程序代写案例-CHEM 120
时间:2021-03-08
CHEM 120 Module 3: Multielectron atoms and the periodic table page 1 of 17

3.1 Multielectron Atoms

The results obtained for the H atom help us understand how one e behaves when it
interacts with a single nucleus. We will extend these ideas to understand and explain the
properties of multielectron atoms. Recall:
 the state of the e is characterized using the quantum numbers (n, , m ,ms)
 the e occupies a certain region of space (i.e. it occupies a particular “orbital”)
In principle, it is an “easy” matter to extend the approach of the last module from the
hydrogen atom (with only one electron) to the helium atom (with two electrons). The
nuclear charge increases by one, from Z=+1 to Z=+2, and the number of electrons
increases by one. There are only three interactions to worry about: electron 1 interacting
with the nucleus; electron 2 interacting with the nucleus; and electron 1 interacting with
electron 2. The Schrodinger equation for the system can be written down immediately, as
it is a simple generalization of the Schrodinger equation given previously for the hydrogen
atom. (There are more coordinates because there are more particles, but that’s about it.)
All we need to do is solve the Schrodinger equation and analyze the results. Writing down
the Schrodinger equation for the helium atom (or any multi-electron atom) is not that hard.
The hard part is solving it, and the difficulty arises from the electron-electron interactions.
Fortunately, we do not have to solve the Schrodinger equation for every atom to
understand how the electrons in a multielectron atom “pack around” the nucleus.
A conceptual model for multi-electron atoms




Some fundamental questions that we must answer include:
 How many electrons can occupy the same orbital?
 What arrangement of electrons gives the atom the lowest possible energy?
We know intuitively that each electron would like to be close to the nucleus but at the
same time, electrons would like to avoid each other. Before we consider how the
electrons in an atom “distribute” themselves amongst the various possible orbitals, it is
instructive to consider how the electrons in a multielectron atom influence each other.
A many-electron atom has a full set of hydrogen-like orbitals at its disposal (i.e. 1s, 2s, 2p, 3s, …)
governed by the same quantum numbers as before, (n, , m ,ms). Into these orbitals will go the
electrons of the atom, each electron occupying a certain region of space and experiencing an
average “attraction” towards the nucleus and an average “repulsion” from the other electrons.
CHEM 120 Module 3: Multielectron atoms and the periodic table page 2 of 17

Penetration and shielding
Penetration is a measure of how close a particular electron gets to the nucleus. Shielding indicates how
thoroughly a particular electron is blocked (or shielded) from the nucleus. In a multielectron atom, the
electrons shield each other from feeling the full nuclear charge. Careful examination of the RDF’s below
shows that:
 a 1s e is more likely to be found close to the nucleus than is a 2s or 2p e. Consequently, a 1s e−
“shields” the 2s and 2p electrons from feeling the full nuclear charge.
 a 2p electron partially shields a 2s electron from the feeling the full nuclear charge. (For the H atoms,
we see that rmp ≈ 200 pm for 2p and rmp ≈ 300 pm for 2s)
 a 2s e “penetrates” towards the nucleus more effectively than does a 2p e. Therefore, a 2s electron
partially shields a 2p electron. Since a 2s electron penetrates more effectively than does a 2p
electron, a 2s electron will actually feel more of the nuclear charge than will a 2p electron.













The position of the maximum in a plot of the RDF versus r tells us the most probable radial distance
between the electron and the nucleus. The area under the curve between two distances, say r1 and r2,
provides a measure of the amount of time the electron spends between r1 and r2. For example, a 1s
electron in the H atom spends most of its time between 0 and 150 pm and very little time at distances
greater than 150 pm. A 2s electron spends most of its time between 100 pm and 600 pm but it does
spend some of its time between 0 and 100 pm.





r r
RDF
1s 2s 2p
There is a reasonable probability that a
2s electron will penetrate close to the
nucleus (e.g. r < 50 pm).
RDF
CHEM 120 Module 3: Multielectron atoms and the periodic table page 3 of 17

Orbital energy diagram for multielectron atoms

















Some terminology: Shells and subshells
n defines a “shell” and so we have an “n = 1 shell”, an “n = 2 shell”, etc. etc.
defines a subshell. For each shell, we have n different subshells because for a given value of n,
the quantum number can have one of n different values: 0, 1, … n – 1. For example, for the n = 3
shell, we have three subshells: a “3s subshell”, “a 3p subshell” and a “3d subshell”.

