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Unformatted text preview: Chapter Seven QUANTUM THEORY AND THE ELECTRONIC
STRUCTURE OF ATOMS
• Electromagnetic Radiation
Bohr's Model of the Hydrogen Atom
The Dual Nature of the Electron
Quantum Mechanics and Quantum Numbers
Electron Configurations and the Aufbau Principle ELECTROMAGNETIC RADIATION
3. Use the equations relating wavelength, frequency, and the speed of light.
Describe a quantum and be able to calculate the energy of a quantum emitted by an atom.
Explain how Einstein accounted for the frequency dependence of the photoelectric effect. Electromagnetic Radiation. To understand some of the experiments that illuminated the behavior of
electrons in atoms requires some appreciation of the nature of electromagnetic radiation. Radiation is energy that
is transmitted through space in the form of waves. Electromagnetic radiation consists of an electric field
component and a magnetic field component that are perpendicular to each other and to their direction of
propagation (see Figure 7.3 of the text). The distance between identical points on successive waves, such as the
wave crests, is called the w avelength , λ (lambda). All electromagnetic waves travel at the same speed through
a vacuum. This speed, known as the speed of light, is 3.00 × 10 8 m/s and has the symbol c. The f requency of
a wave is the number of waves that pass a given point per second. Its symbol is ν (nu), and the unit of
frequency is cycles per second which is written as /s or s–1 . The SI unit for one cycle per second is called a hertz
(Hz). These three quantities, c, λ, and ν , are related by the equation:
c = λν
Figure 7.4 in the textbook shows that the wavelengths of electromagnetic waves have been observed to vary
from as short as 5 × 10 –4 nm for some γ rays to more than 100 m for radio waves. The visible spectrum ranges
from 400 to 700 nm and is only a tiny fraction of the complete span of electromagnetic waves.
According to this equation, the frequency increases as the wavelength decreases. Therefore, radio waves with
their long wavelengths have low frequencies, and γ rays with their short wavelengths have extremely high
frequencies. The radio waves of an AM station broadcasting at a frequency of 1000 kilohertz (1000 kHz), or
106 s –1 , emits electromagnetic waves with a wavelength of 300 m. Quantum Theory. In 1900 Max Planck solved a long-standing riddle when he provided an explanation
for the energy radiated by an object at a certain temperature. Heated bodies emit radiation usually in the infrared.
However, a body will emit visible radiation if it is hot enough. According to classical physics, an object can
acquire or lose any amount of energy no matter how small. But all attempts by scientists to explain the
wavelengths of electromagnetic radiation emitted by heated bodies using classical theory have failed. In Planck's
quantum theory, however, there is a lower limit to the smallest increment of energy (a quantum) that an atom 1 26
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can gain or lose. In addition, the total amount of energy gained or lost by the atom is some multiple of this
increment of energy. A quantum is the smallest quantity of energy that can be emitted in the form of
electromagnetic radiation. The energy of a quantum (E) is given by Planck's equation:
E = hν
where h is Planck's constant, 6.63 × 10 –34 J·s (joule·second) and ν is the frequency of the radiation. With this
equation, Planck was able to explain the entire spectrum of radiation emitted by an object at a certain
temperature. The Photoelectric Effect. In 1905 Albert Einstein used Planck's quantum hypothesis to explain the
photoelectric effect. When ultraviolet light is directed onto a clean metal electrode it causes electrons to be
ejected from the metal surface. If the frequency of the light is varied, the significant finding is that below a
certain frequency no photoelectrons are observed. However, above this frequency, called the threshold frequency,
the number of photoelectrons and their kinetic energy increases with increasing frequency of the radiation.
Therefore, the energy absorbed from a light beam is proportional to its frequency, E ∝ ν . To Einstein, the
significance of the light frequency was that it was a measure of the energy of particle-like entities of light that
are now called photons. The energy of a photon is given by Planck's equation.
Ephotons = h ν
Einstein's interpretation of the existence of the threshold frequency was that a photon of light would interact
with one electron in the surface of the metal. This electron is held to the metal by a force called the binding
energy of the metal (BE). Below the threshold frequency, a photon does not have enough kinetic energy to
overcome the binding energy and eject an electron from the metal surface, but photons associated with light
above the threshold frequency do. As the frequency of light is increased, the energy of a photon increases and
more energy can be transferred to the electrons. Since the binding energy is constant, then as the photon energy
increases, more of its energy is available to become kinetic energy (KE) of the photoelectrons.
KE = hν – BE
Planck's and Einstein's use of the quantum concept demonstrated that light, in addition to having properties of a
wave, has properties of particles as well. This behavior is known as the dual nature of light.
EXAMPLE 7.1 Wavelength and Frequency
Domestic microwave ovens generate microwaves with a frequency of 2.450 GHz. What is the wavelength of this
•Method of Solution
The equation relating the wavelength to the frequency is
c = λν
where c is the speed of light.
Rearranging and substituting, we get:
3 .0 × 1 0 8 m /s
1 G Hz
= 0.122 m
λ= Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 28 / Quantum Theory and the Electronic Structure of Atoms
EXAMPLE 7.2 Planck's Equation
The orange light given off by a sodium vapor lamp has a wavelength of 589 nm. What is the energy of a single
photon of this radiation?
