South Pasadena • Honors Chemistry
Name
9 • Atomic Structure
Period
9.3
NOTES
–
Date
PERIODIC TRENDS
Observations of Patterns
Atomic radius – the size of the atom, from the nucleus to the outer electron. When an outer electron is more
attracted to the nucleus, the atomic radius is smaller. In general, as we go down a family on the Periodic Table,
the atomic radius increases; as we go across a period, the atomic radius decreases.
First Ionization Energy – Energy required to remove an electron from a neutral atom: A + energy  A+ + e‒
The more attracted the outer electron is to the nucleus, the more energy is required to pull it off, resulting in a
higher ionization energy. In general, the first ionization energy decreases going down a family while it increases
going across a period. The elements with the highest ionization energies are in the upper right while the elements
with the lowest ionization energies are in the lower left.
Electron Affinity – Energy released when adding an electron to a neutral atom:
A + e‒  A‒ + energy
Atoms with greater Zeff tend to attract electrons more strongly, and have greater absolute electron affinity.
Similar to ionization energy, elements in the upper right of the Periodic Table have the greatest electron affinity,
while those in the lower left have the lowest electron affinity. However, the trend for electron affinity tends to be
less consistent than that for other properties.
Electronegativity – A value from 0 to 4 created by Linus Pauling to describe how much an atom “wants” an
electron from another atom. It is a combination of an atom’s ionization energy and electron affinity. Elements in
the upper right of the Periodic Table have the highest electronegativity values ‒ they pull strongly on an electron
from another atom ‒ and those in the lower left have the lowest electronegativity values ‒ they don’t pull very
strongly on an electron in a bond. Since noble gases don’t typically bond with other atoms, they do not have
electronegativity values.
H
2.1
Li
1.0
Na
0.9
K
0.8
Rb
0.8
Cs
0.7
Fr
0.7
Be
1.5
Mg
1.2
Ca
1.0
Sr
1.0
Ba
0.9
Ra
0.9
Sc
1.3
Y
1.2
La
1.0
Ac
1.1
Ti
1.5
Zr
1.4
Hf
1.3
Th
1.3
V
1.6
Nb
1.6
Ta
1.5
Pa
1.4
Cr
1.6
Mo
1.8
W
1.7
U
1.4
Mn
1.5
Tc
1.9
Re
1.9
Np
1.4
Fe
1.8
Ru
2.2
Os
2.2
Co
1.9
Rh
2.2
Ir
2.2
Ni
1.9
Pd
2.2
Pt
2.2
Cu
1.9
Ag
1.9
Au
2.4
Zn
1.6
Cd
1.7
Hg
1.9
B
2.0
Al
1.5
Ga
1.6
In
1.7
Tl
1.8
C
2.5
Si
1.8
Ge
1.8
Sn
1.8
Pb
1.9
N
3.0
P
2.1
As
2.0
Sb
1.9
Bi
1.9
O
3.5
S
2.5
Se
2.4
Te
2.1
Po
2.0
F
4.0
Cl
3.0
Br
2.8
I
2.5
At
2.2
He
–
Ne
–
Ar
–
Kr
–
Xe
–
Rn
–
Understanding the Patterns
Sodium
positively-charged
nucleus
core electron shells
“shield” nuclear charge
Protons (Z):
Core Electrons:
Effective Nuclear Charge (Zeff):
Potassium
11
10
11 – 10 = 1
Protons (Z):
Core Electrons:
Effective Nuclear Charge (Zeff):
19
18
19 – 18 = 1
outer electron
attracted to the
nucleus
Calcium
Protons (Z):
Core Electrons:
Effective Nuclear Charge (Zeff):
20
18
20 – 18 = 2
Explaining the Patterns
Element
Sodium
Potassium
Atomic Radius
1.66 Å
2.03 Å
Ionization Energy 495.8 kJ/mol 418.8 kJ/mol
Down a Family: Look at core shells
 The outer electron of Na is in the 3s orbital while that of K is in the 4s orbital.
