922 for his theory explaining the spectrum of the hydrogen atom.
‡ Johannes Robert Rydberg (1854 – 1919). Swedish physicist. Rydberg’s major contribution to physics was his study of the
line spectra of many elements. TOC Study Guide TOC Textbook Website MHHE Website 252 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS Photon n=1
n=2 n=3 FIGURE 7.8 The emission
process in an excited hydrogen
atom, according to Bohr’s theory.
An electron originally in a higherenergy orbit (n 3) falls back to
a lower-energy orbit (n 2). As
a result, a photon with energy h
is given off. The value of h is
equal to the difference in energies of the two orbits occupied
by the electron in the emission
process. For simplicity, only three
orbits are shown. integer called the principal quantum number; it has the values n 1, 2, 3, . . . .
The negative sign in Equation (7.4) is an arbitrary convention, signifying that the
energy of the electron in the atom is lower than the energy of a free electron, which is
an electron that is infinitely far from the nucleus. The energy of a free electron is arbitrarily assigned a value of zero. Mathematically, this corresponds to setting n equal
to infinity in Equation (7.4), so that E∞ 0. As the electron gets closer to the nucleus
(as n decreases), En becomes larger in absolute value, but also more negative. The most
negative value, then, is reached when n 1, which corresponds to the most stable energy state. We call this the ground state, or the ground level, which refers to the lowest energy state of a system (which is an atom in our discussion). The stability of the
electron diminishes for n 2, 3, . . . . Each of these levels is called an excited state,
or excited level, which is higher in energy than the ground state. A hydrogen electron
for which n is greater than 1 is said to be in an excited state. The radius of each circular orbit in Bohr ’s model depends on n2. Thus, as n increases from 1 to 2 to 3, the
orbit radius increases very rapidly. The higher the excited state, the farther away the
electron is from the nucleus (and the less tightly it is held by the nucleus).
Bohr ’s theory enables us to explain the line spectrum of the hydrogen atom.
Radiant energy absorbed by the atom causes the electron to move from a lower-energy
state (characterized by a smaller n value) to a higher-energy state (characterized by a
larger n value). Conversely, radiant energy (in the form of a photon) is emitted when
the electron moves from a higher-energy state to a lower-energy state. The quantized
movement of the electron from one energy state to another is analogous to the movement of a tennis ball either up or down a set of stairs (Figure 7.9). The ball can be on
any of several steps but never between steps. The journey from a lower step to a higher
one is an energy-requiring process, whereas movement from a higher step to a lower
step is an energy-releasing process. The quantity of energy involved in either type of
change is determined by the distance between the beginning and ending steps. Similarly,
the amount of energy needed to move an electron in the Bohr atom depends on the difference in energy levels between the initial and final states.
To apply Equation (7.4) to the emission process in a hydrogen atom, let us suppose that the electron is initially in an excited state characterized by the principal quantum number ni. During emission, the electron drops to a lower energy state characterized by the principal quantum number nf (the subscripts i and f denote the initial and
final states, respectively). This lower energy state may be either another excited state
or the ground state. The difference between the energies of the initial and final states
is
E Ef Ei Ef RH 1
nf2 Ei RH 1
n2
i From Equation (7.4), and
Therefore
FIGURE 7.9 A mechanical
analogy for the emission
processes. The ball can rest on
any step but not between steps. Back Forward E RH
n2
f
1
RH 2
ni RH
n2
i
1
n2
f Because this transition results in the emission of a photon of frequency
h (Figure 7.8), we can write Main Menu TOC Study Guide TOC Textbook Website and energy MHHE Website 7.3 BOHR’S THEORY OF THE HYDROGEN ATOM 253 TABLE 7.1 The Various Series in Atomic Hydrogen
Emission Spectrum
SERIES nf ni Lyman
Balmer
Paschen
Brackett 1
2
3
4 2,
3,
4,
5, SPECTRUM REGION 3,
4,
5,
6, 4, . . .
5, . . .
6, . . .
7, . . . Ultraviolet
Visible and ultraviolet
Infrared
Infrared E h RH 1
n2
i 1
n2
f (7.5) When a photon is emitted, ni nf. Consequently the term in parentheses is negative
and E is negative (energy is lost to the surroundings). When energy is absorbed, ni
nf and the term in parentheses is positive, so E is positive. Each spectral line in the
emission spectrum corresponds to a particular transition in a hydrogen atom. When we
study a large number of hydrogen atoms, we observe all possible transitions and hence
the corresponding spectral lines. The brightness of a spectral line depends on how many
photons of the same wavelength are emitted.
The emission spectrum of hydrogen includes a wide range of wavelengths from
the infrared to the ultraviolet. Table 7.1 lists the series of transitions in the hydrogen
spectrum; they are named after their discoverers. The Balmer series was particularly
easy to study because a number of its lines fall in the visible range.
Figure 7.8 shows a single transition. However, it is more informative to express
transitions as shown in Figure 7.10. Each horizontal line represents an allowed energy
level for the electron in a hydrogen atom. The energy levels are labeled with their principal quantum numbers. FIGURE 7.10 The energy levels in the hydrogen atom and the
various emission series. Each energy level corresponds to the energy associated with an allowed
energy state for an orbit, as postulated by Bohr and shown in
Figure 7.8. The emission lines
are labeled according to the
scheme in Table 7.1. ∞
7
6
5 4 Energy 3 Forward Main Menu Paschen
series 2 n=1 Back Brackett
series TOC Balmer
series Lyman
series Study Guide TOC Textbook Website MHHE Website 254 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS Example 7.4 illustrates the use of Equation (7.5).
EXAMPLE 7.4 What is the wavelength of a photon emitted during a transition from the ni
to the nf 2 state in the hydrogen atom? 5 state Answer Since nf
2, this transition gives rise to a spectral line in the Balmer series (see Figure 7.10). From Equation (7.5) we write E RH 1
n2
i 2.18 1
n2
f
10 4.58 18 10 J 1
52 19 1
22 J The negative sign indicates that this is energy associated with an emission process.
To calculate the wavelength we will omit the minus sign for E because the wavelength of the photon must be positive. Since E h or
E/h, we can calculate the wavelength of the photon by writing The negative sign is in accord with
our convention that energy is given
off to the surroundings. c
ch
E
(3.00 108 m/s)(6.63 4.34 10 7 10 7 m 34 J s) m 4.34 10 1 109 nm
1m 434 nm which is in the visible region of the electromagnetic region (see Figure 7.4). Similar problems: 7.31, 7.32. PRACTICE EXERCISE What is the wavelength (in namometers) of a photon emitted during a transition
from ni 6 to nf 4 state in the H atom?
The Chemistry in Action essays on page 255 and page 256 discuss atomic emission and lasers. 7.4 THE DUAL NATURE OF THE ELECTRON Physicists were both mystified and intrigued by Bohr ’s theory. They questioned why
the energies of the hydrogen electron are quantized. Or, phrasing the question in a more
concrete way, Why is the electron in a Bohr atom restricted to orbiting the nucleus at
certain fixed distances? For a decade no one, not even Bohr himself, had a logical explanation. In 1924 Louis de Broglie† provided a solution to this puzzle. De Broglie † Louis Victor Pierre Raymond Duc de Broglie (1892 – 1977). French physicist. Member of an old and noble family in France,
he held the title of a prince. In his doctoral dissertation, he proposed that matter and radiation have the properties of both
wave and particle. For this work, de Broglie was awarded the Nobel Prize in Physics in 1929. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 7.4 THE DUAL NATURE OF THE ELECTRON 255 Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Element from the Sun
Scientists know that our sun and other stars contain
certain elements. How was this information obtained?
In the early nineteenth century, the German physicist Josef Fraunhofer studied the emission spectrum of
the sun and noticed certain dark lines at specific wavelengths. We interpret the appearance of these lines by
supposing that originally a continuous band of color
was radiated and that, as the emitted light moves outward from the sun, some of the radiation is reabsorbed
at those wavelengths by the atoms in space. These
dark lines are therefore absorption lines. For atoms,
the emission and absorption of light occur at the same
wavelengths, but they differ in appearance — colored
lines for emission and dark lines for absorption. By
matching the absorption lines in the emission spectra Fraunhofer’s original drawing,
showing the dark absorption
lines in the sun’s emission spectrum. reasoned that if light waves can behave like a stream of particles (photons), then perhaps particles such as electrons can possess wave properties. According to de Broglie,
an electron bound to the nucleus behaves like a standing wave. Standing waves can be
generated by plucking, say, a guitar string (Figure 7.11). The waves are described as
standing, or stationary, because they do not travel along the string. Some points on the
string, called nodes, do not move at all; that is, the amplitude of the wave at these
points is zero. There is a node at each end, and there may be nodes between the ends.
The greater the frequency of vibration, the shorter the wavelength of the standing wave
and the greater the number of nodes. As Figure 7.11 shows, there can be only certain
wavelengths in any of the allowed motions of the string. FIGURE 7.11 The standing
waves generated by plucking a
guitar string. Each dot represents
a node. The length of the string
( ) must be equal to a whole
number times one-half the wavelength ( /2). Back Forward of distant stars with the emission spectra of known elements, scientists have been able to deduce the types
of elements present in those stars.
In 1868 the French physicist Pierre Janssen detected a new dark line in the solar emission spectrum
that did not match the emission lines of known elements. The name helium (from the Greek helios, meaning the sun) was given to the element responsible for
the absorption line. Twenty-seven years later, helium
was discovered on Earth by the British chemist William
Ramsay in a mineral of uranium. On Earth, the only
source of helium is through radioactive decay
processes — particles emitted during nuclear decay
are eventually converted to helium atoms. Main Menu l= TOC λ
–
2 –
l = 2λ
2 Study Guide TOC –
l = 3λ
2 Textbook Website MHHE Website 256 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Back Laser — The Splendid Light
Laser is an acronym for light amplification by stimulated emission of radiation. It is a special type of emission that involves either atoms or molecules. Since the
discovery of laser in 1960, it has been used in numerous systems designed to operate in the gas, liquid, and solid states. These systems emit radiation with
wavelengths ranging from infrared through visible and
UV. The advent of laser has truly revolutionized science, medicine, and technology.
Ruby laser was the first known laser. Ruby is a
deep-red mineral containing corundum, Al2O3, in
which some of the Al3 ions have been replaced by
Cr3 ions. A flashlamp is used to excite the chromium
atoms to a higher energy level. The excited atoms are
unstable, so at a given instant some of them will return to the ground state by emitting a photon in the
red region of the spectrum. The photon bounces back
and forth many times between mirrors at opposite ends
of the laser tube. This photon can stimulate the emission of photons of exactly the same wavelength from
other excited chromium atoms; these photons in turn
can stimulate the emission of more photons, and so
on. Because the light waves are in phase — that is,
their maxima and minima coincide — the photons enhance one another, increasing their power with each
Totally reflecting mirror passage between the mirrors. One of the mirrors is
only partially reflecting, so that when the light reaches
a certain intensity it emerges from the mirror as a laser
beam. Depending on the mode of operation, the laser
light may be emitted in pulses (as in the ruby laser
case) or in continuous waves.
Laser light is characterized by three properties: It
is intense, it has precisely known wavelength and
hence energy, and it is coherent. By coherent we mean
that the light waves are all in phase. The applications
of lasers are quite numerous. Their high intensity and
ease of focus make them suitable for doing eye
surgery, for drilling holes in metals and welding, and
for carrying out nuclear fusion. The fact that they are
highly directional and have precisely known wavelengths makes them very useful for telecommunications. Lasers are also used in isotope separation, in
holography (three-dimensional photography), in compact disc players, and in supermarket scanners. Lasers
have played an important role in the spectroscopic investigation of molecular properties and of many chemical and biological processes. Laser lights are increasingly being used to probe the details of chemical
reactions.
