7 Electron Configura
and the Properties
of Atoms
Unit Outline
7.1 Electron Spin and Magnetism
7.2 Orbital Energy
7.3 Electron Configuration of Elements
7.4 Properties of Atoms
7.5 Formation and Electron Configuration of Ions
In This Unit…
In Electromagnetic Radiation and the Electronic Structure of the Atom
(Unit 6) we introduced and explored the concept of orbitals, which
define the shapes electrons take around the nucleus of an atom. In this
unit we expand this description to atoms that contain more than one
electron and compare atoms that differ in their numbers of protons in the
nucleus and electrons surrounding that nucleus. Much of what we know
and can predict about the properties of an atom, such as its size and the
number and types of bonds it will form, can be derived from the number
and arrangement of the atom’s electrons and the energies of the atom’s
orbitals.
ations
s
m
h
s
e
e
w
e
r
s
AlbertSmirnov/iStockphoto.com
7.1 Electron Spin and Magnetism
7.1a Electron Spin and the Spin Quantum Number, m
Although electrons are too small to observe directly, we can detect the m
they exert. This magnetic field is generated by electron spin, the negativ
tron spinning on an axis (Interactive Figure 7.1.1).
The magnetic field produced by an electron occurs in only one o
indicating that electron spin is quantized. That is, an electron has o
spin states. In one spin state, the electron produces a magnetic field
pole in one direction. In the other spin state, the North pole is in the op
(Figure 7.1.2).
Spin states are defined by a fourth quantum number, the spin quant
Because there are two different spin states, ms has two possible values:
sign of ms is used to indicate the fact that the two spin states are in oppos
should not be confused with the negative charge on the electron. The ele
a negative charge, regardless of its spin.
Symbolizing Electron Spin
We saw in Electromagnetic Radiation and the Electronic Structure of th
that lines or boxes can be used to depict orbitals. Electrons in orbitals ar
rows, where the direction of the arrow indicates the spin state of the electr
in the diagram that follows, the upward (c) and downward (T) arrows ind
different spin states.
1s 2s
We will arbitrarily assign ms = +½ to electrons represented with
(also called “spin up” electrons) and ms = −½ to electrons represented
arrow (also called “spin down” electrons).
7.1b Types of Magnetic Materials
Magnetic materials derive their magnetic behavior from the magnetic p
electrons. Because all electrons produce a magnetic field, you might a
Why aren’t all materials magnetic? The answer lies in the fact that the
Unit 7 Electron Configurations and the Properties of Atoms
ms Interactive Figure 7.1.1 © 2013 Cengage Learning
Relate electron spin and magnetic
magnetic field that properties.
vely charged elec-
N
of two directions,
only two possible S
d with the North Electron spin and magnetic field
pposite direction
tum number, ms.
: +½ or −½. The
site directions and
ectron always has
he Atom (Unit 6)
re shown using ar-
ron. For example,
dicate electrons in
an upward arrow
with a downward
properties of their
ask the question,
e magnetic fields
194
© 2013 Cengage LearningInteractive Figure 7.1.3
Use spin states to predict magnetic properties.
(a) (b) (c)
Electron spin representations of a material with (a) diamagnetic,
(b) paramagnetic, and (c) ferromagnetic properties
generated by electrons with opposite spin (in a single atom, molecule
counteract and cancel each other. Therefore, any atom or molecule wit
of spin up and spin down electrons will have a net magnetic field of zer
with opposite spin are said to be spin paired and produce no net magn
even number of electrons leaves unpaired electrons. Materials with un
are magnetic.
The magnetism of most materials can be categorized as being diama
netic, or ferromagnetic. In a diamagnetic material (Interactive Figure 7.1
are spin paired and the material does not have a net magnetic field. Th
slightly repelled by the magnetic field of a strong magnet.
Paramagnetic materials (Interactive Figure 7.1.3b) contain atoms, m
with unpaired electrons. In the absence of an external, strong magnetic fi
fields generated by the individual particles are arranged in random directi
netism produced by each atom or molecule can be cancelled by the magn
it. This results in a magnetic material, but one with a weak net magnetic fi
presence of a strong, external magnet causes the individual spins to align
rial is attracted to the magnet.
Ferromagnetic materials (Interactive Figure 7.1.3c), like parama
contain particles with unpaired electrons. In these materials, however, th
netic fields align naturally and produce a strong, permanent magnetic fie
magnets you are familiar with are ferromagnets.
Unit 7 Electron Configurations and the Properties of Atoms
N Electrons S
SN © 2013 Cengage Learning
Figure 7.1.2 Electron spin and magnetic
e, or ion) directly fields
th equal numbers
ro. Two electrons Section 7.1 Mastery
netic field. An un-
npaired electrons
agnetic, paramag-
1.3a), all electrons
hese materials are
molecules, or ions
field, the magnetic
ions and the mag-
netic fields around
field. However, the
so that the mate-
agnetic materials,
he individual mag-
eld. The common
195
7.2 Orbital Energy
7.2a Orbital Energies in Single- and Multielectron Sp
The relationship between the principal quantum number, n, and orbital e
an orbital energy diagram (Figure 7.2.1). For a single-electron species su
atom, the energy of the atomic orbitals depends only on the value of n. F
orbital in a hydrogen atom has the same energy as a 2s orbital.
an orbital in a single-electron system depends only on the degree of attrac
electron in that orbital and the nucleus. This is mainly a function of the
of the electron from the nucleus, which is controlled by the principal qua
In multielectron species such as helium atoms or sodium ions, the
depend both on the principal quantum number, n, and the type of orb
angular momentum quantum number, ℓ (Interactive Figure 7.2.2). Multie
more complex because the energy of the electron depends on both how
is to the nucleus and the degree to which it experiences repulsive force
electrons present in the atom or ion.
