Download Ionization Energy

Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 4
Periodic Trends
of the Elements
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
4
Periodic Trends of the Elements
4.1 Development of the Periodic Table
4.2 The Modern Periodic Table
Classification of Elements
4.3 Effective Nuclear Charge
4.4 Periodic Trends in Properties of Elements
Atomic Radius
Ionization Energy
Electron Affinity
Metallic Character
4.5 Electron Configuration of Ions
Ions of Main Group Elements
Ions of d-Block Elements
4.6 Ionic Radius
Comparing Ionic Radius with Atomic Radius
Isoelectronic Series
4.1
Development of the Periodic Table
In 1864, John Newlands noted that when the elements were
arranged in order of atomic number that every eighth element had
similar properties.
 He referred to this as the law of octaves.
In 1869, Dmitri Mendeleev and
Lothar Meyer independently
proposed the idea of periodicity.
Mendeleev grouped elements
(66) according to properties.
Mendeleev predicted properties
for elements not yet discovered,
such as Ga.
Development of the Periodic Table
However, Mendeleev could not explain inconsistencies such as
argon coming before potassium in the periodic table, despite having
a higher atomic mass.
In 1913, Henry Moseley discovered the correlation between the
number of protons (atomic number) and frequency of X-rays
generated.
Ordering the periodic table by atomic number instead of atomic
mass enabled scientists to make sense of discrepancies.
Entries today include atomic number and symbol; and are arranged
according to electron configuration.
Worked Example 4.1
What elements would you expect to exhibit properties most similar to those of
chlorine?
Strategy Because elements in the same group tend to have similar properties,
you should identify elements in the same group (7A) as chlorine.
Solution Fluorine, bromine, and iodine, the other nonmetals in Group 7A,
should have properties most similar to those of chlorine.
Think About It Astatine (At) is also in Group 7A. Astatine, though, is
classified as a metalloid, and we have to be careful comparing nonmetals to
metalloids (or to metals). As a metalloid, the properties of astatine should be
less similar to those of chlorine than the other members of Group 7A.
4.2
The Modern Periodic Table
Classification of Elements
The main group elements (also called the representative elements)
are the elements in Groups 1A through 7A.
Classification of Elements
The noble gases are found in Group 8A and have completely filled
p subshells.
The Modern Periodic Table
The transition metals are found in Group 1B and 3B through 8B.
 Group 2B have filled d subshells and are not transition metals.
The Modern Periodic Table
The lanthanides and actinides make up the f-block transition
elements.
The Modern Periodic Table
There is a distinct pattern to the electron configurations of the
elements in a particular group.
For Group 1A: [noble gas]ns1
For Group 2A: [noble gas]ns2
The Modern Periodic Table
The outermost electrons of an atom are called the valence electrons.
Valence electrons are involved in the formation of chemical bonds.
Similarity of valence electron configurations help predict chemical
properties.
For Group 1A: [noble gas]ns1
core
For Group 2A: [noble gas]ns2
core
valence
For Group 7A: [noble gas]ns2np5
core
valence
valence
Worked Example 4.2
Without using a periodic table, give the ground-state electron configuration and
block designation (s-, p-, d-, or f-block) of an atom with (a) 17 electrons, (b) 37
electrons, and (c) 22 electrons. Classify each atom as a main group element or
transition metal.
Strategy Use the figure at right to assign electrons to
orbitals in the correct order. Recall that an s subshell has
one orbital, a p subshell has three orbitals, and a d subshell
has five orbitals. Remember, too, that each orbital can
accommodate a maximum of two electrons.
Solution
(a) 1s22s22p63s23p5, p-block, main group
(b) 1s22s22p63s23p64s23d104p65s1, s-block, main group
(c) 1s22s22p63s23p64s23d2, d-block, transition metal
Think About It Consult the figure to confirm your answer.
4.3
Effective Nuclear Charge
Effective nuclear charge (Zeff) is the actual magnitude of positive
charge that is “experienced” by an electron in the atom.
In a multi-electron atom, electrons are simultaneously attracted to
the nucleus and repelled by one another.
 This results in shielding, where an electron is partially shielded
from the positive charge of the nucleus by the other electrons.