A few important points:
 the energy and size of a given orbital (e.g. 1s) both decrease as Z increases
As Z increases, the attraction between the nucleus and any given electron increases. The
increased attraction lowers the energy of the orbital and causes the orbital to contract.
 For a multielectron atom, the orbitals in the same “shell” are not of the same energy
(s < p < d < f, etc.)
Because of electron-electron repulsions, the subshells are not of the same energy in a
multi-electron atom. In a given shell, the “s” subshell is lowest in energy, then “p”, then
“d”, etc.



1s
1s
2s
2s
2p
2p
3s 3p 3d
3s 3p
3d
H atom
(Z = 1)
Multielectron atom
(Z > 1)
Energy
CHEM 120 Module 3: Multielectron atoms and the periodic table page 4 of 17

3.2 Ground state electron configurations for neutral atoms
How do electrons distribute themselves to obtain the lowest possible energy for the atom?
There are three basic rules that help us predict how the electrons in an atom distribute themselves among
the possible states to achieve the lowest possible energy for the atom.

(1) The Pauli Exclusion Principle
No two electrons can have the same set of quantum numbers n, , m and ms.
Note: If two electrons occupy the same orbital, then n1 = n2 , 1 = 2 and m 1
= m
2
. Therefore, we
must have ms1 ≠ ms2 . i.e., If two electrons occupy the same orbital, they must have different spins.

subshell # orbitals? ( = 2 + 1) max # electrons?
ns 1 2
np 3 6
nd 5 10


(2) The Aufbau Procedure
For neutral atoms, orbitals are filled according to the n + rule:
filling order

1s 2s 2p 3s 3p 4s 3d 4p 5s 4d
(n + ) = 1 2 3 3 4 4 5 5 5 6
max. # e’s? 2 2 6 2 6 2 10 6 2 10
The triangle shown below can be used to
help you remember the orbital filling order.
There are only two exceptions to the (n+ )
rule for Z  36. The exceptions are Cr and
Cu. You should know these two exceptions.
For Z  36, there are many exceptions! (You
don’t need to know, nor could you possibly
know, all of the exceptions for Z > 36.)

1s

2s 2p

3s 3p 3d

4s 4p 4d 4f

5s 5p 5d 5f 5g
n

Did you know? The orbital energies themselves do not follow
the n rule. The rule corresponds simply to the order in which
orbitals must be filled to give the lowest possible energy for the
atom. The orbital filling rule has been established by experiment.
CHEM 120 Module 3: Multielectron atoms and the periodic table page 5 of 17

(3) Hund’s Rule

If there are not enough electrons to fill completely a set of energetically degenerate orbitals,
the lowest energy arrangement is the one which has the maximum number of parallel spins.

For example, we have learned that a set of 2p orbitals can accommodate a total of 6 electrons.
(There are three 2p orbitals and each one can accommodate two electrons.) The three 2p orbitals
are of equal energy. We say that the three 2p orbitals are energetically degenerate.




What are the possible arrangements for two electrons in these three orbitals? There are several
possibilities. Here are a few possible arrangements.
















Q. Which one of the arrangements above is the lowest in energy?
Q. In arrangement “B”, does it matter whether both spins are up or both spins are down?

When electrons have parallel spins, they avoid each other to a greater extent. Thus, they shield each
other less and this increases the attraction each electron feels towards the nucleus. The increased
electron-nucleus attraction results in a lower total energy for the atom.
2px 2py 2pz
2px 2py 2pz
2px 2py 2pz
2px 2py 2pz
We can generate four more arrangements
like this one by putting the two electrons in
2px and 2pz or in 2px and 2pz instead.
We can generate other arrangements that are
equivalent to this one (e.g. put one electron in
2py with its spin up and one electron in 2px with
its spin down; or, put one electron in 2pz with its
spin down and one electron in 2py with its spin
up; etc. etc.)
We can generate two more arrangements
like this one by putting the two electrons in
2py or 2pz instead.
A
B
C
CHEM 120 Module 3: Multielectron atoms and the periodic table page 6 of 17

Example 3-1: What is the ground-state electron configuration for an Ar
atom? What is the ground-state electron configuration for a Ni atom?
Determine the number of unpaired electrons. (The ground-state configuration
is the configuration of lowest energy.)
