As shown in Interactive Figure 7.2.2, subshell energies increase
ℓ (in a given energy level, s < p < d < f ). Using Interactive Figure 7.2.2 a
can make several generalizations about orbital energies in multielectron s
● As n increases, orbital energy increases for orbitals of the same type
A 4s orbital is higher in energy than a 3s orbital.
● As ℓ increases, orbital energy increases.
In the n = 3 shell, 3s < 3p < 3d.
● As n increases, the subshell energies become more closely spaced
occurs.
The 4f orbital is higher in energy than the 5s orbital, despite its lowe
7.3 Electron Configuration of Elements
7.3a The Pauli Exclusion Principle
The electron configuration of an element shows how electrons are distrib
which ones are filled and which ones remain vacant. We can predict the elect
of most elements, and we can use electron configurations to predict phys
properties of the elements.
Unit 7 Electron Configurations and the Properties of Atoms
4s 4p 4d 4f
3s 3p 3d
pecies
2s 2p
energy is shown in Energy
uch as a hydrogen
For example, a 2p © 2013 Cengage Learning
. The energy of 1s
ction between the
Figure 7.2.1 Orbital energies (n = 1 to n = 4)
average distance in a single-electron species
antum number, n.
e orbital energies Section 7.2 Mastery
bital, given by the
electron atoms are
close an electron
es with the other
e with increasing
as a reference, we
species.
e.
d and overlapping
er n value.
buted in orbitals—
tron configuration
sical and chemical
196
To predict the electron configuration for an atom’s ground state, t
state for an atom, electrons are put into the orbitals with the lowest energy
no more than two electrons in an orbital.
The order of subshell filling is related to n, the principal quantum nu
angular momentum quantum number. In general,
● electrons fill orbitals in order of increasing (n + ℓ) and
● when two or more subshells have the same (n + ℓ) value, electrons fi
the lower n value.
These general rules result in the following orbital filling order:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7
The Pauli exclusion principle states that no two electrons within a
the same set of four quantum numbers (n, ℓ, mℓ, and ms). The limits on p
the four quantum numbers means that a single orbital can accommodate n
electrons, and when an orbital contains two electrons, those electrons mu
spins (Figure 7.3.1).
Example Problem 7.3.1 Apply the Pauli exclusion principle.
a. Identify the orbitals in the following list that fill before the 4d orbitals: 3d, 4
b. What is the maximum number of electrons in the 4f subshell?
c. An electron in an orbital has the following quantum numbers: n = 3, ℓ = 1
ms = +½. Identify the orbital and the direction of electron spin (up or dow
Solution:
You are asked to answer questions about orbital filling and to apply the Paul
principle.
You are given information about specific orbitals or electrons.
a. 3d, 4s, and 5s. In general, orbitals fill in order of increasing (n + ℓ). When
subshells have the same (n + ℓ) value, electrons fill the orbital with the low
The order of filling is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p,
7p, 8s, ....
b. 14. There are 7 orbitals in the f subshell, and each orbital can hold two elec
c. 3p, up. The set of three quantum numbers n = 3, ℓ = 1, mℓ = +1 together
orbital. The electrons in the 3p orbital are oriented “spin up” (ms = +½) o
(ms = –½), resulting in two unique sets of four quantum numbers for the t
in this orbital.
Unit 7 Electron Configurations and the Properties of Atoms
the lowest energy Interactive Figure 7.2.2
y possible, placing
umber, and ℓ, the Identify orbital energies in multielectron
species.
fill the orbital with
7s 6d 5f
7p, 8s, … 6p 5d 4f
an atom can have 4d
possible values for 6sOrder of lling3d
no more than two 5p
ust have opposite © 2013 Cengage Learning
5s
4s, 5s, 6p. 4p
1, mℓ = +1, and
wn). 4s
3p
li exclusion
3s
two or more
wer n value. 2p
, 7s, 5f, 6d, 2s
ctrons.
r specifies a 3p 1s
or “spin down” Orbital energies in a multielectron species
two electrons
(a) (b)
Figure 7.3.1 Electron arrangements in an or-
bital that are (a) allowed and (b) not allowed
Video Solution
Tutored Practice
Problem 7.3.1
197
7.3b Electron Configurations for Elements in Periods
Hydrogen and Helium
Hydrogen has a single electron that occupies the orbital with the lowest ene
Two methods are used to represent this electron configuration. The spdf not
spectroscopic notation) has the general format nℓ#, where subshells are list
which they are filled and the number of electrons occupying each subshell is
of the subshell as a superscript. The spdf notation for hydrogen is
H: 1s1 (pronounced “one-ess-one”)
Orbital box notation uses boxes or horizontal lines to represent or
to represent electrons. The electron configuration of hydrogen in orbital
H:
1s
Helium has two electrons, and both occupy the lowest-energy 1s orb
configuration of helium in spdf notation and orbital box notation is theref
He: 1s2 (pronounced “one-ess-two”) n 5 1, < 5 0, m< 5 0,
1s n 5 1, < 5 0, m< 5 0,
Each electron in helium has a unique set of four quantum numbers, as req
exclusion principle. Notice that hydrogen and helium are in the first ro
table and both elements fill orbitals in the first energy level (1s).
Orbital box notations provide information about the number of pair
electrons in an atom, and that information can be used to determine whet
paramagnetic or diamagnetic. Hydrogen has one unpaired electron and i
species, whereas helium’s electrons are paired and it is diamagnetic.
