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Chapter 3 – Atomic Structure and Properties

Introduction

The nuclear atom and quantum theory are the accepted theories for the atom. In this chapter, we demonstrate their utility by using them to explain trends in atomic properties.

3.1 Valence Electrons

Introduction

Most of the properties of an atom are related to the nature of its electron cloud and how strongly the electrons interact with the nucleus. In this section, we identify the electrons that are most important in determining atomic properties.

Prerequisites

Objectives

3.1-1. Valence Electrons Video

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3.1-2. Valence Electron Definition

Valence electrons consist of those electrons in the outermost s sublevel and those in any unfilled sublevel.
The electrons in an atom can be divided into two groups:
The number of valence electrons in an atom equals the group number of the atom.

3.1-3. Valence Electron Configurations of Main Group Elements

All of the elements in a group have the same number of valence electrons (Group number), and their valence electron configurations are the same except for the value of the n quantum number (the period). Thus, the periodicity of chemical properties is due to the periodicity of the valence electron configurations. The table shows the generic valence electron configurations of the elements in each group of the main group elements and the specific configurations of the second period.

Group 1A 2A 3A 4A 5A 6A 7A
Configuration ns1 ns2 ns2np1 ns2np2 ns2np3 ns2np4 ns2np5
2nd Period Element Li Be B C N O F
Configuration 2s1 2s2 2s22p1 2s22p2 2s22p3 2s22p4 2s22p5
Table 3.1: Valence Electron Configurations

3.1-4. Valence Electron Configurations of Transition Elements

The valence electron configuration of the first row transition elements has the form 4s23db, where b is the position of the element in the d block. Cr and Cu are the two exceptions because they each promote one of their 4s electrons into the 3d sublevel to obtain half-filled and completely filled 3d sublevels. Zinc is often considered a transition element because of its location, but its d sublevel is full, and its chemistry is more like that of a Group 2A metal than a transition metal. Thus, its valence electron configuration is 4s2, in keeping with its Group number of 2B. Similarly, Cu is 4s13d10, and the fact that it is a 1B element would lead you to think that its valence electron configuration should be 4s1. However, unlike zinc, copper does use its d orbitals in bonding. Indeed, the most common ion formed by copper is the Cu2+ ion. Thus, copper's chemistry is dictated by its d electrons, and its valence electron configuration is 4s13d10.

3.1-5. Practice with Valence Electrons

Valence electrons and valence electron configurations are very important in chemistry. Practice determining the valence electron configurations for atoms and then check your answer by clicking on the element in the periodic table below.
1
1A
2
2A
3
3B
4
4B
5
5B
6
6B
7
7B
8
8B
9
8B
10
8B
11
1B
12
2B
13
3A
14
4A
15
5A
16
6A
17
7A
18
8A
H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
Si
P
S
Cl
Ar
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
Rb
Sr

3.1-6. Core vs. Valence Electrons Exercise

Exercise 3.1:

Use the facts that all of the electrons in an atom are either core or valence electrons, and the number of valence equals the group number of the element to determine the number of valence and core electrons in each of the following atoms.
O
Valence = 6___ Oxygen is 1s2 2s2 2p4. The valence electrons are the four in the unfilled sublevel (2p) and the two in the outermost s sublevel (2s), so oxygen has six valence electrons: 2 + 4. Alternatively, oxygen is a 6A nonmetal, so it contains six valence electrons.
Core = 2___ Only the two 1s electrons are core electrons.
Ca
Valence = 2___ The electron configuration of calcium is [Ar] 4s2. It contains no unfilled sublevels, so only the outermost s electrons are valence electrons. Thus, calcium has two valence electrons, the two 4s electrons. Alternatively, calcium is a 2A metal, so it contains two valence electrons.
Core = 18___ The electron configuration of calcium is [Ar] 4s2. The argon core, [Ar], is composed of the core electrons in calcium. Argon contains 18 electrons, so calcium contains 18 core electrons. Note that the valence electrons are often all of those after the noble gas core, but not always.
Sn
Valence = 4___ The electron configuration of tin is [Kr] 5s2 4d10 5p2. The 5p sublevel is unfilled, so the 5p and 5s sublevels contain the valence electrons. Thus, tin has four valence electrons. Alternatively, tin is a 4A metal, so it must contain four valence electrons.
Core = 46___ The core electrons come from the filled 4d sublevel and the krypton core, so there are 46 (10 + 36) core electrons in Sn. This is an example of an atom whose valence electrons are not all of those beyond the noble gas core.

