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Chapter 6
Electronic
Structure of Atoms
Electronic
Structure
of Atoms
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Electronic Structure of Atoms
Laser: step-like energy transition
6.1 The Wave Nature of Light
6.2 Quantized Energy and Photons
6.3 Line Spectra and the Bohr Model
6.4 The Wave Behavior of Matter
6.5 Quantum Mechanics and Atomic Orbitals
6.6 Representations of Orbitals
6.7 Many-Electron Atoms
6.8 Electron Configurations
6.9 Electron Configurations and the Periodic Table
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Electronic Structure
• This chapter is all about electronic structure—the
arrangement and energy of electrons.
• It may seem odd to start by talking about waves.
However, extremely small particles have properties
that can only be explained in this manner!
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Waves
• To understand the electronic structure of atoms, one must
understand the nature of electromagnetic radiation.
• The distance between corresponding points on adjacent
Electronic
waves is the wavelength ().
Structure
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Waves
• The number of waves passing a
given point per unit of time is the
frequency ().
• For waves traveling at the same
velocity, the longer the
wavelength, the smaller the
frequency.
• If the time associated with the
lines to the left is one second,
then the frequencies would be 2
s–1 and 4 s–1, respectively.
• s–1: hertz (Hz)
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Electromagnetic Radiation
• All electromagnetic radiation travels at the same
velocity: The speed of light (c) is 2.998  108 m/s.
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c = 
Electronic
Structure
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Sample Exercise 6.1 Concepts of Wavelength and Frequency
Two electromagnetic waves are represented in the margin. (a) Which wave has
the higher frequency? (b) If one wave represents visible light and the other
represents infrared radiation, which wave is which?
Solution
(a) Wave 1 has a longer wavelength (greater distance between peaks). The longer the wavelength, the lower the
frequency (ν = c ⁄ λ). Thus, Wave 1 has the lower frequency, and Wave 2 has the higher frequency.
(b) The electromagnetic spectrum (Figure 6.4) indicates that infrared radiation has a longer wavelength than visible
light. Thus, Wave 1 would be the infrared radiation.
Chemistry: The Central Science, 13th Edition
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Sample Exercise 6.2 Calculating Frequency from Wavelength
The yellow light given off by a sodium vapor lamp used for public lighting has a wavelength of 589 nm. What is
the frequency of this radiation?
Solution
Analyze We are given the wavelength, λ, of the radiation and asked to calculate its frequency, ν.
Plan The relationship between the wavelength and the frequency is given by Equation 6.1. We can solve for ν and
use the values of λ and c to obtain a numerical answer. (The speed of light, c, is 3.00 × 108 m ⁄ s to three significant
figures.)
Solve Solving Equation 6.1 for frequency gives (ν = c ⁄ λ). When we insert the values for c and λ, we note that the
units of length in these two quantities are different. We can convert the Wavelength from nanometers to meters, so
the units cancel:
Check The high frequency is reasonable because of the short wavelength. The units are proper because frequency
has units of “per second,” or s–1.
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The Nature of Energy
The wave nature of light
does not explain how an
object can glow when its
temperature increases.
E = h, 2h, 3h
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The Nature of Energy—Quanta
Max Planck explained it by
assuming that energy comes
in packets called quanta
(singular: quantum).
E = h, 2h, 3h
1918 Nobel Prize in Physics
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The Photoelectric Effect
• Einstein used quanta to explain the
photoelectric effect.
• Each metal has a different energy
at which it ejects electrons (work
function). At lower energy,
electrons are not emitted.
• He concluded that energy is
proportional to frequency:
E = h
where h is Planck’s constant,
6.626  10−34 J·s.
• 1921 Nobel Prize in Physics
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Electronic
Structure
of Atoms
Sample Exercise 6.3 Energy of a Photon
Calculate the energy of one photon of yellow light that has a wavelength of 589 nm.
Solution
Analyze Our task is to calculate the energy, E, of a photon, given λ = 589 nm.
