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Periodic Table and Periodic Trends
Mendeleev’s (1869) Periodic Law stated that elements arranged in order of increasing atomic mass (total # of protons and neutrons) show a periodic
recurrence of properties at regular intervals. The modern periodic law that was revised from Mendeleev’s based on H.G.J. Moseley’s (1887-1915)
discovery of the atomic number (Z) stated that elements arranged in order of increasing atomic number (total # of protons) show a periodic recurrence of
properties at regular intervals. In accordance with this periodic law, elements are arranged in a row-and-column table called the Periodic Table:
• A group – elements with similar chemical properties in a vertical column in the main part of the table, numbered from 1 to 18.
• A period – elements, arranged in a horizontal row, whose properties change from metallic on the left to nonmetallic on the right, numbered from
1 to 7 (top to bottom).
• The “staircase line” – a zigzag line that separates metals (to the left) from nonmetals (to the right).
A knowledge of the electron configurations of elements provides an explanation for the periodic law. Chemical properties repeat
themselves in a regular manner among the elements because electron configurations repeat themselves in a regular manner among the elements.
In general, elements with similar outer shell electron configurations have similar chemical properties. The distinguishing electron for an element is the
last electron that is added to its electron configuration when the configuration is written using the Aufbau Principle.
Periodic Table:
(1) Areas or Regions of the Periodic Table - The periodic table can be divided into 4 areas: s area, p area, d area, and the f area.
(2) No. of Columns in each Region -The number of columns in each area is the same as the maximum number of electrons that the various
types of subshells can accommodate: s area – 2, p area – 6, d area – 10, and f area – 14. for the elements located in the s area, the
distinguishing electron is always found in an s subshell, p area elements the distinguishing electron is always found in an p subshell, and so
on. The extent that the subshell containing an element’s distinguishing electron is filled can also be determined from the element’s position
in the periodic table.
(3) The Row Number (Period Number) of the Periodic Table - The periodic table can also be used to determine the shell in which the
distinguishing electron is located because there is a definite correspondence between them:
(i) In the s and p areas, the shell number equals period number.
(ii) In the d area, the shell number equals period number minus one.
(iii) In the f area, the shell number equals period number minus two.
(4) The Groups’ Roman Numbers of the Periodic Table – The Roman numeral periodic table group number is the same as the number of
valence electrons for the representative elements in a group. Valence electrons are the electrons in the outermost electron shell, which
is the shell with the highest shell number (n). This definition applies only to the representative elements and noble gases in the s and p areas
(The presence of incompletely filled inner d or f subshells is the complicating factor in the definitions involving elements in the d- and fareas.)
Explaining the Periodic Table
The modern view of the atom based on the four quantum numbers was developed using experimental studies of atomic spectra and the
experimentally determined arrangements of elements in the periodic table. It is no coincidence that the maximum number of electrons in the s, p, d,
and f orbitals (Table 5) corresponds exactly to the number of columns of elements in the s, p, d, and f blocks in the periodic table (Table 6). This by itself
is a significant accomplishment that the original Bohr model could not adequately explain. Groups or families in the periodic table were originally
created by Mendeleev to reflect the similar properties of elements in a particular group. The noble gas family, Group 18, is a group of gases that are
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generally nonreactive. The electron configurations for noble gas atoms show that each of them has a filled ns np outer shell of electrons (Table 6). The
original idea from the Bohr theory - filled energy levels as stable (nonreactive) arrangements - still holds, but is more precisely defined. Similar outer
shell or valence electron configurations also apply to most families, in particular, the representative elements. Similarly, the transition elements can
now be explained by our new theory as elements that are filling the d energy sublevel with electrons. The transition elements are sometimes referred
to as the d block of elements. The 5d orbitals can accommodate 10 electrons, and there are 10 elements in each transition-metal period (Table 6).
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Using the same test of the theory on the lanthanides and the actinides, we can explain these series of elements as filling an f energy level. The f
block of elements is 14 elements wide, as expected by filling 7 f orbitals with 14 electrons. The success of the quantum-number and s-p-d-f
theories in explaining the long-established periodic table led to these approaches being widely accepted in the scientific community.
Representative elements – s-blocks - (1) Alkali metals
(2) Alkali-earth metals
p-blocks - (3) Elements that are filling the p energy sublevel with electrons
(4) Halogens
(5) Noble Gases – very stable with filled ns 2 np 6 outer shell of electrons.
Transition Metals –
d-blocks - elements that are filling the d energy sublevel with electrons
Inner-Transition Metals - f-blocks - Lanthanides and actinides – elements that are filling 7 f orbitals with 14 electrons.
Notice that as you progress down the periodic table, the periods contain an increasing number of blocks: the first period contains only the sblocks, the second and third periods contain the s- and p-blocks, the fourth and fifth periods contain the s- , p -, and d-blocks, and the sixth and
seventh periods include all four blocks. This pattern arises from the principles of the quantum theory. The shape of the periodic table is a
result of the way electrons fill the s, p, d, and f orbitals of different energy levels.
Explaining Ion Charges
Transition-metal ions and multiple ions formed by heavy representative metals can now be explained. For example, Zn : [Ar ]4 s 2 3d 10 . If another
atom or ion removes the two 4s electrons this would leave zinc with filled 3d orbitals – a relatively stable state: Zn 2 + : [Ar ]3d 10 .
