Download Byron High School

Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chapter 6
Electronic Structure
of Atoms
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.
Electronic
Structure
of Atoms
Light & Waves
• Light: Crash Course Astronomy #24
Watch this video as an introduction to this
unit.
Light, Energy, Wavelength & Spectroscopy
Electronic
Structure
of Atoms
Waves
• To understand the electronic structure of
atoms, one must understand the nature of
electromagnetic radiation.
• The distance between corresponding points
on adjacent waves is the wavelength () and
Electronic
is measured in meters (m).
Structure
of Atoms
Waves
• The number of waves
passing a given point per
unit of time is the
frequency () and is
measured in hertz (Hz)
or “per second” (s-1).
• For waves traveling at
the same velocity, the
longer the wavelength,
the smaller the
Electronic
Structure
frequency.
of Atoms
Electromagnetic Radiation
• All electromagnetic radiation travels
at the same velocity: the speed of
light (c), 3.00  108 m/s.
• Therefore,
c = 
Electronic
Structure
of Atoms
SAMPLE EXERCISE 6.1 Concepts of Wavelength and Frequency
Two electromagnetic waves are represented in the image below. (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) The lower wave has a longer wavelength (greater distance
between peaks). The longer the wavelength, the lower the frequency ( = c/ ).
Thus, the lower wave has the lower frequency, and the upper one has the
higher frequency.
(b) The electromagnetic spectrum (Figure 6.4) indicates that infrared
radiation has a longer wavelength than visible light. Thus, the lower wave
would be the infrared radiation.
Electronic
Structure
of Atoms
SAMPLE EXERCISE 6.1 Concepts of Wavelength and Frequency
PRACTICE EXERCISE
If one of the waves in the image above represents blue light and the other red
light, which is which?
Answer: The expanded visible-light portion of Figure 6.4 tells you that red
light has a longer wavelength than blue light. The lower wave has the longer
wavelength (lower frequency) and would be the red light.
Electronic
Structure
of Atoms
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?
Solve: Solving Equation 6.1 for frequency gives  = c/ . When we insert the
values for c and , 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:
PRACTICE EXERCISE
(a) A laser used in eye surgery to fuse detached retinas produces radiation with
a wavelength of 640.0 nm. Calculate the frequency of this radiation. (b) An
FM radio station broadcasts electromagnetic radiation at a frequency of 103.4
MHz (megahertz; MHz = 106 s–1). Calculate the wavelength of this radiation.
Answers: (a) 4.688 1014 s–1, (b) 2.901 m
Electronic
Structure
of Atoms
The Nature of Energy
• The wave nature of light
does not explain how
an object can glow
when its temperature
increases.
• Max Planck explained it
by assuming that
energy comes in
packets called quanta.
Electronic
Structure
of Atoms
The Nature of Energy
• Einstein used this
assumption to explain the
photoelectric effect.
• He concluded that energy
is proportional to
frequency:
E = h
where h is Planck’s
constant, 6.63  10−34 J-s.
Electronic
Structure
of Atoms
The Nature of Energy
• Therefore, if one knows the
wavelength of light, one
can calculate the energy in
one photon, or packet, of
that light:
c = 
E = h
Electronic
Structure
of Atoms
SAMPLE EXERCISE 6.3 Energy of a Photon
Calculate the energy of one photon of yellow light whose wavelength is 589 nm.
Analyze: Our task is to calculate the energy, E, of a photon, given 589 nm.
Plan: We can use Equation 6.1 to convert the wavelength to frequency:
We can then use Equation 6.3 to calculate energy:
Solve: The frequency, , is calculated from the given wavelength, as shown in
Sample Exercise 6.2:
The value of Planck’s constant, h, is given both in the text and in the table of
physical constants on the inside front cover of the text, and so we can easily
calculate E:
Comment: If one photon of radiant energy supplies 3.37  10–19J, then one
mole of these photons will supply
This is the magnitude of enthalpies of reactions (Section 5.4), so
radiation can break chemical bonds, producing what are called
photochemical reactions.