Plan We can use Equation 6.1 to convert the wavelength to frequency:
ν=c⁄λ
We can then use Equation 6.3 to calculate energy:
E = hν
Solve The frequency, ν, is calculated from the given wavelength, as shown in Sample Exercise 6.2:
ν = c ⁄ λ = 5.09 × 1014 s–1
The value of Planck constant, h, is given both in the text and in the table of physical constants on the inside back
cover of the text, and so we can easily calculate E:
E = (6.626 × 10–34 J-s)(5.09 × 1014 s–1) = 3.37 × 10–19 J
Comment If one photon of radiant energy supplies 3.37 × 10–19 J, we calculate that one mole of these photons will
supply:
(6.02 × 1023 photons ⁄ mol)(3.37 × 10–19 J ⁄ photon) = 2.03 × 105 J ⁄ mol
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Atomic Emissions
Another mystery in the early twentieth century involved
the emission spectra observed from energy emitted by
atoms and molecules.
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Continuous vs. Line Spectra
• For atoms and molecules,
one does not observe a
continuous spectrum (the
“rainbow”), as one gets
from a white light source.
• Only a line spectrum of
discrete wavelengths is
observed. Each element
has a unique line spectrum.
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The Hydrogen Spectrum
• Johann Balmer (1885) discovered a simple formula
relating the four lines to integers.
• Johannes Rydberg advanced this formula.
• Neils Bohr explained why this mathematical relationship
works.
Electronic
• 1922 Nobel Prize in Physics
Structure
of Atoms
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The Bohr Model
Niels Bohr adopted Planck’s assumption and explained these
phenomena in this way:
1. Only orbits of certain radii, corresponding to certain specific
energies, are permitted for the electron in a hydrogen atom.
2. An electron in a permitted orbit is in an “allowed” energy state. An
electron in an allowed energy state does not radiate energy and,
therefore, does not spiral into the nucleus.
3. Energy is emitted or absorbed by the electron only as the electron
changes from one allowed energy state to another. This energy is
emitted or absorbed as a photon that has energy
E = h
Electronic
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The Bohr Model
The energy absorbed or emitted from
the process of electron promotion or
demotion can be calculated by the
equation
1
E = −hcRH n2
Excited state
E = −hcRH (
1
1
– 2
2
nf
ni
)
where RH is the Rydberg constant,
1.096776  107 m−1, and ni and nf are
the initial and final energy levels of the
electron.
n: principal quantum number
Ground state
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Electronic
Structure
of Atoms
Sample Exercise 6.4 Electronic Transitions in the Hydrogen Atom
Using Figure 6.12, predict which of these electronic transitions
produces the spectral line having the longest wavelength: n = 2
to n = 1, n = 3 to n = 2, or n = 4 to n = 3.
E = −hcRH (
1
1
–
nf2
ni2
)
Solution
The wavelength increases as frequency decreases (λ = c ⁄ ν). Hence, the longest wavelength will be associated with
the lowest frequency. According to Planck’s equation, E = hv, the lowest frequency is associated with the lowest
energy. In Figure 6.12 the shortest vertical line represents the smallest energy change. Thus, the n = 4 to n = 3
transition produces the longest wavelength (lowest frequency) line.
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Important Ideas from the Bohr Model
 Points that are incorporated into the current atomic model
include the following:
1. Electrons exist only in certain discrete energy levels.
2. Energy is involved in the transition of an electron from one
level to another.
Limitations
• It only works for hydrogen.
• Classical physics would result in an electron falling into the
positively charged nucleus. Bohr simply assumed it would
not!
Electronic
• Circular motion is not wave-like in nature.
Structure
of Atoms
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The Wave Nature of Matter
• Louis de Broglie theorized that if
light can have material properties,
matter should exhibit wave properties
(matter waves).
