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Homework Problems
Chapter 7 Homework Problems: 5, 6, 12, 18, 27, 28, 30, 34, 42, 46, 51,
58 a-e, 60 a-e, 65, 76, 82, 84, 96, 116
CHAPTER 7
Electron Configuration and the
Periodic Table
Electron Configuration and the
Periodic Table
The organization of the periodic table reflects the electron
configurations for the elements. Elements with the same (or similar)
valence shell electron configurations have similar chemical and physical
properties, and are grouped together.
Example: Halogens
F
1s2 2s2 2p5 = [He] 2s2 2p5 ( [He] 2s2 2p5 3s0 )
Cl
1s2 2s2 2p6 3s2 3p5 = [Ne] 3s2 3p5
Br
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
= [Ar] 4s2 3d10 4p5
Note that all of the halogens have the ns2 np5 configuration for their
valence electrons, resulting in similarities in their properties.
Organization of the Periodic Table
Main Group Elements
The valence electrons for the first 18 elements of the periodic
table are given below. Notice how elements in the same group have the
same number and type of valence electrons.
Valence Electron Configuration of the Elements
Free Elements in Chemical Equations
Metals - These do not exist as individual “molecules” but (in the
solid state) as crystals. Metals are therefore represented by the symbol
for the element (Fe(s), Cu(s), Hg(l))
Metalloids - Like metals, these elements do not exist as
individual molecules, and so they are represented in the same way as
metals (B(s), Si(s), Ge(s)).
Nonmetals - Representation can vary depending on the element
Noble gases - He(g), Ne(g), Ar(g)…
Diatomic molecules - H2(g), N2(g), O2(g), F2(g), Cl2(g), Br2(l), I2(s)
Other nonmetals - C(s), P(s), S(s),Se(s). We assume the most stable solid
crystal structure. Technically some of the above nonmetals exist as
polyatomic molecules (P4, S8), but we generally ignore this.
Effective Nuclear Charge (Zeff)
It is useful in making predictions concerning the properties of
atoms and ions (called periodic properties) to use a concept called the
effective nuclear charge.
effective nuclear charge - The nuclear charge seen by the valence
electrons. This is different than the charge of the nucleus, because some
of the nuclear charge is screened (blocked) by the core electrons.
Approximate Value for
Effective Nuclear Charge (Zeff)
There is way of determining values for the effective nuclear
charge experimentally. For our purposes, we will use an approximation
method for finding Zeff
Zeff  (atomic number) - (number of core electrons)
The above assumes that each core electron reduces the charge
seen by the valence electrons by one unit. While crude, this
approximation is sufficient for use in determining trends in periodic
properties.
Effective nuclear charge is most useful in discussions involving
the main group elements.
Finding Zeff
If we know the electron configuration for an atom we can
determine the number of core and valence electrons. From this, we can
find the effective nuclear charge.
Example:
# core # valence
O(8 e-) 1s2 2s2 2p4
2
6
Zeff = 8 - 2 = +6
Br(35 e-) [Ar] 4s2 3d10 4p5
28
7
Zeff = 35 - 28 = +7
Notice that the effective nuclear charge is just equal to the
number of valence electrons.
Zeff and Coulomb’s Law
The attractive force acting between particles of opposite charge is
described by Coulomb’s Law:
F ~ Q1 Q2
d2
Q1, Q2 = charge of particles
d = distance between particles
When Q1 and Q2 are of opposite sign, the force is attractive.
If we apply this to the attraction of a valence electron by the
effective nuclear charge, then
F ~ - (Zeff) (1)
d2
The attractive force is stronger as Zeff increases in size and d decreases in
size.
Sizes of Atoms
There are several ways in which values for size can be assigned
for an atom.
Nonbonding atomic radius (van der Waals radius) - Found from
distance between nuclei of adjacent atoms in nonmetallic atomic solids.
Bonding atomic radius - Found from distance between nuclei in
diatomic molecules (nonmetals) or between adjacent atoms in metallic
solids.
Trends in Sizes of Atoms
There are two general cases for which we can make predictions
about the relative sizes of atoms.
1) Atoms in the same group of the periodic table. Since these
atoms all have the same number of valence electrons (except for He in
group 8A) the size depends on the value of n for the valence electrons.