For lead, Pb : [Xe ]6 s 2 4 f 14 5d 10 6 p 2 the electron configuration shows filled 4f, 5d, 6s orbitals, and a partially filled 6p orbitals. The lead atom could
lose the two 6p electrons to form 2+ ion or lose four electrons from the 6s and 6p orbitals to form a 4+ ion.
Explaining Magnetism
The evidence from the electron configurations of iron, cobalt and nickel indicates that unpaired electrons may account for some magnetism.
But it cannot account for the strong ferromagnetism, because ruthenium, rhodium, and palladium, immediately below iron, cobalt, and nickel in
the group are only paramagnetic (weakly magnetic). The theory to explain the strong ferromagnetism for iron, cobalt, and nickel is that their
atoms are able to orient themselves in a magnetic field, acting like a little magnet. These atoms influence each other to form groups (called
domain) in which all of the atoms are oriented with their north pole in the same direction. If most of the domains are then oriented in the same
direction by an external magnetic field of, for example, a strong bar magnet, the ferromagnetic metal becomes a “permanent” magnet.
Ferromagnetism is based on the properties of a collection of atoms, rather than just one atom.
Periodic Trends (more details on pages 49-61):
1.The process of removing an electron from an atom is called ionization, and the energy needed to do this called ionization energy. The energy
needed to remove only one electron is called the first ionization energy. Removal of each successive electron from an Atom requires an
increased amount of energy because the removal of the electron results in a stronger attraction between the nucleuses and the remaining
electrons. The ionization energies increase from left to right across a period in the periodic table, and increase from the bottom to the top of a
group, creating a diagonal relationship in the table.
2. The ionization energy is one of the factors involved in determining the electronegativity, which follows the same diagonal trend as the
ionization energy does.
3. The electron affinity is defined as the energy needed to add an electron to an atom. It has the same diagonal relationship as the ionization
energy in the periodic table.
4. Atomic sizes (radii) decrease from left to right across a period in the periodic table and decrease from bottom to top in a group. In a period,
each succeeding atom has an additional electron and an additional proton. The electrons are added to the same shell and are located at a
relatively constant distance from the nucleus. The increasing nuclear charge has a stronger effect than the electrons’ repelling force. It attracts
the electron clouds closer to the nucleus, thereby decreasing the overall size. From the top each row adds an additional shell to the atom’s
structure. This added outer shell of electrons must be further from the nucleus than the preceding shell. In addition, the added distance means
that the attractive force of the nucleus is reduced. Cations are positively charged ions, and anions are negatively charged. Cations are always
smaller than their neutral atoms, and anions are always larger.
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Concept Map of Periodic Table
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Periodic Trends – Further Explanations
Periodic trends are specific patterns that are present in the periodic table, which illustrate different aspects of a certain element, including its size and its
properties with electrons. The main periodic trends include: electronegativity, ionization energy, electron affinity, atomic radius, melting point, and
metallic character. The periodic trends that arise from the arrangement of the periodic table provide chemists with an invaluable tool to quickly predict an
element's properties. These trends exist because of the similar atomic structure of the elements within their respective group families or period and the
periodic nature of the elements.
1. (1)Periodic Trends for Electronegativity
Ele Electronegativity can be understood as chemical property describing an atom's ability to attract and bind to electrons. Because electronegativity is a
qualitative property, there is not a standardized method for calculating electronegativity. However, the scale that most chemists use in quantifying
electronegativity is the Pauling Scale, named after the chemist Linus Pauling. The numbers assigned by the Pauling scale are dimensionless due to
electronegativity being largely qualitative. Electronegativity values for each element can be found on certain periodic tables. An example is provided
below.
Figure 1. Periodic Table of Electronegativity values
Electronegativity measures an atom's strength to attract and form bonds with electrons. This property exists due to the electronic configuration of atoms.
Most atoms prefer to fulfilling the octet rule (having the valence, or outer, shell comprise of 8 electrons). Since elements on the left side of the periodic
table have less than a half-full valence shell, the energy required to gain electrons is significantly higher compared to the energy required to lose electrons.
As a result, the elements on the left side of the periodic table generally lose electrons in forming bonds. Conversely, elements on the right side of the
periodic table are more energy-efficient in gaining electrons to create a complete valence shell of 8 electrons. This effectively describes the nature of
electronegativity: the more inclined an atom is to gain electrons, the more likely that atom will pull electrons toward itself.
•
As you move to the right across a period of elements, electronegativity increases. When the valence shell of an atom is less than half full, it
requires less energy to lose an electron than gain one and thus, it is easier to lose an electron. Conversely, when the valence shell is more than
half full, it is easier to pull an electron into the valence shell than to donate one.
•
As you move down a group, electronegativity decreases. This is because the atomic number increases down a group and thus there is an
increased distance between the valence electrons and nucleus, or a greater atomic radius.
•
Important exceptions of the above rules include the noble gases, lanthanides, and actinides. The noble gases possess a complete valence shell
and do not usually attract electrons. The lanthanides and actinides possess a more complicated chemistry that does not generally follow any
trends. Therefore, noble gases, lanthanides, and actinides do not have electronegativity values.
•
As for the transition metals, while they have values, there is little variance among them as you move across the period and up and down a
group. This is because of their metallic properties that affect their ability to attract electrons as easily as the other elements.
With these two general trends in mind, we can deduce that the most electronegative element is fluorine, which weighs in at a hefty 3.98 Pauling units.