Lithium to Neon
Lithium has three electrons, two in the 1s orbital and one that is in an orb
energy level. As shown previously, the 2s orbital is lower in energy than the
electron configuration of lithium in spdf notation and orbital box notation i
Li: 1s22s1
1s 2s
Unit 7 Electron Configurations and the Properties of Atoms
s 1–3
ergy, the 1s orbital.
tation (also called
ted in the order in
shown to the right
rbitals and arrows
box notation is
bital. The electron
fore
ms 5 1½
ms 5 2½
quired by the Pauli
ow of the periodic
red and unpaired
ther the atoms are
is a paramagnetic
bital in the second
2p orbitals, so the
is
198
Notice that it would be more correct to draw the orbital box notation el
tion of lithium as shown below because the 2s orbital is higher in en
orbital.
2s
1s
However, it is common to show all orbitals on a horizontal line when w
notation electron configurations in order to make more efficient use of sp
Beryllium has two electrons in the 1s and the 2s orbitals.
Be: 1s22s2
1s 2s
Boron has five electrons. Four electrons fill the 1s and 2s orbitals, and
is in a 2p orbital. Notice that the orbital box diagram shows all three 2p orb
only one of the 2p orbitals is occupied.
B: 1s22s22p1 2p
1s 2s
Carbon has six electrons, four in the 1s and 2s orbitals and two in
When electrons occupy a subshell with multiple orbitals such as 2p,
maximum multiplicity applies. This rule states that the lowest-energy
ration is the one where the maximum number of electrons is unpaire
carbon, this means that the two 2p electrons each occupy a diff
(Interactive Figure 7.3.2).
The electron configurations of nitrogen, oxygen, fluorine, and neon
Notice that the highest-energy orbital for all second-row elements is in th
level (2s or 2p).
N: 1s22s22p3 2p
1s 2s 2p
O: 1s22s22p4
1s 2s
Unit 7 Electron Configurations and the Properties of Atoms
lectron configura-
nergy than the 1s
writing orbital box
pace.
d the fifth electron
bitals even though
n the 2p orbitals. Interactive Figure 7.3.2
Hund’s rule of
electron configu- Apply Hund’s rule of maximum
ed. In the case of multiplicity.
ferent 2p orbital
C: 1s22s22p2 2p
n are shown here. 1s 2s
he second energy
Carbon’s electron configuration has two unpaired
electrons in the 2p orbitals.
199
F: 1s22s22p5 2p
1s 2s
Ne: 1s22s22p6 2p
1s 2s
Sodium to Argon
The elements in the third row of the periodic table have electrons that o
the third energy level, as shown.
Na: 1s22s22p63s1
1s 2s 2p 3s
Mg: 1s22s22p63s2
1s 2s 2p 3s
Al: 1s22s22p63s23p1
1s 2s 2p 3s 3p
Si: 1s22s22p63s23p2
1s 2s 2p 3s 3p
P: 1s22s22p63s23p3
1s 2s 2p 3s 3p
S: 1s22s22p63s23p4
1s 2s 2p 3s 3p
Cl: 1s22s22p63s23p5
1s 2s 2p 3s 3p
Ar: 1s22s22p63s23p6
1s 2s 2p 3s 3p
When electron configurations for multielectron species are written, nob
is often used to represent filled shells (these filled shells are also called co
noble gas notation, the symbol for a noble gas is written within square brack
spdf or orbital box notation representing additional, noncore electrons.
electron configuration for chlorine is written using noble gas notation as sh
Unit 7 Electron Configurations and the Properties of Atoms
occupy orbitals in
ble gas notation
ore electrons). In
kets in front of the
For example, the
hown here.
200
Cl: [Ne]3s23p5 [Ne] 3p
3s
The symbol [Ne] represents the 10 lowest-energy electrons (1s22s22p6
configuration.
Example Problem 7.3.2 Write electron configurations for period 1–3 e
Write the electron configuration for phosphorus using orbital box notation.
Solution:
You are asked to write the electron configuration for an element in orbital bo
You are given the identity of the element.
P:
1s 2s 2p 3s 3p
7.3c Electron Configurations for Elements in Periods
Both potassium and calcium have electron configurations similar to those
in Groups 1A and 2A.
K: [Ar]4s1 [Ar]
4s
Ca: [Ar]4s2 [Ar]
4s
Scandium is the first transition element, and it is the first element to fi
The electron configurations of the transition elements follow Hund’s rule:
Sc: [Ar]4s23d1 [Ar] 3d
4s
Ti: [Ar]4s23d2 [Ar] 3d
4s
Unit 7 Electron Configurations and the Properties of Atoms
6) in the electron Video Solution
elements.
ox notation. Tutored Practice
Problem 7.3.2
s 4–7
of other elements
fill the 3d orbitals.
:
201
V: [Ar]4s23d3 [Ar] 3d
4s
Cr: [Ar]4s13d5 [Ar] 3d
4s
Mn: [Ar]4s23d5 [Ar] 3d
4s
Fe: [Ar]4s23d6 [Ar] 3d
4s
Co: [Ar]4s23d7 [Ar] 3d
4s
Ni: [Ar]4s23d8 [Ar] 3d
4s
Cu: [Ar]4s13d10 [Ar] 3d
4s
Zn: [Ar]4s23d10 [Ar] 3d
4s
Both chromium and copper have electron configurations that do not f
filling order, for reasons that are complex and related to the similar energ
3d orbitals in multielectron atoms. The electron configurations for these el
memorized. The electron configurations for all elements are shown in Intera
There are a few exceptions to the general filling order in the heavier ele
elements follow the general guidelines that we have used to write electron
Unit 7 Electron Configurations and the Properties of Atoms
follow the general
gies of the 4s and
lements should be
active Table 7.3.1.
ements, but most
n configurations.
