Chapter 8
Electron Configuration and
Chemical Periodicity
.
Factors Affecting Atomic Orbital Energies
The Effect of Nuclear Charge (Zeffective)
Higher nuclear charge lowers orbital energy (stabilizes the
system) by increasing nucleus-electron attractions.
The Effect of Electron Repulsions (Shielding)
Additional electron in the same orbital
An additional electron raises the orbital energy through
electron-electron repulsions.
Additional electrons in inner orbitals
Inner electrons shield outer electrons more effectively than
do electrons in the same sublevel.
ELECTRON SPIN & MAGNETISM
Electron Spin: you need to understand magnetism & ionization energy to
understand electron spin and quantum number ms.
Magnetism and what you already know: earth has a magnetic field, with a
north pole and south pole, as do metal magnets
Electromagnetism = magnetic field created by electrical current
Three definitions for magnetic substances:
paramagnetic = attracted to magnetic field
diamagnetic = not attracted to magnetic field
ferromagnetic = substance that retains its magnetism after being
placed in magnetic field (like Fe, Co & Ni)
Figure 8.19
Apparatus for measuring the magnetic behavior of a sample.
ELECTRON SPIN & MAGNETISM
Electrons are spinning charged particles which generate
a tiny magnetic field
Only two orientations are possible and they are called
spin up or spin down, with clockwise being up
This is why the 4th Quantum number, ms, is assigned +½
or -½
ELECTRON SPIN
Data showed that the He atom was not affected by
magnetic field - why?
There are 2 e-s present, must be that one is up and
one is down, cancelling each others magnetic
fields - we say they are "paired"
It turns out that only an atom with one or more
unpaired e-s exhibits paramagnetism
Figure 8.1
Observing the Effect of Electron Spin
The Stern-Gerlach experiment.
ELECTRON SPIN
The lowest energy state for an e- in H is principal level n = 1
There is only one type of orbital at n = 1, because l = 0, which
is an s or a spherical orbital
For H, the 1 e- is generally in the 1s orbital
In He, there are 2 e-s, and it turns out they are both in the 1s
orbital & they are paired up, or coupled with one spin up
and one spin down
Li or Be: 3rd and 4th e-s are in n = 2, but which type of
orbital, s or p?
FOR THIS WE NEED TO LOOK AT DATA FOR
IONIZATION ENERGIES
IONIZATION ENERGIES
Ionization energy (IE): energy required to take an eaway from an atom
First IE removes the e- furthest away from the nucleus
Examples: IE1 = 1.31 x 106 J/mol for H, 1.68 x 106 J/mol for
Fe, and 0.50 x 106 J/mol for Na
IONIZATION ENERGIES
IE in MJ/mol
e-#
1 2
3 4 5 6 7 8 9 10 11
Na
.5 4.6 6.9 9.5 13 17 20 25 29 141 178
Log
5.7 6.7 6.8 7.0 7.1 7.2 7.3 7.4 7.9 8.2 8.3
F
Log
1.7 3.4 6.1 8.4 11 15 18 92 106
6.2 6.5 6.8 6.9 7.0 7.2 7.3 8.0 8.03
Na: three energy levels present as shown by big changes in IE
(from 1 to 2, 9 to 10)
F: two energy levels with slight difference within level because of s & p orbital differences (7 to 8)
IONIZATION ENERGIES
If all the IEs are mathematically adjusted for
increasing force of attraction between protons and
remaining electrons:
-then we find that IE1 to IE5 in F are almost equal,
meaning these 5 e-s are "alike"
- And that IE6 = IE7 but > IE1 to IE5, so these two e-s
are different “levels”
- And that IE8 = IE9 but >>> IE7, so these two elevels are really different
IONIZATION ENERGIES
What does all this mean?
IE data define the energy states and orbitals in the atom
At n = 1, there's one orbital with e-s that are very hard to
remove
At n = 2, there are 4 orbitals, but they're different
because s has different shape than p
IONIZATION ENERGIES
Combine the IE data with pairing of e-s into 2e- per orbital
from magnetic properties and we determine that:
s orbital has up to 2 e-s
and each p orbital has up to 2 e-s for a total of 6
If F has 5 e- of similar IE – they must be in p orbitals and they
are the easiest to remove so they must be outermost
Next 2 e-s are close to same energy, n = 2, but only 2 e-s, so
they are in an s orbital
The last 2 e-s are very different & have much higher IE, must
be close to nucleus, n = 1, 2 e-s in s orbital
This data reveals basic e- configuration of F: 1s has 2 e-s, 2s
has 2 e-s, 2p has 5 e-s.