The general trend is that atomic size increases from top to bottom within
a group.
element*
configuration
radius (nm)
O
[He] 2s2 2p4
0.073
S
[Ne] 3s2 3p4
0.103
Se
[Ar] 4s2 3d10 4p4
0.117
Te
[Kr] 5s2 4d10 5p4
0.143
* Note that for all of these atoms Z = +6.
eff
2) Atoms in the same period (row) in the periodic table. In this
case, the atoms all have the same value of n for the valence electrons.
The determining factor is the value for Zeff, the effective nuclear charge.
The larger the value for Zeff, the smaller the atom.
Example: Consider the elements in the second period.
element configuration
Zeff
radius (nm)
Li
1s2 2s1
+1
0.152
Be
1s2 2s2
+2
0.112
B
1s2 2s2 2p1
+3
0.085
C
1s2 2s2 2p2
+4
0.077
N
1s2 2s2 2p3
+5
0.075
O
1s2 2s2 2p4
+6
0.073
F
1s2 2s2 2p5
+7
0.072
Ne
1s2 2s2 2p6
+8
0.070
Sizes of Main Group Atoms
Sizes of Ions
There are three general cases we will consider.
1) Ions with the same charge within a group. In this case, Zeff
will be the same for all the ions, and so the size of the ion will be
controlled by the largest value of n for the electrons present. The general
trend will be that size (radius) increases from top to bottom.
Be2+
0.031 nm
1s2 = [He]
Mg2+ 0.065 nm
1s22s22p6 = [Ne]
Ca2+
0.099 nm
1s22s22p63s23p6 = [Ar]
F-
0.136 nm
1s22s22p6 = [Ne]
Cl-
0.181 nm
1s22s22p63s23p6 = [Ar]
Br-
0.195 nm
1s22s22p63s23p64s23d104p6 = [Kr]
2) Different ions/atoms of the same element. In this case, the
smaller the number of electrons (more positive the ion) the smaller the
ion.
Cu (29 e-)
0.128 nm
1s22s22p63s23p64s13d10
Cu+ (28 e-)
0.096 nm
1s22s22p63s23p63d10
Cu2+ (27 e-)
0.072 nm
1s22s22p63s23p63d9
Cl (17 e-)
0.099 nm
1s22s22p63s23p5
Cl- (18 e-)
0.181 nm
1s22s22p63s23p6
3) Ions/atoms with the same number of electrons. In this case,
the larger the charge of the nucleus (atomic number) the smaller the ion.
atom/ion
Z
radius (nm)
N3-
+7
0.171
O2-
+8
0.140
F-
+9
0.136
Ne
+10
-*
Na+
+11
0.095
Mg2+
+12
0.065
Al3+
+13
0.050
1s22s22p6
In the above example all the atoms/ions have 10 electrons.
Species with the same number and configuration of electrons are called
isoelectronic.
* Ion sizes are determined from studies of binary ionic compounds, and so cannot be
directly compared to the size for Ne, which is determined in the gas phase or the condensed solid.
Ionization Energy
The ionization energy is defined as the energy required to remove
one electron from an atom, ion, or molecule in the gas phase.
X(g)  X+(g) + e-
IE1
Since atoms in general have a large number of electrons, we can define
the first ionization energy as the energy required to remove the first
electron, the second ionization energy as the energy required to remove
the second electron, and so forth.
O(g)  O+(g) + e-
1st ionization enegy
O+(g)  O2+(g) + e-
2nd ionization energy IE2
O2+(g)  O3+(g) + e-
3rd ionization energy IE3
:
:
:
O7+(g)  O8+(g) + e-
:
:
:
:
IE1
:
8th ionization energy IE8
Note that it will always be true that IE < IE < IE < ...
Trends in 1st Ionization Energy
1) Atoms in the same group of the periodic table. Since all atoms
in the same group have the same effective nuclear charge for the valence
electrons, the determining factor for the ionization energy is the value of
n for the valence electrons. The larger the value of n the further the
valence electrons are from the nucleus, and so the easier to remove a
valence electron. So 1st ionization energy decreases in going from top to
bottom within a group.
element*
configuration
IE1 (kJ/mol)
Li
[He] 2s1
520.
Na
[Ne] 3s1
496.
K
[Ar] 4s1
419.