Electronic
Structure
of Atoms
Do Problem Sets
Electronic
Structure
of Atoms
SAMPLE EXERCISE 6.3 continued
PRACTICE EXERCISE
(a) A laser emits light with a frequency of 4.69  1014 s–1. What is the energy
of one photon of the radiation from this laser? (b) If the laser emits a pulse of
energy containing 5.0  1017 photons of this radiation, what is the total energy
of that pulse? (c) If the laser emits 1.3  10–2 J of energy during a pulse, how
many photons are emitted during the pulse?
Answers: (a) 3.11  10–19 J, (b) 0.16 J, (c) 4.2  1016 photons
Electronic
Structure
of Atoms
The Nature of Energy
• Another mystery involved
the emission spectra
observed from energy
emitted by atoms and
molecules.
Electronic
Structure
of Atoms
The Nature of Energy
• One does not observe
a continuous
spectrum, as one gets
from a white light
source.
• Only a line spectrum of
discrete wavelengths
is observed.
• Watch this video: Spectral Lines Video
• Demonstrate Spectral Tubes in the back store room
Electronic
Structure
of Atoms
The Nature of Energy
•
Niels Bohr adopted Planck’s
assumption and explained
these phenomena in this
way:
1. Electrons in an atom can only
occupy certain orbits
(corresponding to certain
energies).
Bohr Atom Video
link for review
Electronic
Structure
of Atoms
The Nature of Energy
•
Niels Bohr adopted Planck’s
assumption and explained
these phenomena in this
way:
1. Electrons in an atom can only
occupy certain orbits
(corresponding to certain
energies).
Electronic
Structure
of Atoms
The Nature of Energy
•
Niels Bohr adopted Planck’s
assumption and explained
these phenomena in this
way:
2. Electrons in permitted orbits
have specific, “allowed”
energies; these energies will
not be radiated from the atom.
Electronic
Structure
of Atoms
The Nature of Energy
•
Niels Bohr adopted
Planck’s assumption and
explained these
phenomena in this way:
3. Energy is only absorbed or
emitted in such a way as to
move an electron from one
“allowed” energy state to
another; the energy is
defined by
E = h
Electronic
Structure
of Atoms
The Nature of Energy
The energy absorbed or emitted
from the process of electron
promotion or demotion can be
calculated by the equation:
E = −RH (
1
1
- 2
nf2
ni
)
where RH is the Rydberg
constant, 2.18  10−18 J, and ni
and nf are the initial and final
energy levels of the electron. Electronic
Structure
of Atoms
The Nature of Energy
E = −RH (
1
1
nf2
ni2
)
• Electrons can go from
the 2nd level to the 3rd or
4th .
• Electrons can go from
the 1st energy level to
the 2nd, 3rd or even 6th.
Electronic
Structure
of Atoms
The Nature of Energy
The energy absorbed or emitted
from the process of electron
promotion or demotion can be
calculated by the equation:
E = −RH (
1
1
- 2
nf2
ni
)
Note how “n” correlates to
the number in front of the
Electronic
equation; 1
Structure
nf2
of Atoms
SAMPLE EXERCISE 6.4 Electronic Transitions in the Hydrogen Atom
Predict which of the following
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.
Solution The wavelength increases
as frequency decreases ( = c/.
Hence the longest wavelength will
be associated with the lowest
frequency. According to Planck’s
equation, E = h, the lowest
frequency is associated with the
lowest energy. In Figure 6.13 the
shortest vertical line represents the
smallest energy change. Thus, the n
= 4 to n = 3 transition produces the
longest wavelength (lowest
frequency) line.
Electronic
Structure
of Atoms
SAMPLE EXERCISE 6.4 Electronic Transitions in the Hydrogen Atom
PRACTICE EXERCISE
Indicate whether each of
the following electronic
transitions emits energy or
requires the absorption of
energy: (a) n = 3 to n = 1;
(b) n = 2 to n = 4 .