(2) Periodic Trends for Ionization Energy
Ionization Energy is the amount of energy required to remove an electron from a neutral atom in its gaseous phase. Conceptually, ionization energy is
considered the opposite of electronegativity. (i) The lower this energy is, the more readily the atom becomes a cation. Therefore, the higher this energy is,
the more unlikely the atom becomes a cation. Generally, elements on the right side of the periodic table have a higher ionization energy because their
valence shell is nearly filled. Elements on the left side of the periodic table have low ionization energies because of their willingness to lose electrons and
become cations. Thus, ionization energy increases from left to right on the periodic table.
(ii) Another factor that affects ionization energy is electron shielding. Electron shielding describes the ability of an atom's inner electrons to shield its
positively-charged nucleus from its valence electrons. When moving to the right on a period of elements, the number of electrons increases and the
strength of shielding increases. As a result, it is easier for valence shell electrons to ionize and thus the ionization energy decreases when going down a
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group. In certain texts, electron shielding may also be known as screening.
1) The ionization energy of the elements within a period generally increases from left to right. This is due to valence shell stability.
2) The ionization energy of the elements within a group generally decreases from top to bottom. This is due to electron shielding.
The noble gases possess very high ionization energies because of their full valence shell as indicated in the graph. Note that Helium has the highest
ionization energy of all the elements.
Exceptions to the trend:
1. Beryllium vs Boron
2. Nitrogen vs Oxygen
3. Magnesium vs Aluminum
4. Phosphorus vs Sulfur
Figure 3. Graph showing the Ionization Energy of the Elements from Hydrogen to Argon
Some elements can have several ionization energies, so we refer to these varying energies as the first ionization energy, the second ionization energy,
third ionization energy, etc. The first ionization energy is to the energy needed to remove the outermost, or highest, energy electron and the second
ionization energy is the energy required to remove any subsequent high-energy electron from a gaseous cation. Below are the formulas for calculating the
first and second ionization energies.
First Ionization Energy:
Second Ionization Energy:
Generally, any subsequent ionization energies (2nd, 3rd, etc.) follow the same periodic trend as the first ionization energy.
Figure 4. Periodic Table Showing Ionization Energy Trend
Ionization energies decrease as atomic radii increase. This observation is affected by n (the principle quantum number) and Zeff (based on the atomic
number and shows how many protons are seen in the atom) on the ionization energy (I). Given by the following equation:
•
•
Going across a period, the Zeff increases and n (principal quantum number) remains the same, so that the ionization energy increases.
Going down a group, the n increases and Zeff increases slightly, the ionization energy decreases.
Why does the ionization energy increase going across a period? It has to do with two factors. (i) One factor is that the atomic size decreases. (ii) The
second factor is that the effective nuclear charge increases. The effective nuclear charge is the charge experienced by a specific electron within an atom.
Recall that the nuclear charge was used to describe why the atomic size decreased going across a period. Table 9.7 shows the effective nuclear charge
along with the ionization energy for the elements in period 2.
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Element
Lithium (Li)
Beryllium (Be)
Boron (B)
Carbon (C)
Nitrogen (N)
Oxygen (O)
Fluorine (F)
TA B L E 9.7: Effective Nuclear Charge for Period 2 Main Group Elements
Number of
Effective
Ionization Energy
Electron
Number of
Protons
Core Electrons
Configuration
Nuclear
Charge
3
2
1
520 kJ/mol
[He]2s1
2
4
2
2
899 kJ/mol
[He]2s
[He]2s2 2 p1
[He]2s2 2 p2
[He]2s2 2 p3
5
2
3
801 kJ/mol
6
7
2
2
4
5
1086 kJ/mol
1400 kJ/mol
[He]2s2 2 p4
[He]2s2 2 p5
8
2
6
1314 kJ/mol
9
2
7
1680 kJ/mol
2
The electrons that are shielding the nuclear charge are the core electrons, which are the 1s electrons for period 2. The effective nuclear charge is
approximately the difference between the total nuclear charge and the number of core electrons. Notice that as the effective nuclear charge increases,
the ionization energy also increases.
Example:
What would be the effective nuclear charge for chlorine? Would you predict the ionization energy to be higher or lower than the ionization energy for
fluorine?
Solution:
2
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Chlorine has the electron configuration: Cl = [Ne]3s 3p . The effective nuclear charge is 7, which is the same as the nuclear charge for fluorine. Predicting
the ionization energy with just this information would be difficult. The atomic size, however, is larger for chlorine than it is for fluorine because chlorine
has three energy levels (chlorine is in period 3). Now we can conclude that the ionization energy for chlorine should be lower than that of fluorine because
the electron would be easier to pull off when it is further away from the nucleus. (Indeed, the value for the first ionization energy of chlorine is 1251
kJ/mol, compared to 1680 kJ/mol for fluorine.)
The process by which the first ionization energy of hydrogen is measured would be represented by the following equation.
H(g)
+
-
H (g) + e
o
H = -1312.0 kJ/mol
Practice Problem 3:
Use the Bohr model to calculate the wavelength and energy of the photon that would have to be absorbed to ionize a neutral hydrogen atom in the gas
phase.
The magnitude of the first ionization energy of hydrogen can be brought into perspective by comparing it with the energy given off in a chemical reaction.
When we burn natural gas, about 800 kJ of energy is released per mole of methane consumed.