See problems 8-77 and 8-79
from Petrucci (11th edition).
CHEM 120 Module 3: Multielectron atoms and the periodic table page 7 of 17

3.3 The Periodic Table






























The rows of the periodic table are called periods. There are seven periods.
Each period starts a new shell.

The columns are called groups. The groups are numbered 1-18.

Remarks

 The group number provides info about the number of valence
electrons in the atom.





 Atoms in the same group have similar spdf configurations.

e.g. F has ground state config. [He] 2s2 2p5
Cl has ground state config. [Ne] 3s2 3p5

 The periodic table can be divided into s-, p-, d- and f-blocks

e.g. Ca is in the s-block because the last e− added goes into an s orbital
V is in the d-block because the last e− added goes into a d orbital

Valence electrons are electrons
in the outermost shell of an atom.
They are the electrons that
participate in bond formation.




group #, if group # < 12
# valence e
group # 10, if group # 12



In the table above, * is used to identify
elements that have a ground state
configuration that is different from that
expected from its position in the
periodic table. For example, the
configuration expected for Cr, based
on its position, is [Ar] 4s2 3d4. The
actual configuration is [Ar] 4s1 3d5.


G = 1


18
1
H
1s1


G =2


G = 13


14


15


16


17
2
He
1s2
3
Li
2s1
4
Be
2s2
5
B
2s2 2p1
6
C
2s2 2p2
7
N
2s2 2p3
8
O
2s2 2p4
9
F
2s2 2p5
10
Ne
2s2 2p6
11
Na
3s1
12
Mg
3s2


G = 3



4


5


6



7


8



9


10



11


12
13
Al
3s2 3p1
14
Si
3s2 3p2
15
P
3s2 3p3
16
S
3s2 3p4
17
Cl
3s2 3p5
18
Ar
3s2 3p6
19
K
4s1
20
Ca
4s2
21
Sc
4s2 3d1
22
Ti
4s2 3d2
23
V
4s2 3d3
24
Cr*

25
Mn
4s2 3d5
26
Fe
4s2 3d6
27
Co
4s2 3d7
28
Ni
4s2 3d8
29
Cu*

30
Zn
4s2 3d10
31
Ga

32
Ge

33
As

34
Se

35
Br

36
Kr

37
Rb
5s1
38
Sr
5s2
39
Y
5s2 4d1
40
Zr
5s2 4d2
41
Nb*

42
Mo*

43
Tc
5s2 4d5
44
Ru*

45
Rh*

46
Pd*

47
Ag*

48
Cd
5s2 4d10
49
In

50
Sn

51
Sb

52
Te

53
I

54
Xe

55
Cs
6s1
56
Ba
6s2
57-71
La-Lu
72
Hf

73
Ta

74
W

75
Re

76
Os

77
Ir

78
Pt*

79
Au*

80
Hg

81
Tl

82
Pb

83
Bi

84
Po

85
At

86
Rn

87
Fr
7s1
88
Ra
7s2
89-103
Ac-Lr
104
Rf
105
Db
106
Sg
107
Bh
108
Hs
109
Mt
110
Ds
111
Rg
112
Cn
113
Nh
114
Fl
115
Mc
116
Lv
117
Ts
118
Og

57
La*

58
Ce

59
Pr

60
Nd

61
Pm

62
Sm

63
Eu

64
Gd*

65
Tb

66
Dy

67
Ho

68
Er

69
Tm

70
Yb

71
Lu

89
Ac*

90
Th*

91
Pa*

92
U*

93
Np*

94
Pu

95
Am

96
Cm*

97
Bk

98
Cf

99
Es

100
Fm

101
Md

102
No

103
Lr



d block
(n−1)d G−2 ns2
s block
nsG
p block (He is part of the s block.)
ns2 npG−12 for n  3
(n−1)d10 ns2 npG−12 for n > 4

f block

CHEM 120 Module 3: Multielectron atoms and the periodic table page 8 of 17

Learn to use the periodic table to write abbreviated e configurations quickly!

Example 3-2: Use the periodic table to write the abbreviated ground state electron
configurations for As and Fe. How many unpaired electrons does each atom have?























Example: Based on their positions in the periodic table, what do you
expect for the ground-state configurations of Cr and Cu?
















