Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 8
Periodic
Properties of
the Elements
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2007, Prentice Hall
Mendeleev
• order elements by atomic mass
• saw a repeating pattern of properties
• Periodic Law – When the elements are arranged in
•
•
•
order of increasing atomic mass, certain sets of
properties recur periodically
put elements with similar properties in the same
column
used pattern to predict properties of undiscovered
elements
where atomic mass order did not fit other properties,
he re-ordered by other properties
 Te & I
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Periodic Pattern
nm H2O
a/b
H
1
H2
m Li2O m/nm BeOnm B2O3 nm CO2 nm N2O5 nm
O2 nm
Li b
Be a/b B a
C a N a
O
F
7 LiH 9 BeH2 11 ( BH3)n 12 CH4 14 NH3 16 H2O 19 HF
m Na2O m MgO m Al2O3 nm/m SiO2nm P4O10nm SO3 nm Cl2O7
Na b Mg b Al a/b Si a P a
S a Cl a
23 NaH24 MgH2 27 (AlH3) 28 SiH4 31 PH3 32 H2S 35.5 HCl
m = metal, nm = nonmetal, m/nm = metalloid
a = acidic oxide, b = basic oxide, a/b = amphoteric oxide
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Mendeleev's Predictions
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What vs. Why
• Mendeleev’s Periodic Law allows us to predict
what the properties of an element will be based
on its position on the table
• it doesn’t explain why the pattern exists
• Quantum Mechanics is a theory that explains
why the periodic trends in the properties exist
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Electron Spin
• experiments by Stern and Gerlach showed a beam
of silver atoms is split in two by a magnetic field
• the experiment reveals that the electrons spin on
their axis
• as they spin, they generate a magnetic field
spinning charged particles generate a magnetic field
• if there is an even number of electrons, about half
the atoms will have a net magnetic field pointing
“North” and the other half will have a net
magnetic field pointing “South”
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Electron Spin Experiment
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Spin Quantum Number, ms
• spin quantum number describes how the
electron spins on its axis
clockwise or counterclockwise
spin up or spin down
• spins must cancel in an orbital
paired
• ms can have values of ±½
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Pauli Exclusion Principle
• no two electrons in an atom may have the same set of
•
•
4 quantum numbers
therefore no orbital may have more than 2 electrons,
and they must have with opposite spins
knowing the number orbitals in a sublevel allows us to
determine the maximum number of electrons in the
sublevel
 s sublevel has 1 orbital, therefore it can hold 2 electrons
 p sublevel has 3 orbitals, therefore it can hold 6 electrons
 d sublevel has 5 orbitals, therefore it can hold 10 electrons
 f sublevel has 7 orbitals, therefore it can hold 14 electrons
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Allowed Quantum Numbers
Quantum
Number
Principal, n
Values
Number of Significance
Values
1, 2, 3, ...
distance from
nucleus
Azimuthal, l 0, 1, 2, ..., n-1
n
shape of
orbital
Magnetic, ml -l,...,0,...+l
2l + 1
orientation of
orbital
Spin, ms
-½, +½
2
direction of
electron spin
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Quantum Numbers of
Helium’s Electrons
•
•
•
•
helium has two electrons
both electrons are in the first energy level
both electrons are in the s orbital of the first energy level
since they are in the same orbital, they must have opposite spins
first
electron
second
electron
n
l
ml
ms
1
0
0
+½
1
0
0
-½
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Electron Configurations
• the ground state of the electron is the lowest
•
•
•
•
•
energy orbital it can occupy
the distribution of electrons into the various
orbitals in an atom in its ground state is called its
electron configuration
the number designates the principal energy level
the letter designates the sublevel and type of orbital