The emission of light from a
ruby laser. Flash lamp Laser beam Ruby rod Partially reflecting mirror
The stimulated emission of one
photon by another photon in a
cascade event that leads to the
emission of laser light. The synchronization of the light waves
produces an intensely penetrating
laser beam. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 7.4 THE DUAL NATURE OF THE ELECTRON 257 Chemistry in Action Chemistry in Action State-of-the-art lasers used in the research laboratory of Dr. A. H. Zewail at the California
Institute of Technology. De Broglie argued that if an electron does behave like a standing wave in the hydrogen atom, the length of the wave must fit the circumference of the orbit exactly
(Figure 7.12). Otherwise the wave would partially cancel itself on each successive orbit. Eventually the amplitude of the wave would be reduced to zero, and the wave
would not exist.
The relation between the circumference of an allowed orbit (2 r) and the wavelength ( ) of the electron is given by
2r n where r is the radius of the orbit, is the wavelength of the electron wave, and n
1, 2, 3, … . Because n is an integer, it follows that r can have only certain values as n
increases from 1 to 2 to 3 and so on. And since the energy of the electron depends on
the size of the orbit (or the value of r), its value must be quantized.
De Broglie’s reasoning led to the conclusion that waves can behave like particles
and particles can exhibit wavelike properties. De Broglie deduced that the particle and
wave properties are related by the expression (a) h
m (7.7) where , m, and are the wavelengths associated with a moving particle, its mass, and
its velocity, respectively. Equation (7.7) implies that a particle in motion can be treated
as a wave, and a wave can exhibit the properties of a particle.
(b)
FIGURE 7.12 (a) The circumference of the orbit is equal to an
integral number of wavelengths.
This is an allowed orbit. (b) The
circumference of the orbit is not
equal to an integral number of
wavelengths. As a result, the electron wave does not close on itself
evenly. This is a nonallowed orbit. Back Forward Main Menu EXAMPLE 7.5 Calculate the wavelength of the “particle” in the following two cases: (a) The fastest
serve in tennis is about 140 miles per hour, or 62 m/s. Calculate the wavelength associated with a 6.0 10 2 kg tennis ball traveling at this velocity. (b) Calculate the
wavelength associated with an electron moving at 62 m/s.
Answer (a) Using Equation (7.7) we write
h
m TOC Study Guide TOC Textbook Website MHHE Website 258 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS 6.63 The conversion factor is 1 J 10 34 Js 1 kg m2/s2 (see Section 5.7). Therefore
1.8 10 34 m This is an exceedingly small wavelength, since the size of an atom itself is on the
order of 1 10 10 m. For this reason, the wave properties of such a tennis ball cannot be detected by any existing measuring device.
(b) In this case FIGURE 7.13 Patterns made
by a beam of electrons passing
through aluminum. The patterns
are similar to those obtained with
X rays, which are waves. h
m
6.63 10 34 Js where 9.1095 10 31 kg is the mass of an electron. Proceeding as in (a) we obtain
1.2 10 5 m, or 1.3 104 nm, which is in the infrared region. Similar problems: 7.41, 7.42. PRACTICE EXERCISE Calculate the wavelength (in nanometers) of a H atom (mass
moving at 7.00 102 cm/s. 1.674 10 27 kg) Example 7.5 shows that although de Broglie’s equation can be applied to diverse
systems, the wave properties become observable only for submicroscopic objects. This
distinction is due to the smallness of Planck’s constant, h, which appears in the numerator in Equation (7.7).
Shortly after de Broglie introduced his equation, Clinton Davisson† and Lester
Germer‡ in the United States and G. P. Thomson§ in England demonstrated that electrons do indeed possess wavelike properties. By directing a beam of electrons through
a thin piece of gold foil, Thomson obtained a set of concentric rings on a screen, similar to the pattern observed when X rays (which are waves) were used. Figure 7.13
shows such a pattern for aluminum.
The Chemistry in Action essay on p. 259 describes electron microscopy. 7.5
In reality, Bohr ’s theory accounted
for the observed emission spectra of
He and Li2 ions, as well as that
of hydrogen. However, all three systems have one feature in common —
each contains a single electron.
Thus the Bohr model worked successfully only for the hydrogen
atom and for hydrogenlike ions. QUANTUM MECHANICS The spectacular success of Bohr ’s theory was followed by a series of disappointments.
Bohr ’s approach did not account for the emission spectra of atoms containing more
than one electron, such as atoms of helium and lithium. Nor did it explain why extra
lines appear in the hydrogen emission spectrum when a magnetic field is applied.
Another problem arose with the discovery that electrons are wavelike: How can the
“position” of a wave be specified? We cannot define the precise location of a wave because a wave extends in space. † Clinton Joseph Davisson (1881 – 1958). American physicist. He and G. P. Thomson shared the Nobel Prize in Physics in
1937 for demonstrating wave properties of electrons.
‡ Lester Halbert Germer (1896 – 1972). American physicist. Discoverer (with Davisson) of the wave properties of electrons. § George Paget Thomson (1892 – 1975). English physicist. Son of J. J. Thomson, he received the Nobel Prize in Physics in
1937, along with Clinton Davisson, for demonstrating wave properties of electrons. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 7.5 QUANTUM MECHANICS 259 Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Back Electron Microscopy
The electron microscope is an extremely valuable application of the wavelike properties of electrons because it produces images of objects that cannot be
seen with the naked eye or with light microscopes.
According to the laws of optics, it is impossible to form
an image of an object that is smaller than half the
wavelength of the light used for the observation. Since
the range of visible light wavelengths starts at around
400 nm, or 4 10 5 cm, we cannot see anything
smaller than 2 10 5 cm. In principle, we can see
objects on the atomic and molecular scale by using
X rays, whose wavelengths range from about 0.01 nm
to 10 nm. However, X rays cannot be focused, so they
do not produce well-formed images. Electrons, on the
other hand, are charged particles, which can be focused in the same way the image on a TV screen is
focused, that is, by applying an electric field or a magnetic field. According to Equation (7.7), the wavelength of an electron is inversely proportional to its velocity. By accelerating electrons to very high velocities,
we can obtain wavelengths as short as 0.004 nm.
A different type of electron microscope, called the
scanning tunneling microscope (STM), makes use of another quantum mechanical property of the electron
to produce an image of the atoms on the surface of
a sample. Because of its extremely small mass, an electron is able to move or “tunnel” through an energy
barrier (instead of going over it). The STM consists of
a tungsten metal needle with a very fine point, the
source of the tunneling electrons. A voltage is maintained between the needle and the surface of the sample to induce electrons to tunnel through space to the
sample. As the needle moves over the sample, at a
distance of a few atomic diameters from the surface,
the tunneling current is measured. This current decreases with increasing distance from the sample. By
using a feedback loop, the vertical position of the tip
can be adjusted to a constant distance from the surface. The extent of these adjustments, which profile
the sample, is recorded and displayed as a threedimensional false-colored image.
Both the electron microscope and the STM are
among the most powerful tools in chemical and biological research. An electron micrograph showing a normal red blood cell
and a sickled red blood cell from the same person. Forward STM image of iron atoms arranged to display the Chinese
characters for atom on a copper surface. Main Menu TOC Study Guide TOC Textbook Website MHHE Website 260 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS To describe the problem of trying to locate a subatomic particle that behaves like
a wave, Werner Heisenberg† formulated what is now known as the Heisenberg uncertainty principle: it is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certainty. Stated mathematically,
The sign means that the product
x p can be greater than or equal
to h/4 , but it can never be smaller
than h/4 . xp h
4 (7.8) where x and p are the uncertainties in measuring the position and momentum, respectively. Equation (7.8) says that if we make the measurement of the momentum of
a particle more precise (that is, if we make p a small quantity), our knowledge of the
position will become correspondingly less precise (that is, x will become larger).
Similarly, if the position of the particle is known more precisely, then its momentum
measurement must be less precise. Applying the Heisenberg uncertainty principle to
the hydrogen atom, we see that in reality the electron does not orbit the nucleus in a
well-defined path, as Bohr thought. If it did, we could determine precisely both the position of the electron (from the radius of the orbit) and its momentum (from its kinetic
energy) at the same time, a violation of the uncertainty principle.
To be sure, Bohr made a significant contribution to our understanding of atoms,
and his suggestion that the energy of an electron in an atom is quantized remains unchallenged. But his theory did not provide a complete description of electronic behavior in atoms. In 1926 the Austrian physicist Erwin Schrödinger,‡ using a complicated mathematical technique, formulated an equation that describes the behavior and
energies of submicroscopic particles in general, an equation analogous to Newton’s
laws of motion for macroscopic objects. The Schrödinger equation requires advanced
calculus to solve, and we will not discuss it here. It is important to know, however,
that the equation incorporates both particle behavior, in terms of mass m, and wave behavior, in terms of a wave function (psi), which depends on the location in space of
the system (such as an electron in an atom).
The wave function itself has no direct physical meaning. However, the probability of finding the electron in a certain region in space is proportional to the square of
the wave function, 2. The idea of relating 2 to probability stemmed from a wave theory analogy. According to wave theory, the intensity of light is proportional to the
square of the amplitude of the wave, or 2. The most likely place to find a photon is
where the intensity is greatest, that is, where the value of 2 is greatest. A similar argument associates 2 with the likelihood of finding an electron in regions surrounding
the nucleus.
Schrödinger ’s equation began a new era in physics and chemistry, for it launched
a new field, quantum mechanics (also called wave mechanics). We now refer to the
developments in quantum theory from 1913 — the time Bohr presented his analysis for
the hydrogen atom — to 1926 as “old quantum theory.”
THE QUANTUM MECHANICAL DESCRIPTION OF THE HYDROGEN ATOM The Schrödinger equation specifies the possible energy states the electron can occupy
in a hydrogen atom and identifies the corresponding wave functions ( ). These energy
† Werner Karl Heisenberg (1901 – 1976). German physicist. One of the founders of modern quantum theory, Heisenberg received the Nobel Prize in Physics in 1932.
‡
Erwin Schrödinger (1887 – 1961). Austrian physicist. Schrödinger formulated wave mechanics, which laid the foundation
for modern quantum theory. He received the Nobel Prize in Physics in 1933. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 7.6 FIGURE 7.14 A representation
of the electron density distribution
surrounding the nucleus in the hydrogen atom. It shows a high
probability of finding the electron
closer to the nucleus. Although the helium atom has only
two electrons, in quantum mechanics it is regarded as a many-electron
atom. 7.6 QUANTUM NUMBERS 261 states and wave functions are characterized by a set of quantum numbers (to be discussed shortly), with which we can construct a comprehensive model of the hydrogen
atom.
Although quantum mechanics tells us that we cannot pinpoint an electron in an
atom, it does define the region where the electron might be at a given time. The concept of electron density gives the probability that an electron will be found in a particular region of an atom. The square of the wave function, 2, defines the distribution of electron density in three-dimensional space around the nucleus. Regions of high
electron density represent a high probability of locating the electron, whereas the opposite holds for regions of low electron density (Figure 7.14).
To distinguish the quantum mechanical description of an atom from Bohr ’s model,
we speak of an atomic orbital, rather than an orbit. An atomic orbital can be thought
of as the wave function of an electron in an atom. When we say that an electron is in
a certain orbital, we mean that the distribution of the electron density or the probability of locating the electron in space is described by the square of the wave function associated with that orbital. An atomic orbital, therefore, has a characteristic energy, as
well as a characteristic distribution of electron density.