What conclusions can we draw from this example? Not many – except for the points already made
above. A number of textbooks refer to a supposed “special stability” of half-filled subshells. This is a
“cop-out” because it begs the question “What is the origin of this special stability?” The mythical concept
of a special stability of half-filled subshells and filled shells has been demolished but it still lingers in some
books. (Note: The ground-state electron configuration of W is [Xe] 6s2 4f14 5d4 not [Xe] 6s1 4f14 5d5. The
special stability of half-filled subshells is obviously a myth!)
This example illustrates that the Aufbau
procedure does not always give the
correct result.

We expect you to know these two
exceptions (Cr and Cu) but not any
other exceptions to the building up
principle. By their very nature,
exceptions are difficult to predict with
certainty. However, please be aware
that exceptions do exist. The
exceptions are numerous for Z > 36,
and the majority of exceptions involve
atoms from the f-block.
Cr
Aufbau  [Ar] 4s2 3d4
Expt  [Ar] 4s1 3d5
Cu
Aufbau  [Ar] 4s2 3d9
Expt  [Ar] 4s1 3d10
See problems 8-75, 8-85
and 8-86 from Petrucci
(11th edition).
CHEM 120 Module 3: Multielectron atoms and the periodic table page 9 of 17

3.4 Ground-state electron configurations for monatomic ions
When dealing with monatomic ions (e.g. O, Cl, S+, Ni2+, etc.), the following rules are used to predict the
correct ground-state electron configuration.
(i) negative monatomic ions (monatomic anions)
Write the e configuration for the neutral atom first. Then use the (n + ) rule to add e’s to
the appropriate orbitals.
(ii) positive monatomic ions (monatomic cations)
Write the e configuration for the neutral atom first. Then remove electrons from orbitals with
the highest value of n first. If we must choose between removing two e’s that have the same
value of n, then remove the e with the highest value of first.
Rule (ii), has particular significance for cations derived from atoms of the d-block elements.
For atoms in the d block, the valence shell electron configuration is usually ns2 (n−1)d G−2,
where G is the group number. Experiment shows that when electrons are removed from
atoms of the d block, the ns electrons are removed first.

Example 3-3: What is the ground-state electron configuration for
each of the following ions?
(a) S The ground state e− configuration of S is [Ne] 3s2 3p4

So, for S− the ground state e− configuration is …

(b) Br+ The ground state e− configuration of Br is [Ar] 4s2 3d 10 4p5

So, for Br+, the ground state e− configuration is …

(c) Mn2+ The ground state e− configuration of Mn is [Ar] 4s2 3d 5.

So, for Mn2+, the ground state e− configuration is …


Notice that Mn2+ has 23 electrons. If you had “blindly” applied the (n+ )-rule to Mn2+, then you would
predict the ground-state configuration [Ar] 4s2 3d3, which is wrong!! That is, when you use the (n+ )-rule
with 23 electrons, you get the ground electronic state for the neutral atom that has 23 electrons (i.e.
vanadium, V) and not the electron configuration for Mn2+!!
See problem B14 from the
Extra Problems for CHEM 120.
CHEM 120 Module 3: Multielectron atoms and the periodic table page 10 of 17

3.5 Atomic Properties & Periodic Trends

The electron configurations of atoms change in a predictable way as we
move across a period and down a group. It should not be too surprising
that there are systematic and predictable variations in atomic properties.
We shall consider the magnetic properties of atoms as well as trends in
atomic radii, ionization energies and electron affinities.


A. Paramagnetic and Diamagnetic Atoms

As a result of electron spin, each electron produces a magnetic field.

If all of the electrons in an atom are paired up, then the magnetic fields
cancel so that there is no net magnetic field for the atom.

If there are unpaired electrons, then the magnetic fields do not cancel.


diamagnetic atoms 





paramagnetic atoms 






Example 3-4: Which groups of atoms are diamagnetic in their ground
electronic states?

group 2: ns2

group 12: ns2 (n−1)d10

group 18: 1s2 ns2 np6 4s2 3d10 4p6 ns2 (n−2)f14 (n−1)d10 np6
He Ne, Ar Kr, Xe Rn


FYI: There are two elements in the f-block (Yb and No) that are also diamagnetic, but we don’t
expect you to remember these two special cases.