•Method of Solution
The energy of a quantum is rproportional to its frequency. Since wavelength is given here, then substitute c/λ
into Planck's equation for ν .
Ephoton = h ν = hc
λ • Calculation
Introduce values for Planck's constant and the speed of light into the equation. The units will not cancel unless
the wavelength is converted from nanometers into meters.
(6.63 × 1 0 –34 J . s)(3.0 × 1 0 8 m /s)
(589 × 1 0 –9 m )
E = 3.38 × 10 –19 J
We can compare this energy to that of a mole of substance. The energy of a mole of photons is the product of
Ephoton × NA, where NA is Avogadro's number.
E = (3.38 × 10 –19 J)(6.02 × 10 23 /mol)
= 203,000 J/mol = 203 kJ/mol
2. What is the frequency of light that has a wavelength of 750 nm? 3. What is the energy of a photon of radiation having a frequency of 6.2 × 10 14 s –1 ? 4. The red line in the spectrum of lithium occurs at 670.8 nm. What is the energy of a photon of this light?
What is the energy of 1 mole of these photons? 5. A photon has an energy of 2.05 × 10 –19 J. Calculate the wavelength of the radiation. 6. Back What is the wavelength of electromagnetic radiation with frequency 5.0 × 10 14 s –1 ? The binding energy for a electron on a metallic sodium surface is 3.9 × 10 –19 J. Find the kinetic energy of
the photoelectrons when sodium is illuminated with UV light of 305 nm. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Quantum Theory and the Electronic Structure of Atoms / 1 29 BOHR'S MODEL OF THE HYDROGEN ATOM
3. Distinguish between a line spectrum and a continuous spectrum.
List the two assumptions made in Bohr's model of the hydrogen atom.
Calculate the wavelength of radiation emitted by a specific electron transition in the hydrogen atom. Emission Spectra. Elements in the gaseous state that are excited by an electrical current give off
electromagnetic radiation in a process called emission. By passing emitted light through a prism, an emission
spectrum consisting of the wavelengths of emitted electromagnetic radiation is produced. A c ontinuous
spectrum contains all the wavelengths between the longest and shortest emitted by that substance. "White" light
results from the emission of all wavelengths in the visible part of the spectrum.
The light emitted from excited atoms in the gaseous state produces a line spectrum rather than continuous
spectrum. A line spectrum consists of several discrete wavelengths. The observed wavelengths are characteristic
of the element and can serve to identify the presence of the element in a sample.
Emission Spectrum of Hydrogen. A portion of the line spectrum of hydrogen falls in the visible
region, and is called the Balmer series. Figure 7.8(b) of the text shows that the prominent visible lines are at
656 nm, 486 nm, and 434 nm. In all, five spectral series are known for hydrogen. The Lyman series is in the
ultraviolet region, and the Paschen, Brackett, and Pfund series are in the infrared region
Bohr's Model. In 1913 Niels Bohr produced his famous model of the hydrogen atom. The great
achievement of this model was that it explained the source and the observed wavelengths of lines in the
spectrum of the H atom. Bohr postulated a "solar system" model for the H atom in which the electron traveled
in circular orbits about the proton. The basic assumptions of Bohr's model were:
1. 2. The electron moves in a circular orbit about the nucleus. Of the infinite number of possible orbits only
certain orbits with distinct radii are allowed. As long as the electron remains in an orbit, its energy remains
The energy of the electron increases the farther its orbit lies from the nucleus. When energy is absorbed by
the atom, the electron must jump to a higher energy orbit. When an electron transition occurs from a higher
to a lower energy orbit, radiation is emitted by the atom. Bohr was able to calculate the radii of the allowed orbits and their energies. The energies of the H atom are
En = –R H 1 n2 where RH is the Rydberg constant, which in units of joules has the value 2.18 × 10 –18 J. The minus sign in
the equation simply means that the energy of the hydrogen atom is lower than that of a completely separated
proton and electron for which the force of attraction is zero. It does not signify negative energy. The quantity n
is called the principal quantum number, and is an integer; n = 1, 2, 3,…. Inserting values for n into the energy
equation gives the energies of all allowed orbits shown in Figure 7.11 of the text.
The radius of each electron orbit in the hydrogen atom is proportional to n2 . As n increases, the orbit radius
increases rapidly. The farther the electron is from the nucleus, the higher its energy.
Bohr explained that the emission of light, when an electrical current passes through a gas, is due to the
formation of excited atoms. He proposed that in the emission process an electron drops from a higher to a lower
orbit. During this transition the atom emits radiation. For energy to be conserved, the energy lost as radiation
must equal the difference in energy between the initial energy level Ei and the final energy level Ef. For the Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 30 / Quantum Theory and the Electronic Structure of Atoms
emission process, n i represents the higher energy level and nf lower level. Then the change in energy of the
atom ∆ E is
∆ E = Ef – E i
where, according to Bohr's equation, the energy change for the atom is
∆ Eatom = R H 2 – 2 n i nf 1 1 Bohr's stroke of genius was to equate ∆ Eatom to the energy of a photon of emitted light.