3.1-7. Main Group Valence Configurations Exercise

Exercise 3.2:

Write the valence electron configuration of each of the following main group elements.(Separate all terms by a single space and indicate all superscripts with a carat (^). For example, 2s^2 2p^2 for 2s2 2p2.)
Si (Z = 14) o_3s^2 3p^2_s Silicon is a Group 4A element, so it has four valence electrons. It is in the third period, so its valence electron configuration is 3s2 3p2.
K (Z = 19) o_4s^1_s Potassium is a Group 1A element, so it has only one valence electron. It is in the fourth period, so its valence electron configuration is 4s1.
Tl (Z = 81) o_6s^2 6p^1_s Thallium is a Group 3A element, so it has three valence electrons. It is in the sixth period, so its valence electron configuration is 6s2 6p1. Note that the 5d and 4f sublevels are full, so those electrons are not valence electrons.
Br (Z = 35) o_4s^2 4p^5_s Bromine is a Group 7A element, so it has seven valence electrons. It is in the fourth period, so its valence electron configuration is 4s2 4p5.

3.1-8. Transition Metal Valence Configurations Exercise

Exercise 3.3:

Write the valence electron configuration for each of the following transition elements.(Separate all terms by a single space and indicate all superscripts with a carat (^). For example, 2s^2 2p^2 for 2s2 2p2.)
Ti o_4s^2 3d^2_s Titanium is the second member of the d block, so its valence electron configuration is 4s2 3d2.
    Fe o_4s^2 3d^6_s Iron is the sixth member of the d block, so its valence electron configuration is 4s2 3d6.
    Cu o_4s^1 3d^10_s Copper is the ninth member of the d block, but it is one of the two exceptions. Its valence electron configuration is 4s1 3d10.

3.1-9. More Valence Electron Configuration Exercises

Exercise 3.4:

Use the periodic table to help you write valence electron configurations for the following atoms. (Separate all terms by a single space and indicate all superscripts with a carat (^). For example, 2s^2 2p^2 for 2s2 2p2.)
periodic table
Sc o_4s^2 3d^1_s The valence electron configuration of scandium is 4s2 3d1.
N o_2s^2 2p^3_s The valence electron configuration of nitrogen is 2s2 2p3.
Co o_4s^2 3d^7_s The valence electron configuration of cobalt is 4s2 3d7.
Bi o_6s^2 6p^3_s The 4f and 5d blocks are filled, so they are not valence electrons. The valence electron configuration of bismuth is 6s2 6p3. Note the configuration shows five valence electrons, consistent with the fact that bismuth is in Group 5A.
Sr o_5s^2_s Strontium is 5s2.
Mn o_4s^2 3d^5_s Manganese is 4s2 3d5.

3.2 Shielding and Effective Nuclear Charge

Introduction

In this section, you will learn how to predict the relative strengths with which the valence electrons interact with the nucleus.

3.2-1. Effective Nuclear Charge Video

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3.2-2. Shielding and Effective Nuclear Charge

Core electrons shield valence electrons better than do other valence electrons.
The nuclear charge experienced by an electron affects the size and energy of its orbitals, so it is an important factor in determining the properties of the valence electrons and orbitals. However, a valence electron is not attracted by the full positive charge of the nucleus because it is shielded by the other electrons, mainly the core electrons. The nuclear charge that is actually experienced by a valence electron is called the effective nuclear charge, Zeff. The effective nuclear charge experienced by an electron is equal to the charge of the nucleus (Z) minus that portion of the nuclear charge that is shielded by the other electrons (σ).
( 3.1 )
Zeff = Zσ
Effective Nuclear Charge
σ is the shielding of the other electrons. A valence electron is screened both by the core electrons and other valence electrons. However, core electrons are closer to the nucleus, so core electrons shield valence electrons better than do other valence electrons. This important fact will be used to explain the trends of atomic properties within a period.
nucleus
Figure 3.1a
In the absence of any electrons, the full nuclear charge (as shown by the red color) is observed. The nuclear charge is +Z, where Z is the atomic number (the number of protons in the nucleus). If this were a calcium nucleus, the resulting charge would be +20.
core electrons
Figure 3.1b
The nuclear charge is shielded by the negative charge of the core electrons (yellow sphere). Thus, only a portion of the nuclear charge can be felt through the core electrons as shown by the fact that the red color is greatly reduced by the yellow sphere. If this were a Group 2A element, the charge that would result would be +2 as the two valence electrons are not included. Thus, the charge has been reduced from +40 to +2 by the core electrons.
core electrons
Figure 3.1c
Addition of the valence electrons (green sphere) results in the atom. Essentially none of the nuclear charge is felt outside the neutral atom.