the superscript designates the number of electrons
in that sublevel
He = 1s2
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Orbital Diagrams
• we often represent an orbital as a square and the
electrons in that orbital as arrows
 the direction of the arrow represents the spin of the
electron
unoccupied
orbital
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orbital with
1 electron
orbital with
2 electrons
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Sublevel Splitting in
Multielectron Atoms
• the sublevels in each principal energy level of
Hydrogen all have the same energy – we call orbitals
with the same energy degenerate
 or other single electron systems
• for multielectron atoms, the energies of the sublevels
are split
 caused by electron-electron repulsion
• the lower the value of the l quantum number, the less
energy the sublevel has
 s (l = 0) < p (l = 1) < d (l = 2) < f (l = 3)
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Penetrating and Shielding
• the radial distribution function shows that
•
•
•
the 2s orbital penetrates more deeply into
the 1s orbital than does the 2p
the weaker penetration of the 2p sublevel
means that electrons in the 2p sublevel
experience more repulsive force, they are
more shielded from the attractive force of
the nucleus
the deeper penetration of the 2s electrons
means electrons in the 2s sublevel
experience a greater attractive force to the
nucleus and are not shielded as
effectively
the result is that the electrons in the 2s
sublevel are lower in energy than the
electrons in the 2p
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Penetration & Shielding
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7s
6s
Energy
5s
4s
3s
2s
1s
6p
5p
6
d
5d
5f
4f
4d
4p
3d
3p Notice the following:
1. because of penetration, sublevels within
an energy level are not degenerate
2. penetration of the 4th and higher energy
2p
levels is so strong that their s sublevel is
lower in energy than the d sublevel of the
previous energy level
3. the energy difference between levels
becomes smaller for higher energy levels
Order of Subshell Filling
in Ground State Electron Configurations
start by drawing a diagram
putting each energy shell on
a row and listing the subshells,
(s, p, d, f), for that shell in
order of energy, (left-to-right)
next, draw arrows through
the diagonals, looping back
to the next diagonal
each time
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1s
2s
2p
3s
3p
3d
4s
4p
4d
4f
5s
5p
5d
5f
6s
6p
6d
7s
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Filling the Orbitals with Electrons
• energy shells fill from lowest energy to high
• subshells fill from lowest energy to high
s → p → d → f
 Aufbau Principle
• orbitals that are in the same subshell have the same
•
energy
no more than 2 electrons per orbital
 Pauli Exclusion Principle
• when filling orbitals that have the same energy, place
one electron in each before completing pairs
 Hund’s Rule
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Example 8.1 – Write the Ground State
Electron Configuration and Orbital Diagram
and of Magnesium.
1. Determine the atomic number of the element
from the Periodic Table
 This gives the number of protons and electrons in
the atom
Mg Z = 12, so Mg has 12 protons and 12 electrons
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Example 8.1 – Write the Ground State
Electron Configuration and Orbital
Diagram and of Magnesium.
2. Draw 9 boxes to represent the first 3 energy
levels s and p orbitals
a) since there are only 12 electrons, 9 should be
plenty
1s
2s
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2p
3s
3p
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Example 8.1 – Write the Ground State
Electron Configuration and Orbital
Diagram and of Magnesium.
3. Add one electron to each box in a set, then
pair the electrons before going to the next set
until you use all the electrons
 When pair, put in opposite arrows