Ephoton = ∆ Eatom
Since Ephoton = hν = hc
hν = ∆ Eatom
and the frequency of light emitted should be given by
The wavelengths of emitted light can be calculated from
ν= λ= c
∆ Eatom With this model of the hydrogen atom, Bohr was able to predict the wavelengths of all the literally hundreds of
observed lines in the spectrum of hydrogen!
EXAMPLE 7.3 The Hydrogen Atom
b. What amount of energy in joules is lost by a hydrogen atom when an electron transition from n = 3 to n =
2 occurs in the hydrogen atom?
What is the wavelength of the light emitted when the electron transition n = 3 → n = 2 occurs? •Method of Solution for (a)
The energy of the atom is quantized and depends on the orbit of the electron. Each orbit is assigned a principal
quantum number n:
En = –R H 1 n2 For this transition ni = 3 and nf = 2:
∆ E = Ef – E i = E2 – E 3 = 2 – 2 2 3 • Calculation for (a)
Substituting for RH gives:
∆ E = E2 – E 3 = –2.18 × 1 0 –18 J – –2.18 × 1 0 –18 J 4 9 ∆ E = –0.545 × 10 –18 J + 0.242 × 10 –18 J
∆ E = –0.303 × 10 –18 J = –3.03 × 10 –19 J Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Quantum Theory and the Electronic Structure of Atoms / 1 31
•Method of Solution for (b)
The energy lost by the atom appears as a photon of radiation with its respective frequency and wavelength. Since
the energy of the photon is positive we will drop the negative sign from the calculation of the wavelength.
∆ Eatom = Ephoton =
• Calculation for (b)
3.03 × 1 0 –19 J
( 6.63 × 1 0 –34 J . s)(3.00 × 1 0 8 m s –1 )
3.03 × 1 0 –19 J
λ = 656 × 10 –9 m
= 656 nm
7. Consider the following energy levels for the hydrogen atom.
Considering only these levels:
a. How many emission lines are possible?
b. Which transition produces photons of the greatest energy?
c. Which transition for the H atom produces the emission line with the longest wavelength?
8. What wavelength of radiation will be emitted when an electron in a hydrogen atom jumps from the n = 5 to
the n = 1 principal energy level? Name the region of the electromagnetic spectrum corresponding to this
9. A hydrogen emission line in the ultraviolet region of the spectrum at 95.2 nm corresponds to a transition
from a higher energy level n to the n = 1 level. What is the value of n for the higher energy level?
10. The second line in the Balmer series occurs at 486.1 nm. What is the energy difference between the initial
and final energy levels involved in the electron transition? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 32 / Quantum Theory and the Electronic Structure of Atoms T HE DUAL NATURE OF THE ELECTRON
1. Calculate the wavelength of a particle, given its momentum. Matter Waves. In 1925 Louis de Broglie had the novel idea that since light had been found to have a dual
nature with properties of both waves and particles, then particles of matter should have a dual nature and exhibit
wave properties as well. He deduced that the wavelength λ of a particle depends on its mass m and velocity u:
Experimental evidence of the wave properties of small particles was provided in 1927 by diffraction experiments
with electron beams. Diffraction is a phenomenon that can be explained only by wave motion. Since electrons
were observed to produce diffraction patterns, then electrons must behave as waves. According to de Broglie's
equation, the wavelength of a particle increases as its mass decreases, so wave properties (diffraction) are most
prominent for the electron, and are the basic principle behind the electron microscope. Diffraction patterns of
neutron beams have also been observed and serve as a means to learn about the structures of molecules. As the
mass of a particle reaches macroscopic size (say, > 10–12 g) its wavelength becomes extremely short and wave
properties cannot be observed.
EXAMPLE 7.4 De Broglie Wavelength
When an atom of Th-232 undergoes radioactive decay, an alpha particle which has a mass of 4.0 amu is ejected
from the Th nucleus with a velocity of 1.4 × 10 7 m/s. What is the de Broglie wavelength of the alpha particle?
•Method of Solution
The de Broglie wavelength depends on the mass and velocity of the particle. The equation is
Because Planck's constant has units of J·s, and 1 J = 1 kg·m2 ·s –2 , the mass of the alpha particle must be
expressed in kg. Since 1 mole of alpha particles has a mass of 4.0 g, then the mass of one particle is:
4 .0 × 1 0 –3 k g/mol
= 6.6 × 10 –27 kg/particle
6.02 × 1 0 23 particles/mol
The wavelength of this alpha particle is:
mu ( 6.63 × 1 0 –34 J · s ) × 1 kg·m 2 /s2 1J (6.6 × 1 0 –27 k g)(1.4 × 1 0 7 m /s)
Notice that all the units cancel except meters. The wavelength is:
λ = 7.2 × 10 –15 m •Comment
The wavelength is smaller than the diameter of the thorium nucleus, which is about 2 × 10 –14 m .
_______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Quantum Theory and the Electronic Structure of Atoms / 1 33
11. Calculate the wavelength in nanometers of a proton (mass = 1.6725 × 10 –27 kg) that is moving at 10% of
the speed of light. QUANTUM MECHANICS AND QUANTUM NUMBERS
4. Describe the use of the terms probability, electron density, and orbital in the wave mechanical model of the
Describe the significance of the three quantum numbers that correspond to an orbital in an atom.