3.2-3. Trends in Zeff

Effective nuclear charge increases going from left to right in a period.
Trends within a period: The charge of the nucleus increases by a full unit of positive charge in going from one element to the next in a period, but the additional electrons are valence electrons, which do not shield with a full negative charge. Consequently, effective nuclear charge increases going from left to right in a period. This effect is very important because it explains many trends within a period. Consider the following comparison of Li and F: Although F has five more electrons than Li, they increase the shielding by only 2 (from 2 in Li to 4 in F). This is because they are valence electrons, which do not shield very well.
line graph Z_eff versus group number
Figure 3.2
Effective Nuclear Charges of the Elements in the 2nd and 3rd Periods
Trends within a group: The shielding ability of the core electrons decreases as their n quantum number increases. As a result, the effective nuclear charge experienced by the valence electrons within a group increases with the n quantum number of the valence level. Consequently, effective nuclear charge increases going down a group. For example, the 2s electron on Li experiences an effective nuclear charge of ~1, while the 3s electron on Na experiences a charge of ~2, which shows that the
n = 2
electrons do not shield as well as the
n = 1
electrons.
This effect opposes that of changing the n quantum number because increasing n increases the energy of the valence levels, but increasing the effective nuclear charge lowers the energy of the valence levels. The effect of increasing n is the more important because the energies of the valence levels do increase going down a group.

3.2-4. Effective Nuclear Charge Exercise

Exercise 3.5:

Select the electron in each pair that experiences the greater effective nuclear charge.
  • 2p in Cl
  • 2p in F The Cl nucleus has eight more protons and eight more electrons, but most of the additional electrons are in the
    n = 3
    level, so they do not shield the 2p very will. Thus, the 2p in Cl experiences a higher nuclear charge than do the 2p in F.
  • 3p in Cl
  • 3p in P Effective nuclear charge increases going across a period.
  • 4p in Br The p orbitals have a nodal plane at the nucleus, but s orbitals do not. Thus, s orbitals screen the p orbitals better than the p orbitals screen the s.
  • 4s in Br
  • 5s in Rb
  • 4s in K Effective nuclear charge experienced by the valence electrons increases slightly going down a group.

3.3 Relative Atomic Size

Introduction

Atoms are not hard spheres with well defined boundaries, so the term "atomic radius" is somewhat vague and there are several definitions of what the atomic radius is. Consequently, atomic radii are not measured directly. Rather, they are inferred from the distances between atoms in molecules, which can be readily determined with several techniques. However, a discussion of how atomic radii are defined requires an understanding of chemical bonding and the solid state, so a detailed discussion of the different types of radii and their values is postponed until Chapter 8. In this section, we restrict our discussion to trends in the relative sizes of atoms.

Prerequisites

Objectives

3.3-1. The Bohr Model and Atomic Size

The size of the atom is given by the size of its valence electron clouds. Although the orbits of fixed radii suggested in the Bohr model are not correct, the conclusions of the model can still be used to gain a qualitative understanding of trends in atomic radii by substituting the effective nuclear charge for the atomic number in Equation 2.4 to obtain the following:
rn
n2
Zeff

3.3-2. Atomic Size

Atomic radii increase going down a group and decrease going across a period.
The size of an atom is defined by the size of the valence orbitals. Using the Bohr model as discussed above, we conclude that Note that the atomic radii of H and Rb are
0.37 Å
and
2.11 Å,
respectively.
representation of atom sizes
Figure 3.3
Relative Atomic Radii

3.3-3. Atomic Size Exercise

Exercise 3.6:

Use only a periodic table to determine the largest atom in each group.
  • Cl Atomic radii increase going down a group.
  • F Atomic radii increase going down a group.
  • I
  • Na
  • S Atomic radii decrease going across a period.
  • Al Atomic radii decrease going across a period.
  • Br Atomic radii decrease going across a period.
  • K
  • Ge Atomic radii decrease going across a period.