all electrons are paired; the atom does not possess a magnetic moment;
the atom interacts only weakly with an external magnetic field
We say the atom
possesses a
magnetic moment.
one or more unpaired electrons; the atom possesses a magnetic
moment; the atom interacts strongly with an external magnetic field
See problem 9-35 from
Petrucci (11th edition).
CHEM 120 Module 3: Multielectron atoms and the periodic table page 11 of 17

B. Atomic Radii

The radius of an atom is a little more difficult to define than you might
think. There is no precise outer boundary to an atom and even if there
was, we couldn’t really measure the size of a single atom anyway. The
size of an atom is determined by measuring the distance between
nuclei in a certain environment.


covalent radius
(of X)

The covalent radius is especially useful for elements that form
diatomic molecules.


metallic radius


The metallic radius is especially useful for elements that are metals.
Most elements in the periodic table are metals at room temperature
and atmospheric pressure.


Remarks:


 atomic radii decrease across a period

As we move from left to right across a period, we are adding electrons into
vacancies in an existing shell. The nuclear charge increases from left to
right, and because electrons in the same shell do not shield each other
effectively (only partially), the increased nuclear charge brings all the
electrons closer to the nucleus.

 atomic radii increase down a group

As we move from top to
bottom within a group, we
begin a new shell and each
new shell has its maximum
electron density a greater
distance from the nucleus.











= ½ of the diatomic bond length for the X2 molecule
= ½ of the distance between “nearest neighbours” in a metallic solid
For you to think about …
In order to answer the question
“How big is an atom?”, one
must first answer the questions
“How are we going to measure
the size of the atom?” and
“How hard are we going to
push on the atom when we
measure its size?” If we are
going to measure the size of
an overripe peach using a pair
of calipers, for example, then
the value we get depends on
how hard we squeeze. The
size we measure depends on
the environment of the atom
and how we plan to make the
measurement!!
CHEM 120 Module 3: Multielectron atoms and the periodic table page 12 of 17

C. Ionization Energy







e.g. Consider the ionization energies of the magnesium atom, Mg.

Mg(g) → Mg+(g) + e− ∆H = 738 kJ mol−1  IE(1) = 738 kJ mol−1

Mg+(g) → Mg2+(g) + e− ∆H = 1451 kJ mol−1  IE(2) = 1451 kJ mol−1

overall: Mg(g) → Mg2+(g) + 2e− ∆Htot = IE(1) + IE(2) = 2189 kJ mol−1


Remarks:

 It becomes increasingly difficult to remove electrons if one or more
electrons have already been removed.

So, IE(1) < IE(2) < IE(3) < …

 Ionization energies decrease down a group

The valence electrons of larger atoms occupy orbitals with higher n values and
these electrons have higher energies. The valence electrons of larger atoms
are more easily removed.

 Ionization energies generally increase across a period (but there are
easily explained exceptions!)

The atomic radius decreases as we move left to right across a period. The
outer electrons are closer to the nucleus and held more strongly, so more
energy is required to remove them from the atom.

















ionization
energy
(IE)
energy required to remove an electron from a gas-phase atom
(often reported as an enthalpy change, ΔH)
It always takes energy to remove an
electron from an atom. Therefore,
ionization is always an endothermic
process and ionization energies are
always positive.
Li
Be
B
C
N
O
F
Ne
IE decreases as we go from
group 2 to group 13 and from
group 15 to group 16.
CHEM 120 Module 3: Multielectron atoms and the periodic table page 13 of 17

Why does IE decrease as we move from Be to B?

Be has configuration [He] 2s2

B has configuration [He] 2s2 2p1


Why does IE decrease as we move from N to O?

This may not be the appropriate question to ask. As suggested by the diagram on the preceding page,
not only is the first ionization of O lower than that of N, but the first ionization energies of O, F and Ne are
all lower than expected. That is, the first ionization energies of O, F and Ne are considerably lower than
the values predicted by extrapolating the first ionization energies of B, C and N. Thus, it may be more
appropriate to ask: Why are the first ionization energies of O, F and Ne lower than expected? We
shall consider two explanations, both of which focus on the on the fact that the ionization of O, F and Ne
involves the removal of a paired 2p electron whereas the ionization of B, C or N involves the removal of
an unpaired 2p electron.

According to one explanation, electron-electron repulsions are the key consideration. Paired electrons
occupy the same orbital and are, on average, closer together than electrons in separate orbitals. Thus,
they experience “extra” repulsion and are more easily removed. Although the electron-electron repulsions
increase with the number of electrons in the 2p orbitals, there is a significant increase in the electronic
repulsions, and a corresponding decrease in ionization energy, once orbital sharing begins, that is, as we
proceed from N([He] 2s2 2p3) to O([He] 2s2 2p4).