Describe the shapes of s, p, and d orbitals.
State the Pauli exclusion principle. The dual nature of the electron is further described by the H eisenberg uncertainty principle which
states that it is impossible to know simultaneously both the momentum and the position of an electron with
certainty. The principle can be interpreted as follows: If we want to know the energy of an electron in an atom
with only a small uncertainty, then we will have a large uncertainty in knowing the position of the electron in
some region of the atom. The result of this principle is that we can only know the probabilty of finding the
electron in a certain region of the atom. Quantum Mechanics. In 1926 Erwin Schrödinger incorporated all the wave and particle properties into a
form of mechanics called wave mechanics or quantum mechanics. He formulated the now famous Schrödinger
wave equation and obtained a set of mathematical functions called wave functions, ψ. For any set of coordinates
within the atom, ψ has a numerical value. The square of the wave function ψ2 is related to the probability that
an electron will be found in a small region (volume) in space. Where ψ2 is large, the probability of finding the
electron is high. Where ψ2 is small, the probability of finding the electron is low.
The term electron density (Figure 7.22 in the text) is used to represent the probability concept. The term
electron density refers to the probability of finding the electron in a certain region of the atom. The electron
density must not be interpreted to mean that one electron occupies the whole of the space covered by the
probability distribution at one time . Rather, the electron is considered a point charge, and the electron density
refers to the probability of finding the electron in a specific volume of the atom.
As a result of the probability concept, we no longer speak of distinct Bohr orbits. In the wave mechanical
model of an atom, the Bohr orbit is replaced with an orbital or atomic orbital. Orbitals are regions in an atom
within which electrons have a high probability, usually 90 percent, of being found. Each atomic orbital has a
characteristic shape and energy.
Atomic Orbitals. An orbital is a graphical representation of the electron probability around the nucleus.
The density of the probability cloud at each region in space (see Figure 7.22 text) represents the probability of
finding the electron there. Orbitals are often drawn using equal-probability contours. If we go far enough from
the nucleus in a particular direction, we come to a point at which 90 percent of the time the electron, when it is
in that direction, will be inside that point. Connecting all such points produces an equal-probability boundary
surface such that 90 percent of the time the electron will be inside that surface no matter what direction from the
nucleus we consider. This graphical representation gives the shape of the orbital.
The different types of orbitals can be recognized by their shapes. While there are more than three types of
orbitals only the characteristics of s, p, and d orbitals are given in Table 7.1. The shapes of s and p orbitals are
shown in Figure 7.18 and 7.19 of the text. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 34 / Quantum Theory and the Electronic Structure of Atoms
T able 7.1 Symbols and Shapes of Orbitals
Shape of 90 percent
dumbbell (2 lobes)
__________________________________________________________________ Quantum Numbers. Both the energy and the probability distribution are described by a set of numbers
called quantum numbers. The p rincipal quantum number n determines the energy of a hydrogen atom
according to the formula En = –R H (1/n 2 ). This number can have any integral value, n = 1, 2, 3,…. The larger
the value of n, the greater the energy, and the farther the electron is from the nucleus on the average. When n =
1, the electron in the hydrogen atom is said to be in the ground state. All other values of n correspond to excited
states. When n = ∞ , the highest possible energy, the atom has lost the electron and exists as a H+ ion.
The a ngular momentum quantum number l is related to the shape of the orbital. Within each
principal energy level one or more orbital types will exist. Thus any one principle quantum level may have
more than one value for l. For a given n, l has integral values from 0 to (n – 1); l = 1, 2, 3,…, (n – 1). Each
value of l defines a group of orbitals composing a specific sublevel or subshell.
The shape of atomic orbitals will be discussed in the next section. For now each orbital is specified by a
letter designation. Orbitals for which l = 0 are also referred to as s orbitals. Electrons in s orbitals have zero
units of angular momentum. The letters used to designate orbitals with different l values are
The m agnetic quantum number m l describes the orientation of an orbital in space. The value of m l
may have integral values from –l t o + l, including zero.
ml = – l, (–l + 1), …, 0, …, (l – 2), (l – 1), +l
The number of ml values is very important because it designates the number of orbitals within a subshell. For
example, when l = 2, then ml = –2, –1, 0, 1, 2. We see there are five ml values. d orbitals always occur in
groups of five orbitals. These five d orbitals all have the same energy, but have five different orientations in
space. They make up the d subshell. A s ubshell is a group of orbitals designated by a particular l value within
a shell. Table 7.2 shows the number of orbitals in the most common subshells.
Table 7.2 The Number of Orbitals in Various Subshells
–1, 0, 1
–2, –1, 0, 1, 2
–3, –2, –1, 0, 1, 2, 3
___________________________________________________________ The e lectron spin quantum number , ms , designates one of two possible spin directions for electrons.
The two possible values of ms are +1/2 and –1/2 . The P auli exclusion principle states that no two Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Quantum Theory and the Electronic Structure of Atoms / 1 35
electrons in an atom can have all four quantum numbers the same. Accordingly, this principle limits the
numbers of electrons each atomic orbital can hold to no more than two electrons, one electron will have m s =
+ 1/2 and the other ms = – 1/2 .