IONIZATION ENERGIES
Now look at Na data:
First IE is < 2nd IE
but IE2 to IE9 are about the same
then big jump to IE10 and IE11
Means that 1 e- is outermost at n = 3
then 8 e-s in n = 2 (notice slight diff for 2s & 2p)
then 2 e-s in n=1
IONIZATION ENERGIES
Now we look at first IE vs. atomic number
Noble gases have high IE
All of Group IA atoms have low IE, Group IIA has fairly
low IE
Transition metals into same IE - these are d e-s
Periodic trend of IE - highest at He, lowest at Fr
Figure 8.10
Periodicity of first ionization energy (IE1)
Figure 8.11
First ionization energies of the main-group elements.
Table 8.1 Summary of Quantum Numbers of Electrons in Atoms
Name
Symbol
Permitted Values
Property
principal
n
positive integers(1,2,3,…) orbital energy (size)
angular
momentum
l
integers from 0 to n-1
orbital shape (The l values
0, 1, 2, and 3 correspond to
s, p, d, and f orbitals,
respectively.)
magnetic
ml
integers from -l to 0 to +l
orbital orientation
spin
ms
+1/2 or -1/2
direction of e- spin
QUANTUM NUMBERS
We know that e-s pair up into two per orbital
maximum
Pauli Exclusion Principle - a statement of the facts:
no two e-s in an atom can have the exact same set
of four quantum numbers
He 2 e-s: n
l
ml
ms
1
0
0
+½
1
0
0
-½
ELECTRON CONFIGURATIONS
The best way to figure out quantum numbers is to know
electron configuration, so we will do that first!
There are several “rules” or physical laws based on data like
Ionization Energies
If 2p is at higher energy level than 2s, then 3p is higher than
3s
Also find that 3d is slightly higher than 3p
In multi-electron atoms:
- 4s slightly lower energy than 3d, so fill 4s before 3d
- always start at 1s, fill in according to increasing energy
levels
Figure 8.3 Order for filling energy
sublevels with electrons
Order for filling energy sublevels with
electrons
Illustrating Orbital Occupancies
The electron configuration
#
of electrons in the sublevel
n l
as s,p,d,f
The orbital diagram (box or circle, label
with orbital name)
1s
1s
ELECTRON
CONFIGURATIONS
Hund's Rule: max number of unpaired e-s will occur
in ground state
Two methods as seen in previous slide: orbital box
(next slide) or spectroscopic (spdf)
1s2
Figure
page 251
A vertical orbital diagram for the Li ground state
no color-empty
light - half-filled
dark - filled, spin-paired
You will do horizontal orbital box
notation.
ELECTRON CONFIG USING PERIODIC TABLE
Remembering all the rules and the order for filling
orbitals looks difficult!
It turns out the periodic table is layed out in blocks:
s block is groups 1 & 2
p block is groups 13 to 18
d block is transition elements groups 3 - 12
f block is inner transition
You can figure out the electron config for the last e- in
any element by looking at the periodic table. Then fill
in starting from H or nearest noble gas.
Figure 8.5
A periodic table of partial
ground-state electron
configurations
Figure 8.6
The relation between orbital filling and the
periodic table
Practice
 Pair up and use a plain (not large) periodic table to do
the spdf and orbital box notation for B, Ne and Mg.
SAMPLE PROBLEM 8.2
PROBLEM:
Determining Electron Configuration
Give the full and condensed electrons configurations, partial orbital
diagrams showing valence electrons, and number of inner electrons
for the following elements:
(a) potassium
(b) molybdenum
(c) lead
SOLUTION:
(a) for K
1s22s22p63s23p64s1
full configuration
condensed configuration [Ar] 4s1
partial orbital diagram
4s1
There are 18 inner electrons.
3d
4p
SAMPLE PROBLEM 8.2
continued
(b) for Mo
full configuration 1s22s22p63s23p64s23d104p65s14d5
condensed configuration [Kr] 5s14d5
partial orbital diagram
There are 36 inner electrons
and 6 valence electrons.
5s1
4d5
5p
(c) for Pb
full configuration
1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2
condensed configuration
[Xe] 6s24f145d106p2
partial orbital diagram
There are 78 inner electrons
and 4 valence electrons.