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
o
H = -802.4 kJ/mol
The thermite reaction, which is used to weld iron rails, gives off about 850 kJ of energy per mole of iron oxide consumed.
Fe2O3(s) + 2 Al(s)
Al2O3(s) + 2 Fe(l)
o
H = -851.5 kJ/mol
The first ionization energy of hydrogen is half again as large as the energy given off in either of these reactions.
A few anomalies exist with respect to the ionization energy trends. Going across a period as we noted before there are two ways in which the ionization
energy may be affected by the electron configuration. When we look at period 3, we can see that there is an anomaly as we move from the 3s sublevels to
the 3p sublevel. The table below shows the electron configurations and first ionization energy for the main group elements in period 3.
In the table, we see that when we compare magnesium to aluminum, the IE1 decreases instead of increases. Why is this? Magnesium has its outermost
electrons in the 3s sub-level. The aluminum atom has its outermost electron in the 3 p sublevel. Since p electrons have just slightly more energy than s
2
1
electrons, it takes a little less energy to remove that electron from aluminum. One other factor is that the electrons in 3s shield the electron in 3 p . These
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two factors allow the IE1 for aluminum to be less than IE1 for magnesium.
When we look again at the table, we can see that the ionization energies for beryllium and nitrogen also do not follow the general trend. The first
ionization energy of boron is smaller than beryllium, and the first ionization energy of oxygen is smaller than nitrogen.
These observations can be explained by looking at the electron configurations of these elements. The electron removed when a beryllium atom is ionized
comes from the 2s orbital, but a 2p electron is removed when boron is ionized.
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Be: [He] 2s
2
B: [He] 2s 2p
1
The electrons removed when nitrogen and oxygen are ionized also come from 2p orbitals.
2
3
2
4
N: [He] 2s 2p
O: [He] 2s 2p
But there is an important difference in the way electrons are distributed in these atoms. Hund's rules predict that the three electrons in the 2p orbitals of
a nitrogen atom all have the same spin, but electrons are paired in one of the 2p orbitals on an oxygen atom. While nitrogen has one electron occupying
each of the three p orbitals in the second sub-level, oxygen has an additional electron in one of the three 2p orbitals. The presence of two electrons in an
orbital lead to greater electron-electron repulsion experienced by these 2 p electrons, which lowers the amount of energy needed to remove one of these
electrons. Therefore, IE1 for oxygen is less than that for nitrogen.
Hund's rules can be understood by assuming that electrons try to stay as far apart as possible to minimize the force of repulsion between these particles.
The three electrons in the 2p orbitals on nitrogen therefore enter different orbitals with their spins aligned in the same direction. In oxygen, two electrons
must occupy one of the 2p orbitals. The force of repulsion between these electrons is minimized to some extent by pairing the electrons. There is still
some residual repulsion between these electrons, however, which makes it slightly easier to remove an electron from a neutral oxygen atom than we
would expect from the number of protons in the nucleus of the atom.
Practice Problem 4:
Predict which element in each of the following pairs has the larger first ionization energy.
(a) Na or Mg
(b) Mg or Al
(c) F or Cl
Second, Third, Fourth, and Higher Ionization Energies
+
2+
3+
By now you know that sodium forms Na ions, magnesium forms Mg ions, and aluminum forms Al ions. But have you ever wondered why sodium
2+
3+
doesn't form Na ions, or even Na ions? The answer can be obtained from data for the second, third, and higher ionization energies of the element.
The first ionization energy of sodium, for example, is the energy it takes to remove one electron from a neutral atom.
Na(g) + energy
+
-
Na (g) + e
2+
The second ionization energy is the energy it takes to remove another electron to form an Na ion in the gas phase.
+
Na (g) + energy
2+
-
Na (g) + e
The third ionization energy can be represented by the following equation.
2+
Na (g) + energy
3+
-
Na (g) + e
3+
The energy required to form a Na ion in the gas phase is the sum of the first, second, and third ionization energies of the element.
First, Second, Third, and Fourth Ionization Energies of Sodium, Magnesium, and Aluminum (kJ/mol)
+
It doesn't take much energy to remove one electron from a sodium atom to form an Na ion with a filled-shell electron configuration. Once this is done,
however, it takes almost 10 times as much energy to break into this filled-shell configuration to remove a second electron. Because it takes more energy
+
to remove the second electron than is given off in any chemical reaction, sodium can react with other elements to form compounds that contain Na ions
2+
3+
but not Na or Na ions.
A similar pattern is observed when the ionization energies of magnesium are analyzed. The first ionization energy of magnesium is larger than sodium
because magnesium has one more proton in its nucleus to hold on to the electrons in the 3s orbital.
Mg: [Ne] 3s
2
The second ionization energy of Mg is larger than the first because it always takes more energy to remove an electron from a positively charged ion than
2+
from a neutral atom. The third ionization energy of magnesium is enormous, however, because the Mg ion has a filled-shell electron configuration.
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The same pattern can be seen in the ionization energies of aluminum. The first ionization energy of aluminum is smaller than magnesium. The second
ionization energy of aluminum is larger than the first, and the third ionization energy is even larger. Although it takes a considerable amount of energy to
3+
3+
remove three electrons from an aluminum atom to form an Al ion, the energy needed to break into the filled-shell configuration of the Al ion is
4+
astronomical. Thus, it would be a mistake to look for an Al ion as the product of a chemical reaction.