202
Interactive Table 7.3.1
Electron Configurations of Atoms in the Ground State
Z Element Configuration Z Element Configuration
1H 1s1 41 Nb [Kr]5s14d4
2 He 1s2 42 Mo [Kr]5s14d5
3 Li [He]2s1 43 Tc [Kr]5s24d5
4 Be [He]2s2 44 Ru [Kr]5s14d7
5B [He]2s22p1 45 Rh [Kr]5s14d8
6C [He]2s22p2 46 Pd [Kr]4d10
7N [He]2s22p3 47 Ag [Kr]5s14d10
8O [He]2s22p4 48 Cd [Kr]5s24d10
9F [He]2s22p5 49 In [Kr]5s24d105p1
10 Ne [He]2s22p6 50 Sn [Kr]5s24d105p2
11 Na [Ne]3s1 51 Sb [Kr]5s24d105p3
12 Mg [Ne]3s2 52 Te [Kr]5s24d105p4
13 Al [Ne]3s23p1 53 I [Kr]5s24d105p5
14 Si [Ne]3s23p2 54 Xe [Kr]5s24d105p6
15 P [Ne]3s23p3 55 Cs [Xe]6s1
16 S [Ne]3s23p4 56 Ba [Xe]6s2
17 Cl [Ne]3s23p5 57 La [Xe]6s25d1
18 Ar [Ne]3s23p6 58 Ce [Xe]6s25d14f 1
19 K [Ar]4s1 59 Pr [Xe]6s24f 3
20 Ca [Ar]4s2 60 Nd [Xe]6s24f 4
21 Sc [Ar]4s23d1 61 Pm [Xe]6s24f 5
22 Ti [Ar]4s23d2 62 Sm [Xe]6s24f 6
23 V [Ar]4s23d3 63 Eu [Xe]6s24f 7
24 Cr [Ar]4s13d5 64 Gd [Xe]6s25d14f 7
Unit 7 Electron Configurations and the Properties of Atoms
Z Element Configuration
81 Tl [Xe]6s25d104f 146p1 c
82 Pb [Xe]6s25d104f 146p2
83 Bi [Xe]6s25d104f 146p3
84 Po [Xe]6s25d104f 146p4
85 At [Xe]6s25d104f 146p5
86 Rn [Xe]6s25d104f 146p6
87 Fr [Rn]7s1
88 Ra [Rn]7s2
89 Ac [Rn]7s26d1
90 Th [Rn]7s26d2
91 Pa [Rn]7s26d15f 2
92 U [Rn]7s26d15f 3
93 Np [Rn]7s26d15f 4
94 Pu [Rn]7s25f 6
95 Am [Rn]7s25f 7
96 Cm [Rn]7s26d15f 7
97 Bk [Rn]7s25f 9
98 Cf [Rn]7s25f 10
99 Es [Rn]7s25f 11
100 Fm [Rn]7s25f 12
101 Md [Rn]7s25f 13
102 No [Rn]7s25f 14
103 Lr [Rn]7s26d15f 14
104 Rf [Rn]7s26d25f 14
203
b
Table 7.3.1 (continued)
Z Element Configuration Z Element Configuration
25 Mn [Ar]4s23d5 65 Tb [Xe]6s24f 9
26 Fe [Ar]4s23d6 66 Dy [Xe]6s24f 10
27 Co [Ar]4s23d7 67 Ho [Xe]6s24f 11
28 Ni [Ar]4s23d8 68 Er [Xe]6s24f 12
29 Cu [Ar]4s13d10 69 Tm [Xe]6s24f 13
30 Zn [Ar]4s23d10 70 Yb [Xe]6s24f 14
31 Ga [Ar]4s23d104p1 71 Lu [Xe]6s25d14f 14
32 Ge [Ar]4s23d104p2 72 Hf [Xe]6s25d24f 14
33 As [Ar]4s23d104p3 73 Ta [Xe]6s25d34f 14
34 Se [Ar]4s23d104p4 74 W [Xe]6s25d44f 14
35 Br [Ar]4s23d104p5 75 Re [Xe]6s25d54f 14
36 Kr [Ar]4s23d104p6 76 Os [Xe]6s25d64f 14
37 Rb [Kr]5s1 77 Ir [Xe]6s25d74f 14
38 Sr [Kr]5s2 78 Pt [Xe]6s15d94f 14
39 Y [Kr]5s24d1 79 Au [Xe]6s15d104f 14
40 Zr [Kr]5s24d2 80 Hg [Xe]6s25d104f 14
Example Problem 7.3.3 Write electron configurations for period 4–7 e
Write electron configurations for the following elements, in spdf notation and
tion. Identify the element as paramagnetic or diamagnetic.
a. Zn (Do not use the noble gas notation.)
b. Sm (Use the noble gas notation.)
Solution:
You are asked to write electron configurations in both spdf and orbital box n
identify the element as paramagnetic or diamagnetic.
You are given the identity of the element.