The Schrödinger equation works nicely for the simple hydrogen atom with its one
proton and one electron, but it turns out that it cannot be solved exactly for any atom
containing more than one electron! Fortunately, chemists and physicists have learned
to get around this kind of difficulty by approximation. For example, although the behavior of electrons in many-electron atoms (that is, atoms containing two or more electrons) is not the same as in the hydrogen atom, we assume that the difference is probably not too great. Thus we can use the energies and wave functions obtained from the
hydrogen atom as good approximations of the behavior of electrons in more complex
atoms. In fact, this approach provides fairly good descriptions of electronic behavior
in many-electron atoms. QUANTUM NUMBERS In quantum mechanics, three quantum numbers are required to describe the distribution of electrons in hydrogen and other atoms. These numbers are derived from the
mathematical solution of the Schrödinger equation for the hydrogen atom. They are
called the principal quantum number, the angular momentum quantum number, and the
magnetic quantum number. These quantum numbers will be used to describe atomic
orbitals and to label electrons that reside in them. A fourth quantum number — the spin
quantum number — describes the behavior of a specific electron and completes the description of electrons in atoms. THE PRINCIPAL QUANTUM NUMBER (n)
Equation (7.4) holds only for the
hydrogen atom. Back Forward Main Menu The principal quantum number (n) can have integral values 1, 2, 3, and so forth; it corresponds to the quantum number in Equation (7.4). In a hydrogen atom, the value of
n determines the energy of an orbital. As we will see shortly, this is not the case for a
many-electron atom. The principal quantum number also relates to the average distance of the electron from the nucleus in a particular orbital. The larger n is, the greater
the average distance of an electron in the orbital from the nucleus and therefore the
larger (and less stable) the orbital. TOC Study Guide TOC Textbook Website MHHE Website 262 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS THE ANGULAR MOMENTUM QUANTUM NUMBER ( ) The angular momentum quantum number ( ) tells us the “shape” of the orbitals (see
Section 7.7). The values of depend on the value of the principal quantum number, n.
For a given value of n, has possible integral values from 0 to (n 1). If n 1, there
is only one possible value of ; that is,
n 1 1 1 0. If n 2, there are two
values of , given by 0 and 1. If n 3, there are three values of , given by 0, 1, and
2. The value of is generally designated by the letters s, p, d, . . . as follows:
0
Name of orbital Remember that the “2” in 2s refers
to the value of n and the “s” symbolizes the value of . 1 2 3 4 5 s p d f g h Thus if
0, we have an s orbital; if
1, we have a p orbital; and so on.
The unusual sequence of letters (s, p, and d ) has a historical origin. Physicists
who studied atomic emission spectra tried to correlate the observed spectral lines with
the particular energy states involved in the transitions. They noted that some of the
lines were sharp; some were rather spread out, or diffuse; and some were very strong
and hence referred to as principal lines. Subsequently, the initial letters of each adjective were assigned to those energy states. However, after the letter d and starting with
the letter f (for fundamental), the orbital designations follow alphabetical order.
A collection of orbitals with the same value of n is frequently called a shell. One
or more orbitals with the same n and values are referred to as subshell. For example, the shell with n 2 is composed of two subshells,
0 and 1 (the allowed values for n 2). These subshells are called the 2s and 2p subshells where 2 denotes the
value of n, and s and p denote the values of .
THE MAGNETIC QUANTUM NUMBER (m ) The magnetic quantum number (m ) describes the orientation of the orbital in space
(to be discussed in Section 7.7). Within a subshell, the value of m depends on the
value of the angular momentum quantum number, . For a certain value of , there are
(2
1) integral values of m as follows:
,( 1), . . . 0, . . . ( 1), If
0, then m
0. If
1, then there are [(2 1) 1], or three values of m ,
namely, 1, 0, and 1. If
2, there are [(2 2) 1], or five values of m , namely,
2, 1, 0, 1, and 2. The number of m values indicates the number of orbitals in a
subshell with a particular value.
To conclude our discussion of these three quantum numbers, let us consider a situation in which n 2 and
1. The values of n and indicate that we have a 2p
subshell, and in this subshell we have three 2p orbitals (because there are three values
of m , given by 1, 0, and 1).
THE ELECTRON SPIN QUANTUM NUMBER (ms) Experiments on the emission spectra of hydrogen and sodium atoms indicated that lines
in the emission spectra could be split by the application of an external magnetic field.
The only way physicists could explain these results was to assume that electrons act
like tiny magnets. If electrons are thought of as spinning on their own axes, as Earth Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 7.7 ATOMIC ORBITALS 263 Oven
Atom beam
N S S N Detecting screen
Magnet (a) Slit screen (b) FIGURE 7.15 The (a) counterclockwise and (b) clockwise spins
of an electron. The magnetic
fields generated by these two
spinning motions are analogous
to those from the two magnets.
The upward and downward arrows are used to denote the direction of spin. 7.7
That the wave function for an orbital theoretically has no outer limit
as one moves outward from the nucleus raises interesting philosophical
questions regarding the sizes of
atoms. Chemists have agreed on an
operational definition of atomic size,
as we will see in later chapters. FIGURE 7.16 Experimental arrangement for demonstrating the spinning motion of electrons. A
beam of atoms is directed through a magnetic field. For example, when a hydrogen atom with a
single electron passes through the field, it is deflected in one direction or the other, depending on
the direction of the spin. In a stream consisting of many atoms, there will be equal distributions of
the two kinds of spins, so that two spots of equal intensity are detected on the screen. does, their magnetic properties can be accounted for. According to electromagnetic theory, a spinning charge generates a magnetic field, and it is this motion that causes an
electron to behave like a magnet. Figure 7.15 shows the two possible spinning motions
of an electron, one clockwise and the other counterclockwise. To take the electron spin
into account, it is necessary to introduce a fourth quantum number, called the electron
spin quantum number (ms), which has a value of 1 or 1 .
2
2
Conclusive proof of electron spin was provided by Otto Stern† and Walther
Gerlach‡ in 1924. Figure 7.16 shows the basic experimental arrangement. A beam of
gaseous atoms generated in a hot furnace passes through a nonhomogeneous magnetic
field. The interaction between an electron and the magnetic field causes the atom to
be deflected from its straight-line path. Because the spinning motion is completely random, the electrons in half of the atoms will be spinning in one direction, and those
atoms will be deflected in one way; the electrons in the other half of the atoms will be
spinning in the opposite direction, and those atoms will be deflected in the other direction. Thus, two spots of equal intensity are observed on the detecting screen. ATOMIC ORBITALS Table 7.2 shows the relation between quantum numbers and atomic orbitals. We see
that when
0, (2
1) 1 and there is only one value of m , thus we have an s
orbital. When
1, (2
1) 3, so there are three values of m or three p orbitals,
labeled px, py, and pz. When
2, (2
1) 5 and there are five values of m , and
the corresponding five d orbitals are labeled with more elaborate subscripts. In the following sections we will consider the s, p, and d orbitals separately.
† Otto Stern (1888 – 1969). German physicist. He made important contributions to the study of magnetic properties of atoms
and the kinetic theory of gases. Stern was awarded the Nobel Prize in Physics in 1943.
‡ Walther Gerlach (1889 – 1979). German physicist. Gerlach’s main area of research was in quantum theory. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 264 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS TABLE 7.2 Relation Between Quantum Numbers and Atomic Orbitals n
An s subshell has one orbital, a p
subshell has three orbitals, and a d
subshell has five orbitals. 1
2
3 .
.
. m 0
0
1
0
1
2 NUMBER OF ORBITALS ATOMIC ORBITAL DESIGNATIONS 1
1
3
1
3
5 1s
2s
2px, 2py, 2pz
3s
3px, 3py, 3pz
3dxy, 3dyz, 3dxz,
3dx2 y2, 3dz2
.
.
. 0
0
1, 0, 1
0
1, 0, 1
2, 1, 0, 1, 2 .
.
. .
.
. .
.
. Electron density FIGURE 7.17 (a) Plot of electron density in the hydrogen 1s
orbital as a function of the distance from the nucleus. The electron density falls off rapidly as
the distance from the nucleus increases. (b) Boundary surface diagram of the hydrogen 1s orbital. Distance from nucleus
(a) (b) One of the important questions we ask when studying the properties of
atomic orbitals is, What are the shapes of the orbitals? Strictly speaking, an orbital does
not have a well-defined shape because the wave function characterizing the orbital extends from the nucleus to infinity. In that sense it is difficult to say what an orbital
looks like. On the other hand, it is certainly convenient to think of orbitals as having
specific shapes, particularly in discussing the formation of chemical bonds between
atoms, as we will do in Chapters 9 and 10.
Although in principle an electron can be found anywhere, we know that most of
the time it is quite close to the nucleus. Figure 7.17(a) shows a plot of the electron
density in a hydrogen atomic 1s orbital versus the distance from the nucleus. As you
can see, the electron density falls off rapidly as the distance from the nucleus increases.
Roughly speaking, there is about a 90 percent probability of finding the electron within
a sphere of radius 100 pm (1 pm 1 10 12 m) surrounding the nucleus. Thus, we
can represent the 1s orbital by drawing a boundary surface diagram that encloses
about 90 percent of the total electron density in an orbital, as shown in Figure 7.17(b).
A 1s orbital represented in this manner is merely a sphere.
Figure 7.18 shows boundary surface diagrams for the 1s, 2s, and 3s hydrogen
atomic orbitals. All s orbitals are spherical in shape but differ in size, which increases
as the principal quantum number increases. Although the details of electron density
variation within each boundary surface are lost, there is no serious disadvantage. For
us the most important features of atomic orbitals are their shapes and relative sizes,
which are adequately represented by boundary surface diagrams.
s Orbitals. 1s
2s
3s
FIGURE 7.18 Boundary surface diagrams of the hydrogen
1s, 2s, and 3s orbitals. Each
sphere contains about 90 percent
of the total electron density. All s
orbitals are spherical. Roughly
speaking, the size of an orbital is
proportional to n2, where n is the
principal quantum number. Back Forward It should be clear that the p orbitals start with the principal quantum num2. If n 1, then the angular momentum quantum number can assume only p Orbitals. ber n Main Menu TOC Study Guide TOC Textbook Website MHHE Website 7.7 z FIGURE 7.19 The boundary
surface diagrams of the three 2p
orbitals. These orbitals are identical in shape and energy, but
their orientations are different.
The p orbitals of higher principal
quantum numbers have a similar
shape. y x z x 2py 2pz z y x
3dx 2 – y 2 y x y z z x y
y 265 z z 2px
z ATOMIC ORBITALS x y 3dxy 3dz 2 FIGURE 7.20 Boundary surface diagrams of the five 3d orbitals. Although the 3d z 2 orbital
looks different, it is equivalent to
the other four orbitals in all other
respects. The d orbitals of higher
principal quantum numbers have
similar shapes. x y 3d xz x
3d yz the value of zero; therefore, there is only a 1s orbital. As we saw earlier, when
1,
the magnetic quantum number m can have values of 1, 0, 1. Starting with n 2 and
1, we therefore have three 2p orbitals: 2px, 2py, and 2pz (Figure 7.19). The letter
subscripts indicate the axes along which the orbitals are oriented. These three p orbitals are identical in size, shape, and energy; they differ from one another only in orientation. Note, however, that there is no simple relation between the values of m and
the x, y, and z directions. For our purpose, you need only remember that because there
are three possible values of m , there are three p orbitals with different orientations.
The boundary surface diagrams of p orbitals in Figure 7.19 show that each p orbital can be thought of as two lobes on opposite sides of the nucleus. Like s orbitals,
p orbitals increase in size from 2p to 3p to 4p orbital and so on.
When
2, there are five values of
m , which correspond to five d orbitals. The lowest value of n for a d orbital is 3.