1s
2s
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  
2p

3s
3p
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Example 8.1 – Write the Ground State
Electron Configuration and Orbital
Diagram and of Magnesium.
4. Use the diagram to write the electron
configuration
 Write the number of electrons in each set as a
superscript next to the name of the orbital set
1s22s22p63s2 = [Ne]3s2


1s
2s
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  
2p

3s
3p
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Valence Electrons
• the electrons in all the subshells with the
highest principal energy shell are called the
valence electrons
• electrons in lower energy shells are called
core electrons
• chemists have observed that one of the most
important factors in the way an atom
behaves, both chemically and physically, is
the number of valence electrons
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Electron Configuration of Atoms
in their Ground State
• Kr = 36 electrons
1s22s22p63s23p64s23d104p6
 there are 28 core electrons and 8 valence electrons
• Rb = 37 electrons
•
•
1s22s22p63s23p64s23d104p65s1
[Kr]5s1
for the 5s1 electron in Rb the set of quantum numbers is
n = 5, l = 0, ml = 0, ms = +½
for an electron in the 2p sublevel, the set of quantum
numbers is n = 2, l = 1, ml = -1 or (0,+1), and ms = - ½
or (+½)
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Electron Configurations
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Electron Configuration & the
Periodic Table
• the Group number corresponds to the number of
valence electrons
• the length of each “block” is the maximum
number of electrons the sublevel can hold
• the Period number corresponds to the principal
energy level of the valence electrons
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s1
1
2
3
4
5
6
7
s2
p 1 p 2 p 3 p 4 p 5 s2
p6
d1 d2 d3 d4 d5 d6 d7 d8 d9 d10
f2 f3 f4 f5 f6 f7 f8 f9 f10 f11 f12 f13 f14 f14d1
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Electron Configuration from
the Periodic Table
8A
1A
1
2
3
4
5
6
7
3A 4A 5A 6A 7A
2A
Ne
P
3s2
3p3
P = [Ne]3s23p3
P has 5 valence electrons
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Transition Elements
• for the d block metals, the principal energy level is one less than
valence shell
 one less than the Period number
 sometimes s electron “promoted” to d sublevel
Zn
Z = 30, Period 4, Group 2B
[Ar]4s23d10
4s
3d
• for the f block metals, the principal energy level is two
less than valence shell
 two less than the Period number they really belong to
 sometimes d electron in configuration
Eu
Z = 63, Period 6
6s
4f
[Xe]6s24f 7
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Electron Configuration from
the Periodic Table
8A
1A
1
2
3
4
5
6
7
3A 4A 5A 6A 7A
2A
3d10
Ar
As
4s2
4p3
As = [Ar]4s23d104p3
As has 5 valence electrons
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Practice – Use the Periodic Table to write the short
electron configuration and orbital diagram for each of the
following
• Na (at. no. 11)
• Te (at. no. 52)
• Tc (at. no. 43)
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Practice – Use the Periodic Table to write the short
electron configuration and orbital diagram for each of the
following
• Na (at. no. 11) [Ne]3s1
3s
• Te (at. no. 52) [Kr]5s24d105p4
5s
4d
5p
• Tc (at. no. 43) [Kr]5s24d5
5s
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4d
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Properties & Electron Configuration
• elements in the same
column have similar
chemical and physical
properties because they
have the same number of
valence electrons in the
same kinds of orbitals
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Electron Configuration &
Element Properties
• the number of valence electrons largely determines the behavior
of an element
 chemical and some physical
• since the number of valence electrons follows a Periodic pattern,
•
•
the properties of the elements should also be periodic
quantum mechanical calculations show that 8 valence electrons
should result in a very unreactive atom, an atom that is very
stable – and the noble gases, that have 8 valence electrons are all
very stable and unreactive
conversely, elements that have either one more or one less
electron should be very reactive – and the halogens are the most
reactive nonmetals and alkali metals the most reactive metals
 as a group
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Electron Configuration &
Ion Charge
• we have seen that many metals and nonmetals
form one ion, and that the charge on that ion is
predictable based on its position on the Periodic
Table
Group 1A = +1, Group 2A = +2, Group 7A = -1,
Group 6A = -2, etc.
• these atoms form ions that will result in an
electron configuration that is the same as the
nearest noble gas
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Electron Configuration of Anions
in their Ground State
• anions are formed when atoms gain enough
electrons to have 8 valence electrons
filling the s and p sublevels of the valence shell
• the sulfur atom has 6 valence electrons
S atom = 1s22s22p63s23p4
• in order to have 8 valence electrons, it must gain
2 more
S2- anion = 1s22s22p63s23p6
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Electron Configuration of
Cations in their Ground State
• cations are formed when an atom loses all its
valence electrons
resulting in a new lower energy level valence shell
however the process is always endothermic
• the magnesium atom has 2 valence electrons
Mg atom
= 1s22s22p63s2
• when it forms a cation, it loses its valence electrons
Mg2+ cation = 1s22s22p6