Another explanation focuses instead on the extent of shielding and the strength of the electron-nucleus
attractions. Unpaired electrons with parallel spins tend to avoid each other, screen each other less,
experience a higher effective nuclear charge, interact more strongly with the nucleus, and are harder to
remove. According to this line of reasoning, the “extra” electron-nuclear attraction causes the ionization
energy for N([He] 2s2 2p3) to be greater than expected by the backward-extrapolation of the first ionization
energies of O, F, and Ne. Conversely, because paired electrons do not avoid each other to the same
extent, they shield each other to a greater extent, interact less strongly with the nucleus and are more
easily removed.

Is the observed “dip” in ionization energy that occurs as we move from group 15 to 16 caused by
increased electron-electron repulsions or by decreased electron-nucleus attractions? It is difficult to
answer this question with complete certainty, but it is clear that the dip occurs once orbital sharing begins.
The difficulty of providing an unambiguous explanation for the observed dip in IE(1) is not totally
unexpected. As we have already pointed out, the energy of an atom is a delicate balance of electron-
electron repulsions and electron-nucleus attractions. The rationalization of ionization energies is further
complicated by the fact that, for example, the ionization energy of N depends on the energies of both
The 2p electron in B is in a higher energy orbital and
is easier to remove than the 2s electron in Be.
CHEM 120 Module 3: Multielectron atoms and the periodic table page 14 of 17

N([He] 2s2 2p3) and N+([He] 2s2 2p2). Similarly, the ionization energy for the O atom depends on the
energies of both O([He] 2s2 2p4) and O+([He] 2s2 2p3). Thus, the drop in ionization energy that occurs as
we move from N to O depends upon a delicate balance of electron-electron repulsions and electron-
nucleus attractions in four different species.




B has configuration [He] 2s2 2p1



C has configuration [He] 2s2 2p2



N has configuration [He] 2s2 2p3



O has configuration [He] 2s2 2p4



F has configuration [He] 2s2 2p5



Ne has configuration [He] 2s2 2p6







Although we have not provided a definitive answer to the question “Why does the ionization
energy decrease as we move from N to O?”, it is important to remember that the decrease in
ionization energy occurs once the sharing of 2p orbitals begins.





2p 2s
Ionization of B, C or N
involves the removal
of an unpaired 2p
electron.
Ionization of O, F or Ne
involves the removal
of a paired 2p electron.
CHEM 120 Module 3: Multielectron atoms and the periodic table page 15 of 17

D. Electron Affinities





e.g. Consider the electron affinities of C, O, and F.

C(g) + e− → C−(g) EA(1) = −122 kJ mol−1 is the 1st electron affinity of C

O(g) + e− → O−(g) EA(1) = −141 kJ mol−1 is the 1st electron affinity of O

O−(g) + e− → O2−(g) EA(2) = +744 kJ mol−1 is the 2nd electron affinity of O

F(g) + e− → F−(g) EA(1) = −328 kJ mol−1 is the 1st electron affinity of F


Remarks: (focusing on the “main group” elements, i.e. the “s” and “p” blocks)

 Most atoms (except for N and those in groups 2 and 18) release energy
when they acquire an electron.

group 2: ns2 filled subshell

group 18: ns2 np6 filled shell


 For atoms near the right-hand end of a period (except for the noble
gases in group 18), a large quantity of energy is released when an
electron is added.




 For atoms near the left-hand end of a period, only a small quantity of
energy is released. (For group 2, energy is required to add an
electron to the atom!)











electron
affinity
(EA)
energy change that accompanies the addition of an electron
to a gas-phase atom
These atoms do not “want” more e−’s.
For O and F, the EA’s are quite large.
For Li, Na, K, etc. the EA’s are
quite small.