6s2
6p2
ELECTRON
CONFIGURATIONS
For many e- atoms we can use a shorthand for either
method called "noble gas core designation" or
condensed version
Try for examples: Cl and As
Cl: [Ne]3s23p5
As: [Ar]4s23d104p3
ELECTRON CONFIGURATIONS:
VALENCE ELECTRONS
Electrons beyond noble gas core are valence electrons:
e-s in outermost principal quantum level of atom
Practice with Na, As, Mn, and Pu
Then determine the principal quantum number of the
last electron in each of the above
ELECTRON
CONFIGURATIONS
Hund's Rule: max number of unpaired e-s will occur in
ground state
Why Cr & Cu don't exactly follow the filling rules:
Cr is more stable with 1 e- per orbital including 4s
Cu is more stable with full d shell and 1 e- in 4s
BACK TO QUANTUM NUMBERS
 If you have the electron configuration, you have the
first two quantum numbers for each electron, n and l.
 An s subshell has only one orbital with only one
ml quantum number, 0. A p subshell has three
orbitals, so you have to list each one’s ml: -1, 0, +1.
A d subshell has ml ranging from -2.,,,0,…+2; etc.
 Within each orbital, the ms for the first e- is by
definition +½. The second e- is assigned –½.
SAMPLE PROBLEM 8.1
PROBLEM:
Determining Quantum Numbers
from Orbital Diagrams
Write a set of quantum numbers for the third electron and a set
for the eighth electron of the F atom.
PLAN: Use the orbital diagram to find the third and eighth electrons.
Within 2p, you have
-1, 0, +1.
9F
1s
2s
2p
SOLUTION: The third electron is in the 2s orbital. Its quantum numbers are
n= 2
l= 0
ml = 0
ms= +1/2
Write: 2, 0, 0, +1/2
The eighth electron is in a 2p orbital. Its quantum numbers are
n= 2
l= 1
ml = -1
So write
ms= -1/2
2, 1, -1, -1/2
ELECTRON
CONFIGURATIONS
Make a table of all four quantum numbers for every electron in vanadium.
n
l
ml
ms
n
l
ml
ms
Table 8.2 in new ed.
8.2
Figuree 8.4 Condensed ground-state electron configurations in
the first three periods.
Table 8.3 in new ed.
Table 8.3
ELECTRON CONFIGURATIONS
Table to remember energy levels IF you don’t have a
periodic table handy!
1s
2s 2p
3s 3p 3d
4s 4d 4p 4f
5s 5p 5d 5f (5g)
6s 6p 6d 6f (6g)
7s 7p 7d 7f
Follow arrows down and to left to fill in electron
configuration.
A video on how to write electron
configurations and orbital diagrams
 YouTube - How to Write Electron Configurations and
Orbital Diagrams
HISTORY OF PERIODIC TABLE
Origin is based on "periodic properties" and relative masses
Johann Dobereiner grouped triads of elements with similar
properties and increasing relative mass
In 1864, John Newlands conceived the idea of octaves, since
the chem prop's seemed to repeat for every eighth element
Current table: Julius Meyer and Dmitri Mendeleev
 YouTube - The Periodic Table: Groups and Trends
PERIODIC TRENDS
Trends are the result of atom's e- configuration - # of e-s or
really # of protons since its arranged by atomic number
Look at Argon's e- density vs. distance from nucleus
Not like a billiard ball!
Radius is "soft" and is affected by covalent bonding, since it
can overlap
Cl by itself is 132 pm, but in Cl2 radius is 100 pm
Trend: smallest radii are upper right, largest to lower left in
general
Figure 8.7
Defining metallic and covalent radii
Figure 8.8
Atomic radii of the maingroup and transition
elements.
Figure 8.9
Periodicity of atomic radius
SAMPLE PROBLEM 8.3
PROBLEM:
Ranking Elements by Atomic Size
Using only the periodic table (not Figure 8.15)m rank each set of
main group elements in order of decreasing atomic size:
(a) Ca, Mg, Sr
(b) K, Ga, Ca
(c) Br, Rb, Kr
(d) Sr, Ca, Rb
SOLUTION:
(a) Sr > Ca > Mg
These elements are in Group 2A(2).
(b) K > Ca > Ga
These elements are in Period 4.
(c) Rb > Br > Kr
Rb has a higher energy level and is far to the left.
Br is to the left of Kr.
(d) Rb > Sr > Ca
Ca is one energy level smaller than Rb and Sr.
Rb is to the left of Sr.
PERIODIC TRENDS
IE: discussed previously, but trend is for highest to
upper right, lowest to lower left. Always endothermic
process to remove eElectron Affinity: Trend is most negative EA at F, least
likely is Fr, worst are noble gases
Table 8.4
Figure 8.13
Electron affinities of the main-group elements.
Figure 8.14
Trends in three atomic properties.
Figure 8.15
Trends in metallic behavior.