Because can never be greater than n 1, when n 3 and
2, we have five 3d
orbitals (3dxy, 3dyz, 3dxz, 3dx2 y2, and 3dz2), shown in Figure 7.20. As in the case of the
p orbitals, the different orientations of the d orbitals correspond to the different values
of m , but again there is no direct correspondence between a given orientation and a
particular m value. All the 3d orbitals in an atom are identical in energy. The d orbitals for which n is greater than 3 (4d, 5d, . . .) have similar shapes.
Orbitals having higher energy than d orbitals are labeled f, g, . . . and so on. The
f orbitals are important in accounting for the behavior of elements with atomic numbers greater than 57, but their shapes are difficult to represent. In general chemistry we
are not concerned with orbitals having values greater than 3 (the g orbitals and beyond).
The following examples illustrate the labeling of orbitals with quantum numbers
and the calculation of total number of orbitals associated with a given principal quantum number.
d Orbitals and Other Higher-Energy Orbitals. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 266 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS EXAMPLE 7.6 List the values of n, , and m for orbitals in the 4d subshell.
As we saw earlier, the number given in the designation of the subshell is
the principal quantum number, so in this case n 4. Since we are dealing with d
orbitals,
2. The values of m can vary from
to . Therefore, m can be 2,
1, 0, 1, 2 (which correspond to the five d orbitals). Answer Similar problem: 7.55. PRACTICE EXERCISE Give the values of the quantum numbers associated with the orbitals in the 3p subshell.
EXAMPLE 7.7 What is the total number of orbitals associated with the principal quantum number
n 3?
For n 3, the possible values of are 0, 1, and 2. Thus there is one 3s
orbital (n 3,
0, and m
0); there are three 3p orbitals (n 3,
1, and
m
1, 0, 1); there are five 3d orbitals (n 3,
2, and m
2, 1, 0, 1,
2). Therefore the total number of orbitals is 1 3 5 9.
Answer Similar problem: 7.60. PRACTICE EXERCISE What is the total number of orbitals associated with the principal quantum number
n 4? THE ENERGIES OF ORBITALS Now that we have some understanding of the shapes and sizes of atomic orbitals, we
are ready to inquire into their relative energies and look at how energy levels affect
the actual arrangement of electrons in atoms. FIGURE 7.21 Orbital energy
levels in the hydrogen atom.
Each short horizontal line represents one orbital. Orbitals with
the same principal quantum number (n) all have the same energy. 4p 4d 3s 3p 3d 2s 4f 2p Energy 4s 1s Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 7.8 FIGURE 7.22 Orbital energy
levels in a many-electron atom.
Note that the energy level depends on both n and values. 5s Energy 2s 267 4d
4p
3d 4s
3s ELECTRON CONFIGURATION 3p 2p 1s According to Equation (7.4), the energy of an electron in a hydrogen atom is determined solely by its principal quantum number. Thus the energies of hydrogen orbitals increase as follows (Figure 7.21):
1s
1s
2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p FIGURE 7.23 The order in
which atomic subshells are filled
in a many-electron atom. Start
with the 1s orbital and move
downward, following the direction of the arrows. Thus the order
goes as follows: 1s 2s 2p
3s 3p 4s 3d . . . . 7.8 2s 2p 3s 3p 3d 4s 4p 4f . Although the electron density distributions are different in the 2s and 2p orbitals, hydrogen’s electron has the same energy whether it is in the 2s orbital or a 2p orbital.
The 1s orbital in a hydrogen atom corresponds to the most stable condition, the ground
state. An electron residing in this orbital is most strongly held by the nucleus because
it is closest to the nucleus. An electron in the 2s, 2p, or higher orbitals in a hydrogen
atom is in an excited state.
The energy picture is more complex for many-electron atoms than for hydrogen.
The energy of an electron in such an atom depends on its angular momentum quantum number as well as on its principal quantum number (Figure 7.22). For many-electron atoms, the 3d energy level is very close to the 4s energy level. The total energy
of an atom, however, depends not only on the sum of the orbital energies but also on
the energy of repulsion between the electrons in these orbitals (each orbital can accommodate up to two electrons, as we will see in Section 7.8). It turns out that the total energy of an atom is lower when the 4s subshell is filled before a 3d subshell. Figure
7.23 depicts the order in which atomic orbitals are filled in a many-electron atom. We
will consider specific examples in the next section. ELECTRON CONFIGURATION The four quantum numbers n, , m , and ms enable us to label completely an electron
in any orbital in any atom. In a sense, we can regard the set of four quantum numbers
as the “address” of an electron in an atom, somewhat in the same way that a street address, city, state, and postal ZIP code specify the address of an individual. For example, the four quantum numbers for a 2s orbital electron are n 2,
0, m
0, and
1
ms
or 1 . It is inconvenient to write out all the individual quantum numbers,
2
2
and so we use the simplified notation (n, , m , ms). For the example above, the quan- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 268 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS tum numbers are either (2, 0, 0, 1 ) or (2, 0, 0, 1 ). The value of ms has no effect on
2
2
the energy, size, shape, or orientation of an orbital, but it determines how electrons are
arranged in an orbital.
Example 7.8 shows how quantum numbers of an electron in an orbital are assigned. EXAMPLE 7.8 Write the four quantum numbers for an electron in a 3p orbital.
Answer To start with, we know that the principal quantum number n is 3 and the
angular momentum quantum number must be 1 (because we are dealing with a p
orbital). For
1, there are three values of m given by 1, 0, 1. Since the electron spin quantum number ms can be either 1 or 1 , we conclude that there are
2
2
six possible ways to designate the electron: (3, 1, 1, 1 )
2
(3, 1, 0, 1 )
2
(3, 1, 1, 1 )
2 Similar problems: 7.55, 7.56. (3, 1, 1, 1 )
2
(3, 1, 0, 1 )
2
(3, 1, 1, 1 )
2 PRACTICE EXERCISE Write the four quantum numbers for an electron in a 5p orbital.
The hydrogen atom is a particularly simple system because it contains only one
electron. The electron may reside in the 1s orbital (the ground state), or it may be found
in some higher-energy orbital (an excited state). For many-electron atoms, however,
we must know the electron configuration of the atom, that is, how the electrons are
distributed among the various atomic orbitals, in order to understand electronic behavior. We will use the first ten elements (hydrogen to neon) to illustrate the rules for
writing electron configurations for atoms in the ground state. (Section 7.9 will describe
how these rules can be applied to the remainder of the elements in the periodic table.)
For this discussion, recall that the number of electrons in an atom is equal to its atomic
number Z.
Figure 7.21 indicates that the electron in a ground-state hydrogen atom must be
in the 1s orbital, so its electron configuration is 1s1:
denotes the number of electrons
in the orbital or subshell
8 88
m8 denotes the principal
quantum number n 8
m8
888 1s1m denotes the angular momentum
quantum number The electron configuration can also be represented by an orbital diagram that
shows the spin of the electron (see Figure 7.15):
H h
1s1 Remember that the direction of electron spin has no effect on the energy
of the electron. Back Forward The upward arrow denotes one of the two possible spinning motions of the electron.
(Alternatively, we could have represented the electron with a downward arrow.) The
box represents an atomic orbital. Main Menu TOC Study Guide TOC Textbook Website MHHE Website 7.8 ELECTRON CONFIGURATION 269 THE PAULI EXCLUSION PRINCIPLE For many-electron atoms we use the Pauli† exclusion principle to determine electron
configurations. This principle states that no two electrons in an atom can have the same
four quantum numbers. If two electrons in an atom should have the same n, , and m
values (that is, these two electrons are in the same atomic orbital), then they must have
different values of ms. In other words, only two electrons may occupy the same atomic
orbital, and these electrons must have opposite spins. Consider the helium atom, which
has two electrons. The three possible ways of placing two electrons in the 1s orbital
are as follows:
hh gg hg 1s2 1s2 1s2 (a) He (b) (c) Diagrams (a) and (b) are ruled out by the Pauli exclusion principle. In (a), both electrons have the same upward spin and would have the quantum numbers (1, 0, 0, 1 );
2
in (b), both electrons have downward spins and would have the quantum numbers (1,
0, 0, 1 ). Only the configuration in (c) is physically acceptable, because one electron
2
has the quantum numbers (1, 0, 0, 1 ) and the other has (1, 0, 0, 1 ). Thus the he2
2
lium atom has the following configuration:
Electrons that have opposite spins
are said to be paired. In helium,
1
1
ms
2 for one electron; ms
2
for the other. He hg
1s2 Note that 1s2 is read “one s two,” not “one s squared.”
DIAMAGNETISM AND PARAMAGNETISM The Pauli exclusion principle is one of the fundamental principles of quantum mechanics. It can be tested by a simple observation. If the two electrons in the 1s orbital
of a helium atom had the same, or parallel, spins (hh or gg), their net magnetic fields
would reinforce each other. Such an arrangement would make the helium atom paramagnetic [Figure 7.24(a)]. Paramagnetic substances are those that are attracted by a
† Wolfgang Pauli (1900 – 1958). Austrian physicist. One of the founders of quantum mechanics, Pauli was awarded the Nobel
Prize in Physics in 1945. FIGURE 7.24 The (a) parallel
and (b) antiparallel spins of two
electrons. In (a) the two magnetic
fields reinforce each other, and
the atom is paramagnetic. In
(b) the two magnetic fields cancel
each other, and the atom is diamagnetic. N N N S S S S N (a) Back Forward Main Menu TOC (b) Study Guide TOC Textbook Website MHHE Website 270 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS magnet. On the other hand, if the electron spins are paired, or antiparallel to each other
(hg or gh), the magnetic effects cancel out and the atom is diamagnetic [Figure 7.24(b)].
Diamagnetic substances are slightly repelled by a magnet.
Measurements of magnetic properties provide the most direct evidence for specific electron configurations of elements. Advances in instrument design during the last
twenty years or so enable us to determine not only whether an atom is paramagnetic
but also how many unpaired electrons are present. By experiment we find that the helium atom is diamagnetic in the ground state, in accord with the Pauli exclusion principle. A useful general rule to keep in mind is that any atom with an odd number of
electrons must be paramagnetic, because we need an even number of electrons for complete pairing. On the other hand, atoms containing an even number of electrons may
be either diamagnetic or paramagnetic. We will see the reason for this behavior shortly.
As another example, consider the lithium atom (Z 3) which has three electrons.
The third electron cannot go into the 1s orbital because it would inevitably have the
same four quantum numbers as one of the first two electrons. Therefore, this electron
“enters” the next (energetically) higher orbital, which is the 2s orbital (see Figure 7.22).
The electron configuration of lithium is 1s22s1, and its orbital diagram is
hg h 1s2 Li 2s1 The lithium atom contains one unpaired electron and is therefore paramagnetic. THE SHIELDING EFFECT IN MANY-ELECTRON ATOMS Experimentally we find that the 2s orbital lies at a lower energy level than the 2p orbital in a many-electron atom. Why? In comparing the electron configurations of 1s22s1
and 1s22p1, we note that, in both cases, the 1s orbital is filled with two electrons.
Because the 2s and 2p orbitals are larger than the 1s orbital, an electron in either of
these orbitals will spend more time away from the nucleus (on the average) than an
electron in the 1s orbital. Thus, we can speak of a 2s or 2p electron being partly
“shielded” from the attractive force of the nucleus by the 1s electrons. The important
consequence of the shielding effect is that it reduces the electrostatic attraction between
protons in the nucleus and the electron in the 2s or 2p orbital.