• He demonstrated that the relationship
between mass and wavelength was
The wave nature of light
is used to produce this
electron micrograph.
h
 = mv
p
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Electronic
Structure
of Atoms
Sample Exercise 6.5 Matter Waves
What is the wavelength of an electron moving with a speed of 5.97 × 106 m ⁄ s? The mass of the electron
is 9.11 × 10–31 kg.
Solution
Analyze We are given the mass, m, and velocity, v, of the electron, and we must calculate its de Broglie wavelength, λ.
Plan The wavelength of a moving particle is given by Equation 6.8, so λ is calculated by inserting the known quantities
h, m, and v. In doing so, however, we must pay attention to units.
Solve Using the value of Planck constant:
h = 6.626 × 10–34 J-s
we have the following:
Chemistry: The Central Science, 13th Edition
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The Uncertainty Principle
Heisenberg showed that
the more precisely the
momentum of a particle is
known, the less precisely
is its position is known:
h
(x) (mv) 
4
Electron (5 x 106 m/s, 1% uncertainty)
Electronic
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Quantum Mechanics
• Erwin Schrödinger developed a
mathematical treatment into which both
the wave and particle nature of matter
could be incorporated.
• This is known as quantum mechanics.
• The solution of Schrödinger’s wave
equation is designated with a
lowercase Greek psi ().
• The square of the wave equation, 2,
gives the electron density, or
probability of where an electron is likely
to be at any given time.
Electronic
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Quantum Numbers
• Solving the wave equation gives a set of wave
functions, or orbitals, and their corresponding
energies.
• Each orbital describes a spatial distribution of
electron density.
• An orbital is described by a set of three quantum
numbers (n, l, ml).
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Principal Quantum Number (n)
• The principal quantum number, n, describes the
energy level on which the orbital resides.
• The values of n are integers ≥ 1.
• These correspond to the values in the Bohr model.
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Angular Momentum Quantum Number (l)
• This quantum number defines the shape of the orbital.
• Allowed values of l are integers ranging from 0 to n − 1.
• We use letter designations to communicate the different
values of l and, therefore, the shapes and types of
orbitals.
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Magnetic Quantum Number (ml)
• The magnetic quantum number describes the threedimensional orientation of the orbital.
• Allowed values of ml are integers ranging from −l to l:
−l ≤ ml ≤ l
• Therefore, on any given energy level, there can be up
to 1 s orbital, 3 p orbitals, 5 d orbitals, 7 f orbitals, and
so forth.
Electronic
Structure
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Shell and Subshell
1.
2.
3.
4.
Orbitals with the same value of n form an electron shell: different
orbital types within a shell are subshells.
The shell with principal quantum number n consists of exactly n
subshells.
Each subshell consists of a specific number of orbitals (−l ≤ ml ≤ l).
The total number of orbitals in a shell is n2.
Electronic
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Sample Exercise 6.6 Subshells of the Hydrogen Atom
(a) Without referring to Table 6.2, predict the number of subshells in the fourth shell, that is, for n = 4. (b) Give the
label for each of these subshells. (c) How many orbitals are in each of these subshells?
Solution
Analyze and Plan We are given the value of the principal quantum number, n. We need to determine the allowed
values of l and ml for this given value of n and then count the number of orbitals in each subshell.
Solve There are four subshells in the fourth shell, corresponding to the four possible values of l (0, 1, 2, and 3).
These subshells are labeled 4s, 4p, 4d, and 4f. The number given in the designation of a subshell is the principal
quantum number, n; the letter designates the value of the angular momentum quantum number, l: for l = 0, s; for l = 1,
p; for l = 2, d; for l = 3, f.
There is one 4s orbital (when l = 0, there is only one possible value of ml: 0). There are three 4p orbitals (when
l = 1, there are three possible values of ml: 1, 0, −1). There are five 4d orbitals (when l = 2, there are five allowed
values of ml: 2, 1, 0, −1, −2). There are seven 4f orbitals (when l = 3, there are seven permitted values of ml: 3, 2, 1, 0,
−1, −2, −3).
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s Orbitals
• The value of l for s orbitals is 0.