3.4 Relative Orbital Energies

Introduction

The relative energy of the valence orbitals in an atom is an important characteristic of the atom as it dictates both properties of the atom and the manner and strength of its interaction with other atoms. Indeed, we will invoke relative orbital energies in making predictions about chemical processes throughout the remainder of this course. In this section, we use the Bohr model to predict the relative energies of the valence orbitals in some small atoms.

Prerequisites

Objectives

3.4-1. Orbital Energies Video

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3.4-2. The Bohr Model and Orbital/Electron Energy

The ease with which an electron is lost from an atom is given by how strongly the electron is bound to the atom, which, in turn, is given by the energy of the electron in the atom. We again use the conclusions of the Bohr model in a qualitative way to understand trends in valence electron energies by substituting the effective nuclear charge for the atomic number in Equation 2.5 to obtain Equation 3.2.
( 3.2 )
Ex ∝ −
Zeff2
n2
Electron/Orbital Energy Approximation
Ex is the energy of a valence orbital on atom X, and n is the principle quantum number of its valence shell. While Equation 3.2 is only a very rough approximation, it is useful in demonstrating how the relative orbital energies of the valence electrons in the atoms of the first three periods are related. We conclude that

3.4-3. Relative Energies, an Example

The valence orbitals of nonmetals are relatively low in energy, while those of metals are relatively high.

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We now determine the relative valence orbital energies of Li, C, and F as shown in Figure 3.4. We use the fact that they are each in the second period to get n and Figure 3.2 to obtain values of Zeff and construct the table shown in Figure 3.4. We use Equation 3.2
Ex ∝ −
Zeff2
n2
and the values in the table to conclude the following.
This figure contains a table with 4 columns with 3 rows of data on the right.  The first column contains the element, the second contains the value of Z_eff, the third contains the value of n, and the fourth contains the value of the formula -(Z_eff)^2 / n^2.  The first row is Li; 1.3; 2; -0.4.  The second row is C; 3.3; 2; -2.7.  The third row is F; 5.2; 2; -6.8.  On the left side of this figure is an energy level diagram that shows energy increasing up.  At the top of the diagram is an energy level of 0.  Slightly below that is the 2s orbital of Li containing one electron.  Below that are the three 2p orbitals of C containing 2 electrons in separate orbitals.  Below that are three 2p orbitals of F with 5 total electrons.  Two of the F orbitals contain two electrons and one orbital contains a single electron.
Figure 3.4

3.4-4. Predicting Relative Energy Exercise

Exercise 3.7:

Use the effective nuclear charges (Figure 3.2) and the n quantum number of the valence electrons of chlorine, oxygen, sulfur, and silicon in Equation 3.2
Ex ∝ −
Zeff2
n2
to predict the relative energies of their valence orbitals. Compare to the values of
Zeff2
n2
to those for Li (0.4), C (2.7), and F (6.8).
    Chlorine (Zeff = 6.1)
  • above Li
    Zeff2
    n2
    =
    6.12
    32
    = 4.1,
    which places it between F and C. As expected, the n quantum number dominates the trends in a group to place Cl above F.
  • Li - - - - - - -
  • between Li and C
    Zeff2
    n2
    =
    6.12
    32
    = 4.1,
    which places it between F and C. As expected, the n quantum number dominates the trends in a group to place Cl above F.
  • C - - - - - - -
  • between C and F
  • F - - - - - - -
  • below F
    Zeff2
    n2
    =
    6.12
    32
    = 4.1,
    which places it between F and C. As expected, the n quantum number dominates the trends in a group to place Cl above F.
    Oxygen (Zeff = 4.6)
  • above Li
    Zeff2
    n2
    =
    4.62
    32
    = 5.3,
    which places it between F and C. Note that the valence orbitals of oxygen are lower than those of chlorine.
  • Li - - - - - - -
  • between Li and C
    Zeff2
    n2
    =
    4.62
    32
    = 5.3,
    which places it between F and C. Note that the valence orbitals of oxygen are lower than those of chlorine.
  • C - - - - - - -
  • between C and F
  • F - - - - - - -
  • below F
    Zeff2
    n2
    =
    4.62
    32
    = 5.3,
    which places it between F and C. Note that the valence orbitals of oxygen are lower than those of chlorine.
    Sulfur (Zeff = 5.5)
  • above Li
    Zeff2
    n2
    =
    5.52
    32
    = 3.4,
    which places it between F and C. As expected, the n quantum number dominates the trends in a group to place S above O.
  • Li - - - - - - -
  • between Li and C
    Zeff2
    n2
    =
    5.52
    32
    = 3.4,
    which places it between F and C. As expected, the n quantum number dominates the trends in a group to place S above O.
  • C - - - - - - -
  • between C and F
  • F - - - - - - -
  • below F
    Zeff2
    n2
    =
    5.52
    32
    = 3.4,
    which places it between F and C. As expected, the n quantum number dominates the trends in a group to place S above O.
    Silicon (Zeff = 4.3)
  • above Li
    Zeff2
    n2
    =
    4.32
    32
    = 2.1,
    which places it between C and Li. Note that this is also above C as expected for two elements in the same group.
  • Li - - - - - - -
  • between Li and C
  • C - - - - - - -
  • between C and F
    Zeff2
    n2
    =
    4.32
    32
    = 2.1,
    which places it between C and Li. Note that this is also above C as expected for two elements in the same group.
  • F - - - - - - -
  • below F
    Zeff2
    n2
    =
    4.32
    32
    = 2.1,
    which places it between C and Li. Note that this is also above C as expected for two elements in the same group.