2
1
In the case of magnesium it has the electronic stability of a full 2s shell whilst Aluminum has 3 electrons in its secondary energy level, 2s & 2p , hence it
will readily lose its electron in the p shell in order to gain the stability of a full 2s orbital and hence its first ionization energy is less than that of magnesium,
thus bucking the trend of increasing first ionization energy from left to right in the same period in the periodic table
Practice Problem 5:
Practice Problem 6:
Predict the group in the periodic table in which an element with the
following ionization energies would most likely be found.
Use the trends in the ionization energies of the elements to explain
the following observations.
1st IE = 786 kJ/mol
(a) Elements on the left side of the periodic table are more likely
than those on the right to form positive ions.
2nd IE = 1577
3rd IE = 3232
4th IE = 4355
5th IE = 16,091
6th IE = 19,784
(b) The maximum positive charge on an ion is equal to the group
number of the element
Summary - Ionization energy is the energy required to remove the most loosely held electron from a gaseous atom or ion. Ionization energy generally
increases across a period and decreases down a group.
• Once one electron has been removed, a second electron can be removed, but IE1 < IE2. If a third electron is removed, IE1 < IE2 < IE3, and so on.
• The effective nuclear charge is the charge of the nucleus felt by the valence electrons.
• The effective nuclear charge and the atomic size help explain the trend of ionization energy. Going down a group, the atomic size gets larger and the
electrons can be more readily removed. Therefore, ionization energy decreases down a group. Going across a period, both the effective nuclear charge
and the ionization energy increases, because the electrons are harder to remove.
(3) Periodic Trends for Electron Affinity
Like the name suggests, electron affinity describes the ability of an atom to accept an electron. Unlike electronegativity, electron affinity is a quantitative
measure that measures the energy change that occurs when an electron is added to a neutral gas atom. When measuring electron affinity, the more
negative the value, the more of an affinity to electrons that atom has.
Electron affinity generally decreases down a group of elements because each atom is larger than the atom above it (this is the atomic radius trend, which
will be discussed later in this text). This means that an added electron is further away from the atom's nucleus compared to its position in the smaller
atom. With a larger distance between the negatively-charged electron and the positively-charged nucleus, the force of attraction is relatively weaker.
Therefore, electron affinity decreases. Moving from left to right across a period, atoms become smaller as the forces of attraction become stronger. This
causes the electron to move closer to the nucleus, thus increasing the electron affinity from left to right across a period.
•
•
Electron affinity increases from left to right within a period. This is caused by the decrease in atomic radius.
Electron affinity decreases from top to bottom within a group. This is caused by the increase in atomic radius.
Figure 5. Periodic Table showing Electron Affinity Trend
Ionization energies measure the tendency of a neutral atom to resist the loss of electrons. It takes a considerable amount of energy, for example, to
remove an electron from a neutral fluorine atom to form a positively charged ion.
F(g)
F+(g) + e-
Ho = 1681.0 kJ/mol
The electron affinity of an element is the energy given off when a neutral atom in the gas phase gains an extra electron to form a negatively charged ion.
A fluorine atom in the gas phase, for example, gives off energy when it gains an electron to form a fluoride ion.
F(g) + e-
F-(g)
Ho = -328.0 kJ/mol
Electron affinities are more difficult to measure than ionization energies and are usually known to fewer significant figures. The electron affinities of the
main group elements are shown in the figure below.
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Several patterns can be found in these data.
• Electron affinities generally become smaller as we go down a column of the periodic table for two reasons. First, the electron being added to the
atom is placed in larger orbitals, where it spends less time near the nucleus of the atom. Second, the number of electrons on an atom increases
as we go down a column, so the force of repulsion between the electron being added and the electrons already present on a neutral atom
becomes larger.
•
Electron affinity data are complicated by the fact that the repulsion between the electron being added to the atom and the electrons already
present on the atom depends on the volume of the atom. Among the nonmetals in Groups VIA and VIIA, this force of repulsion is largest for the
very smallest atoms in these columns: oxygen and fluorine. As a result, these elements have a smaller electron affinity than the elements below
them in these columns as shown in the figure below. From that point on, however, the electron affinities decrease as we continue down these
columns.
At first glance, there appears to be no pattern in electron affinity across a row of the periodic table, as shown in the figure below. When these data are
listed along with the electron configurations of these elements, however, they make sense. These data can be explained by noting that electron affinities
are much smaller than ionization energies. As a result, elements such as helium, beryllium, nitrogen, and neon, which have unusually stable electron
configurations, have such small affinities for extra electrons that no energy is given off when a neutral atom of these elements picks up an electron. These
configurations are so stable that it actually takes energy to force one of these elements to pick up an extra electron to form a negative ion.
Electron Affinities and Electron Configurations for the First 10 Elements in the Periodic Table
Element
Electron Affinity (kJ/mol)
Electron Configuration
H
72.8
1s1
He
<0
1s2
Li
59.8
[He] 2s1
Be
<0
[He] 2s2
B
27
[He] 2s2 2p1
C
122.3
[He] 2s2 2p2
N
<0
[He] 2s2 2p3
O
141.1
[He] 2s2 2p4
F
328.0
[He] 2s2 2p5
Ne
<0
[He] 2s2 2p6
Consequences of the Relative Size of Ionization Energies and Electron Affinities
+
-
Students often believe that sodium reacts with chlorine to form Na and Cl ions because chlorine atoms "like" electrons more than sodium atoms do.