Let's take an electron in a particular orbital, say, a 3px orbital. The three quantum numbers associated with
this orbital are n = 3, l = 1, and m l = –1. Any electrons in this orbital must have these identical n, l, and ml
quantum numbers. The two sets of quantum numbers for electrons in this orbital differ in their ms values:
electron 2: n = 3, l = 1, ml = –1, ms = + 1/2
n = 3, l = 1, ml = –1, ms = – 1/2 T he result of this qualitative treatment of wave mechanics gives several essential pieces
o f information about electrons in atoms:
1. Quantum numbers (n, l, m) and their relationships to each other.
2. Electron probability density and atomic orbitals.
3. The maximum number of electrons per orbital.
4. An electron shell is a collection of orbitals with the same value of n.
5. A subshell is a set of orbitals with the same values of n and l. _______________________________________________________________________________
EXAMPLE 7.5 Quantum Numbers
If the principal quantum number of an electron is n = 2, what values could the following have?
a. Its l quantum number
b. Its m l quantum number
c. Its m s quantum number
•Method of Solution
c. Recall that the angular momentum quantum number l has values that depend on the value of the principal
quantum number n; l = 0, 1, 2, …, (n – 1). Answer: When n = 2, l = 0, 1.
The magnetic quantum number ml depends on the value of l where ml = – l , – l + 1, …, 0,…, l + 1 , l .
Answer: When l = 1, ml = –1, 0, 1; and when l = 0, ml = 0.
Answer: The ms quantum number for a single electron could be either +1/2 or – 1/2 . _______________________________________________________________________________
EXAMPLE 7.6 Quantum Numbers and Orbitals
List all the possible types of orbitals associated with the principal energy level n = 5.
•Method of Solution
The type of orbital is given by the angular momentum quantum number. When n = 5; l = 0, 1, 2, 3, and 4.
Answer: These correspond to subshells consising of s, p, d, and f orbitals.
EXAMPLE 7.7 Quantum Numbers
For the following sets of quantum numbers for electrons, indicate which quantum numbers — n, l, m l — could
not occur and state why.
a. 3, 2, 2
b. 2, 2, 2
c. 2, 0, –1 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 36 / Quantum Theory and the Electronic Structure of Atoms
• Method of Solution
b. When n = 3, l = 0, 1, 2. When l = 2; ml = –2, –1, 0, 1, 2. Answer: Thus set (a) can occur.
When n = 2, l = 0, 1. Answer: The set in (b) cannot occur because l ≠ 2 when n = 2. This imeans that the
n = 2 energy level cannot have d orbitals.
c. When n = 2, l = 0, 1. When l = 0, ml = 0. Answer: The set in (c) cannot occur because ml ≠ –1 when
l = 0.
EXAMPLE 7.8 Quantum Numbers and Orbitals
What values can ml take for
a. a 3d orbital?
b. a 2s orbital?
•Method of Solution
The value of ml depends only on l and so first convert each orbital designation to the corresponding value of l
a. For a d orbital l = 2; therefore, possible m l values are –2, –1, 0, 1, and 2.
b. For an s orbital l = 0; therefore, ml = 0.
12. Which of the following sets of quantum numbers are not allowed for describing an electron in an orbital?
13. Which choice is a possible set of quantum numbers for the last electron added to make up an atom of
gallium (Ga) in its ground state? a.
1 /2 –1/2
1 /2 –1/2 14. Which of the following are incorrect designations for an atomic orbital?
a. 3f b. 4s c. 2d d. 4f
15. How many orbitals in an atom can have the following designations?
d. n = 3
16. For each of the following give the subshell designation, the ml values, and the number of possible orbitals.
a. n = 3 l = 2
b. n = 4 l = 3
c. n = 5 l = 1
17. For each of the following subshells give the n and l values, and the number of possible orbitals.
c. 4f Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Quantum Theory and the Electronic Structure of Atoms / 1 37 ELECTRON CONFIGURATIONS AND THE AUFBAU PRINCIPLE
3. Compare the order of orbital energies in the hydrogen atom to the order in a many-electron atom.
Use the Aufbau principle to write the electron configuration of an element.
Use Hund's rule to construct orbital diagrams for the electron configurations of elements. The Energies of Orbitals. The way electrons are distributed among the orbitals of an atom is its
electron configuration and electronic structure. Each of the principal energy levels (called shells) of an atom can
be divided into subshells, which are sets of orbitals. The s subshell has one s orbital, the p subshell has three p
orbitals, the d subshell has five d orbitals, and so on. In addition each subshell has an electron capacity based on
a maximum of two electrons per orbital. The orbitals available to a shell and the subshell capacities are
summarized in Table 7.3.
Allowed Numbers of Orbitals per Energy Level
Orbitals per Subshell
In a hydrogen atom all of the orbitals in a particular energy level (same n value) have the same energy. In a
hydrogen atom the electron resides in the 1s orbital because it has the lowest possible energy in that orbital. The
ground state electron configuration of H is 1s1 , as shown below. The number of electrons per subshell
1 Principal electron shell
(value of n) 1s Type of orbital occupied Atoms of elements other than hyrogen have more than one electron and are called m any-electron atoms .