Unit 7 Electron Configurations and the Properties of Atoms
Z Element Configuration
105 Db [Rn]7s26d35f 14
106 Sg [Rn]7s26d45f 14
107 Bh [Rn]7s26d55f 14
108 Hs [Rn]7s26d65f 14
109 Mt [Rn]7s26d75f 14
110 Ds [Rn]7s26d85f 14
111 Rg [Rn]7s26d95f 14
112 Cn [Rn]7s26d105f 14
113 — [Rn]7s26d105f 147p1
114 — [Rn]7s26d105f 147p2
115 — [Rn]7s26d105f 147p3
116 — [Rn]7s26d105f 147p4
117 — [Rn]7s26d105f 147p5
118 — [Rn]7s26d105f 147p6
elements.
orbital box nota-
notation and to
c
204
b
Example Problem 7.3.3 (continued)
a. Zinc is element 30. Use the filling order to write the electron configuration,
the maximum number of electrons that can be accommodated in a subshell
spdf notation: 1s22s22p63s23p64s23d10
orbital box notation:
1s 2s 2p 3s 3p 4s
Zinc is diamagnetic because its electron configuration shows no unpaired
b. Samarium is element 62. Use the symbol [Xe] to represent the first 54 elect
tron configuration.
spdf notation: [Xe]6s24f 6
orbital box notation: [Xe] 4f
6s
Samarium is paramagnetic because its electron configuration shows six unp
7.3d Electron Configurations and the Periodic Table
Interactive Figure 7.3.3 shows the electron configurations for the elemen
written using spdf notation and noble gas notation.
Interactive Figure 7.3.3
Relate electron configuration and the periodic table.
1A
1
H
1s1 2A 3A 4A 5A 6A 7
34 5 67 8
Li Be BC NOF
[He]2s1 [He]2s2 [He]2s22p1 [He]2s22p2 [He]2s22p3 [He]2s22p4 [He]2
11 12 13 14 15 16
Na Mg S
Al Si P
[Ne]3s1 [Ne]3s2 [Ne]3s23p1 [Ne]3s23p2 [Ne]3s23p3 [Ne]3s23p4 [Ne]3
Electron configurations, periods 1–3.
Unit 7 Electron Configurations and the Properties of Atoms
, keeping in mind Video Solution
l.
Tutored Practice
3d Problem 7.3.3
electrons.
trons in the elec-
paired electrons.
nts in periods 1–3,
8A © 2013 Cengage Learning
2
He
7A 1s2
9 10
F Ne
2s22p5 [He]2s22p6
17 18
Cl Ar
3s23p5 [Ne]3s23p6
205
Notice that elements within a group have the following in common:
● They have the same number of electrons beyond the core electrons
noble gas notation) and similar electron configurations. For example
have one electron in addition to the core electrons, and both elements
electron configuration [noble gas]ns1.
● For the main group elements, the number of electrons beyond the
equal to the group number (with the exception of He).
The electrons beyond the core electrons are the valence electron
The valence electrons are the highest-energy electrons and are the electr
attracted to the nucleus. It is these electrons that are involved in chemi
the formation of chemical bonds. As shown in Interactive Figure 7.3.3, for
ments, the number of valence electrons is equal to the group number of
fact that elements within a group have similar electron configurations and
of valence electrons suggests that elements within a group have similar p
thing we will investigate later in this unit.
The orbital filling order is related to the structure of the periodic
in Interactive Figure 7.3.4. The periodic table can be divided into four
Interactive Figure 7.3.4
Identify electron configuration and valence electrons
for elements using the periodic table.
1s 3d 1s
2s 4d 2p
3s 5d 3p
4s 6d 4p
5s 5p
6s 6p
7s
4f © 2013 Cengage Learning
5f
s–block elements d–block elements (transition metals)
p–block elements f–block elements: lanthanides (4f )
and actinides (5f )
Subshells and the periodic table
Unit 7 Electron Configurations and the Properties of Atoms
s (represented by
e, both Li and Na
s have the general
core electrons is
ns for an element.
rons least strongly
ical reactions and
r the Group A ele-
the element. The
the same number
properties, some-
c table, as shown
blocks, indicated
206
by color in Interactive Figure 7.3.4, each of which represents the type o
filled with the highest-energy electrons for the elements in that blo
Groups 1A and 2A constitute the ns-block because the highest-energ
signed to the ns orbital. Groups 3A through 8A make up the np-block, wh
are filled last. The s-block and p-block elements are the Group A elemen
table and are commonly called the main-group elements. The transit
within the (n − 1)d-block, and the lanthanide and actinide elements mak
f-block.
The number of elements in each horizontal row within a block is relat
of orbitals within that block and the number of electrons that can fill t
example, the d-block is 10 elements across because the d subshell conta
and can accommodate a maximum of 10 electrons.
The periodic table can be used to generate electron configurations
from hydrogen to the desired element. Each element represents the
electron to an orbital. For example, arsenic, As, is a p-block element.
As, you begin with the first-row elements, H and He, each of which rep
tron in the 1s orbital (1s2). Next, Li and Be represent 2s2 and B throu
2p6. Similarly, Na and Mg represent 3s2 and Al through Ar represent 3
row of the periodic table, K and Ca represent 4s2, Sc through Zn repres
through As represent three electrons in the 4p orbital, 4p3. The electr
of arsenic is
As: 1s22s22p63s23p64s23d104p3 [Ar]4s23d104p3
7.4 Properties of Atoms
7.4a Trends in Orbital Energies
As shown in the previous section, the organization of the periodic table is
electron configurations. The energy of atomic orbitals is also related to th
periodic table.
As you move down within a group, the energy of highest-energy occ
creases. Likewise, the elements have greater numbers of electrons, and the
with higher n. As the value of n increases from element to element, the o
the electrons in these orbitals are farther from the nucleus, and the electro
energy orbitals experience weaker electron–nucleus attractions.