1
EA values for the main group elements
18
H
−73

2

13

14

15

16

17
He
> 0
Li
−60
Be
> 0
B
−27
C
−122
N
+7
O
−141
F
−328
Ne
> 0
Na
−53
Mg
> 0
Al
−43
Si
−134
P
−72
S
−200
Cl
−349
Ar
> 0
K
−48
Ca
−2
Ga
−29
Ge
−119
As
−78
Se
−195
Br
−325
Kr
> 0
Rb
−47
Sr
−5
In
−29
Sn
−107
Sb
−103
Te
−190
I
−295
Xe
> 0
Cs
−46
Ba
−14
Tl
−19
Pb
−35
Bi
−91
Po
−186
At
−270
Rn
> 0

CHEM 120 Module 3: Multielectron atoms and the periodic table page 16 of 17

Summary of Periodic Trends

















































See problem 9-42 from
Petrucci (11th edition).
A summary of trends in atomic radius, first ionization energy, electron affinity, electronegativity and atomic
polarizability. Atomic radiusa and polarizabilityb decrease from left to right in a given a period and increase from top to
bottom in a given group. Ionization energy, electron affinity, and electronegativity increase from left to right in a given
period and decrease from top to bottom in a given group.

aAtomic radius refers to metallic radius for metals and covalent radius for nonmetals.
bPolarizability will be discussed in Module 6. It provides a measure of the extent to which the electron cloud of an atom
can be distorted.
cIonization energy refers to the first ionization energy. It is the energy change for the process X(g) → X+(g) + e−.
dElectron affinity refers to the first electron affinity. It is the energy change for the process X(g) + e− → X−(g). An atom
has a large electron affinity if the addition of an electron leads to the release of a large quantity of energy (and the
formation of an anion that is much lower in energy than the atom from which it was formed). An atom has a small
electron affinity if the addition of an electron leads to the release of a small quantity of energy or requires the input of
energy.
eElectronegativity will be discussed in more detail in Module 4. It provides a measure of the “power” of an atom to
attract electrons to itself.
1
ns1
18
ns2np6
n=1 H
2
ns2
13
ns2np1
14
ns2np2
15
ns2np3
16
ns2np4
17
ns2np5
He
n=2 Li
Be

B C N O F Ne
n=3 Na
Mg

Al Si P S Cl Ar
n=4 K
Ca

Ga Ge As Se Br Kr
n=5 Rb
Sr

In Sn Sb Te I Xe
n=6 Cs
Ba

Tl Pb Bi Po At Rn
n=7 Fr
Ra


Ionization Energy (IE), Electron Affinity (EA), Electronegativity (EN)
Atomic Radius, Polarizability
A
to
m
ic
R
a
d
iu
s
, P
o
la
riz
a
b
ility

IE
, E
A
, E
N

CHEM 120 Module 3: Multielectron atoms and the periodic table page 17 of 17

Study questions for Module 3

 Which three quantum numbers are used to characterize an atomic orbital?
 Which two quantum numbers are used to characterize an electron’s spin?
 What is the conceptual model used for multi-electron atoms?
 What is the difference between “penetration” and “shielding”?
 In a multi-electron atom, we say that electrons in “inner shells” shield electrons in outer shells. What
does this mean?
 How does the orbital energy diagram for a multi-electron atom differ from that of the hydrogen atom?
(There are two important differences.)
 Why, in a multielectron atom, is a 3s orbital lower in energy than 3p? Why is 3p lower in energy than 3d?
(Answer: In a multi-electron atom, an electron in an 3s orbital penetrates towards the nucleus more
effectively than does an electron in a 3p orbital, and an electron in a 3p orbital penetrates more effectively
than does an electron in a 3d orbital.)
 What is the Pauli Exclusion Principle and what does it say about the maximum number of electrons that
can occupy a given orbital?
 What is the Aufbau procedure?
 Orbitals in the same subshell are “energetically degenerate”? What does “energetically degenerate” mean?
 What is Hund’s rule?
 How do we use the periodic table for quickly writing abbreviated electron configurations?
 How do we write the electron configuration for a positive monatomic ion, such as Fe3+ or Mn2+?
 What must be true of an atom that is diamagnetic?
 What must be true of an atom that is paramagnetic?
 Are most atoms, in their ground electronic states, diamagnetic or paramagnetic? Which groups of
atoms are diamagnetic?
 What is the difference between “covalent radius” and “metallic radius”?
 What are the observed trends in atomic radius?
 To what does “ionization energy” refer?
 What are the observed trends in ionization energy?
 Generally speaking, ionization energies increase from left to right across a period. However, there are
two notable exceptions to this general trend? When do they occur and why?
 To what does “electron affinity” refer?
 What are the observed trends in electron affinity?





























































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































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