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Trend in Atomic Radius – Main Group
• Different methods for measuring the radius of an
atom, and they give slightly different trends
 van der Waals radius = nonbonding
 covalent radius = bonding radius
 atomic radius is an average radius of an atom based on
measuring large numbers of elements and compounds
• Atomic Radius Increases down group
 valence shell farther from nucleus
 effective nuclear charge fairly close
• Atomic Radius Decreases across period (left to right)
 adding electrons to same valence shell
 effective nuclear charge increases
 valence shell held closer
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Effective Nuclear Charge
• in a multi-electron system, electrons are simultaneously
•
•
•
attracted to the nucleus and repelled by each other
outer electrons are shielded from full strength of nucleus
screening effect
effective nuclear charge is net positive charge that is
attracting a particular electron
Z is nuclear charge, S is electrons in lower energy levels
 electrons in same energy level contribute to screening, but
very little
 effective nuclear charge on sublevels trend, s > p > d > f
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Zeffective = Z - S
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Screening & Effective Nuclear Charge
43
Trends in Atomic Radius
Transition Metals
• increase in size down the Group
• atomic radii of transition metals roughly the
same size across the d block
must less difference than across main group
elements
valence shell ns2, not the d electrons
effective nuclear charge on the ns2 electrons
approximately the same
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Example 8.5 – Choose the
Larger Atom in Each Pair
1)
2)
3)
4)
N or F
F, N is further left
C or Ge
Ge, Ge is further down
N or Al,
Al Al is further down & left
Al or Ge? opposing trends
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Electron Configuration of
Cations in their Ground State
• cations form when the atom loses electrons
from the valence shell
• for transition metals electrons, may be removed
from the sublevel closest to the valence shell
Al atom =
Al+3 ion =
Fe atom =
Fe+2 ion =
Fe+3 ion =
Cu atom =
Cu+1 ion =
1s22s22p63s23p1
1s22s22p6
1s22s22p63s23p64s23d6
1s22s22p63s23p63d6
1s22s22p63s23p63d5
1s22s22p63s23p64s13d10
1s22s22p63s23p63d10
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Magnetic Properties of
Transition Metal Atoms & Ions
• electron configurations that result in unpaired electrons
mean that the atom or ion will have a net magnetic field –
this is called paramagnetism
 will be attracted to a magnetic field
• electron configurations that result in all paired electrons
mean that the atom or ion will have no magnetic field –
this is called diamagnetism
 slightly repelled by a magnetic field
• both Zn atoms and Zn2+ ions are diamagnetic, showing
that the two 4s electrons are lost before the 3d
 Zn atoms [Ar]4s23d10
 Zn2+ ions [Ar]4s03d10
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Example 8.6 – Write the Electron Configuration
and Determine whether the Fe atom and Fe3+ ion
are Paramagnetic or Diamagnetic
• Fe Z = 26
• previous noble gas = Ar
18 electrons
• Fe3+atom
ion = [Ar]4s023d56
• unpaired electrons
• paramagnetic
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4s
3d
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Trends in Ionic Radius
• Ions in same group have same charge
• Ion size increases down the group
 higher valence shell, larger
• Cations smaller than neutral atom; Anions bigger
•
than neutral atom
Cations smaller than anions
 except Rb+1 & Cs+1 bigger or same size as F-1 and O-2
• Larger positive charge = smaller cation
 for isoelectronic species
 isoelectronic = same electron configuration
• Larger negative charge = larger anion
 for isoelectronic series
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1A
-1
Periodic Pattern - Ionic Radius (Å)
+1
2A
H
+1
+2
3A
4A
+3
+2
K 1.33 Ca 0.99
0.62 +3
Ga +1
0.81 +3 0.71 +4
Rb 1.47 Sr 1.13 In +1 Sn +2
+1
-2
-3
P 2.12 S
-3
-1
-1
1.84 Cl 1.81
-2
-1
As 2.22 Se 1.98 Br 1.96
+2
+2 0.95 +3 0.84 +4
Cs 1.69 Ba 1.35 Tl +1 Pb +2
-2
7A
N 1.71 O 1.40 F 1.33
-4
Ge
6A
-3
-4
Li 0.68 Be 0.31 B 0.23 C
+1
+2
+3
-4
Na 0.97 Mg 0.66 Al 0.51 Si
+1
5A
-2
Sb
+1
Bi
Te 2.21
-1
I 2.20
53
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Ionization Energy
• minimum energy needed to remove an electron
from an atom
gas state
endothermic process
valence electron easiest to remove
M(g) + IE1  M1+(g) + 1 eM+1(g) + IE2  M2+(g) + 1 efirst ionization energy = energy to remove electron from
neutral atom; 2nd IE = energy to remove from +1 ion; etc.
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General Trends in 1st Ionization Energy
• larger the effective nuclear charge on the
electron, the more energy it takes to remove it
• the farther the most probable distance the
electron is from the nucleus, the less energy it
takes to remove it
• 1st IE decreases down the group
valence electron farther from nucleus
• 1st IE generally increases across the period
effective nuclear charge increases
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58
Example 8.8 – Choose the Atom in Each
Pair with the Higher First Ionization Energy
1)
2)
3)
4)
Al or S
S, Al is further left
As or Sb
Sb, Sb is further down
N or Si
Si, Si is further down & left
O or Cl? opposing trends
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Irregularities in the Trend
• Ionization Energy generally increases from left
to right across a Period
• except from 2A to 3A, 5A to 6A
Be 
1s