Answers: (a) emits
energy, (b) requires
absorption of energy
Electronic
Structure
of Atoms
Do Problem Sets
Electronic
Structure
of Atoms
The Wave Nature of Matter
• Louis de Broglie posited that if light can
have material properties, matter should
exhibit wave properties.
• He demonstrated that the relationship
between mass and wavelength was
h
 = mv
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  10–28 g.)
Solve: Using the value of Planck’s constant,
and recalling that
we have the following:
Comment: By comparing this value with the wavelengths of electromagnetic
Electronic
radiation shown in Figure 6.4, we see that the wavelength of this electron is about theStructure
of Atoms
same as that of X rays.
SAMPLE EXERCISE 6.5 continued
PRACTICE EXERCISE
Calculate the velocity of a neutron whose de Broglie wavelength is 500 pm.
The mass of a neutron is given in the table on the back inside cover of the text.
Answer: 7.92 102 m/s
Electronic
Structure
of Atoms
The Uncertainty Principle
• Heisenberg showed that the more precisely
the momentum of a particle is known, the less
precisely is its position known:
(x) (mv) 
h
4
• In many cases, our uncertainty of the
whereabouts of an electron is greater than the
size of the atom itself!
Electronic
Structure
of Atoms
The Uncertainty Calculation
(x) 
h
4 m (v)
v = velocity x uncertainty
• Use this & page 211 to calculate the
uncertainty for problem sets #’s 37 & 38.
• Here is a video that will help with your
calculation:
• Uncertainty Calculation video
 Stop watching at the 9:50 minute mark.
Electronic
Structure
of Atoms
Quantum Mechanics
• Erwin Schrödinger
developed a
mathematical treatment
into which both the
wave and particle nature
of matter could be
incorporated.
• It is known as quantum
mechanics.
Electronic
Structure
of Atoms
Quantum Mechanics
• Erwin Schrödinger
developed a
mathematical treatment
into which both the
wave and particle nature
of matter could be
incorporated.
• It is known as quantum
mechanics.
Electronic
Structure
of Atoms
Quantum Mechanics
• The wave equation is
designated with a lower
case Greek psi ().
• The square of the wave
equation, 2, gives a
probability density map of
where an electron has a
certain statistical likelihood
of being at any given instant
in time.
Electronic
Structure
of Atoms
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.
Electronic
Structure
of Atoms
Principal Quantum Number, n
• The principal quantum number, n,
describes the energy level on which the
orbital resides.
• The values of n are integers ≥ 0.
Electronic
Structure
of Atoms
Azimuthal Quantum Number, l
• Also called “sublevels”; s, p, d, f
• 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.
Electronic
Structure
of Atoms
Azimuthal Quantum Number, l
Value of l
0
1
2
3
Type of orbital
s
p
d
f
MEMORIZE THESE VALUES!!!
Electronic
Structure
of Atoms
Do Problem Sets
Electronic
Structure
of Atoms
Review:
Principal Quantum Number, n
• The principal quantum number, n,
describes the energy level on which the
orbital resides.
• The values of n are integers ≥ 0.
Electronic
Structure
of Atoms
Azimuthal Quantum Number, l
• Also called “sublevels”; s, p, d, f
• 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.
Electronic
Structure
of Atoms
Azimuthal Quantum Number, l
Value of l
0
1
2
3
Type of orbital
s
p
d
f
MEMORIZE THESE VALUES!!!
Electronic
Structure
of Atoms
Magnetic Quantum Number, ml
• Psst: That is read as “m sub ell”
• We called them “orbitals”
• Describes the three-dimensional
orientation of the orbital.
• These were the “lines” we used to place
the electrons;
• __, __, __ __ __
• 1s, 2s,
2p
Electronic
Structure
of Atoms
Magnetic Quantum Number, ml
• Therefore, on any given energy level,
there can be a max of 1s orbital, 3 p
orbitals, 5 d orbitals, 7 f orbitals, etc.