Rb
[Kr] 5s1
403.
* In all cases Zeff = +1.
2) Atoms in the same period (row) in the periodic table. The
valence electrons all have the same value for n, and so it is Zeff that
determines the ionization energy. The larger the value for Zeff the more
difficult it is to remove an electron. So 1st ionization energy generally
increases in going from left to right within a period.
element
configuration Zeff
IE1 (kJ/mol)
Li
[He] 2s1
+1
520.
Be
[He] 2s2
+2
899.
B
[He] 2s2 2p1
+3
801.
C
[He] 2s2 2p2
+4
1086.
N
[He] 2s2 2p3
+5
1402.
O
[He] 2s2 2p4
+6
1314.
F
[He] 2s2 2p5
+7
1681.
Ne
[He] 2s2 2p6
+8
2081.
Higher Ionization Energies
As noted, ionization energies always increase as we go from
removing the first electron to the second, third, and so forth. However,
when we have removed all the valence electrons and begin removing
core electrons there is a big jump in “Zeff”, and so usually a large jump in
ionization energy, as can be seen below.
atom/ion - configuration
“Zeff” IE (kJ/mol)
reaction
Na(g)  Na+(g) + e-
Na 1s2 2s2 2p6 3s1
+1
496.
Na+ 1s2 2s2 2p6
+9
4560.
Na+(g)  Na2+(g) + e-
Na2+ 1s2 2s2 2p5
+9
6910.
Na2+(g)  Na3+(g) + e-
Mg 1s2 2s2 2p6 3s2
+2
738.
Mg(g)  Mg+(g) + e-
Mg+ 1s2 2s2 2p6 3s1
+2
1450.
Mg+(g)  Mg2+(g) + e-
Mg2+1s2 2s2 2p6
+10
7730.
Mg2+(g)  Mg3+(g) + e-
Ionization Energies For Elements 3-11
Notice that while the ionization always increases as more and
more electrons are removed, the biggest jump occurs after all the valence
electrons have been removed.
Electron Affinity
Electron affinity is defined as the energy change when an atom or
molecule adds an electron in the gas phase.
X(g) + e-  X-(g)
EA
Unlike ionization energy, electron affinity can be either positive or
negative.
EA > 0 Must add energy to force the atom to add an electron
EA < 0 Energy is released when atom adds an electron
If EA is negative that means the atom wants to add an electron.
Trends in Electron Affinity
While there are no strong trends for electron affinity we can make
the following observations.
1) EA is large and negative for the halogens.
atom
configuration
EA (kJ/mol)
F
[He] 2s2 2p5
- 328.
Cl
[Ne] 3s2 3p5
- 349.
Br
[Ar] 4s2 3d10 4p5
- 325.
I
[Kr] 5s2 4d10 5p5
- 298.
Explanation - There is space for one more electron in the valence
shell. That electron will see a large effective nuclear charge (Zeff = +7)
and so it is easy to add.
Trends in Electron Affinity
2) EA is large and positive for the noble gases.
Ne(g) + e-  Ne-(g)
Ne 1s2 2s2 2p6
Ne- 1s2 2s2 2p6 3s1
Explanation – The electron being added is going to a higher
energy orbital. In fact, it will see an effective nuclear charge Zeff = 0, and
so there is no strong attraction of the electron towards the atom. In fact,
it takes energy to add the electron due to electron-electron repulsion.
Examples of “Trend Questions”
1) Place the following in order from largest to smallest
a) Ca, Cl, Mg
b) P3-, S2-, Cl-
2) Which of the following has the largest 1st ionization energy?
a) Cl or Br
b) C or O
Examples of “Trend Questions”
1) Place the following in order from largest to smallest
a) Ca, Cl, Mg
Ca > Mg ; Mg > Cl so Ca > Mg > Cl
b) P3-, S2-, Cl-
All have 18 e-, and so P3- > S2- > Cl-
2) Which of the following has the largest 1st ionization energy?
a) Cl or Br
Both are in the same group, so chlorine is larger.
b) C or O
Both are in the same row, so oxygen is larger.
Ion Formation
By knowing the electron configuration of an atom we can often
predict the ion that will form from that atom. That is particularly true for
main group elements.