2s

1s

2s
B
N

1s

2s
  
2p
O

1s

2s
  
2p
2p

2p
Which
Which is
is easier
easier to
to remove
remove an
an electron
electron from
from B
N or
or Be?
O? Why?
Why?
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Irregularities in the
First Ionization Energy Trends
Be  
Be+  
1s
2s
2p
1s
2s
2p
To ionize Be you must break up a full sublevel, cost extra energy
  
B+  
1s
2s
2p
1s
2s
2p
When you ionize B you get a full sublevel, costs less energy
B
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Irregularities in the
First Ionization Energy Trends
    
N+    
1s
2s
2p
1s
2s
2p
To ionize N you must break up a half-full sublevel, cost extra energy
N
O

1s

2s
  
2p
O+ 
1s

2s
  
2p
When you ionize O you get a half-full sublevel, costs less energy
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Trends in Successive
Ionization Energies
• removal of each successive
electron costs more energy
 shrinkage in size due to having more
protons than electrons
 outer electrons closer to the nucleus,
therefore harder to remove
• regular increase in energy for each
•
successive valence electron
large increase in energy when start
removing core electrons
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Trends in Electron Affinity
• energy released when an neutral atom gains an electron
 gas state
 M(g) + 1e-  M-1(g) + EA
• defined as exothermic (-), but may actually be
endothermic (+)
 alkali earth metals & noble gases endothermic, WHY?
• more energy released (more -); the larger the EA
• generally increases across period
 becomes more negative from left to right
 not absolute
 lowest EA in period = alkali earth metal or noble gas
 highest EA in period = halogen
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• Metals
Metallic Character
 malleable & ductile
 shiny, lusterous, reflect light
 conduct heat and electricity
 most oxides basic and ionic
 form cations in solution
 lose electrons in reactions - oxidized
• Nonmetals
 brittle in solid state
 dull
 electrical and thermal insulators
 most oxides are acidic and molecular
 form anions and polyatomic anions
 gain electrons in reactions - reduced
• metallic character increases left
• metallic character increase down
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Example 8.9 – Choose the
More Metallic Element in Each Pair
1)
2)
3)
4)
Sn or Te
Te, Sn is further left
P or Sb,
Sb Sb is further down
Ge or In
In, In is further down & left
S or Br? opposing trends
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Trends in the Alkali Metals
• atomic radius increases down the column
• ionization energy decreases down the column
• very low ionization energies
 good reducing agents, easy to oxidize
 very reactive, not found uncombined in nature
 react with nonmetals to form salts
 compounds generally soluble in water  found in seawater
• electron affinity decreases down the column
• melting point decreases down the column
 all very low MP for metals
• density increases down the column
 except K
 in general, the increase in mass is greater than the increase
in volume
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2 Na(s) + 2 H2O(l)  2 NaOH(aq) + H2(g)
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Trends in the Halogens
• atomic radius increases down the column
• ionization energy decreases down the column
• very high electron affinities
 good oxidizing agents, easy to reduce
 very reactive, not found uncombined in nature
 react with metals to form salts
 compounds generally soluble in water  found in seawater
• reactivity increases down the column
• react with hydrogen to form HX, acids
• melting point and boiling point increases down the
•
column
density increases down the column
 in general, the increase in mass is greater than the increase in
volume
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Example 8.10 – Write a balanced chemical
reaction for the following.
• reaction between potassium metal and bromine
gas
K(s) + Br2(g) 
K(s) + Br2(g)  K+ Br
2 K(s) + Br2(g)  2 KBr(s)
(ionic compounds are all solids at room temperature)
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Example 8.10 – Write a balanced chemical
reaction for the following.
• reaction between rubidium metal and liquid
water
Rb(s) + H2O(l) 
Rb(s) + H2O(l)  Rb+(aq) + OH(aq) + H2(g)
2 Rb(s) + 2 H2O(l)  2 Rb+(aq) + 2 OH(aq) + H2(g)
(alkali metal ionic compounds are soluble in water)
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Example 8.10 – Write a balanced chemical
reaction for the following.
• reaction between chlorine gas and solid iodine
Cl2(g) + I2(s) 
Cl2(g) + I2(s)  ICl
write the halogen lower in the column first
assume 1:1 ratio, though others also exist
2 Cl2(g) + I2(s)  2 ICl(g)
(molecular compounds found in all states at room
temperature)
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Trends in the Noble Gases
• atomic radius increases down the column
• ionization energy decreases down the column
 very high IE
• very unreactive
 only found uncombined in nature
 used as “inert” atmosphere when reactions with other gases
would be undersirable
• melting point and boiling point increases down the
column
 all gases at room temperature
 very low boiling points
• density increases down the column
 in general, the increase in mass is greater than the increase in
volume
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