The manner in which the electron density varies as we move from the nucleus
outward depends on the type of orbital. The density near the nucleus is greater for the
2s electron than for the 2p electron. In other words, a 2s electron spends more time
near the nucleus than a 2p electron does (on the average). For this reason, the 2s orbital is said to be more “penetrating” than the 2p orbital, and it is less shielded by the
1s electrons. In fact, for the same principal quantum number n, the penetrating power
decreases as the angular momentum quantum number increases, or
s p d f Since the stability of an electron is determined by the strength of its attraction to the
nucleus, it follows that a 2s electron will be lower in energy than a 2p electron. To put
it another way, less energy is required to remove a 2p electron than a 2s electron because a 2p electron is not held quite as strongly by the nucleus. The hydrogen atom
has only one electron and, therefore, is without such a shielding effect. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 7.8 ELECTRON CONFIGURATION 271 Continuing our discussion of atoms of the first ten elements, we go next to beryllium (Z 4). The ground-state electron configuration of beryllium is 1s22s2, or
hg hg 1s2 Be 2s2 Beryllium atoms are diamagnetic, as we would expect.
The electron configuration of boron (Z 5) is 1s22s22p1, or
hg hg 1s2 B h 2s2 2p1 Note that the unpaired electron can be in the 2px, 2py, or 2pz orbital. The choice is
completely arbitrary because the three p orbitals are equivalent in energy. As the diagram shows, boron atoms are paramagnetic. HUND’S RULE The electron configuration of carbon (Z 6) is 1s22s22p2. The following are different
ways of distributing two electrons among three p orbitals:
hg hg hh 2px 2py 2pz 2px 2py 2pz 2px 2py 2pz (a) (b) (c) None of the three arrangements violates the Pauli exclusion principle, so we must determine which one will give the greatest stability. The answer is provided by Hund’s
rule,† which states that the most stable arrangement of electrons in subshells is the one
with the greatest number of parallel spins. The arrangement shown in (c) satisfies this
condition. In both (a) and (b) the two spins cancel each other. Thus the orbital diagram
for carbon is
hg hg 1s2 C 2s2 hh
2px 2py 2pz Qualitatively, we can understand why (c) is preferred to (a). In (a), the two electrons are in the same 2px orbital, and their proximity results in a greater mutual repulsion than when they occupy two separate orbitals, say 2px and 2py. The choice of
(c) over (b) is more subtle but can be justified on theoretical grounds. The fact that
carbon atoms are paramagnetic, each containing two unpaired electrons, is in accord
with Hund’s rule.
The electron configuration of nitrogen (Z 7) is 1s22s22p3:
hg hg hhh 1s2 N 2s2 2p3 †
Frederick Hund (1896 – 0000) German physicist. Hund’s work was mainly in quantum mechanics. He also helped to develop the molecular orbital theory of chemical bonding. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 272 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS Again, Hund’s rule dictates that all three 2p electrons have spins parallel to one another; the nitrogen atom is therefore paramagnetic, containing three unpaired electrons.
The electron configuration of oxygen (Z 8) is 1s22s22p4. An oxygen atom is
paramagnetic because it has two unpaired electrons:
hg hg hg h h 1s2 O 2s2 2p4 The electron configuration of fluorine (Z
arranged as follows: 9) is 1s22s22p5. The nine electrons are hg hg hg hg h 1s2 F 2s2 2p5 The fluorine atom is thus paramagnetic, having one unpaired electron.
In neon (Z 10), the 2p orbitals are completely filled. The electron configuration
of neon is 1s22s22p6, and all the electrons are paired, as follows:
hg hg hg hg hg 1s2 Ne 2s2 2p6 The neon atom thus should be diamagnetic, and experimental observation bears out
this prediction.
GENERAL RULES FOR ASSIGNING ELECTRONS TO ATOMIC ORBITALS Based on the preceding examples we can formulate some general rules for determining the maximum number of electrons that can be assigned to the various subshells
and orbitals for a given value of n:
Each shell or principal level of quantum number n contains n subshells. For example, if n 2, then there are two subshells (two values of ) of angular momentum
quantum numbers 0 and 1.
• Each subshell of quantum number contains 2
1 orbitals. For example, if
1,
then there are three p orbitals.
• No more than two electrons can be placed in each orbital. Therefore, the maximum
number of electrons is simply twice the number of orbitals that are employed.
• A quick way to determine the maximum number of electrons that an atom can have
is to use the formula 2n2.
• The following examples illustrate the procedure for calculating the number of electrons in orbitals and labeling electrons with the four quantum numbers. EXAMPLE 7.9 What is the maximum number of electrons that can be present in the principal level
for which n 3?
When n
is given by Answer of Back Forward Main Menu TOC 3, then 0, 1, and 2. The number of orbitals for each value Study Guide TOC Textbook Website MHHE Website 7.8 Value of 273 Number of Orbitals
(2
1) 0
1
2 Similar problems: 7.61, 7.62, 7.63. ELECTRON CONFIGURATION 1
3
5 The total number of orbitals is nine. Since each orbital can accommodate two electrons, the maximum number of electrons that can reside in the orbitals is 2 9, or
18. Or, using the formula 2n2, we have n 3, so 2(32) 18.
PRACTICE EXERCISE Calculate the total number of electrons that can be present in the principal level for
which n 4. EXAMPLE 7.10 An oxygen atom has a total of eight electrons. Write the four quantum numbers for
each of the eight electrons in the ground state.
We start with n 1, so
0, a subshell corresponding to the 1s orbital.
This orbital can accommodate a total of two electrons. Next, n 2, and may be
either 0 or 1. The
0 subshell contains one 2s orbital, which can accommodate
two electrons. The remaining four electrons are placed in the
1 subshell, which
contains three 2p orbitals. The orbital diagram is
Answer hg hg hg h h 1s2 2s2 2p4 O The results are summarized in the following table:
ELECTRON 1 n 1 m 0 ms ORBITAL 0 1
2 2 1 0 0 1
2 3 2 0 0 1
2 4 0 0 5 2 1 1 1
2 6 2 1 0 1
2 7 2 1 1 1
2 8 Similar problem: 7.85. 2 1
2 2 1 1 1
2 1s 2s 2px, 2py, 2pz 1 is comOf course, the placement of the eighth electron in the orbital labeled m
pletely arbitrary. It would be equally correct to assign it to m
0 or m
1.
PRACTICE EXERCISE Write a complete set of quantum numbers for each of the electrons in boron (B). Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 274 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS 7.9 THE BUILDING-UP PRINCIPLE Here we will extend the rules used in writing electron configurations for the first ten
elements to the rest of the elements. This process is based on the Aufbau principle.
(The German word “Aufbau” means “building up.”) The Aufbau principle dictates that
as protons are added one by one to the nucleus to build up the elements, electrons are
similarly added to the atomic orbitals. Through this process we gain a detailed knowledge of the ground-state electron configurations of the elements. As we will see later,
knowledge of electron configurations helps us to understand and predict the properties
of the elements; it also explains why the periodic table works so well.
Table 7.3 gives the ground-state electron configurations of elements from H
(Z 1) through Mt (Z 109). The electron configurations of all elements except hydrogen and helium are represented by a noble gas core, which shows in brackets the
noble gas element that most nearly precedes the element being considered, followed
by the symbol for the highest filled subshells in the outermost shells. Notice that the
electron configurations of the highest filled subshells in the outermost shells for the elements sodium (Z 11) through argon (Z 18) follow a pattern similar to those of
lithium (Z 3) through neon (Z 10).
As mentioned in Section 7.7, the 4s subshell is filled before the 3d subshell in a
many-electron atom (see Figure 7.23). Thus the electron configuration of potassium
(Z 19) is 1s22s22p63s23p64s1. Since 1s22s22p63s23p6 is the electron configuration of
argon, we can simplify the electron configuration of potassium by writing [Ar]4s1,
where [Ar] denotes the “argon core.” Similarly, we can write the electron configuration of calcium (Z 20) as [Ar]4s2. The placement of the outermost electron in the 4s
orbital (rather than in the 3d orbital) of potassium is strongly supported by experimental evidence. The following comparison also suggests that this is the correct configuration. The chemistry of potassium is very similar to that of lithium and sodium,
the first two alkali metals. The outermost electron of both lithium and sodium is in an
s orbital (there is no ambiguity in assigning their electron configurations); therefore,
we expect the last electron in potassium to occupy the 4s rather than the 3d orbital.
The elements from scandium (Z 21) to copper (Z 29) are transition metals.
Transition metals either have incompletely filled d subshells or readily give rise to
cations that have incompletely filled d subshells. Consider the first transition metal series, from scandium through copper. In this series additional electrons are placed in the
3d orbitals, according to Hund’s rule. However, there are two irregularities. The electron configuration of chromium (Z 24) is [Ar]4s13d 5 and not [Ar]4s23d 4, as we might
expect. A similar break in the pattern is observed for copper, whose electron configuration is [Ar]4s13d10 rather than [Ar]4s23d 9. The reason for these irregularities is that
a slightly greater stability is associated with the half-filled (3d 5) and completely filled
(3d10) subshells. Electrons in the same subshells (in this case, the d orbitals) have equal
energy but different spatial distributions. Consequently, their shielding of one another
is relatively small, and the electrons are more strongly attracted by the nucleus when
they have the 3d 5 configuration. According to Hund’s rule, the orbital diagram for
Cr is
[Ar] h hhhhh 4s1 Cr 3d 5 Thus, Cr has a total of six unpaired electrons. The orbital diagram for copper is Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 7.9 TABLE 7.3 THE BUILDING-UP PRINCIPLE 275 The Ground-State Electron Configurations of the Elements* ATOMIC ELECTRON ATOMIC ELECTRON ATOMIC ELECTRON NUMBER SYMBOL CONFIGURATION NUMBER SYMBOL CONFIGURATION NUMBER SYMBOL CONFIGURATION 1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36 H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
Si
P
S
Cl
Ar
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr 1s1
1s2
[He]2s1
[He]2s2
[He]2s22p1
[He]2s22p2
[He]2s22p3
[He]2s22p4
[He]2s22p5
[He]2s22p6
[Ne]3s1
[Ne]3s2
[Ne]3s23p1
[Ne]3s23p2
[Ne]3s23p3
[Ne]3s23p4
[Ne]3s23p5
[Ne]3s23p6
[Ar]4s1
[Ar]4s2
[Ar]4s23d1
[Ar]4s23d 2
[Ar]4s23d 3
[Ar]4s13d 5
[Ar]4s23d 5
[Ar]4s23d 6
[Ar]4s23d 7
[Ar]4s23d 8
[Ar]4s13d10
[Ar]4s23d10
[Ar]4s23d104p1
[Ar]4s23d104p2
[Ar]4s23d104p3
[Ar]4s23d104p4
[Ar]4s23d104p5
[Ar]4s23d104p6 37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72 [Kr]5s1
[Kr]5s2
[Kr]5s24d1
[Kr]5s24d2
[Kr]5s14d 4
[Kr]5s14d 5
[Kr]5s24d 5
[Kr]5s14d 7
[Kr]5s14d 8
[Kr]4d10
[Kr]5s14d10
[Kr]5s24d10
[Kr]5s24d105p1
[Kr]5s24d105p2
[Kr]5s24d105p3
[Kr]5s24d105p4
[Kr]5s24d105p5
[Kr]5s24d105p6
[Xe]6s1
[Xe]6s2
[Xe]6s25d1
[Xe]6s24f 15d1
[Xe]6s24f 3
[Xe]6s24f 4
[Xe]6s24f 5
[Xe]6s24f 6
[Xe]6s24f 7
[Xe]6s24f 75d 1
[Xe]6s24f 9
[Xe]6s24f 10
[Xe]6s24f 11
[Xe]6s24f 12
[Xe]6s24f 13
[Xe]6s24f 14
[Xe]6s24f 145d 1
[Xe]6s24f 145d 2 Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
Cs
Ba
La
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
Hf 73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109 Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
Fr
Ra
Ac
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
Rf
Ha
Sg
Ns
Hs
Mt [Xe]6s24f 145d 3
[Xe]6s24f 145d 4
[Xe]6s24f 145d 5
[Xe]6s24f 145d 6
[Xe]6s24f 145d 7
[Xe]6s14f 145d 9
[Xe]6s14f 145d10
[Xe]6s24f 145d10
[Xe]6s24f 145d106p1
[Xe]6s24f 145d106p2
[Xe]6s24f 145d106p3
[Xe]6s24f 145d106p4
[Xe]6s24f 145d106p5
[Xe]6s24f 145d106p6
[Rn]7s1
[Rn]7s2
[Rn]7s26d1
[Rn]7s26d 2
[Rn]7s25f 26d1
[Rn]7s25f 36d1
[Rn]7s25f 46d1
[Rn]7s25f 6
[Rn]7s25f 7
[Rn]7s25f 76d1
[Rn]7s25f 9
[Rn]7s25f 10
[Rn]7s25f 11
[Rn]7s25f 12
[Rn]7s25f 13
[Rn]7s25f 14
[Rn]7s25f 146d1
[Rn]7s25f 146d 2
[Rn]7s25f 146d 3
[Rn]7s25f 146d 4
[Rn]7s25f 146d 5
[Rn]7s25f 146d 6
[Rn]7s25f 146d7 *The symbol [He] is called the helium core and represents 1s2. [Ne] is called the neon core and represents 1s22s22p6. [Ar] is called the argon core and represents [Ne]3s23p6.