3.4-5. Relative Valence Orbital Energies for Several Atoms

Before leaving our discussion of relative orbital energies, we add H (Zeff = 1) and O (Zeff = 4.5) to our energy diagram to obtain the following diagram that contains eight elements of the first three periods.
Figure 3.5
The relative energies of the valence orbitals of some atoms studied in the first three periods. Values of
Zeff2
n2
are given to the right.
We draw the following two conclusions based on the diagram:

3.4-6. Orbital Energy Exercise

Exercise 3.8:

Consider the following valence orbital energy diagram that we have constructed to this point.

a
Li (–0.4)
b
H (–1.0)
c
Si (–2.0)
d
C (–2.5)
e
S (–3.3)
f
Cl (–4.2)
g
O (–5.0)
h
F (–6.5)
Where would the valence orbitals of the following atoms be placed? Give the letter of the location.
    (a)    N (Zeff = 3.83)
  • a
    Zeff2/n2 = −3.7
  • b
    Zeff2/n2 = −3.7
  • c
    Zeff2/n2 = −3.7
  • d
    Zeff2/n2 = −3.7
  • e
    Zeff2/n2 = −3.7
  • f
  • g
    Zeff2/n2 = −3.7
  • h
    Zeff2/n2 = −3.7
    (b)    P (Zeff = 4.89)
  • a
    Zeff2/n2 = −2.7
  • b
    Zeff2/n2 = −2.7
  • c
    Zeff2/n2 = −2.7
  • d
    Zeff2/n2 = −2.7
  • e
  • f
    Zeff2/n2 = −2.7
  • g
    Zeff2/n2 = −2.7
  • h
    Zeff2/n2 = −2.7

3.4-7. Core vs. Valence Orbital Energies Exercise

The energies of the valence orbitals of all atoms lie in a relatively narrow range due to the periodicity in Zeff and an increasing n quantum number. Core electrons, on the other hand, continue to drop in energy as the number of protons increases because they are not shielded very efficiently by the valence electrons. Thus, the valence 2p orbitals of oxygen are at lower energy than the valence 3p valence orbitals on sulfur because valence orbital energies increase going down a group, but the 2p electrons in oxygen are much higher in energy than the 2p electrons on sulfur because sulfur has 16 protons while oxygen has only eight.
Exercise 3.9:

Select the orbital that is at lower energy in each pair.
  • 2p on N The orbitals are on the same atom, so use
    n + l,
    which is 3 for a 2p and 2 for a 2s.
  • 2s on N
  • 2p on Br
  • 2p on F
    n = 2
    for both orbitals, but Zeff is much greater in the vicinity of the 35 protons of a Br atom than around the nine protons of a F atom. The 2p electrons in Br are core electrons, while those in F are valence electrons, and core electrons are always lower in energy than valence electrons.
  • 2p on N Zeff increases going across a period, while n remains the same.
  • 2p on O

3.5 Ionization Energy

Introduction

One of the properties of an atom that is important in dictating the chemical properties of the atom is the ease with which the atom loses one or more of its valence electrons. This property of an atom is given by the atom's ionization energy or ionization potential, the topic of this section.