 Although all electrons shield one another to some extent, the
most effective are the core electrons.
 As a result, the value of Zeff increases steadily from left to right
because the core electrons remain the same but Z increases.
Z
Zeff
Li
Be
B
C
N
O
F
3
4
5
6
7
8
9
1.28
1.91
2.42
3.14
3.83
4.45
5.10
4.3
Effective Nuclear Charge
In general, the effective nuclear charge is given by
Zeff = Z – σ
 Z is the nuclear charge or simply the number of protons in the
nucleus.
 σ is the shielding constant.
Zeff increases from left to right across a period; changes very little
down a column.
Z
Zeff
Li
Be
B
C
N
O
F
3
4
5
6
7
8
9
1.28
1.91
2.42
3.14
3.83
4.45
5.10
4.4
Periodic Trends in Properties of Elements
Atomic radius is the distance between the nucleus of an atom and
its valence shell.
(a) Atomic radius in metals, or metallic radius,
is half the distance between the nuclei of two
adjacent, identical metal atoms.
(b) Atomic radius in nonmetals, or covalent
radius, is half the distance between adjacent,
identical nuclei connected by a chemical bond.
Atomic Radius
Atomic radii (in picometers)
The atomic radius increases
from top to bottom down a
group.
 Increasing n, so outermost
shell lies farther from the
nucleus
Atomic radius decreases from
left to right across a period.
 Increasing Zeff which
draws the valence shell
closer to the nucleus
Atomic Radius
Atomic radius decreases left to right across a period due to
increased electrostatic attraction between the effective nuclear
charge and the charge on the valence shell.
Worked Example 4.3
Referring only to a periodic table, arrange the elements P, S, and O in order of
increasing atomic radius.
Strategy Use the left-to-right (decreasing) and top-to-bottom (increasing) trends
to compare the atomic radii of two of the three elements at a time.
Solution
O<S<P
Think About It Consult the atomic radii figure to confirm the order. Note
that there are circumstances under which the trends alone will be insufficient
to compare the radii of two elements. Using only a periodic table, for
example, it would not be possible to determine that bromine (r = 114 pm) has
a smaller radius than silicon (r = 118 pm).
Ionization Energy
Ionization energy (IE) is the minimum energy required to remove
an electron from an atom in the gas phase.
The result is an ion, a chemical species with a net charge.
Na(g) → Na+(g) + e−
Sodium has an ionization energy of 495.8 kJ/mol.
Specifically, 495.8 kJ/mol is the first ionization energy of sodium,
IE1(Na), which corresponds to the removal of the most loosely
held electron.
Ionization Energy
Ionization Energy
In general, as Zeff increases, ionization energy also increases.
 Thus, IE1 increases from left to right across a period.
Ionization Energy
Within a given shell, electrons with a higher value of l are higher
in energy and thus, easier to remove.
Ionization Energy
Removing a paired electron is easier because of the repulsive
forces between two electrons in the same orbital.
Ionization Energy
It is possible to remove additional electrons in subsequent
ionizations, giving IE1, IE2, and so on.
Na(g) → Na+(g) + e−
IE1(Na) = 496 kJ/mol
Na+(g) → Na2+(g) + e−
IE2(Na) = 4562 kJ/mol
Ionization Energy
It takes more energy to remove the 2nd, 3rd, 4th, etc. electrons
because it is harder to remove an electron from a cation than an atom.
It takes much more energy to remove core electrons than valence.
 Core electrons are closer to nucleus.
 Core electrons experience greater Zeff because of fewer filled
shells shielding them from the nucleus.
Worked Example 4.4
Would you expect Na or Mg to have the greater first ionization energy (IE1)?
Which should have the greater second ionization energy (IE2)?
Strategy Consider effective nuclear charge and electron configuration to
compare the ionization energies. Na has one valence electron and Mg has two.
Solution
IE1(Mg) > IE1(Na) because Mg is to the right of Na in the periodic table (i.e., Mg
has the greater Zeff, so it is more difficult to remove its electron).
IE2(Na) > IE2(Mg) because the second ionization of Mg removes a valence
electron, whereas the second ionization of Na removes a core electron.