There is no doubt that sodium reacts vigorously with chlorine to form NaCl.
2 Na(s) + Cl2(g)
2 NaCl(s)
Furthermore, the ease with which solutions of NaCl in water conduct electricity is evidence for the fact that the product of this reaction is a salt, which
+
contains Na and Cl ions.
NaCl(s)
H2O
Na+(aq) + Cl- (aq)
The only question is whether it is legitimate to assume that this reaction occurs because chlorine atoms "like" electrons more than sodium atoms.
The first ionization energy for sodium is one and one-half times larger than the electron affinity for chlorine.
Na: 1st IE = 495.8 kJ/mol
Cl: EA = 328.8 kJ/mol
Thus, it takes more energy to remove an electron from a neutral sodium atom than is given off when the electron is picked up by a neutral chlorine atom.
We will obviously have to find another explanation for why sodium reacts with chlorine to form NaCl. Before we can do this, however, we need to know
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more about the chemistry of ionic compounds.
Lattice Energies and the Strength of the Ionic Bond
The force of attraction between oppositely charged particles is directly proportional to the product of the charges on the two objects (q1 and q2) and
2
inversely proportional to the square of the distance between the objects (r ).
The strength of the bond between the ions of opposite charge in an ionic compound therefore depends on the charges on the ions and the distance
between the centers of the ions when they pack to form a crystal.
An estimate of the strength of the bonds in an ionic compound can be obtained by measuring the lattice energy of the compound, which is the energy
given off when oppositely charged ions in the gas phase come together to form a solid.
+
-
+
Example: The lattice energy of NaCl is the energy given off when Na and Cl ions in the gas phase come together to form the lattice of alternating Na and
Cl ions in the NaCl crystal shown in the figure below.
Na+(g) + Cl-(g)
Ho = -787.3 kJ/mol
NaCl(s)
The lattice energies of ionic compounds are relatively large. The lattice energy of NaCl, for example, is 787.3 kJ/mol, which is only slightly less than the
energy given off when natural gas burns. The bond between ions of opposite charge is strongest when the ions are small. The lattice energies for the
alkali metal halides is therefore largest for LiF and smallest for CsI, as shown in the table below.
Lattice Energies of Alkali Metals Halides (kJ/mol)
Li+
Na
K
+
+
F-
Cl-
Br-
I-
1036
853
807
757
923
787
747
704
821
715
682
649
+
785
689
660
630
+
740
659
631
604
Rb
Cs
The ionic bond should also become stronger as the charge on the ions becomes larger. The data in the table below show that the lattice energies for salts
2of the OH and O ions increase rapidly as the charge on the ion becomes larger.
-
2-
Lattice Energies of Salts of the OH and O Ions (kJ/mol)
OH-
O2-
Na+
900
2481
Mg2+
3006
3791
5627
15,916
Al
3+
Lattice Energies and Solubility
+
-
When a salt, such as NaCl dissolves in water, the crystals disappear on the macroscopic scale. On the atomic scale, the Na and Cl ions in the crystal are
released into solution.
NaCl(s)
H2O
Na+(aq) + Cl-(aq)
The lattice energy of a salt therefore gives a rough indication of the solubility of the salt in water because it reflects the energy needed to separate the
positive and negative ions in a salt. Sodium and potassium salts are soluble in water because they have relatively small lattice energies. Magnesium and
aluminum salts are often much less soluble because it takes more energy to separate the positive and negative ions in these salts. NaOH, for example, is
very soluble in water (420 g/L), but Mg(OH)2 dissolves in water only to the extent of 0.009 g/L, and Al(OH)3 is essentially insoluble in water.
Why Does Sodium Form NaCl?
+
-
Sodium reacts with chlorine to form Na ions and Cl ions in spite of the fact that the first ionization energy of sodium is larger than the electron affinity of
chlorine. To explain this, we need to divide the reaction between sodium and chlorine into a number of hypothetical steps for which we know the amount
of energy given off or absorbed.
The starting materials for this reaction are solid sodium metal and chlorine molecules in the gas phase, and the product of the reaction is solid sodium
chloride.
2 Na(s) + Cl2(g)
2 NaCl(s)
Let's imagine that the reaction takes place by the following sequence of steps.
•
A mole of sodium is converted from the solid to a gas. As might be expected, this reaction is endothermic.
Na(s)
Na(g)
Ho = 107.3 kJ
12
•
+
An electron is then removed from each sodium atom to form a mole of Na ions. The energy consumed in this reaction is equal to the first
ionization energy of sodium.
Na+(g)
Na(g)
•
A mole of chlorine atoms is formed by breaking the bonds in one-half a mole of chlorine molecules. Like the previous steps, this is an
endothermic reaction.
1
•
/2 Cl2(g)
Cl(g)
Ho = 121.7 kJ
-
An electron is then added to each chlorine atom to form a Cl ion. This is the first exothermic step in this process, and the energy released is
equal to the electron affinity of chlorine.
Cl(g) + e-
•
Ho = 495.8 kJ
+
Cl-(g)
Ho = -348.8 kJ
-
The isolated Na and Cl ions in the gas phase then come together to form solid NaCl. This is a strongly exothermic reaction, for which
equal to the lattice energy of NaCl.
Na+(g) + Cl-(g)
NaCl(s)
o
H is
Ho = -787.3 kJ
+
-
If we consider just the first four steps in this reaction, Hess's law suggests that it 5.16 takes 376.0 kJ/mol to form Na and Cl ions from sodium metal and
chlorine gas.