Prerequisites

Objectives

3.5-1. Ionization Energy Video

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3.5-2. Ionization Energy and Orbital Energy

Ionization energies measure the energy of the highest energy electron.
The ionization energy (IE) of an atom is the energy required to remove the highest energy electron. The ionization energy of lithium is 520 kJ/mol, which means that the electrons in a mole of lithium atoms can be removed by the input of 520 kJ. The ionization process is
Li Li1+ + e1−      ΔE = +520 kJ
where ΔE = IE, the ionization energy. The energy change results because the energy of the electron in the initial state is different than in the final state. In the initial state, the electron is in an orbital of energy En, but in the final state, it is a free electron with no potential energy. The relationship between the ionization energy and the orbital energy is obtained as follows:
IE = EfinalEinitial
IE = free electron energy − orbital energy
IE = 0 − En
Thus, IE = –En. The orbital energy of the highest energy electron can be approximated as the negative of the ionization energy of the atom. In the above discussion, we have considered the removal of only the highest energy electron, but as shown in the following, other electrons can also be removed by successive ionizations. Thus, the energy required to remove the first electron is the first ionization energy and that required to remove the second electron is the second ionization energy, and so forth. Consider the case of the metal M.
MM1+ + e1−    First ionization
M1+M2+ + e1−Second ionization
M2+M3+ + e1−Third ionization
M3+M4+ + e1−Fourth ionization
The loss of an electron reduces the screening of the remaining electrons, so the remaining electrons experience greater nuclear charge and are more difficult to remove. Thus, each successive ionization energy is greater than the preceding. We use the term ionization energy to refer to the first ionization energy in the remainder of the chapter.

3.5-3. Putting Numbers to Orbital Energies

The energy required to remove one 2p electron from each atom in a mole of F atoms to produce a mole of F1+ ions is 1681 kJ, so the ionization energy of fluorine is 1681 kJ/mol. In addition, 1681 kJ/mole are released in the reverse process, so we conclude that the 2p orbital energy is –1681 kJ/mol. Nonmetals are characterized by high ionization energies and low valence orbital energies because they have high effective nuclear charges. The ionization energy of carbon is 1086 kJ/mol, so the orbital energy of a 2p orbital in carbon is –1086 kJ/mol. The ionization energy of lithium is only 520 kJ/mol, so the orbital energy of its 2s orbital is –520 kJ/mol. The valence orbitals of metals are characterized by low ionization energies and high valence orbital energies because they have low effective nuclear charges.
Figure 3.6
Ionization Energy and Orbital Energy

3.5-4. Ionization Energy Trends

Ionization energies increase going up a group and going from left to right in a period.
Ionization energy is the energy required to remove the highest energy electron from an atom, which can be estimated with the assumption that IE = –En; i.e., that the ionization energy equals the negative of the orbital energy of the electron. En can be estimated with Equation 3.2
Ex ∝ −
Zeff2
n2
, so we conclude that ionization energies
This trend is shown in Figure 3.7, which shows the ionization energies of the atoms in order of atomic number. However, there are some apparent exceptions that arise because electron configurations in which sublevels are filled (Groups 2A and 8A) or half-filled (Group 5A) are unusually stable, so removing an electron from an element in one of these groups is more difficult and results in deviations from the expected periodicity. For example, the effective nuclear charge of B is greater than that of Be, but the ionization energy of Be is greater than that of B because the electron must be removed from a filled 2s sublevel in Be. Similarly, the ionization energy of N is greater than that of O because the 2p sublevel of N is half-filled.
line graph of ionization energy versus atomic number
Figure 3.7
First Ionization Energies of the Main Group Elements: Circles of the same color represent elements of the same group. For example, all Group 1A elements are shown as red circles.
Summary

3.5-5. Ionization Energy Exercise

Exercise 3.10:

Use only a Periodic Table to determine the atom in each group with the greatest ionization energy.
  • Cl Ionization energies increase going up a group.
  • F
  • I Ionization energies increase going up a group.
  • Na Ionization energies increase going across a period.
  • S
  • Al Ionization energies increase going across a period.
  • Br
  • K Ionization energies increase going across a period.
  • Ge Ionization energies increase going across a period.
  • Si Half-filled sublevels are unusually stable.
  • P
  • S Half-filled sublevels are unusually stable.