The general tendency for the main group elements is to either add
electrons to have completely filled outer s and p orbitals, or to lose
electrons to completely empty the outermost s and p orbitals.
Which of the above will occur depends on which process
involves the fewest electrons.
In terms of a general statement we can say the following:
Main group (representative) elements tend to add or lose
electrons to achieve an electron configuration that is the same as that of
the nearest noble gas.
Examples
Group 2A elements – 2 valence electron
Be
1s2 2s2 = [He]2s2
Mg
1s2 2s22p63s2 = [Ne]3s2
Ca
1s2 2s2 2p6 3s2 3p6 4s2 = [Ar]4s2
In all cases, the outermost s and p orbitals can be emptied by
removing two electron (the valence electrons). So the ions expected to
form are Be2+, Mg2+, and Ca2+.
Group 6A elements – 6 valence electrons
O
1s2 2s2 2p4 = [He]2s2 2p4
S
1s2 2s2 2p6 3s2 3p4 = [Ne]3s2 3p4
Se
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4 = [Ar]4s2 3d10 4p4
In this case we can fill the outermost (valence) s and p orbitals by
adding two electrons. So the ions expected to form are O2-, S2-, and Se2-.
Common Ions For Main Group Elements
The common ions formed from the main group elements are
indicated below. In a few cases main group elements can forms ions
with different charges. For example, thallium (Tl), a group 3A element,
forms ions with a +1 and +3 charge (Tl+, Tl3+). Lead (Pb), a group 4A
element, can for ions wit a +2 and +4 charge.
Electron Configurations For Metal Cations
We may write electron configurations for ions just as we can for
atoms. For anions the same rules used for atoms apply. For cations we
must use a slightly different approach.
1) Write the electron configuration for the atom.
2) Remove electrons to form the cation. Start by removing
electrons from orbitals with the largest value of n. If there are two
orbitals with the same value of n, begin with the orbital with the largest
value of .
Example: What are the electron configurations for Fe, Fe2+ and Fe3+?
Example: What are the electron configurations for Fe2+ and Fe3+?
Fe(26 e-)
[Ar] 4s2 3d6 = 1s2 2s2 2p6 3s2 3p6 4s2 3d6
Fe2+(24 e-) [Ar] 3d6
Fe3+(23 e-) [Ar] 3d5
Transition metal ions often form by creating empty s-orbitals,
empty d-orbitals, or half filled d-orbitals, but there are many exceptions
to this general tendency.
Properties of Metals and Nonmetals
Metals
Are usually solids at room temperature.
Are shiny, lustrous, malleable, and ductile.
Are good conductors of heat and electricity.
Tend to form cations in ionic compounds.
Do not usually form molecular compounds.
Form oxides that are ionic and usually are bases.
Nonmetals
Exhibit a variety of colors, and states (s, l, g).
Are not shiny, and are brittle rather than malleable.
Are poor conductors of heat and electricity.
Tend to form anions in ionic compounds.
Form molecular compounds with other nonmetals.
Form oxides that are molecular compounds and usually
acids.
Examples of Metal and Nonmetal Oxides
Metals
CaO(s) + H2O(l)  Ca(OH)2(aq)
Li2O(s) + H2O(l)  2 LiOH(s)
Nonmetals
SO3(g) + H2O(l)  H2SO4(aq)
N2O5(g) + H2O(l)  2 HNO3(aq)
Metallic Character of Elements
Metals tend to lose electrons to form cations, while nonmetals
tend to gain electrons to form anions. Based on the periodic trends in
ionization energy (which measures how easy it is to remove an electron
from an atom) and electron affinity (which measures how easy it is to
add an electron to an atom) we can make the following prediction about
the “metallic character” of an element.
Metallic character decreases in going from bottom to top and
from left to right in the periodic table.
Metallic character can be related to electronegativity, a concept
we will introduce in the next chapter.
Trends in Metallic Character
End of Chapter 7
“When I first saw the periodic table, it hit me with the force of
revelation - it embodied, I was convinced, eternal truths...I thought of
Mendeleev as a sort of Moses, bearing the tablets of the God-given
periodic law.” - Oliver Sachs
Two ions are talking to each other in a solution. One says: "Are
you a cation or an anion?" The other replies, "Oh, I'm a cation." The first
asks, "Are you sure?" The reply - "I'm positive." - anonymous