Think About It The first ionization energies of Na and Mg are 496 and 738
kJ/mol, respectively. The second ionization energies of Na and Mg are 4562
and 1451 kJ/mol, respectively.
Electron Affinity
Electron affinity (EA) is the energy released when an atom in the
gas phase accepts an electron.
Cl(g) + e− → Cl−(g)
Electron Affinity
Like ionization energy, electron affinity increases from left to right
across a period as Zeff increases.
 Easier to add an electron as the positive charge of the nucleus
increases.
Electron Affinity
It is easier to add an electron to an s orbital than to add one to a p
orbital with the same principal quantum number.
Electron Affinity
Within a p subshell, it is easier to add an electron to an empty
orbital than to add one to an orbital that already contains an
electron.
Electron Affinity
More than one electron may be added to an atom.
Process
Electron Affinity
O(g) + e− → O−(g)
EA1 = 141 kJ/mol
O− (g) + e− → O2−(g)
EA2 = −741 kJ/mol
While many first electron affinities are positive, subsequent
electron affinities are always negative.
 Considerable energy is required to overcome the repulsive
forces between the electron and the negatively charged ion.
Worked Example 4.5
For each pair of elements, indicate which one you would expect to have the
greater first electron affinity, EA1: (a) Al or Si, (b) Si or P.
Strategy Consider effective nuclear charge and electron configuration to
Think
About It energies.
The first (a)
electron
affinities
of Al,
are 4A. Al
compare
the ionization
Al is in
Group 3A
andSi,
Siand
is inPGroup
23p1), and Si has four valence electrons
42.5,
134, and
72.0 kJ/mol,
has three
valence
electrons
([Ne]3srespectively.
([Ne]3s23p2). (b) P is in Group 5A, so it has five valence electrons ([Ne]3s23p3).
Solution
EA1(Si) > EA1(Al) because Si is to the right of Al and therefore has a greater Zeff.
EA1(Si) > EA1(P) because although P is to the right of Si, adding an electron to a
P atom requires placing it in a partially occupied 3p orbital. The energy cost of
pairing electrons outweighs the energy advantage of adding an electron to an
atom with a larger Zeff.
Metallic Character
Metals tend to
 Be shiny, lustrous, malleable, and ductile
 Be good conductors of both heat and electricity
 Have low ionization energies (commonly form cations)
Metallic Character
Nonmetals tend to
 Vary in color and are not shiny
 Be brittle, rather than malleable
 Be poor conductors of both heat and electricity
 Have high electron affinities (commonly form anions)
Metallic Character
Metalloids are elements with properties intermediate between
those of metals and nonmetals.
Metallic Character
Many of the periodic trends of the elements can be explained using
Coulomb’s law, which states that the force (F) between two
charged objects (Q1 and Q2) is directly proportional to the product
of the two charges and inversely proportional to the distance (d)
between the objects squared.
Q1×Q2
Fα
d2
Worked Example 4.6
For carbon and nitrogen, use their effective nuclear charges and atomic radii to
compare the attractive force between the nucleus in each atom and the valence
electron that would be removed by the first ionization.
Strategy Zeff for C and N are 3.14 and 3.83, respectively; the radii of C and N
are 77 pm and 75 pm, respectively. IE1 are 1086 kJ/mol (C) and 1402 kJ/mol (N).
The charge on the valence electron in each case is -1.
Think About It The negative sign indicates that the force is
attractive rather than repulsive. The calculated number for
Solution
(3.14)×(-1)
-4 than for carbon.
nitrogen
is about 28 percent
larger
For C: F α
= -5.3×10
2
(77 pm)
(3.83)×(-1)
-4
For N: F α
=
-6.8×10
2
(75 pm)
Note that it doesn’t matter what units we use for the distance between the charges.
We are not trying to calculate a particular attractive force, only compare the
magnitude of these two attractive forces.
4.5
Electron Configurations of Ions
To write the electron configuration of an ion formed by a main
group element:
1) Write the configuration for the atom.
2) Add or remove the appropriate number of electrons.