In these atoms each principal energy level is split into several energies. These energies correspond to the orbitals
of one type having a slightly different energy than orbitals of another type. This splitting is shown in Figure
7.1. The energy of an orbital in a many-electron atom depends on both n and l. For a given value of n, the Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 38 / Quantum Theory and the Electronic Structure of Atoms
energy of orbitals increases as l increases. Therefore, for atoms other than hydrogen atoms, the 2p orbital has a
higher energy than the 2s orbital.
3d Energy 4s
Figure 7.1 The orbital energies in a many-electron atom. Within a shell the orbital
energies increase as the l quantum number increases. The energy of an orbital depends on
its n and l values. Here the 4s subshell is shown to be lower than the 3d subshell. The order of increasing energy for orbitals within a given principal energy level is
This trend results from the fact that s orbitals have a greater electron density near the nucleus than p orbitals, and
p orbitals have greater electron density near the nucleus than d orbitals. In a many-electron atom, the energies of
the orbitals increase as follows (Figure 7.1):
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s …
Note also that the higher principal energy levels (n values) have more subshells, and as these split in energy
some of these subshells can actually be lower in energy than subshells from a lower principal energy level.
Thus the energy of a 4s orbital is very close to the energy of a 3d orbital in a many-electron atom. In potassium
and calcium atoms the energy of the 4s orbital is less than the energy of the 3d orbital. Electron Configurations. To write the electron configuration for the ground state of a many-electron
atom, recall that the order in which subshells are occupied is from lowest energy to highest energy. The order
shown in Figure 7.1 can easily be remembered by drawing Figure 7.27 of the textbook. The general approach of
building up the atoms of the elements by adding a proton to the nucleus and an electron to an orbital is called
the Aufbau or building up principle.
Helium has two electrons both of which can occupy the 1s orbital. The electron configuration is: He 1s2 .
A lithium atom has three electrons. According to the Pauli principle, we can place only two electrons in the 1s
orbital; the third electron must enter the next higher orbital, the 2s orbital. Its electron configuration is: Li
1s2 2s1 . The electron configurations of the second period elements are:
Be Back Forward 1s2 2s2 Main Menu TOC Study Guide TOC Textbook Website MHHE Website Quantum Theory and the Electronic Structure of Atoms / 1 39
With Be the 1s and 2s orbitals are filled, and so additional electrons must enter the 2p orbitals. The 2p subshell
is being filled over the next six elements.
Ne 1s2 2s2 2p1
1s2 2s2 2p2
1s2 2s2 2p3
1s2 2s2 2p4
1s2 2s2 2p5
1s2 2s2 2p6 Orbital Diagrams. The above electron configurations tend to hide information about the number of
electrons in any one outer orbital. The 2p subshell consists of three 2p orbitals: 2px , 2p y , and 2pz. Thus its
capacity is 6 electrons. If we use an orbital diagram, a choice arises about where to place the electrons. In
carbon, for example, are the 2 electrons in the p subshell both in the 2px orbital, or are they distributed so that
the 2px and the 2py each have one electron? To make the choice first we need to describe an orbital diagram.
An orbital diagram uses boxes to designate individual orbitals, and groups of boxes to designate subshells. ] ] 1s 2s 2p An arrow pointing up ↑ stands for an electron spinning in one direction, and an arrow pointing down ↓ stands
for an electron spinning in the opposite direction.
The orbital diagrams for carbon and nitrogen are drawn as follows. Electrons are placed into orbitals
according to H und's rule , which states that electrons entering a subshell containing more than one orbital will
have the most stable arrangement if the electrons occupy the orbitals singly rather than in pairs. Thus carbon
has one electron in the 2p x and one in the 2p y , rather than two electrons in the 2px . [↑↓ ]
2s [↑↓ ] [↑↓ ] 1s N [↑↓ ]
1s C [↑ ] ↑ ] ] 2s 2p [↑ ] ↑ ] ↑ ]
2p According to Hund's rule, nitrogen atoms have the electrons in the 2p subshell distributed with one each in the
2px , 2p y , and 2pz.
The orbital diagram indicates the number of unpaired electrons in an atom. The presence, or absence, of
unpaired electrons is detected experimentally by the behavior of a substance when placed in a magnetic field.
Paramagnetism is the property of attraction to a magnetic field. Diamagnetism is the property of repulsion by a
magnetic field. The atoms of paramagnetic substances contain unpaired electrons. Those of diamagnetic
substance contain only paired electrons. Using the three principles that have been given, the electron
configurations of the ground states of most atoms can be estimated. Noble Gas Core Abbreviations. The electron configurations of the noble gas elements can be used as
abbreviations when writing electron configurations. Neon has the configuration Ne 1s2 2s2 2p6 . All elements
beyond neon have this configuration for the first 10 electrons. We use the symbol [Ne] to represent the
configuration of the 10 electrons in neon and call it a neon core. Similarly 18 electrons arranged in the ground
state of an argon atom (1s 2 2s2 2p6 3s2 3p6 ) are called an argon core, written [Ar]. [Kr] and [Xe] cores can also be
used. In practice we select the noble gas that most nearly precedes the element being considered. Therefore,
calcium and strontium could be written in full or abbreviated as follows:
Ca 1s 2 2s2 2p6 3s2 3p6 4s2
Sr 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 Back Forward Main Menu TOC [Ar]4s2
or [Kr]5s2 Study Guide TOC Textbook Website MHHE Website 1 40 / Quantum Theory and the Electronic Structure of Atoms
EXAMPLE 7.9 Electron Configurations
Which of the following electron configurations would correspond to ground states and which to excited states?