Na(s) + 1/2 Cl2(g)
Na+(g) + Cl-(g)
Ho = 376.0 kJ/mol
When we include the last step in the calculation, the lattice energy of NaCl is large enough to compensate for all of the steps in this reaction that consume
energy, as shown in the figure below.
Na(s) + 1/2 Cl2(g)
NaCl(s)
Ho = -411.3 kJ/mol
+
-
The primary driving force behind this reaction is therefore the force of attraction between the Na and Cl ions formed in the reaction, not the affinity of a
chlorine atom for electrons.
Why does the reaction stop at NaCl? Why doesn't it keep going to form NaCl2or NaCl3? The lattice energy would increase as the charge on the sodium
+
2+
3+
2+
atom increased from Na to Na or Na . But to form an Na ion, we have to remove a second electron from the sodium atom, and the second ionization
energy of sodium (4562.4 kJ/mol) is almost 10 times as large as the first ionization energy. The increase in the lattice energy that would result from
2+
+
forming an Na ion can't begin to compensate for the energy needed to break into the filled-shell configuration of the Na ion to remove a second
electron. The reaction between sodium and chlorine therefore stops at NaCl.
Why Does Magnesium Form MgCl2?
If the reaction between sodium and chlorine stops at NaCl, why does the reaction between magnesium and chlorine go on to form MgCl2? To answer this
question, let's break the reaction into the following steps.
•
A mole of magnesium is converted from the solid to a gas.
13
Mg(s)
•
+
An electron is removed from each magnesium atom to form a mole of Mg ions.
Mg+(g)
Mg(g)
•
2+
Mg2+(g)
Ho = 243.4 kJ
2 Cl(g)
-
An electron is then added to each chlorine atom to form Cl ions.
2 Cl(g) + e-
•
Ho = 1450.6 kJ
Two moles of chlorine atoms are formed by breaking the bonds in a mole of chlorine molecules.
Cl2(g)
•
Ho = 737.7 kJ
A second electron is then removed to form a mole of Mg ions.
Mg+(g)
•
Ho = 147.7 kJ
Mg(g)
2+
2 Cl-(g)
Ho = -697.4 kJ
-
The isolated Mg and Cl ions in the gas phase then come together to form solid MgCl2.
Mg2+(g) + 2 Cl-(g)
Ho = -2526 kJ
MgCl2(s)
The lattice energy for MgCl2 is large enough to compensate for the energy it takes to remove the second electron from a magnesium atom because we
2+
don't have to break into a filled-shell configuration to form Mg ions.
Mg(s) + Cl2(g)
Ho = -644 kJ/mol
MgCl2(s)
2+
The reaction stops at MgCl2, however, because it would take an enormous amount of energy to break into the filled-shell configuration of the Mg ion to
remove another electron.
Why Do Semimetals Exist?
Metals have some or all of the following properties.
• They have a metallic shine or luster.
• They are typically solids at room temperature.
• They are malleable and ductile.
• They conduct heat and electricity.
• They exist as extended planes of atoms.
• They combine with other metals to form alloys, which behave like metals.
•
+
2+
3+
2+
They form positive ions, such as the Na , Mg , Fe , and Cu ions.
Nonmetals have the opposite properties.
• They seldom have a metallic luster.
• They are often gases at room temperature.
• They are neither malleable nor ductile.
• They are poor conductors of both heat and electricity.
• They often form molecules in their elemental form.
• They combine with other nonmetals to form covalent compounds.
• They tend to form negative ions, such as the F-, Cl-, P3-, SO42-, and PO43- ions.
The differences in the chemical and physical properties of metals and nonmetals can be traced to differences in their electron configurations, their
ionization energies, their electron affinities, and the radii of their atoms and ions. As a rule, metals have relatively few electrons in their outermost shell
of orbitals, lower ionization energies, smaller electron affinities, and larger atoms than nonmetals.
There are no abrupt changes in the physical properties discussed in this chapter as we go across a row of the periodic table or down a column. As a result,
the change from metal to nonmetal must be gradual. Instead of arbitrarily dividing elements into metals and nonmetals, it might be better to describe
some elements as being more metallic and other elements as more nonmetallic. Metallic character decreases as we go across a row of the periodic table
from left to right, while nonmetallic character increases.
Metallic character decreases
Na
Mg
Al
Si
P
S
Cl
Ar
Nonmetallic character increases
For the same reasons, elements toward the top of a column of the periodic table are the most nonmetallic, while elements toward the bottom of a
column are the most metallic. This is evident in Group IVA, where a gradual transition is seen from a nonmetallic element (carbon) toward the well-known
metals tin and lead.
C
metallic
Si nonmetallic
character
character
Ge
increases
increases
Sn
14
Pb
If elements become less metallic and more nonmetallic as we go across a row of the periodic table from left to right, we should encounter elements along
each row that have properties that lie between the extremes of metals and nonmetals. The eight elements in this class (B, Si, Ge, As, Sb, Te, Po, and At)
often look metallic, but they are brittle, like nonmetals. They are neither good conductors nor good insulators but serve as the basis for the semiconductor
industry. These eight elements are often known as metalloids, or semimetals
(4) Periodic Trends for Atomic Radius
The actual trends that are observed with atomic size have to do with three factors. These factors are:
1) the number of protons in the nucleus (called the nuclear charge).