• They are spherical in shape.
• The radius of the sphere increases with the value of n.
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s Orbitals
• For an ns orbital, the
number of peaks is n.
• For an ns orbital, the
number of nodes (where
there is zero probability of
finding an electron) is n – 1.
• As n increases, the electron
density is more spread out
and there is a greater
probability of finding an
electron further from the
nucleus.
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p Orbitals
• The value of l for p orbitals is 1.
• ml : -1, 0, 1
• They have two lobes with a node between them.
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d Orbitals
• The value of l for a d orbital is 2.
• ml : -2, -1, 0, 1, 2
• Four of the five d orbitals have four lobes; the other Electronic
Structure
resembles a p orbital with a doughnut around the center.
of Atoms
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f Orbitals
•
•
•
•
Very complicated shapes
Seven equivalent orbitals in a sublevel
l=3
ml : -3, -2, -1, 0, 1, 2, 3
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Electronic
Structure
of Atoms
Energies of Orbitals—Hydrogen
• For a one-electron
hydrogen atom, orbitals
on the same energy
level have the same
energy.
• Chemists call them
degenerate orbitals.
Electronic
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Energies of Orbitals—Many-electron Atoms
• As the number of electrons
increases, so does the
repulsion between them.
• Therefore, in atoms with more
than one electron, not all
orbitals on the same energy
level are degenerate.
• Orbital sets in the same
sublevel are still degenerate.
• Energy levels start to overlap
in energy (e.g., 4s is lower
in energy than 3d.)
Electronic
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Spin Magnetic Quantum Number, ms
• In the 1920s, it was discovered that two
electrons in the same orbital do not have
exactly the same energy.
• The “spin” of an electron describes its
magnetic field, which affects its energy.
• This led to the spin quantum number, ms.
• The spin quantum number has only two
allowed values, +½ and –½.
Electronic
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Pauli Exclusion Principle
• No two electrons in the same atom can have exactly
the same energy.
• Therefore, no two electrons in the same atom can
have identical sets of quantum numbers.
• This means that every electron in an atom must
differ by at least one of the four quantum number
values: n, l, ml, and ms.
• An orbital can hold a maximum of two electrons and
they must have opposite spins.
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Electron Configurations
5
4p
• The way electrons are distributed in an atom
is called its electron configuration.
• The most stable organization is the lowest
possible energy, called the ground state.
• Each component consists of
– a number denoting the energy level;
– a letter denoting the type of orbital;
– a superscript denoting the number of
electrons in those orbitals.
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Orbital Diagrams
• Each box in the diagram
represents one orbital.
• Half-arrows represent the
electrons.
• The direction of the arrow
represents the relative spin
of the electron.
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Hund’s Rule
“For degenerate orbitals, the lowest energy is attained when
the number of electrons with the same spin is maximized.”
 This means that, for a set of orbitals in the same sublevel, there
must be one electron in each orbital before pairing and the Electronic
Structure
electrons have the same spin, as much as possible.
of Atoms
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Sample Exercise 6.7 Orbital Diagrams and Electron Configurations
Draw the orbital diagram for the electron configuration of oxygen, atomic number 8. How many unpaired electrons
does an oxygen atom possess?
Solution
Analyze and Plan Because oxygen has an atomic number of 8,
each oxygen atom has eight electrons. Figure 6.25 shows the ordering
of orbitals. The electrons (represented as half arrows) are placed in
the orbitals (represented as boxes) beginning with the lowest-energy
orbital, the 1s. Each orbital can hold a maximum of two electrons
(the Pauli exclusion principle). Because the 2p orbitals are degenerate,
we place one electron in each of these orbitals (spin-up) before
pairing any electrons (Hund’s rule).
Solve Two electrons each go into the 1s and 2s orbitals with their
spins paired. This leaves four electrons for the three degenerate
2p orbitals. Following Hund’s rule, we put one electron into each
2p orbital until all three orbitals have one electron each. The fourth
electron is then paired up with one of the three electrons already in
a 2p orbital, so that the orbital diagram is
The corresponding electron configuration is written 1s22s22p4. The atom has two unpaired electrons.