[Kr] is called the krypton core and represents [Ar]4s23d104p6. [Xe] is called the xenon core and represents [Kr]5s24d105p6. [Rn] is called the radon core and represents
[Xe]6s24f 145d106p6. [Ar] h hg hg hg hg hg 4s1 Cu 3d10 Again, extra stability is gained in this case by having the 3d orbitals completely filled. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 276 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS FIGURE 7.25 Classification of
groups of elements in the periodic table according to the type
of subshell being filled with electrons. 1s 1s
2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 5d 6p 7s 6d 4f
5f For elements Zn (Z 30) through Kr (Z 36), the 4s and 4p subshells fill in a
straightforward manner. With rubidium (Z 37), electrons begin to enter the n 5
energy level.
The electron configurations in the second transition metal series [yttrium (Z 39)
to silver (Z 47)] are also irregular, but we will not be concerned with the details here.
The sixth period of the periodic table begins with cesium (Z 55) and barium
(Z 56), whose electron configurations are [Xe]6s1 and [Xe]6s2, respectively. Next
we come to lanthanum (Z 57). From Figure 7.23 we would expect that after filling
the 6s orbital we would place the additional electrons in 4f orbitals. In reality, the energies of the 5d and 4f orbitals are very close; in fact, for lanthanum 4f is slightly higher
in energy than 5d. Thus lanthanum’s electron configuration is [Xe]6s25d1 and not
[Xe]6s24f 1.
Following lanthanum are the 14 elements known as the lanthanides, or rare earth
series [cerium (Z 58) to lutetium (Z 71)]. The rare earth metals have incompletely
filled 4f subshells or readily give rise to cations that have incompletely filled 4f subshells. In this series, the added electrons are placed in 4f orbitals. After the 4f subshells
are completely filled, the next electron enters the 5d subshell of lutetium. Note that the
electron configuration of gadolinium (Z 64) is [Xe]6s24f 75d1 rather than [Xe]6s24f 8.
Like chromium, gadolinium gains extra stability by having half-filled subshells (4f 7).
The third transition metal series, including lanthanum and hafnium (Z 72) and
extending through gold (Z 79), is characterized by the filling of the 5d orbitals. The
6s and 6p subshells are filled next, which takes us to radon (Z 86).
The last row of elements is the actinide series, which starts at thorium (Z 90).
Most of these elements are not found in nature but have been synthesized.
With few exceptions, you should be able to write the electron configuration of
any element, using Figure 7.23 as a guide. Elements that require particular care are the
transition metals, the lanthanides, and the actinides. As we noted earlier, at larger values of the principal quantum number n, the order of subshell filling may reverse from
one element to the next. Figure 7.25 groups the elements according to the type of subshell in which the outermost electrons are placed. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website SUMMARY OF KEY EQUATIONS 277 EXAMPLE 7.11 Write the electron configurations for sulfur and for palladium, which is diamagnetic.
Sulfur (Z Answer 16) Sulfur has 16 electrons.
It takes 10 electrons to complete the first and second periods (1s22s22p6). This
leaves us 6 electrons to fill the 3s orbital and partially fill the 3p orbitals. Thus
the electron configuration of S is 1.
2. 1s22s22p63s23p4 or [Ne]3s23p4.
Palladium (Z 46) Palladium has 46 electrons.
We need 36 electrons to complete the fourth period, leaving us with 10 more
electrons to distribute among the 5s and 4d orbitals. The three choices are
(a) 4d10, (b) 4d 95s1, and (c) 4d 85s2. Since atomic palladium is diamagnetic, its
electron configuration must be 1.
2. 1s22s22p63s23p64s23d104p64d10 or simply [Kr]4d10. The configurations in (b) and (c) both give paramagnetic Pd
atoms. Similar problems: 7.91, 7.92. PRACTICE EXERCISE Write the ground-state electron configuration for phosphorus (P). SUMMARY OF
KEY EQUATIONS •u (7.1) •E (7.2) h • En • • • Back Forward Main Menu Relating speed of a wave to its wavelength and
frequency.
Relating energy of a quantum (and of a photon) to the
frequency. 1
n2 RH (7.4) h RH h
m 1
n2
i (7.7) E xp h
4 TOC 1
n2
f Energy of an electron in the nth state in a hydrogen
atom. (7.5) Energy of a photon emitted as the electron undergoes a
transition from the ni level to the nf level.
Relating wavelength of a particle to its mass m and
velocity . (7.8) Calculating the uncertainty in the position or in the
momentum of a particle. Study Guide TOC Textbook Website MHHE Website 278 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS SUMMARY OF FACTS
AND CONCEPTS 1. The quantum theory developed by Planck successfully explains the emission of radiation
by heated solids. The quantum theory states that radiant energy is emitted by atoms and
molecules in small discrete amounts (quanta), rather than over a continuous range. This
behavior is governed by the relationship E h , where E is the energy of the radiation, h
is Planck’s constant, and is the frequency of the radiation. Energy is always emitted in
whole-number multiples of h (1 h , 2 h , 3 h , . . . ).
2. Using quantum theory, Einstein solved another mystery of physics — the photoelectric effect. Einstein proposed that light can behave like a stream of particles (photons).
3. The line spectrum of hydrogen, yet another mystery to nineteenth-century physicists, was
also explained by applying the quantum theory. Bohr developed a model of the hydrogen
atom in which the energy of its single electron is quantized — limited to certain energy
values determined by an integer, the principal quantum number.
4. An electron in its most stable energy state is said to be in the ground state, and an electron at an energy level higher than its most stable state is said to be in an excited state. In
the Bohr model, an electron emits a photon when it drops from a higher-energy state (an
excited state) to a lower-energy state (the ground state or another, less excited state). The
release of specific amounts of energy in the form of photons accounts for the lines in the
hydrogen emission spectrum.
5. De Broglie extended Einstein’s wave-particle description of light to all matter in motion.
The wavelength of a moving particle of mass m and velocity v is given by the de Broglie
equation
h/mv.
6. The Schrödinger equation describes the motions and energies of submicroscopic particles.
This equation launched quantum mechanics and a new era in physics.
7. The Schrödinger equation tells us the possible energy states of the electron in a hydrogen
atom and the probability of its location in a particular region surrounding the nucleus.
These results can be applied with reasonable accuracy to many-electron atoms.
8. An atomic orbital is a function ( ) that defines the distribution of electron density ( 2) in
space. Orbitals are represented by electron density diagrams or boundary surface diagrams.
9. Four quantum numbers characterize each electron in an atom: the principal quantum number n identifies the main energy level, or shell, of the orbital; the angular momentum
quantum number indicates the shape of the orbital; the magnetic quantum number m
specifies the orientation of the orbital in space; and the electron spin quantum number ms
indicates the direction of the electron’s spin on its own axis.
10. The single s orbital for each energy level is spherical and centered on the nucleus. The
three p orbitals present at n 2 and higher each has two lobes, and the pairs of lobes are
arranged at right angles to one another. Starting with n 3, there are five d orbitals, with
more complex shapes and orientations.
11. The energy of the electron in a hydrogen atom is determined solely by its principal quantum number. In many-electron atoms, the principal quantum number and the angular momentum quantum number together determine the energy of an electron.
12. No two electrons in the same atom can have the same four quantum numbers (the Pauli
exclusion principle).
13. The most stable arrangement of electrons in a subshell is the one that has the greatest
number of parallel spins (Hund’s rule). Atoms with one or more unpaired electron spins
are paramagnetic. Atoms in which all electrons are paired are diamagnetic.
14. The Aufbau principle provides the guideline for building up the elements. The periodic
table classifies the elements according to their atomic numbers and thus also by the electronic configurations of their atoms. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 279 KEY WORDS
Actinide series, p. 276
Amplitude, p. 244
Atomic orbital, p. 261
Aufbau principle, p. 274
Boundary surface diagram,
p. 264
Diamagnetic, p. 270
Electromagnetic radiation,
p. 246 Electromagnetic wave, p. 246
Electron configuration, p. 268
Electron density, p. 261
Emission spectrum, p. 250
Excited level (or state), p. 252
Frequency ( ), p. 244
Ground level (or state), p. 252
Ground state, p. 252
Heisenberg uncertainty principle, p. 260
Hund’s rule, p. 271
Lanthanide (rare earth) series,
p. 276
Line spectrum, p. 250
Many-electron atom, p. 261
Noble gas core, p. 274
Node, p. 255
Paramagnetic, p. 269 Pauli exclusion principle,
p. 269
Photon, p. 248
Quantum, p. 248
Quantum numbers, p. 261
Rare earth series, p. 276
Transition metals, p. 274
Wave, p. 244
Wavelength ( ), p. 244 QUESTIONS AND PROBLEMS
QUANTUM THEORY AND
ELECTROMAGNETIC RADIATION
Review Questions 7.1 What is a wave? Explain the following terms associated with waves: wavelength, frequency, amplitude.
7.2 What are the units for wavelength and frequency of
electromagnetic waves? What is the speed of light,
in meters per second and miles per hour?
7.3 List the types of electromagnetic radiation, starting
with the radiation having the longest wavelength and
ending with the radiation having the shortest wavelength.
7.4 Give the high and low wavelength values that define
the visible region of the electromagnetic spectrum.
7.5 Briefly explain Planck’s quantum theory and explain
what a quantum is. What are the units for Planck’s
constant?
7.6 Give two everyday examples that illustrate the concept of quantization. certain emission process in the cesium atom.
Calculate the wavelength of this radiation (to three
significant figures). In which region of the electromagnetic spectrums is this wavelength found?
7.12 The SI unit of length is the meter, which is defined
as the length equal to 1,650,763.73 wavelengths of
the light emitted by a particular energy transition in
krypton atoms. Calculate the frequency of the light
to three significant figures.
THE PHOTOELECTRIC EFFECT
Review Questions 7.13 Explain what is meant by the photoelectric effect.
7.14 What are photons? What role did Einstein’s explanation of the photoelectric effect play in the development of the particle-wave interpretation of the nature
of electromagnetic radiation?
Problems Problems 7.7 Convert 8.6 1013 Hz to nanometers, and 566 nm to
hertz.