• s: __
• p: __ __ __
• d: __ __ __ __ __
• f: __ __ __ __ __ __ __
Electronic
Structure
of Atoms
Magnetic Quantum Number, ml
• Values are integers ranging from -l to l:
−l ≤ ml ≤ l.
• How to read:
ANY s sublevel: value = 0
• Therefore ml value can be from -0 to +0
Will only have a single orbital
• Therefore ml value can only be = 0
__
0
Electronic
Structure
of Atoms
Magnetic Quantum Number, ml
• Values are integers ranging from -l to l:
−l ≤ ml ≤ l.
• How to read:
ANY p sublevel: value = 1
Will have 3 orbitals
• Therefore ml value can be from -1 to +1
• Think number lines: -1, 0, +1
• So for each “p orbital” put “0” in the middle and
-1 to left and +1 to right
__ __ __
Electronic
Structure
-1 0 +1
of Atoms
Magnetic Quantum Number, ml
• Values are integers ranging from -l to l:
−l ≤ ml ≤ l.
• How to read:
ANY d sublevel: value = 2
Will have 5 orbitals
• Therefore ml value can be from -2 to +2
• So for each “d orbital” put “0” in the middle and
work left and right from there
__ __ __ __ __
0
Electronic
Structure
__ __ __ __ __
of Atoms
-2 -1 0 +1 +2
Magnetic Quantum Number, ml
• Orbitals with the same value of n form a
shell/level.
• Different orbital types within a shell are
subshells/sublevel.
Electronic
Structure
of Atoms
s Orbitals
• Value of l = 0.
• Spherical in shape.
• Radius of sphere
increases with
increasing value of n.
Electronic
Structure
of Atoms
s Orbitals
Observing a graph of
probabilities of finding
an electron versus
distance from the
nucleus, we see that s
orbitals possess n−1
nodes/standing waves,
or regions where there
is 0 (ZERO) probability
of finding an electron.
Electronic
Structure
of Atoms
p Orbitals
• Value of l = 1.
• Have two lobes with a node between them.
Electronic
Structure
of Atoms
d Orbitals
• Value of l is 2.
• Four of the
five orbitals
have 4 lobes;
the other
resembles a p
orbital with a
doughnut
around the
center.
Electronic
Structure
of Atoms
SAMPLE EXERCISE 6.6 Subshells of the Hydrogen Atom
(a) 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?
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.
Solution (a) There are four subshells in the fourth shell, corresponding to the four
possible values of l (0, 1, 2, and 3).
Solution (b) 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 azimuthal quantum number, l: for l = 0, s; for l = 1, p; for l = 2, d; for l = 3,
f.
Solution (c) 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,
and –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).
Electronic
Structure
of Atoms
Do Problem Sets
Electronic
Structure
of Atoms
SAMPLE EXERCISE 6.6 Subshells of the Hydrogen Atom
PRACTICE EXERCISE
(a) What is the designation for the subshell with n = 5 and l = 1? (b) How
many orbitals are in this subshell? (c) Indicate the values of ml for each of
these orbitals in letter “b”.
Answers: (a) 5p; (b) 3; (c) 1, 0, –1
Electronic
Structure
of Atoms
Sample Problem
An electron has the quantum
numbers: n=2; l=1; ml=-1; ms=1/2.
a) What is the subshell designation of
this electron?
b) What if it were: 3, 0, 1, ½?
Electronic
Structure
of Atoms
Sample Problem
An electron has the quantum
numbers: n=2; l=1; ml=-1; ms=1/2.
a) What is the subshell designation of
this electron?
n=2 implies 2nd energy level
l=1 implies “p” sublevel
That is all we need; subshell implies
sublevel;
Electronic
Answer = 2p
Structure
of Atoms
Sample Problem
What if it were: 3, 0, 1, ½?