3.6 Electronegativity

Introduction

The electrons in a bond can lower their potential energy by residing closer to the atom in the bond that has the valence orbital at lower energy. The ability of an atom to attract the bonding electrons to itself is called its electronegativity (χ). Atoms are most electronegative when their valence orbitals are low in energy. Electronegativity is the topic of this section.

Prerequisites

Objectives

3.6-1. Electronegativity Video

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3.6-2. Electronegativity and Orbital Energy

Atoms with low-energy valence orbitals are highly electronegative.
Electronegativity (χ) is a measure of an atom's ability to attract bonding electrons, so electron density in a bond accumulates near those atoms with higher electronegativities. Bonding electrons reside in orbitals involving the valence orbitals of the atoms, especially those that are unfilled, and electrons seek to minimize their energy, so an atom that is highly electronegative is simply one whose valence orbitals, especially those that are unfilled, are low in energy.
Example:
En = −
(Zeff)2
n2
,
so the valence orbitals of atoms with high electronegativities have
  • 1
    large Zeff (nonmetals) and
  • 2
    low n quantum numbers (high in the periodic table).
Consider the orbital energies of Li, C, and F shown in Figure 3.8. Li is a metal with a low Zeff, so its orbital energy is high. The electron in it is readily lost (low ionization energy), but bonding electrons are not drawn to the high-energy orbital, so Li has a very low electronegativity. F is a nonmetal with a high Zeff, so its orbital energy is low. Thus, it is very difficult to remove a 2p electron from F (high ionization energy), but bonding electrons are drawn to the low energy unfilled orbital, so F is highly electronegative. The 2p orbital energy of C is about halfway between the valence orbital energies of Li and F, and its electronegativity is also about half way between these two extremes in the period.
Figure 3.8
Electronegativity and Orbital Energy

3.6-3. The Meaning of Electronegativity

The electronegativity difference between two atoms dictates the type of bond that forms between them. Refer to Figure 3.9 for the following discussion.
Figure 3.9
Ionic bonds are the topic of Chapter 4, and covalent bonds are discussed in Chapters 5 and 6.

3.6-4. Electronegativity Trends

Electronegativities decrease going down a group and increase going across a period.
We use the Bohr model to approximate the energy of the lowest energy empty orbital –En and conclude that Consequently,
line graph of electronegativity versus atomic number
Figure 3.10
Electronegativities of the Main Group Elements: Circles of the same color represent elements of the same group. For example, all Group 1A elements are shown as red circles.

3.6-5. Late Metals

An exception to the above generality about the electronegativities of metals arises from the fact that d and f electrons do not shield very well because they contain two and three nodal planes, respectively. Therefore, the effective nuclear charge experienced by the valence orbitals in late metals (metals that lie on the right side of the Periodic Table) can be quite large. For example, Pb has 27 more protons and electrons than does Cs, but 24 of those electrons are d and f electrons, which do not shield the 27 additional protons very well. Thus, the 6p electrons in Tl and Pb experience relatively high Zeff (12.25 and 12.39, respectively), which makes both of these metals fairly electronegative. Indeed, the electronegativity of Pb is much greater than that of Si even though they are in the same Group, and the valence orbitals in Pb have a much higher n quantum number. We conclude that due to their high effective nuclear charges, late metals have unusually high electronegativities (see Table 3.2), which impacts significantly on their chemical properties.

Metal χ
Ag 1.9
Sn 2.0
Hg 2.0
Tl 2.0
Pb 2.3
Table 3.2: Electronegativities of Late Metals

3.6-6. Metals vs. Nonmetals

Metals have low ionization energies and nonmetals have high electronegativities.
Metals are characterized by lower effective nuclear charges, so they have Nonmetals are characterized by high effective nuclear charges, so they have Thus, metals tend to lose electrons (low ionization energy), while nonmetals tend to gain electrons (high electronegativities).