7.8 (a) What is the frequency of light having a wavelength of 456 nm? (b) What is the wavelength (in
nanometers) of radiation having a frequency of
2.45 109 Hz? (This is the type of radiation used in
microwave ovens.)
7.9 The average distance between Mars and Earth is
about 1.3 108 miles. How long would it take TV
pictures transmitted from the Viking space vehicle on
Mars’ surface to reach Earth? (1 mile 1.61 km.)
7.10 How many minutes would it take a radio wave to
travel from the planet Venus to Earth? (Average distance from Venus to Earth 28 million miles.)
7.11 The SI unit of time is the second, which is defined as
9,192,631,770 cycles of radiation associated with a Back Forward Main Menu TOC 7.15 A photon has a wavelength of 624 nm. Calculate the
energy of the photon in joules.
7.16 The blue color of the sky results from the scattering
of sunlight by air molecules. The blue light has a frequency of about 7.5 1014 Hz. (a) Calculate the
wavelength, in nm, associated with this radiation, and
(b) calculate the energy, in joules, of a single photon
associated with this frequency.
7.17 A photon has a frequency of 6.0 104 Hz. (a) Convert
this frequency into wavelength (nm). Does this frequency fall in the visible region? (b) Calculate the energy (in joules) of this photon. (c) Calculate the energy (in joules) of 1 mole of photons all with this
frequency.
7.18 What is the wavelength, in nm, of radiation that has
an energy content of 1.0 103 kJ/mol? In which region of the electromagnetic spectrum is this radiation
found? Study Guide TOC Textbook Website MHHE Website 280 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS 7.19 When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in
joules) associated with the photons if the wavelength
of the X rays is 0.154 nm.
7.20 A particular form of electromagnetic radiation has a
frequency of 8.11 1014 Hz. (a) What is its wavelength in nanometers? In meters? (b) To what region
of the electromagnetic spectrum would you assign it?
(c) What is the energy (in joules) of one quantum of
this radiation?
BOHR’S THEORY OF THE HYDROGEN ATOM
Review Questions 7.21 What are emission spectra? How do line spectra differ from continuous spectra?
7.22 What is an energy level? Explain the difference between ground state and excited state.
7.23 Briefly describe Bohr ’s theory of the hydrogen atom
and how it explains the appearance of an emission
spectrum. How does Bohr ’s theory differ from concepts of classical physics?
7.24 Explain the meaning of the negative sign in Equation
(7.4).
Problems 7.25 Explain why elements produce their own characteristic colors when they emit photons?
7.26 Some copper compounds emit green light when they
are heated in a flame. How would you determine
whether the light is of one wavelength or a mixture
of two or more wavelengths?
7.27 Is it possible for a fluorescent material to emit radiation in the ultraviolet region after absorbing visible
light? Explain your answer.
7.28 Explain how astronomers are able to tell which elements are present in distant stars by analyzing the
electromagnetic radiation emitted by the stars.
7.29 Consider the following energy levels of a hypothetical atom:
1.0 10 19 J
E4
5.0 10 19 J
E3
E2
10 10 19 J
15 10 19 J
E1
(a) What is the wavelength of the photon needed to
excite an electron from E1 to E4? (b) What is the energy (in joules) a photon must have in order to excite an electron from E2 to E3? (c) When an electron
drops from the E3 level to the E1 level, the atom is
said to undergo emission. Calculate the wavelength
of the photon emitted in this process. Back Forward Main Menu TOC 7.30 The first line of the Balmer series occurs at a wavelength of 656.3 nm. What is the energy difference between the two energy levels involved in the emission
that results in this spectral line?
7.31 Calculate the wavelength (in nanometers) of a photon emitted by a hydrogen atom when its electron
drops from the n 5 state to the n 3 state.
7.32 Calculate the frequency (Hz) and wavelength (nm) of
the emitted photon when an electron drops from the
n 4 to the n 2 level in a hydrogen atom.
7.33 Careful spectral analysis shows that the familiar yellow light of sodium lamps (such as street lamps) is
made up of photons of two wavelengths, 589.0 nm
and 589.6 nm. What is the difference in energy (in
joules) between photons with these wavelengths?
7.34 An electron in the hydrogen atom makes a transition
from an energy state of principal quantum numbers
ni to the n 2 state. If the photon emitted has a wavelength of 434 nm, what is the value of ni?
PARTICLE-WAVE DUALITY
Review Questions 7.35 Explain the statement, Matter and radiation have a
“dual nature.”
7.36 How does de Broglie’s hypothesis account for the fact
that the energies of the electron in a hydrogen atom
are quantized?
7.37 Why is Equation (7.7) meaningful only for submicroscopic particles, such as electrons and atoms, and
not for macroscopic objects?
7.38 Does a baseball in flight possess wave properties? If
so, why can we not determine its wave properties?
Problems 7.39 Thermal neutrons are neutrons that move at speeds
comparable to those of air molecules at room temperature. These neutrons are most effective in initiating a nuclear chain reaction among 235U isotopes.
Calculate the wavelength (in nm) associated with a
beam of neutrons moving at 7.00 102 m/s. (Mass
of a neutron 1.675 10 27 kg.)
7.40 Protons can be accelerated to speeds near that of light
in particle accelerators. Estimate the wavelength (in
nm) of such a proton moving at 2.90 108 m/s.
(Mass of a proton 1.673 10 27 kg.)
7.41 What is the de Broglie wavelength, in cm, of a
12.4-g hummingbird flying at 1.20 102 mph?
(1 mile 1.61 km.)
7.42 What is the de Broglie wavelength (in nm) associated with a 2.5-g Ping-Pong ball traveling 35 mph? Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS QUANTUM MECHANICS
Review Questions 7.43 What are the inadequacies of Bohr ’s theory?
7.44 What is the Heisenberg uncertainty principle? What
is the Schrödinger equation?
7.45 What is the physical significance of the wave function?
7.46 How is the concept of electron density used to describe the position of an electron in the quantum mechanical treatment of an atom?
7.47 What is an atomic orbital? How does an atomic orbital differ from an orbit?
7.48 Describe the characteristics of an s orbital, a p orbital, and a d orbital. Which of the following orbitals
do not exist: 1p, 2s, 2d, 3p, 3d, 3f, 4g?
7.49 Why is a boundary surface diagram useful in representing an atomic orbital?
7.50 Describe the four quantum numbers used to characterize an electron in an atom.
7.51 Which quantum number defines a shell? Which quantum numbers define a subshell?
7.52 Which of the four quantum numbers (n, , m , ms)
determine (a) the energy of an electron in a hydrogen atom and in a many-electron atom, (b) the size
of an orbital, (c) the shape of an orbital, (d) the orientation of an orbital in space? Problems 7.53 An electron in a certain atom is in the n 2 quantum level. List the possible values of and m that
it can have.
7.54 An electron in an atom is in the n 3 quantum level.
List the possible values of and m that it can have.
7.55 Give the values of the quantum numbers associated
with the following orbitals: (a) 2p, (b) 3s, (c) 5d.
7.56 For the following subshells give the values of the
quantum numbers (n, , and m ) and the number of
orbitals in each subshell: (a) 4p, (b) 3d, (c) 3s, (d) 5f.
7.57 Discuss the similarities and differences between a 1s
and a 2s orbital.
7.58 What is the difference between a 2px and a 2py orbital?
7.59 List all the possible subshells and orbitals associated
with the principal quantum number n, if n 5.
7.60 List all the possible subshells and orbitals associated
with the principal quantum number n, if n 6.
7.61 Calculate the total number of electrons that can occupy (a) one s orbital, (b) three p orbitals, (c) five d
orbitals, (d) seven f orbitals. Back Forward Main Menu TOC 281 7.62 What is the total number of electrons that can be held
in all orbitals having the same principal quantum
number n?
7.63 Determine the maximum number of electrons that can
be found in each of the following subshells: 3s, 3d,
4p, 4f, 5f.
7.64 Indicate the total number of (a) p electrons in N (Z
7); (b) s electrons in Si (Z 14); and (c) 3d electrons
in S (Z 16).
7.65 Make a chart of all allowable orbitals in the first four
principal energy levels of the hydrogen atom.
Designate each by type (for example, s, p) and indicate how many orbitals of each type there are.
7.66 Why do the 3s, 3p, and 3d orbitals have the same energy in a hydrogen atom but different energies in a
many-electron atom?
7.67 For each of the following pairs of hydrogen orbitals,
indicate which is higher in energy: (a) 1s, 2s; (b) 2p,
3p; (c) 3dxy, 3dyz; (d) 3s, 3d; (e) 4f, 5s.
7.68 Which orbital in each of the following pairs is lower
in energy in a many-electron atom? (a) 2s, 2p; (b) 3p,
3d; (c) 3s, 4s; (d) 4d, 5f.
ATOMIC ORBITALS
Review Questions 7.69 Describe the shapes of s, p, and d orbitals. How are
these orbitals related to the quantum numbers n, ,
and m ?
7.70 List the hydrogen orbitals in increasing order of energy.
ELECTRON CONFIGURATION
Review Questions 7.71 What is electron configuration? Describe the roles
that the Pauli exclusion principle and Hund’s rule
play in writing the electron configuration of elements.
7.72 Explain the meaning of the symbol 4d 6.
7.73 Explain the meaning of diamagnetic and paramagnetic. Give an example of an atom that is diamagnetic and one that is paramagnetic. What does it mean
when we say that electrons are paired?
7.74 What is meant by the term “shielding of electrons”
in an atom? Using the Li atom as an example, describe the effect of shielding on the energy of electrons in an atom.
7.75 Define the following terms and give an example of
each: transition metals, lanthanides, actinides.
7.76 Explain why the ground-state electron configurations
of Cr and Cu are different from what we might expect. Study Guide TOC Textbook Website MHHE Website 282 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS 7.77 Explain what is meant by a noble gas core. Write the
electron configuration of a xenon core.
7.78 Comment on the correctness of the following statement: The probability of finding two electrons with
the same four quantum numbers in an atom is zero.
Problems 7.79 Indicate which of the following sets of quantum numbers in an atom are unacceptable and explain why:
(a) (1, 0, 1 , 1 ), (b) (3, 0, 0, 1 ), (c) (2, 2, 1, 1 ),
22
2
2
(d) (4, 3, 2, 1 ), (e) (3, 2, 1, 1).
2
7.80 The ground-state electron configurations listed here
are incorrect. Explain what mistakes have been made
in each and write the correct electron configurations.
Al: 1s22s22p43s23p3
B: 1s22s22p5
F: 1s22s22p6
7.81 The atomic number of an element is 73. Are the atoms
of this element diamagnetic or paramagnetic?
7.82 Indicate the number of unpaired electrons present in
each of the following atoms: B, Ne, P, Sc, Mn, Se,
Kr, Fe, Cd, I, Pb.
7.83 Write the ground-state electron configurations for the
following elements: B, V, Ni, As, I, Au.
7.84 Write the ground-state electron configurations for the
following elements: Ge, Fe, Zn, Ni, W, Tl.
7.85 The electron configuration of a neutral atom is
1s22s22p63s2. Write a complete set of quantum numbers for each of the electrons. Name the element.
7.86 Which of the following species has the most unpaired
electrons? S , S, or S . Explain how you arrive at
your answer.
THE AUFBAU PRINCIPLE
Review Questions 7.87 State the Aufbau principle and explain the role it plays
in classifying the elements in the periodic table.
7.88 Describe the characteristics of the following groups
of elements: transition metals, lanthanides, actinides.
7.89 What is the noble gas core? How does it simplify the
writing of electron configurations?
7.90 What are the group and period of the element osmium?
Problems 7.91 Use the Aufbau principle to obtain the ground-state
electron configuration of selenium.