Na: 1s22s22p63s1
Na+: 1s22s22p6
10 electrons total,
isoelectronic with Ne
Cl: 1s22s22p63s23p5
Cl− : 1s22s22p63s23p6
18 electrons total,
isoelectronic with Ar
Worked Example 4.7
Write electron configurations for the following ions of main group elements:
(a) N3-, (b) Ba2+, and (c) Be2+.
Strategy First write electron configurations for the atoms. Then add electrons
(for anions) and remove electrons (for cations) to account for the charge.
Solution
(a) [He]2s22p6 or [Ne]
(b) [Kr]5s24d105p6 or [Xe]
(c) 1s2 or [He]
Think About It Be sure to add electrons to form an anion, and remove electrons
to form a cation.
Ions of d-Block Elements
Ions of d-block elements are formed by removing electrons first
from the shell with the highest value of n.
For Fe to form Fe2+, two electrons are lost from the 4s subshell not
the 3d.
Fe: [Ar]4s23d 6
Fe2+: [Ar]3d 6
Fe can also form Fe3+, in which case the third electron is removed
from the 3d subshell.
Fe: [Ar]4s23d 6
Fe3+: [Ar]3d 5
Worked Example 4.8
Write electron configurations for the following ions of d-block elements: (a) Zn2+,
(b) Mn2+, and (c) Cr3+.
Strategy First write electron configurations for the atoms. Then add electrons
(for anions) and remove electrons (for cations) to account for the charge. The
electrons removed from a d-block element must come first from the outermost s
subshell, not the partially filled d subshell.
Solution
(a) [Ar]3d10
(b) [Ar]3d5
(c) [Ar]3d3
Think About It Be sure to add electrons to form an anion, and remove
electrons to form a cation. Also, double-check to make sure that electrons
removed from a d-block elment come first from the ns subshell and then, if
necessary, from the (n – 1)d subshell.
4.6
Ionic Radius
The ionic radius is the radius
of a cation or an anion.
When an atom loses an
electron to become a cation,
its radius decreases due in
part to a reduction in
electron-electron repulsions
in the valence shell.
A significant decrease in
radius occurs when all of an
atom’s valence electrons are
removed.
Comparing Ionic Radius with Atomic Radius
When an atom gains one or
more electrons and becomes
an anion, its radius increases
due to increased electronelectron repulsions.
Isoelectronic Series
An isoelectronic series is a series of two or more species that have
identical electron configurations, but different nuclear charges.
O2− : 1s22s22p6
F−: 1s22s22p6
isoelectronic
Ne: 1s22s22p6
Worked Example 4.9
Identify the isoelectronic series in the following group of species, and arrange
them in order of increasing radius: K+, Ne, Ar, Kr, P3-, S2-, and Cl-.
Strategy Isoelectronic series are species with identical electron configurations
but different nuclear charges. Determine the number of electrons in each species.
The radii of isoelectronic series members decreases with increasing nuclear
charge.
Solution
The isoelectronic series includes K+, Ar, P3-, and Cl- (18 electrons each).
In order of increasing radius: K+ < Ar < Cl- < S2- < P3-.
Think About It With identical electron configurations, the attractive force
between the valence electrons and the nucleus will be strongest for the largest
nuclear charge. Thus, the larger the nuclear charge, the closer in the valence
electrons will be pulled and the smaller the radius will be.
4
Chapter Summary: Key Points
Development of Periodic Table
The Modern Periodic Table
Classification of Elements
Effective Nuclear Charge
Periodic Trends in Properties of Elements
Atomic Radius
Ionization Energy
Electron Affinity
Metallic Character
Electron Configuration of Ions
Ions of Main Group Elements
Ions of d-Block Elements
Ionic Radius
Comparing Ionic Radius with Atomic Radius
Isoelectronic Series
Group Quiz #4
• Write electron configurations for the
following atoms/ions.
• K (long-hand)
• S2- (long-hand)
• Cu (short-hand)
• Al3+ (long-hand)
• Fe2+ (short-hand)
48
Group Quiz #5
• For each periodic trend indicated,
identify the atom or ion with a larger
value:
• Radius: Cl or Cl1• Radius: O2- or Na1+
• Ionization Eng: Na or K
• Ionization Eng: Ca or Se
• Electron Affinity: C or F
49