Unit 7 Electron Configurations and the Properties of Atoms
of subshell that is
ock. Elements in
gy electron is as-
here np subshells
nts in the periodic
tion elements sit
ke up the (n − 2)
ted to the number
that subshell. For
ains five d orbitals
by “counting up”
e addition of one
To “count up” to
presents an elec-
ugh Ne represent
3p6. In the fourth
sent 3d10, and Ga
ron configuration
Section 7.3 Mastery
closely related to
he structure of the
cupied orbitals in-
ey occupy orbitals
orbitals are larger,
ons in these high-
207
Interactive Figure 7.4.1
Explore orbital energies.
4s 4s 4s 4s 4s 4s 4s 4s
3p 3p 3p 3p 3p 3p 3p 3p
3s 3s 3sEnergy 3s 3s 3s 3s 3s
2p 2p 2p 2p 2p 2p 2p 2p
© 2013 Cengage Learning
2s 2s 2s 2s 2s 2s 2s 2s
1s 1s 1s 1s 1s
1s 1s 1s Si P S Cl Ar
Na Mg Al
Orbital energies, Na–Ar
Similarly, as you move left to right across the periodic table within a p
of all atomic orbitals decreases. Consider the elements in the third period
for example (Interactive Figure 7.4.1). Comparing Na to Mg, two factors
of the atomic orbitals. First, the Mg nucleus has one more proton than
which increases the nucleus–electron attractive forces. This will lower the
orbitals. Second, Mg has one more electron than Na, which increases the e
repulsions. This will increase the energy of atomic orbitals.
As shown in Interactive Figure 7.4.1, the Mg atomic orbitals are low
the Na atomic orbitals. This suggests that the attractive forces that resu
protons in the nucleus are more important than the repulsive forces fro
electrons. In general, when moving across a period from left to right, the
between the electrons and the nucleus increase and thus the orbital ener
Figure 7.4.2 shows the periodic trend for relative energy of atomic o
in the upper right of the periodic table have the lowest-energy orbitals
lower left have the highest-energy orbitals.
Unit 7 Electron Configurations and the Properties of Atoms
period, the energy Lowest © 2013 Cengage Learning
d, Na through Ar, energy
affect the energy orbitals
n the Na nucleus,
e energy of atomic Highest
electron–electron energy
orbitals
wer in energy than
ult from additional Figure 7.4.2 Relative orbital energy of the
om the additional elements
e attractive forces
rgies decrease. 208
orbitals; elements
and those in the
Effective Nuclear Charge
The combination of attractive forces between electrons and the nucle
electron repulsive forces is called the effective nuclear charge and is
Z*. The effective nuclear charge for the highest-energy electrons in an at
charge felt by those electrons, taking into account the electron–electron
between high-energy electrons and core electrons. A simplified method f
for the highest-energy electrons in an atom is
Z* = Z − [number of core electrons]
Using this simplified method, we can calculate values of Z and Z* for Na t
Na Mg Al Si P S Cl Ar
Z 11 12 13 14 15 16 17 18
Z* 1 2 3 4 5 6 7 8
In general, effective nuclear charge increases moving left to right ac
table. More sophisticated calculation methods for determining the electro
sive forces in an atom or ion result in Z* values that also increase moving l
the periodic table, but these values more accurately represent electron ene
Figure 7.4.3). As Z* increases, the electrons feel a strong attractive force
This has the effect of decreasing the energy of the atomic orbitals. As t
atomic orbitals decreases, the electrons are held closer to the nucleus a
tightly to the nucleus. This general trend in orbital energies controls mos
trends we see in atomic properties, as we will explore in the rest of this u
7.4b Atomic Size
The size of an atom is controlled by the size of its orbitals, but orbitals ar
no defined outer limit. Chemists use many different definitions for the siz
covalent radius of an element is the distance between the nuclei of two
ment when they are held together by a single bond. For example, the dista
bonded Cl atoms in Cl2 is 198 pm (Figure 7.4.4). Therefore, the covalent
atom is 99.0 pm. Other covalent radii are determined from atom distan
Because distances between bonded atoms vary from molecule to molecul
lent radii show average values.
The metallic radius of an element is the distance between the nucle
a metallic crystal. Tables of atomic radii generally report covalent radii fo
metallic radii for metals.
Unit 7 Electron Configurations and the Properties of Atoms
eus and electron–
given the symbol
tom is the nuclear
n repulsive forces
for calculating Z*
through Ar:
cross the periodic Interactive Figure 7.4.3
on-electron repul-
left to right across Explore effective nuclear charge.
ergies (Interactive Higher Z*, electrons
from the nucleus. held more tightly
the energy of the
and are held more © 2013 Cengage Learning
st of the observed
unit. Relative effective nuclear charge for the elements© 2013 Cengage Learning
Cl Cl
re boundless, with
ze of an atom. The 198 pm
atoms of that ele-
ance between two Figure 7.4.4 The distance between
two Cl atoms in Cl2 is 198 pm.
radius of each Cl
nces in molecules.
le, tables of cova-
ei of two atoms in
or nonmetals and
209
The size of an atom is closely related to its electron configuration. Rec
an orbital’s energy, the more tightly an electron in that orbital is held to th
smaller the orbital. Therefore, the trend in atomic size follows that of orbi
the smallest atoms in the top right of the periodic table and the largest at
left (Interactive Figure 7.4.5).
Interactive Figure 7.4.5
Explore atomic size of the elements.