If you need it: a 15 minute
video from the Khan Academy
 YouTube - Other Periodic Table Trends
SAMPLE PROBLEM 8.4
PROBLEM:
Ranking Elements by First Ionization Energy
Using the periodic table only, rank the elements in each of the
following sets in order of decreasing IE1:
(a) Kr, He, Ar
(b) Sb, Te, Sn
(c) K, Ca, Rb
(d) I, Xe, Cs
SOLUTION:
(a) He > Ar > Kr
Group 8A(18) - IE decreases down a group.
(b) Te > Sb > Sn
Period 5 elements - IE increases across a period.
(c) Ca > K > Rb
Ca is to the right of K; Rb is below K.
(d) Xe > I > Cs
I is to the left of Xe; Cs is furtther to the left and
down one period.
SAMPLE PROBLEM 8.5
PROBLEM:
Identifying an Element from Successive
Ionization Energies
Name the Period 3 element with the following ionization energies
(in kJ/mol) and write its electron configuration:
IE1
IE2
IE3
IE4
IE5
1012
1903
2910
4956
6278
IE6
22,230
SOLUTION:
The largest increase occurs after IE5, that is, after the 5th valence
electron has been removed. Five electrons would mean that the
valence configuration is 3s23p3 and the element must be
phosphorous, P (Z = 15).
The complete electron configuration is 1s22s22p63s23p3.
Atoms to Ions
What happens to e- configuration if an atom becomes an
ion?
Ion size: cations get smaller, anions get bigger
More charge is more effect on size
ISOELECTRONIC: iso = same, electrons, so same # of
electrons
We can list isoelectronic series of ions and noble gas atoms
that have the same number of electrons.
Figure 8.17
Main-group ions and the noble gas configurations.
Isoelectronic series: O2-, F-, Ne, Na+, Mg2+
Atoms to Ions
Rank the isoelectronic series on the previous slide by
size smallest to largest:
Do abbrev. e- config: Mg2+, Fe2+ and Fe3+, Cu1+ and Cu2+.
Determine which will be paramagnetic, which is more
strongly paramagnetic, etc. Then rank them by size.
SAMPLE PROBLEM 8.6 Writing Electron Configurations of Main-Group Ions
PROBLEM:
(a) Iodine
Using condensed electron configurations, write reactions for the
formation of the common ions of the following elements:
(b) Potassium
(c) Indium
SOLUTION:
(a) Iodine (Z = 53) is in Group 7A(17) and will gain one electron to be isoelectronic
with Xe: I ([Kr]5s24d105p5) + eI- ([Kr]5s24d105p6)
(b) Potassium (Z = 19) is in Group 1A(1) and will lose one electron to be isoelectronic
with Ar: K ([Ar]4s1)
K+ ([Ar]) + e(c) Indium (Z = 49) is in Group 3A(13) and can lose either one electron or three
electrons: In ([Kr]5s24d105p1)
In+ ([Kr]5s24d10) + e+
In ([Kr]5s24d105p1)
In3+([Kr] 4d10) + 3e-
SAMPLE PROBLEM 8.7
PROBLEM:
Writing Electron Configurations and Predicting
Magnetic Behavior of Transition Metal Ions
Use condensed electron configurations to write the reaction for the
formation of each transition metal ion, and predict whether the ion is
paramagnetic.
(a) Mn2+
(b) Cr3+
(c) Hg2+
SOLUTION:
(a) Mn2+(Z = 25) Mn([Ar]4s23d5)
(b) Cr3+(Z = 24) Cr([Ar]4s13d5)
(c) Hg2+(Z = 80) Hg([Xe]6s24f145d10)
Mn2+ ([Ar] 3d5) + 2eCr3+ ([Ar] 3d3) + 3e-
paramagnetic
paramagnetic
Hg2+ ([Xe] 4f145d10) + 2enot paramagnetic (is diamagnetic)
Figure 8.20
Depicting ionic radius.
Figure 8.21
Ionic vs. atomic radius.
SAMPLE PROBLEM 8.8
PROBLEM:
Ranking Ions by Size
Rank each set of ions in order of decreasing size, and explain your
ranking:
(a) Ca2+, Sr2+, Mg2+
(b) K+, S2-, Cl -
(c) Au+, Au3+
SOLUTION:
(a) Sr2+ > Ca2+ > Mg2+
(b) S2- > Cl - > K+
(c) Au+ > Au3+
These are members of the same Group (2A/2) and
therefore decrease in size going up the group.
The ions are isoelectronic; S2- has the smallest Zeff and
therefore is the largest while K+ is a cation with a large Zeff
and is the smallest.
The higher the + charge, the smaller the ion.