3.6-7.Electronegativity Exercise

Exercise 3.11:

Use only a Periodic Table to determine the atom in each group with the greatest electronegativity.
  • Cl Electronegativity increases going up a group.
  • F
  • I Electronegativity increases going up a group.
  • Na Electronegativity increases going across a period.
  • S
  • Al Electronegativity increases going across a period.
  • Br
  • K Electronegativity increases going across a period.
  • Ge Electronegativity increases going across a period.
  • Al Late metals have unusually high electronegativities.
  • Tl
  • Ga Late metals have unusually high electronegativities.

3.6-8. Orbital Energy and Atomic Properties Exercise

Exercise 3.12:

Use what you have learned about the relationships between orbital energies and ionization energies and electronegativities and the valence orbital occupancies of atoms X and Y given below to answer the following questions about atoms X and Y.
energy diagram
    The atom with the lower ionization energy:
  • X Atoms with lower ionization energies have electrons at higher energy. The electrons in atom Y are higher than those in atom X, therefore, element Y has the lower ionization energy.
  • Y
    The atom with the higher electronegativity:
  • X
  • Y Atoms with higher electronegativities have valence orbitals at lower energy. The valence orbitals of X are lower than those of Y, so atom X is much more electronegative than atom Y.
The group of the periodic table to which each atom belongs:
X o_6A_s Atom X has six valence electrons (ns2np4) so it is a Group 6A element.
Y o_2A_s Atom Y has two valence electrons (ns2), so it is a Group 2A element.

3.7 Magnetic Properties

Introduction

All magnetic properties are due to the magnetic fields caused by electron spin. However, no magnetic field is generated by paired electrons because the two different electron spins are opposed and their magnetic fields cancel. Consequently, the magnetic properties of an atom are due solely to its unpaired electrons. In this section, we give a brief introduction into the magnetic properties of atoms and materials.

Prerequisites

Objectives

3.7-1. Paramagnetism and Diamagnetism

The paramagnetism of an atom increases with the number of unpaired electrons.
Atoms with unpaired electrons are paramagnetic. Paramagnetic atoms align in magnetic fields due to the presence of the unpaired electrons. The more unpaired electrons an atom has, the more paramagnetic it is. Both Li and N have unfilled valence subshells, so both have unpaired electrons and are paramagnetic. N has three unpaired electrons while Li has only one, so N is more paramagnetic than Li.
This figure shows the three 2p orbitals above the one 2s orbital for Li, Be, N, and Ne.  Li has one electron in the 2s orbital.  Be has two electrons in the 2s orbital.  N has two electrons in the 2s orbital and one electron in each of the 2p orbitals.  Ne has two electrons in the 2s orbital and 2 electrons in each of the 2p orbitals.
Figure 3.11
Atoms with no unpaired electrons are diamagnetic. Diamagnetic atoms do not align in magnetic fields because they have no unpaired electrons. Neither Be nor Ne has any unfilled valence shells, so both are diamagnetic.

3.7-2. Ferromagnetism

Ferromagnetism is a measure of the bulk magnetism of a material, while paramagnetism is a measure of the magnetism of individual atoms.
Paramagnetism and diamagnetism are atomic properties, not bulk properties. Thus, N and Li are paramagnetic atoms, but nitrogen gas and lithium metal are not magnetic because the unpaired electrons on the atoms pair with one another to form materials that are not magnetic. The magnetism you are familiar with is called ferromagnetism. It is a bulk property because it requires unpaired electrons in a material, not just in an isolated atom. Iron is the best known example. Fe atoms have four unpaired electrons, so Fe atoms are paramagnetic. When Fe atoms bond to form iron metal, there is electron pairing in the solid, but not all four electrons pair. Consequently, iron metal is magnetic and iron is a ferromagnet.
energy diagram: Fe
Figure 3.12
Valence orbital occupancy of Fe. It is both paramagnetic and ferromagnetic.

3.7-3. Paramagnetism Exercise

Exercise 3.13:

Use a Periodic Table to determine the more paramagnetic atom in each pair.
  • O O is 2s22p4, so it has two unpaired electrons in its 2p sublevel. N is 2s22p3, so it has three unpaired electrons.
  • N
  • Na
  • Mg Na is 3s1 and has one unpaired electron. Mg is 3s2, so it has no unpaired electrons
  • Cr
  • Fe Cr is 4s13d5 (an exception), so it has six unpaired electrons. Fe has only four unpaired electrons.

3.8 Exercises and Solutions

Select the links to view either the end-of-chapter exercises or the solutions to the odd exercises.