1A
H, 37
2A 3A 4A 5A 6A 7A
Li, 152 Be, 113 B, 83 C, 77 N, 71 O, 66 F, 71
Na, 186 Mg, 160 Al, 143 Si, 117 P, 115 S, 104 Cl, 99
K, 227 Ca, 197 Ga, 122 Ge, 123 As, 125 Se, 117 Br, 114
Rb, 248 Sr, 215 In, 163 Sn, 141 Sb, 141 Te, 143 I, 133
Cs, 265 Ba, 217 Tl, 170 Pb, 154 Bi, 155 Po, 167 At, 140
Atom radii (pm)
Unit 7 Electron Configurations and the Properties of Atoms
call that the lower© 2013 Cengage Learning
he nucleus and the
ital energies, with
toms in the lower
8A
He, 32
Ne, 69
Ar, 97
Kr, 110
Xe, 130
Rn, 145
210
7.4c Ionization Energy
Ionization energy is the amount of energy required to remove an electro
atom.
X(g) S X+(g) + e−(g) ∆E = ionization energy
All atoms require energy to remove an electron, and therefore ionization
positive. The trends in ionization energies follow those expected based o
(Interactive Figure 7.4.6).
Notice that the elements farther down within a group, for example, mo
to Na, have decreased ionization energy. This trend is related to orbital ener
removed from sodium occupies a 3s orbital, which is higher in energy than
by lithium (2s), which in turn is higher than the energy of the electron lost
Ionization energy generally increases when moving left to right acro
trend is related both to orbital energies and to effective nuclear charge, Z*
decrease and Z* increases in the elements as you move left to right across a p
the farther right an element is, the more strongly the outermost electrons ar
nucleus and the higher its ionization energy. In general, a large, positive
value indicates that an element is more stable as a neutral atom than as a c
Interactive Figure 7.4.6
Explore ionization energy values.
Period
12 3
2500 He
First ionization energy (kJ/mol)
2000 Ne Ar
1500 © 2013 Cengage Learning Cl
F S
N
H Be C O P
1000
B Mg
500 Li Si
Na Al
0
5 10 15
Atomic number
Ionization energy values for elements in periods 1–3
Unit 7 Electron Configurations and the Properties of Atoms
on from a gaseous
energy values are
on orbital energies
oving from H to Li
rgies. The electron
n the electron lost
by hydrogen (1s).
oss a period. This
*. Orbital energies
period. Therefore,
re attracted to the
ionization energy
cation.
211
Notice that Interactive Figure 7.4.6 shows two areas where there are
general trend in ionization energy values: at the Group 3A elements and
elements. In both cases, elements have ionization energy values that are
be predicted based on the general trend.
The lower-than-expected ionization energy values for the Group 3A
explained by examining electron configurations of the Group 2A and 3A e
Group 2A: [noble gas]ns2 Group 3A: [noble gas]ns2np
The np electron removed from a Group 3A element has higher energy tha
removed from a Group 2A element. The np electron is easier to remove tha
resulting in ionization energies for the Group 3A elements that are lower th
The lower-than-expected ionization energy values for the Group 6A
explained by examining electron configurations of the Group 5A and 6A
Group 5A: [noble gas] np
ns
Group 6A: [noble gas] np
ns
The electron removed from a Group 6A element is in a filled np o
electron repulsive forces make it easier to remove this paired electron t
np electron removed from a Group 5A element; therefore, Group 6A elem
tion energies that are lower than predicted.
7.4d Electron Affinity
Electron affinity is the energy change when a gaseous atom gains an el
X(g) + e− S X−(g) ∆E = electron affinity
Although most elements have negative electron affinity values, some are s
close to zero. Keep in mind that a negative electron affinity value indicat
released when an electron is added to a gaseous atom and that a large,
affinity value indicates that an element is more stable as an anion than a
Not all atoms have a measurable electron affinity. Despite this, the trends
ties generally follow those of the other periodic properties we have examin
numerous exceptions) (Interactive Figure 7.4.7).
Elements lower in a group generally have less negative electron affi
those higher in a group.
Unit 7 Electron Configurations and the Properties of Atoms
exceptions to the
d at the Group 6A
lower than would
A elements can be
elements.
p1
an the ns electron
an the ns electron,
han predicted.
A elements can be
A elements.
orbital. Electron– Interactive Figure 7.4.7
than the unpaired
ments have ioniza- Explore electron affinity values.
lectron. 100 1 Period 3
2
slightly positive or
tes that energy is Electron af nity (kJ/mol)0 HeBe NNe Mg Ar
negative electron B Al
as a neutral atom. © 2013 Cengage Learning
in electron affini- –100 H Li Na P
ned thus far (with
C O Si
finity values than S
–200
–300
F Cl
–400 5 10 15
Atomic number
Electron affinity values for elements in periods 1–3
212
H: −72.8 kJ/mol Li: −59.6 kJ/mol Na: −52.9 kJ/mol K
As atomic size increases down a group, the orbital that is occupied by
increases in energy. As the orbital energy increases, the attractive forces
electron and the nucleus decrease, so electron affinity becomes less neg
adding electrons becomes less favorable the farther down a group the ele
Electron affinity values generally are more negative in elements farther
a period. There are some notable exceptions, however, in the Group 2A and
Group 2A: [noble gas]
ns
Group 5A: [noble gas] np
ns
The Group 2A elements have electron affinities that are very close to
added electron occupies the np orbital. Adding an electron to this higher-
quires energy, so the electron affinity is much lower (less negative) tha
general trends. The Group 5A elements have low (less negative) electron
cause the added electron fills a half-filled np orbital and introduces new
repulsion forces.
Notice that the Group 8A elements also have electron affinity values
zero. An electron added to a noble gas element would occupy the (n +
higher-energy level. Thus, the noble gases generally do not form anions.