a. 1s 2 2s2 2p1
b. 1s2 2p1
c . 1s 2 2s2 2p1 3s1
•Method of Solution
a. In the above configuration electrons occupy the lowest possible energy states. Answer: It corresponds to a
b. In this configuration electrons do not occupy the lowest possible energy states. The 2s orbital which lies
lower than the 2p is vacant, and the last electron is in the 2p orbital. Answer: This is an excited state.
c. In this case the last electron is in the 3s orbital while the lower energy 2p subshell is not completely filled.
Answer: This is an excited state.
EXAMPLE 7.10 Electron Configurations
Write the electron configuration for a potassium atom.
•Method of Solution
Potassium has 19 electrons. The first 10 of these will occupy the same orbitals as neon which is given above:
1s2 2s2 2p6 .
This leaves 9 electrons to account for. Keeping in mind the order of increasing orbital energies, the next 2
electrons fill the 3s orbital. Then 6 electrons can be placed into the 3p subshell. This leaves 1 electron.
According to the order of orbital energies, the 3d does not fill next. Rather, the 4s orbital is lower in energy than
the 3d. The last electron in potassium enters the 4s orbital. The electron configuration of potassium is:
K 1s 2 2s2 2p6 3s2 3p6 4s1
or in terms of the argon core abbreviation:
EXAMPLE 7.11 Orbital Diagrams
c. Write the electron configuration for arsenic.
Draw its orbital diagram.
Are As atoms diamagnetic or paramagnetic? •Method of Solution
a. From the atomic number of As we see there are 33 electrons per arsenic atom. From Figure 7.29 in the
text, the order of filling orbitals is 1s, 2s, 2p, 3s, 3p, 4s, etc. Placing electrons in the lowest energy
orbitals until they are filled, we find that the first 18 electrons are arranged 1s2 2s2 2p6 3s2 3p6 , which
corresponds to an Ar core. The next two enter the 4s, and the next ten enter the 3d. This leaves 3 electrons
for the 4p subshell. Answer: The electron configuration of As is:
As [Ar]4s2 3d10 4p3 b. Back All of the orbitals are filled except for the 4p orbitals. The electrons must be placed into the 4p orbitals in
accordance with Hund's rule. The orbital diagram for the outer orbitals is: Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Quantum Theory and the Electronic Structure of Atoms / 1 41 As [↑↓ ]
4s [↑↓ ]↑↓ ]↑↓ ]↑↓ ]↑↓ ] [↑ ] ↑ ] ↑ ] 3d 4p c. Answer: Arsenic atoms are paramagnetic because they contain 3 unpaired electrons.
18. What element has atoms with the electron configuration [Xe]6s2 4f14 5d10 6p2 ?
19. What third period element has atoms in the ground state with three unpaired electrons?
20. Write the electron configuration for the following atoms: Sb, V, Pb.
21. How many unpaired electrons do oxygen atoms have?
_______________________________________________________________________________ CONCEPTUAL QUESTIONS
1. Briefly describe Bohr's theory of the hydrogen atom and discuss how the theory explains the production of
an emission line spectrum. 2. How is the concept of electron density used to describe the position of an electron in an atom according to
quantum mechanics? 3. Describe in what way Bohr's model of the hydrogen atom violates Heisenberg's uncertainty principle. 4. Which of the following are observable.
a. the position of an electron in an H atom.
b. the wavelength of light emitted by H atoms.
c. the nucleus of an atom.
d. an s orbital 5. Is the electron density distributed evenly inside an s orbital? 6. In many-electron atoms the 2s orbital has a lower energy than the 2p. Explain. PRACTICE TEST
1. A certain AM radio station broadcasts at a frequency of 600 kHz. What is the wavelength of these radio
waves in meters?
2. How long would it take a radio wave of frequency 5.5 × 10 5 Hz to travel from the planet Venus to Earth
(28 million miles)?
3. Calculate the energy required to remove an electron from a hydrogen atom in the ground state.
4. Which of the following has the greatest ionization energy?
b. He+ c . Li 2+ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 42 / Quantum Theory and the Electronic Structure of Atoms
5. The average kinetic energy of a neutron at 25°C is 6.2 × 10 –21 J. What is its de Broglie wavelength? The
mass of a neutron is 1.008 amu. Hint: kinetic energy = 1 /2 mu 2 = (mu)2 /2m.