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Condensed Electron Configurations
• Elements in the same group of the
periodic table have the same number of
electrons in the outer most shell. These
are the valence electrons.
• The filled inner shell electrons are called
core electrons. These include
completely filled d or f sublevels.
• We write a shortened version of an
electron configuration using brackets
around a noble gas symbol and listing
only valence electrons.
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Outer-Shell and Valence Electrons
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Periodic Table
• We fill orbitals in increasing order of energy.
• Different blocks on the periodic table correspond to
different types of orbitals: s = blue, p = pink (main-group
elements), d = orange (transition elements), f = tan
(lanthanides and actinides, or inner transition elements)
Electronic
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Se: [Ar]3d104s24p4
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Sample Exercise 6.8 Electron Configurations for a Group
What is the characteristic valence electron configuration of the group 7A elements, the halogens?
Solution
Analyze and Plan We first locate the halogens in the periodic table, write the electron configurations for the first two
elements, and then determine the general similarity between the configurations.
Solve The first member of the halogen group is fluorine (F, element 9). Moving backward from F, we find that the
noble-gas core is [He]. Moving from He to the element of next higher atomic number brings us to Li, element 3.
Because Li is in the second period of the s block, we add electrons to the 2s subshell. Moving across this block gives
2s2. Continuing to move to the right, we enter the p block. Counting the squares to F gives 2p5. Thus, the condensed
electron configuration for fluorine is
F:
[He]2s22p5
The electron configuration for chlorine, the second halogen, is
Cl:
[Ne]3s23p5
From these two examples, we see that the characteristic valence electron configuration of a halogen is ns2np5, where n
ranges from 2 in the case of fluorine to 6 in the case of astatine.
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Sample Exercise 6.9 Electron Configurations from the Periodic Table
(a) Based on its position in the periodic table, write the condensed electron configuration for bismuth, element 83.
(b) How many unpaired electrons does a bismuth atom have?
Solution
(a) Our first step is to write the noble-gas core. We do this by locating bismuth, element 83, in the periodic table. We
then move backward to the nearest noble gas, which is Xe, element 54. Thus, the noble-gas core is [Xe].
Next, we trace the path in order of increasing atomic numbers from Xe to Bi. Moving from Xe to Cs,
element 55, we find ourselves in period 6 of the s block. Knowing the block and the period identifies the subshell in
which we begin placing outer electrons, 6s. As we move through the s block, we add two electrons: 6s2.
As we move beyond the s block, from element 56 to element 57, the curved arrow below the periodic table
reminds us that we are entering the f block. The first row of the f block corresponds to the 4f subshell. As we move
across this block, we add 14 electrons: 4f 14.
With element 71, we move into the third row of the d block. Because the first row of the d block is 3d, the
second row is 4d and the third row is 5d. Thus, as we move through the ten elements of the d block, from element 71
to element 80, we fill the 5d subshell with ten electrons: 5d10.
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Sample Exercise 6.9 Electron Configurations from the Periodic Table
Continued
Moving from element 80 to element 81 puts us into the p block in the 6p subshell. (Remember that the principal
quantum number in the p block is the same as that in the s block.) Moving across to Bi requires three electrons: 6p3.
The path we have taken is
Chemistry: The Central Science, 13th Edition
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Sample Exercise 6.9 Electron Configurations from the Periodic Table
Continued
Putting the parts together, we obtain the condensed electron configuration: [Xe]6s24f 145d106p3. This configuration
can also be written with the subshells arranged in order of increasing principal quantum number: [Xe]4f 145d106s26p3.
Finally, we check our result to see if the number of electrons equals the atomic number of Bi, 83: Because Xe
has 54 electrons (its atomic number), we have 54 + 2 + 14 + 10 + 3 = 83. (If we had 14 electrons too few, we would
realize that we have missed the f block.)