7.5 Formation and Electron Configuratio
7.5a Cations
A cation forms when an atom loses one or more electrons. Metals have low
values, so metals generally form cations. For example, sodium, a Group 1A e
electron to form the sodium cation, Na+. Examination of the Na electron co
that the highest-energy electron occupies the 3s orbital. This is the sodium
and it is this electron that is lost upon formation of Na+.
Unit 7 Electron Configurations and the Properties of Atoms
K: −48.4 kJ/mol
the new electron
between the new
gative. In general,
ement is.
to the right across
d 5A elements.
o zero because the Section 7.4 Mastery
-energy orbital re-
an predicted from
affinity values be-
electron–electron
s that are close to
+ 1)s orbital in a
on of Ions
w ionization energy
element, loses one
onfiguration shows
valence electron,
213
3s IE1 5 496 kJ
2p
3s
2s
2p 1
2s
1s 1s
Na Na+
1s22s22p63s1 1s22s22p6 1 e–
The energy required to remove a second electron from an atom is the
energy, IE2. For any atom, IE2 > IE1. Sodium does not normally form a +2 c
second electron is removed from the 2p subshell, which is much lower in e
subshell. Removing an electron from an inner energy shell requires a great
this case almost 10 times as much as is required to remove the 3s valence e
3s IE2 5 4562 kJ
2p
3s
2s
2p 1
2s
1s 1s
Na+ Na2+
1s22s22p6 1s22s22p5 1 e–
Magnesium, a Group 2A element, loses both of its valence electrons to
Unit 7 Electron Configurations and the Properties of Atoms
J/mol
second ionization
cation because the
energy than the 3s
t deal of energy, in
electron.
J/mol
form a +2 cation.
214
3s 3s
2p
2p 1
2s
2s
1s 1s
Mg Mg2+
1s22s22p63s2 1s22s22p6 1 2 e–
Like sodium, magnesium is expected to easily lose its valence electrons, bu
from a lower-energy shell. As shown in Table 7.5.1, the first and second io
for Mg are relatively low, but the third ionization energy (IE3) is more than
than IE1.
In general, when metals form cations, they lose electrons from the high
pied orbitals, those with the highest n value. Transition metals have valenc
(n − 1)d, and in the case of lanthanides and actinides, (n − 2)f orbitals.
metal forms a cation, electrons are first lost from the outermost occupied o
with highest n). For example, when iron forms a +2 cation, it loses two 4s
4s 3d 4s 3d
3p 3p
3s 3s
2p 2p
2s 2s
1s 1s
Fe Fe2+
[Ar]4s23d 6 [Ar]3d6 1 2 e
Unit 7 Electron Configurations and the Properties of Atoms
ut not an electron Table 7.5.1 First, Second, and Third IE values
onization energies for Mg
n 10 times greater
Mg(g) S Mg+(g) + e− IE1 = 738 kJ/mol
hest-energy occu- Mg+(g) S Mg2+(g) + e− IE2 = 1451 kJ/mol
ce electrons in ns, Mg2+(g) S Mg3+(g) + e− IE3 = 7733 kJ/mol
When a transition
orbital (the orbital
s electrons.
d
1
e–
215
Recall that the 4s orbital is filled before the 3d orbitals are filled. Howev
removed from the highest-energy orbitals because those electrons feel th
tive “pull” from the positively charged nucleus.
Transition metals often form more than one cation. Iron is commonly
+2 and +3 oxidation states. In the formation of Fe3+, two 4s electrons an
are lost.
4s 3d 4s 3d
3p 3p
3s 3s
2p 2p
2s 2s
1s 1s
Fe2+ Fe3+
[Ar]3d 6 [Ar]3d5 1 e–
Notice that the electron removed to form Fe3+ from Fe2+ is taken from a
not a half-filled orbital. This minimizes the repulsive forces between elect
Example Problem 7.5.1 Write electron configurations for cations.
Write electron configurations for the following ions in spdf and orbital box not
a. Al3+ (Do not use noble gas notation.)
b. Cr2+ (Use noble gas notation.)
Solution:
You are asked to write an electron configuration for a cation using spdf and
notation.
You are given the identity of the cation.
a. Aluminum is element 13. The element loses three electrons from its highes
to form the Al3+ ion.
Al: 1s22s22p63s23p1
1s 2s 2p 3s 3p
Unit 7 Electron Configurations and the Properties of Atoms
ver, electrons are
he weakest attrac-
found in both the
nd one 3d electron
d
1
a filled 3d orbital,
trons.
tation.
orbital box
st-energy orbitals
c
216
b
Example Problem 7.5.1 (continued)
Al31: 1s22s22p6 2p
1s 2s
b. Chromium is element 24. The element loses two electrons from its highest-
to form the Cr2+ ion.
Cr: [Ar]4s13d5 [Ar] 3d
4s
Cr21: [Ar]3d4 [Ar] 3d
4s
7.5b Anions
An anion forms when an atom gains one or more electrons. Nonmetals hav
negative electron affinity values, so nonmetals generally form anions. For e
gains one electron to form the chloride ion, Cl−. Examination of the Cl el
tion shows that the 3p subshell in Cl contains a single vacancy. Cl theref
electron in a 3p orbital to form Cl−.
3p EA1 5 –349 k
3s 1 3p
3s
2p
2p
2s 2s
1s 1s
Cl Cl–
[Ne]3s23p5 1 e– [Ne]3s23p6
Sulfur has two 3p orbital vacancies and can therefore gain two electr
Unit 7 Electron Configurations and the Properties of Atoms
-energy orbitals Video Solution
ve relatively large, Tutored Practice
example, chlorine Problem 7.5.1
lectron configura-
fore gains a single
kJ/mol
rons.
217