7.92 Use the Aufbau principle to obtain the ground-state
electron configuration of technetium. Back Forward Main Menu TOC ADDITIONAL PROBLEMS 7.93 When a compound containing cesium ion is heated
in a Bunsen burner flame, photons with an energy of
4.30 10 19 J are emitted. What color is the cesium
flame?
7.94 Discuss the current view of the correctness of the following statements. (a) The electron in the hydrogen
atom is in an orbit that never brings it closer than
100 pm to the nucleus. (b) Atomic absorption spectra result from transitions of electrons from lower to
higher energy levels. (c) A many-electron atom behaves somewhat like a solar system that has a number of planets.
7.95 Distinguish carefully between the following terms:
(a) wavelength and frequency, (b) wave properties
and particle properties, (c) quantization of energy and
continuous variation in energy.
7.96 What is the maximum number of electrons in an atom
that can have the following quantum numbers?
Specify the orbitals in which the electrons would be
1
found. (a) n 2, ms
4, m
1;
2 ; (b) n
1
(c) n 3,
2; (d) n 2,
0, ms
2;
2.
(e) n 4,
3, m
7.97 Identify the following individuals and their contributions to the development of quantum theory: Bohr, de
Broglie, Einstein, Planck, Heisenberg, Schrödinger.
7.98 What properties of electrons are used in the operation of an electron microscope?
7.99 In a photoelectric experiment a student uses a light
source whose frequency is greater than that needed
to eject electrons from a certain metal. However, after continuously shining the light on the same area
of the metal for a long period of time the student notices that the maximum kinetic energy of ejected electrons begins to decrease, even though the frequency
of the light is held constant. How would you account
for this behavior?
7.100 A certain pitcher ’s fastballs have been clocked at
about 120 mph. (a) Calculate the wavelength of a
0.141-kg baseball (in nm) at this speed. (b) What is
the wavelength of a hydrogen atom at the same
speed? (1 mile 1609 m.)
7.101 Considering only the ground-state electron configuration, are there more elements that have diamagnetic
or paramagnetic atoms? Explain.
7.102 A ruby laser produces radiation of wavelength
633 nm in pulses whose duration is 1.00 10 9 s.
(a) If the laser produces 0.376 J of energy per pulse,
how many photons are produced in each pulse?
(b) Calculate the power (in watts) delivered by the
laser per pulse. (1 W 1 J/s.) Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 7.103 A 368-g sample of water absorbs infrared radiation
at 1.06 104 nm from a carbon dioxide laser.
Suppose all the absorbed radiation is converted to
heat. Calculate the number of photons at this wavelength required to raise the temperature of the water
by 5.00°C.
7.104 Photodissociation of water
H2O( ) h 88n H2(g) 1
2 O2(g) o
has been suggested as a source of hydrogen. The Hrxn
for the reaction, calculated from thermochemical data,
is 285.8 kJ per mole of water decomposed. Calculate
the maximum wavelength (in nm) that would provide
the necessary energy. In principle, is it feasible to use
sunlight as a source of energy for this process? 7.105 Spectral lines of the Lyman and Balmer series do not
overlap. Verify this statement by calculating the
longest wavelength associated with the Lyman series
and the shortest wavelength associated with the
Balmer series (in nm).
7.106 Only a fraction of the electrical energy supplied to a
tungsten light bulb is converted to visible light. The
rest of the energy shows up as infrared radiation (that
is, heat). A 75-W light bulb converts 15.0 percent of
the energy supplied to it into visible light (assume the
wavelength to be 550 nm). How many photons are
emitted by the light bulb per second? (1 W 1 J/s.)
7.107 Certain sunglasses have small crystals of silver chloride (AgCl) incorporated in the lenses. When the
lenses are exposed to light of the appropriate wavelength, the following reaction occurs:
AgCl 88n Ag Cl The Ag atoms formed produce a uniform gray color
that reduces the glare. If H for the reaction is 248 kJ,
calculate the maximum wavelength of light that can
induce this process.
7.108 The He ion contains only one electron and is therefore a hydrogenlike ion. Calculate the wavelengths,
in increasing order, of the first four transitions in the
Balmer series of the He ion. Compare these wavelengths with the same transitions in a H atom.
Comment on the differences. (The Rydberg constant
for He is 8.72 10 18 J.)
7.109 Ozone (O3) in the stratosphere absorbs the harmful
radiation from the sun by undergoing decomposition:
O3 88n O O2. (a) Referring to Table 6.3, calculate the H° for this process. (b) Calculate the maximum wavelength of photons (in nm) that possess this
energy to cause the decomposition of ozone photochemically. Back Forward Main Menu TOC 283 7.110 The retina of a human eye can detect light when radiant energy incident on it is at least 4.0 10 17 J.
For light of 600-nm wavelength, how many photons
does this correspond to?
7.111 An electron in an excited state in a hydrogen atom
can return to the ground state in two different ways:
(a) via a direct transition in which a photon of wavelength 1 is emitted and (b) via an intermediate excited state reached by the emission of a photon of
wavelength 2. This intermediate excited state then
decays to the ground state by emitting another photon of wavelength 3. Derive an equation that relates
1 to 2 and 3.
7.112 A photoelectric experiment was performed by separately shining a laser at 450 nm (blue light) and a laser
at 560 nm (yellow light) on a clean metal surface and
measuring the number and kinetic energy of the
ejected electrons. Which light would generate more
electrons? Which light would eject electrons with
greater kinetic energy? Assume that the same amount
of energy is delivered to the metal surface by each
laser and that the frequencies of the laser lights exceed the threshold frequency.
7.113 Draw the shapes (boundary surfaces) of the following orbitals: (a) 2py, (b) 3dz2, (c) 3dx2 y2. (Show coordinate axes in your sketches.)
7.114 The electron configurations described in this chapter
all refer to gaseous atoms in their ground states. An
atom may absorb a quantum of energy and promote
one of its electrons to a higher-energy orbital. When
this happens, we say that the atom is in an excited
state. The electron configurations of some excited
atoms are given. Identify these atoms and write their
ground-state configurations:
(a) 1s12s1
(b) 1s22s22p23d1
(c) 1s22s22p64s1
(d) [Ar]4s13d104p4
(e) [Ne]3s23p43d1
7.115 Draw orbital diagrams for atoms with the following
electron configurations:
(a) 1s22s22p5
(b) 1s22s22p63s23p3
(c) 1s22s22p63s23p64s23d7
7.116 If Rutherford and his coworkers had used electrons
instead of alpha particles to probe the structure of the
nucleus (see Chapter 2), what might have they discovered?
7.117 Scientists have found interstellar hydrogen atoms
with quantum number n in the hundreds. Calculate
the wavelength of light emitted when a hydrogen
atom undergoes a transition from n 236 to n 235. Study Guide TOC Textbook Website MHHE Website 284 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS 7.118
7.119 7.120 7.121 7.122 7.123 7.124 In what region of the electromagnetic spectrum does
this wavelength fall?
Calculate the wavelength of a helium atom whose
speed is equal to the root-mean-square speed at 20°C.
Ionization energy is the energy required to remove a
ground state (n 1) electron from an atom. It is usually expressed in units of kJ/mol, that is, the energy
in kilojoules required to remove one mole of electrons from one mole of atoms. (a) Calculate the ionization energy (in kJ/mol) for the hydrogen atom.
(b) Repeat the calculation, assuming in this second
case that the electrons are removed from the n 2
state.
An electron in a hydrogen atom is excited from the
ground state to the n 4 state. Comment on the correctness of the following statements (true or false).
(a) n 4 is the first excited state.
(b) It takes more energy to ionize (remove) the electron from n 4 than from the ground state.
(c) The electron is farther from the nucleus (on average) in n 4 than in the ground state.
(d) The wavelength of light emitted when the electron drops from n 4 to n 1 is longer than that
from n 4 to n 2.
(e) The wavelength the atom absorbs in going from
n 1 to n 4 is the same as that emitted as it
goes from n 4 to n 1.
The ionization energy of a certain element is
412 kJ/mol (see Problem 7.119). However, when the
atoms of this element are in the first excited state
(n 2), the ionization energy is only 126 kJ/mol.
Based on this information, calculate the wavelength
of light emitted in a transition from n 2 to n 1.
Alveoli are the tiny sacs of air in the lungs (see
Problem 5.122) whose average diameter is 5.0
10 5 m. Consider an oxygen molecule (5.3
10 26 kg) trapped within a sac. Calculate the uncertainty in the velocity of the oxygen molecule. (Hint:
The maximum uncertainty in the position of the molecule is given by the diameter of the sac.)
How many photons at 660 nm must be absorbed to
melt 5.0 102 g of ice? On average, how many H2O
molecules does one photon convert from ice to water? (Hint: It takes 334 J to melt 1 g of ice at 0°C.)
Shown below are portions of orbital diagrams representing the ground-state electron configurations of
certain elements. Which of them violate the Pauli exclusion principle? Which of them violate Hund’s
rule? h h hh h hg h (a) Back h hg g
(b) (c) Forward Main Menu TOC hg hhh h h h g hg (d) (e) hg hg gg hg hg
(f ) 7.125 The UV light that is responsible for tanning the skin
falls in the 320 – 400 nm region. Calculate the total
energy (in joules) absorbed by a person exposed to
this radiation for 2.0 hours, given that there are 2.0
1016 photons hitting Earth’s surface per square centimeter per second over a 80-nm (320 nm to 400 nm)
range and that the exposed body area is 0.45 m2.
Assume that only half of the radiation is absorbed
and the other half is reflected by the body. (Hint: Use
an average wavelength of 360 nm in calculating the
energy of a photon.)
7.126 The sun is surrounded by a white circle of gaseous
material called corona, which becomes visible during a total eclipse of the sun. The temperature of the
corona is in the millions of degrees Celsius, which is
high enough to break up molecules and remove some
or all of the electrons from atoms. One way astronomers have been able to estimate the temperature
of corona is by studying the emission lines of ions of
certain elements. For example, the emission spectrum
of Fe14 ions has been recorded and analyzed.
Knowing that it takes 3.5 104 kJ/mol to convert
Fe13 to Fe14 , estimate the temperature of the sun’s
corona. (Hint: The average kinetic energy of one
mole of a gas is 3 RT.)
2 7.127 In 1996 physicists created an anti-atom of hydrogen.
In such an atom, which is the antimatter equivalent
of an ordinary atom, the electrical charges of all the
component particles are reversed. Thus the nucleus
of an anti-atom is made of an anti-proton, which has
the same mass as a proton but bears a negative charge,
while the electron is replaced by an anti-electron (also Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS called positron) with the same mass as an electron,
but bearing a positive charge. Would you expect the
energy levels, emission spectra, and atomic orbitals
of an antihydrogen atom to be different from those
of a hydrogen atom? What would happen if an antiatom of hydrogen collided with a hydrogen atom? Back Forward Main Menu TOC 285 Answers to Practice Exercises: 7.1 625 Hz. 7.2 8.24 m. 7.3 3.39 103 nm. 7.4 2.63 103 nm. 7.5 56.6 nm. 7.6 n 3,
1, m
1, 0, 1. 7.7 16. 7.8 (5, 1, 1, 1 ), (5, 1, 0, 1 ),
2
2
1
(5, 1, 1, 2 ), (5, 1, 1, 1 ), (5, 1, 0, 1 ), (5, 1, 1, 1 ). 7.9
2
2
2
32. 7.10 (1, 0, 0, 1 ), (1, 0, 0, 1 ), (2, 0, 0, 1 ), (2, 0, 0, 1 ),
2
2
2
2
(2, 1, 1, 1 ). There are 5 other acceptable ways to write the
2
quantum numbers for the last electron. 7.11 [Ne]3s23p3. Study Guide TOC Textbook Website MHHE Website ...

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