(b) We see from the condensed electron configuration that the only partially occupied subshell is 6p. The orbital
diagram representation for this subshell is
In accordance with Hund’s rule, the three 6p electrons occupy the three 6p orbitals singly, with their spins parallel.
Thus, there are three unpaired electrons in the bismuth atom.
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Chromium as an Anomaly
• For instance, the electron configuration for chromium is
[Ar] 4s1 3d5
rather than the expected
[Ar] 4s2 3d4.
• This occurs because the 4s and 3d orbitals are very
close in energy.
• These anomalies occur in f-block atoms with f and d
orbitals, as well.
Electronic
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Some Anomalies
 Some irregularities
occur when there
are enough
electrons to half-fill
s and d orbitals on
a given row.
•
•
•
•
•
•
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Expected
Cr = [Ar]4s23d4
Cu = [Ar]4s23d9
Mo = [Kr]5s24d4
Ru = [Kr]5s24d6
Pd = [Kr]5s24d8
•
•
•
•
•
•
Found Experimentally
Cr = [Ar]4s13d5
Cu = [Ar]4s13d10
Mo = [Kr]5s14d5
Ru = [Kr]5s14d7
Pd = [Kr]5s04d10
Electronic
Structure
of Atoms
Sample Integrative Exercise Putting Concepts Together
Boron, atomic number 5, occurs naturally as two isotopes, 10B and 11B, with natural abundances of 19.9% and 80.1%,
respectively. (a) In what ways do the two isotopes differ from each other? Does the electronic configuration of 10B
differ from that of 11B? (b) Draw the orbital diagram for an atom of 11B. Which electrons are the valence electrons?
(c) Indicate three major ways in which the 1s electrons in boron differ from its 2s electrons. (d) Elemental boron reacts
with fluorine to form BF3, a gas. Write a balanced chemical equation for the reaction of solid boron with fluorine gas.
(e) ΔHf゜ for BF3(g) is −1135.6 kJ ⁄ mol. Calculate the standard enthalpy change in the reaction of boron with fluorine.
(f) Will the mass percentage of F be the same in 10BF3 and 11BF3? If not, why is that the case?
Solution
(a) The two isotopes of boron differ in the number of neutrons in the nucleus.
(Sections 2.3 and 2.4) Each of
the isotopes contains five protons, but 10B contains five neutrons, whereas 11B contains six neutrons. The two
isotopes of boron have identical electron configurations, 1s22s22p1, because each has five electrons.
(b) The complete orbital diagram is
The valence electrons are the ones in the outermost occupied shell, the 2s2 and 2p1 electrons. The 1s2 electrons
constitute the core electrons, which we represent as [He] when we write the condensed electron configuration,
[He]2s22p1.
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Sample Integrative Exercise Putting Concepts Together
Continued
(c) The 1s and 2s orbitals are both spherical, but they differ in three important respects: First, the 1s orbital is lower
in energy than the 2s orbital. Second, the average distance of the 2s electrons from the nucleus is greater than
that of the 1s electrons, so the 1s orbital is smaller than the 2s. Third, the 2s orbital has one node, whereas the 1s
orbital has no nodes (Figure 6.19).
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Sample Integrative Exercise Putting Concepts Together
Continued
(d) The balanced chemical equation is
2 B(s) + 3 F2(g) → 2 BF3(g)
(e) ΔH ゜ = 2(−1135.6) − [0 + 0] = −2271.2 kJ. The reaction is strongly exothermic.
(f) As we saw in Equation 3.10 (Section 3.3), the mass percentage of an element in a substance depends on the formula
weight of the substance. The formula weights of 10BF3 and 11BF3 are different because of the difference in the
masses of the two isotopes (the isotope masses of 10B and 11B are 10.01294 and 11.00931 amu, respectively). The
denominators in Equation 3.10 would therefore be different for the two isotopes, whereas the numerators would
remain the same.
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