6. Deduce the possible sets of four quantum numbers for an electron that arise when n = 2.
7. What is the maximum number of electrons that can occupy an n = 3 energy level? The n = 4 energy level?
8. Which of the following subshells has a capacity of 10 electrons?
a. 5s b. 2p c. 4p d. 3d e. 6s
9. How many orbitals are occupied by one or more electrons in a germanium atom?
10. Complete the sentence: An electron with l = 2 must
a. have m l = –2.
b. be in an n = 3 energy level
c. be in a p orbital
d. be in a d orbital
11. Write quantum numbers for:
a. an electron in a 2s orbital
b. the outermost electrons in Ge
12. Write the electron configuration for the following atoms: Ar, Se, Ag.
13. How many electrons are represented by the abbreviation [Kr]?
14. Which of the following is the correct orbital diagram for chromium? (b) [↑↓ ]↑↓ ] ] ] ] [↑↓ ] 3d (a) 4s [↑ ]↑ ]↑ ]↑ ] ] [↑↓ ] 3d 4s (c) [↑↓ ]↑↓ ]↑↓ ]↑↓ ] ]
3d (d) 4s [↑ ]↑ ]↑ ]↑ ]↑ ]
3d [↑ ]
4s 15. Which of the following atoms has the greatest number of unpaired electrons?
b. Ag c. O
16. If the energy required to ionize 1 mole of H atoms were used to raise the temperature of water, what mass of
water could have its temperature increased by 50°C?
17. If 5 percent of the energy supplied to an incandescent light bulb is radiated as visible light, how many
"visible" photons per second are emitted by a 100 watt bulb? Assume the wavelength of all visible light to
be 560 nm. Given: 1 watt = 1 J/s.
18. The light-sensitive compound in most photographic films is silver bromide (AgBr). Assume, when film is
exposed, that the light energy absorbed dissociates the molecule into atoms. (The actual process is more
complex.) If the energy of dissociation of AgBr is 100 kJ/mol, find the wavelength of light that is just
able to dissociate AgBr. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Quantum Theory and the Electronic Structure of Atoms / 1 43
21. 600 nm
4.0 × 10 14 Hz
4.11 × 10 –19 J
2.97 × 10 –19 J/photon, 179 kJ/mol
9.70 × 10 –7 m
2.62 × 10 –19 J
a. 6 b. n = 4 to n = 1 c. n = 4 to n = 3
95.2 nm, ultraviolet
1.32 × 10 –14 m
a. 1 b. 5 c. 3 d. 9
a. 3d, m l = –2, –1, 0, 1, 2. 5 orbitals
b. 4f, ml = –3, –2, –1, 0, 1, 2, 3. 7 orbitals
c. 5p, ml = –1, 0, 1. 3 orbitals
a. n = 2 l = 0 b. n = 3 l = 1 c. n = 4 l = 3
Sb [Kr]5s2 4d10 5p3 ; V [Ar]4s 2 3d3 ; Pb [Xe]6s2 4f14 5d10 6p2
2 Conceptual Questions
1. 2. 3. 4.
5. Back See Section 7.3 of the text for details. Essentially each orbit in the H atom has a certain energy associated
with it. The emission of radiation by an excited H atom is explained by the electron dropping from a higher
to a lower energy orbit. Since only a few orbits are permitted, then only a few transitions are possible. Each
transition produces one line in the spectrum. With the number of transitions restricted, one observes a line
spectrum rather than a continuous spectrum.
Electron density gives the probability that an electron will be found in a particular region of an atom.
Regions of high electron density represent a high probability of locating the electron, whereas regions of
low electron density represent a low probability of finding the electron there. Figure 7.23 of the text shows
the plot of electron density in the hydrogen 1s orbital. The darkness of a point is proportional to the
probability of finding the electron at that position.
According to the uncertainty principle, ∆ x ∆ p > h. This means that if the uncertainty in the momentum of
an electron ∆ p is small, then the uncertainty in the position of the electron ∆ x will be large. Momentum is
related to energy, and for the H atom the principle energy levels are known accurately. According to the
uncertainty principle, the position of the electron which is its distance from the nucleus must have a large
uncertainty. It is not appropriate to assume that the electron is moving around the nucleus in a well defined
Only b. The wavelength of light emitted by H atoms can be observed.
No, in the 1s orbital the electron density is more dense near the nucleus. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 44 / Quantum Theory and the Electronic Structure of Atoms
6. In the hydrogen atom the 2s and 2p orbitals have the same energy. The energy of an electron in a manyelectron atom depends not only on its principle quantum number (n) but also on its angular momentum
quantum number (l). Electrons are assigned to subshells in order of increasing "n + l" value. For the 2s
subshell n + l = 2, while for the 2p subshell n + l = 3. If two subshells have the same value of n + l ,
then electrons are assigned to the one with the lower n value first. Practice Test
11. 12. 13.
18. 500 m
2.18 × 10 –18 J
λ = 95.2 nm; ultraviolet
a. n = 2, l = 0, ml = 0, ms = 1 /2
b. n = 4, l = 0, ml = 0, ms = 1 /2
n = 4, l = 0, ml = 0, ms = – 1/2
n = 4, l = 1, ml = -1, ms = 1 /2
n = 4, l = 1, ml = 0, ms = 1 /2
Ar 1s2 2s2 2p6 3s2 3p6
Se [Ar]4s2 3d10 4p4
Ag [Kr]5s1 4d10
1.4 × 10 19 photons/s
1200 nm _____________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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This note was uploaded on 09/15/2009 for the course CHEM 102 taught by Professor Bastos during the Spring '08 term at Adelphi.
- Spring '08