•
3 = 3rd energy level (n)
0 = “s” sublevel
1 = orbital with a value of 1 (_-1_ _0_ _+1_)
BUT the “s” sublevel can only have ONE
orbital and this has 3, so this electron does
not exist!!
Electronic
Structure
of Atoms
Energies of Orbitals
• For a one-electron
hydrogen atom,
orbitals on the same
energy level have
the same energy.
• That is, they are
degenerate.
Electronic
Structure
of Atoms
Energies of Orbitals
• As the number of
electrons increases,
though, so does the
repulsion between
them.
• Therefore, in manyelectron atoms,
orbitals on the same
energy level are no
longer degenerate. Electronic
Structure
of Atoms
Spin 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.
Electronic
Structure
of Atoms
Spin Quantum Number, ms
• This led to a fourth
quantum number, the
spin quantum number,
ms.
• The spin quantum
number has only 2
allowed values: +1/2
and −1/2.
Electronic
Structure
of Atoms
Pauli Exclusion Principle
• No two electrons in the
same atom can have
exactly the same energy.
• For example, no two
electrons in the same
atom can have identical
sets of quantum
numbers.
Electronic
Structure
of Atoms
Do Problem Sets
Electronic
Structure
of Atoms
Electron Configurations
• Review Electronic Configurations
Using The Periodic Table Method
(video link)
Electronic
Structure
of Atoms
Electron Configurations
• Distribution of all
electrons in an atom
• Consist of
 Number denoting the
energy level
Electronic
Structure
of Atoms
Electron Configurations
• Distribution of all
electrons in an atom
• Consist of
 Number denoting the
energy level
 Letter denoting the type
of orbital
Electronic
Structure
of Atoms
Electron Configurations
• Distribution of all
electrons in an atom.
• Consist of
 Number denoting the
energy level.
 Letter denoting the type
of orbital.
 Superscript denoting the
number of electrons in
those orbitals.
Electronic
Structure
of Atoms
Electron Configurations
• Distribution of all
electrons in an atom.
• Consist of
 Number denoting the
energy level.
 Letter denoting the type
of orbital.
 Superscript denoting the
number of electrons in
those orbitals.
Electronic
Structure
of Atoms
SAMPLE EXERCISE 6.7 Orbital Diagrams and Electron
Configurations
PRACTICE EXERCISE
(a) Write the electron configuration for phosphorus, element 15. (b) How
many unpaired electrons does a phosphorus atom possess?
Answers: (a) 1s22s22p63s23p3, (b) three
Electronic
Structure
of Atoms
SAMPLE EXERCISE 6.7 Orbital Diagrams and Electron
Configurations
PRACTICE EXERCISE
(a) Write the electron configuration for a phosphide ion.
Answers: (a) 1s22s22p63s23p6, remember that nonmetals gain e-’s. So
phosphide has 5 valence e-’s and will gain 3 more e-’s giving it a full energy
level.
Electronic
Structure
of Atoms
SAMPLE EXERCISE 6.7 Orbital Diagrams and Electron
Configurations
PRACTICE EXERCISE
(a) Write the electron configuration for Iron, element 26.
(b) Now write the electron configuration of the ion Iron II.
Answers: (a) 1s22s22p63s23p64s23d 5
(b) 1s22s22p63s23p63d 5; Iron II is Fe+2, meaning it lost 2 e-’s from its
outer most level. Since 4s was the outer most level it
loses 2 e-’s from there first.
Electronic
Structure
of Atoms
Orbital Diagrams
• Each box represents
one orbital.
• Half-arrows represent
the electrons.
• The direction of the
arrow represents the
spin of the electron.
Electronic
Structure
of Atoms
Hund’s Rule
“For degenerate
orbitals, the lowest
energy is attained
when the number of
electrons with the
same spin is
maximized.”