 An outer electron in higher shells experiences greater shielding from inner electrons, which cancels out the
increased positive charges in the nucleus. It is further away and is less attracted to the nucleus.
 Because K has more core shells, the outer electron in K is further away from the nucleus than that in Na, so K
has a larger atomic radius and a lower ionization energy.
Element
Potassium
Calcium
Atomic Radius
2.03 Å
1.76 Å
Ionization Energy 418.8 kJ/mol 589.8 kJ/mol
Across a Period: Look at number of protons
 The outer electrons in K and Ca are both in the 4s subshell. However, K has 19 protons, while Ca has 20
protons.
 The outer electron in an atom with more protons has a greater Zeff so it is more attracted to the nucleus.
Atomic vs. Ionic Radii
Atoms can gain or lose electrons to form ions. Because the number of protons and core electron shells are the
same, we want to focus on the number of electrons. If there are more electrons, the outer electrons experience
greater electron-electron repulsions, resulting in a larger radius.
Sodium atom, Na Sodium ion, Na+
Radius
1.54 Å
0.97 Å
Protons
11
11
Electrons
11
10
+
 Both Na and Na have 11 protons. Na has 11
electrons while Na+ has 10 electrons.
 An atom with fewer electrons experience weaker
electron-electron repulsions.
 Because Na+ has fewer electrons, there are
weaker electron-electron repulsions, so it has a
smaller radius.
(An explanation stating that the outer electron of Na is
in the 3s orbital while that of Na+ is in the 2p orbital
is also acceptable.)
Chlorine atom, Cl Chloride ion, Cl‒
Radius
0.99 Å
1.81 Å
Protons
17
17
In summary, a cation has a smaller radius than its
Electrons
17
18
neutral atom, and an anion has a larger radius than its



Both Cl and Cl‒ have 17 protons. Cl has 17
electrons while Cl‒ has 18 electrons.
An atom with more electrons experience stronger
electron-electron repulsions.
Because Cl‒ has more electrons, there are
stronger electron-electron repulsions, so it has a
larger radius.
neutral atom.
Oxide ion
Fluoride ion
Sodium ion
Magnesium ion
2‒
‒
+
O
F
Na
Mg2+
Radius
1.40 Å
1.33 Å
0.97 Å
0.66 Å
Protons
8
9
11
12
Electrons
10
10
10
10
Atoms are isoelectronic if they have the same number of electrons.
When comparing their radii, we want to focus on the number of
protons.
 O2‒, F‒, Ne, Na+, and Mg2+ all have 10 electrons, but they have
O2‒
F‒
Ne
Na+ Mg2+
8, 9, 10, 11, and 12 protons, respectively.
 The outer electron is more attracted to an atom with more protons than to one with fewer protons.
 Because Mg2+ has the most protons, its outer electron is more attracted to the nucleus than the other species,
and it has the smallest radius. Because O2‒ has the fewest protons, its outer electron is least attracted to the
nucleus than the other species, and it has the largest radius.
Successive Ionization Energies
The first ionization energy is the energy for removing the first electron from an atom. We can also look at the
energy required to remove additional electrons.
First Ionization Energy
Second Ionization Energy
Third Ionization Energy
Fourth Ionization Energy
IE1 for Al = 580 kJ/mol
IE2 for Al = 1,815 kJ/mol IE3 for Al = 2,740 kJ/mol IE4 for Al = 11,600 kJ/mol
Equation:
Equation:
Equation:
Al + IE1
 Al+ + e‒
Al+ + IE2  Al2+ + e‒
Al2+ + IE3  Al3+ + e‒
Successive Ionization Energies increase because:
as there are fewer electrons, there is less electron-electron repulsion so the outer
electron is more attracted and is closer to the nucleus (the electron cloud shrinks).
Equation:
Al3+ + IE4  Al4+ + e‒
When a successive IE
makes a large jump:
we are removing an inner
electron in the 2p subshell,
which is much closer to the
nucleus, rather than ones
in the 3s or 3p subshells.
Using successive ionization energies, we can predict that the Al3+ would be the favored and most stable ion
formed for aluminum.