Electronic
Structure
of Atoms
Hund’s Rule
Not This:
This:
“Remember to fill each
orbital with a single
electron with the
SAME SPIN before
you go back and pair
them up, using
OPPOSITE SPIN”
Electronic
Structure
of Atoms
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 8 electrons. The electrons (represented as 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 representation is
The corresponding electron configuration is written 1s22s22p4. The atom has two
unpaired electrons.
Electronic
Structure
of Atoms
Periodic Table
• We fill orbitals in
increasing order of
energy.
• Different blocks on
the periodic table,
then correspond to
different types of
orbitals.
Electronic
Structure
of Atoms
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 them.
Solve: The first member of the halogen group is fluorine, atomic number 9. The
condensed electron configuration for fluorine is
Similarly, that for chlorine, the second halogen, is
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.
Electronic
Structure
of Atoms
SAMPLE EXERCISE 6.8 Electron Configurations for a Group
PRACTICE EXERCISE
Which family of elements is characterized by an ns2np2 electron configuration in the
outermost occupied shell?
Answer: group 4A
Electronic
Structure
of Atoms
SAMPLE EXERCISE 6.9 Electron Configurations from the Periodic Table
(a) Write the full electron configuration for bismuth, element number 83. (b) Write the
condensed electron configuration for this element. (c) How many unpaired electrons
does each atom of bismuth possess?
Solution (a) We write the electron configuration by moving across the periodic table
one row at a time and writing the occupancies of the orbital corresponding to each row
(refer to Figure 6.29).
1s22s22p63s23p64s23d104p65s24d105p66s24ƒ 145d106p3.
Electronic
Structure
of Atoms
SAMPLE EXERCISE 6.9 Electron Configurations from the Periodic Table
(a) Write the electron configuration for bismuth, element number 83. (b) Write the
condensed electron configuration for this element. (c) How many unpaired electrons
does each atom of bismuth possess?
Solution
(b) We write the condensed electron configuration by locating bismuth on the periodic
table and then moving backward to the nearest noble gas, which is Xe, element 54. Thus,
the noble-gas core is [Xe]. The outer electrons are then read from the periodic table as
before. Moving from Xe to Cs, element 55, we find ourselves in the sixth row. Moving
across this row to Bi gives us the outer electrons.
(b Thus, the abbreviated electron configuration is: [Xe]6s24ƒ145d106p3
Electronic
Structure
of Atoms
SAMPLE EXERCISE 6.9 continued
(c) How many unpaired electrons does each atom of bismuth possess?
Solution
c) We can see from the abbreviated electron configuration that the only partially
occupied subshell is the 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 each atom of
bismuth.
Electronic
Structure
of Atoms
SAMPLE EXERCISE 6.9 continued
PRACTICE EXERCISE
Use the periodic table to write the condensed electron configurations for (a)
Co (atomic number 27), (b) Te (atomic number 52).
Answers: (a) [Ar]4s23d7 , (b) [Kr]5s24d105p4
Electronic
Structure
of Atoms
Some Anomalies
Some
irregularities
occur when there
are enough
electrons to halffill s and d
orbitals on a
given row.
Electronic
Structure
of Atoms
Some Anomalies
For instance, the
electron
configuration for
chromium is
[Ar] 4s1 3d 5
rather than the
expected
[Ar] 4s2 3d 4.
This is because a ½ -full “s & d” sublevel are
more stable than a full “s” sublevel
Electronic
Structure
of Atoms
Some Anomalies
Predict: What
other element in
the same period
will do the same
thing?
Did you say
Copper? 
Remember ½ of full
sublevels are more
stable
The electron configuration for copper is
[Ar] 4s1 3d 10 rather than [Ar] 4s2 3d 9.
Electronic
Structure
of Atoms
Some Anomalies
• This occurs because the 4s and 3d orbitals are
very close in energy.
 Others elements to know: Mo & Ag
• Find them and compare their locations to the others
• These anomalies occur in f-block atoms, as well.
 Don’t worry about these. 
Electronic
Structure
of Atoms
Finish your
Problem Sets
Electronic
Structure
of Atoms