2) the number of energy levels holding electrons and the number of electrons in the outer energy level, n.
3) the number of electrons held between the nucleus and its outermost electrons (called the shielding effect).
The atomic radius is one-half the distance between the centers of a homonuclear diatomic molecule, as illustrated below. A diatomic molecule is a
molecule made of exactly two atoms, while homonuclear means both atoms are the same element.
The 3rd contributing factor to atomic size is the shielding effect. The protons in the nucleus attract the valence electrons in the outer energy level, but the
strength of this attraction depends on the size of the charges, the distance between the charges, and the number of electrons between the nucleus and
the valence electrons. The presence of the core electrons weakens the attraction between the nucleus and the valence electrons. This weakening is called
the shielding effect. Note that although valence electrons do participate in shielding, electrons in the same energy level do not shield each other as
effectively as the core electrons do. As a result, the amount of shielding primarily depends on the number of electrons between the nucleus and the
valence electrons. When the nucleus pulls strongly on the valence electrons, the valence shell can be pulled in tighter and closer to the nucleus. When the
attraction is weakened by shielding, the valence shell cannot be pulled in as close. The more shielding that occurs, the further the valence shell can spread
out.
For example, if you are looking at the element sodium, it has the electron configuration: Na 1s22s22 p63s1
The outer energy level is n = 3. There is one valence electron, but the attraction between this lone valence electron and the nucleus, which has 11 protons,
is shielded by the other 10 inner (or core) electrons.
When we compare an atom of sodium to one of cesium, we notice that the number of protons increases, as well as the number of energy levels occupied
by electrons. The increase in the number of protons, however, is also accompanied by the same increase in the number of shielding electrons. Cs 1s22s22
p63s23 p64s23d104 p65s24d105 p66s1
The result is that the valence electron in both atoms feels a similar pull from the nucleus, but the valence electron in the cesium atom is further from the
nucleus because it is in a higher energy level. Compared to the shielding effect, the increase in the number of energy levels has a greater impact on the
atom’s size. Consequently, the size of a cesium atom is larger than that of a sodium atom.
This is true for not only Group 1A metals, but for all of the groups across the periodic table. For any given group, as you move downward in the periodic
table, the size of the atoms increases. For instance, the largest atoms in the halogen family are bromine and iodine (astatine is radioactive and only exists
for short periods of time, so we won’t include it in the discussion). You can imagine that with the increase in the number of energy levels, the size of the
atom must increase. The increase in the number of energy levels in the electron cloud takes up more space.
In order to determine the trend for the periods, we need to look at 1) the number of protons (nuclear charge), 2) the number of energy levels, and 3) the
shielding effect. For a row in the periodic table, the atomic number still increases (as it did for the groups), and thus the number of protons would
increase. For a given period, however, we find that the outermost energy level does not change as the number of electrons increases. In period 2, for
example, each additional electron goes into the second energy level, so the total number of energy levels does not go up. Table 9.2 shows the electron
configuration for the elements in period 2.
Looking at the elements in period 2, the number of protons increases from three (for lithium) to nine (for fluorine). Therefore, the nuclear charge
increases across a period. Meanwhile, the number of energy levels occupied by the electrons remains the same. How will this affect the radius? We know
that every one of the elements in this period has two core electrons in the inner energy level (n = 1). The core electrons shield the outer electrons from
the charge of the nucleus. Since the number of protons attracting the outer electrons increases while the shielding remains the same, the valence
electrons are pulled closer to the nucleus, making the atom smaller.
Consider the elements lithium, beryllium, and fluorine from period 2. With lithium, the two core electrons will shield the one valence electron from three
protons. Beryllium has four protons being shielded by the 1s2 electrons. For fluorine, there are nine protons and nine electrons. All three of these
elements have the same core electrons: the
1s2 electrons. As the number of protons increases, the nuclear charge increases. With an increase in nuclear charge, there is an increase in the pull
between the protons and the outer level, pulling the outer electrons toward the nucleus. The amount of shielding from the nucleus does not increase
because the number of core electrons remains the same. The net result is that the atomic size decreases going across the row. In the graph below, the
values are shown for the atomic radii for period 2.
15
Atomic Radii of Transition Elements - Until now, we have worked solely with the main group elements. Let’s consider how three factors affecting atomic
size affect transition metals. The first row of the transition metals all contain electrons in the 3d sublevel and are referred to as the 3d metals. Table 9.3
shows the electron configuration for the ten elements in this row. The number of protons is increasing, so the nuclear charge is increasing.
TA B L E 9.3: Electron Configuration for 3
Element
Scandium (Sc)
Number of Protons
21
Titanium (Ti)
22
Vanadium (V)
23
Electron Configuration
[Ar]3d 1 4s2
[Ar]3d 2 4s2
[Ar]3d 3 4s2
Chromium (Cr)
24
Manganese (Mn)
25
Iron (Fe)
26
Cobalt (Co)
27
Nickel (Ni)
28
[Ar]3d 7 4s2
[Ar]3d 8 4s2
Copper (Cu)
Zinc (Zn)
29
30
[Ar]3d 10 4s1
[Ar]3d 10 4s2
[Ar]3d 5 4s1
[Ar]3d 5 4s2
[Ar]3d 6 4s2
You may notice that some of these configurations are not what you would expect based on the information presented so far. Both chromium and copper
have one of the 4s electrons moved to a 3d orbital. A simplified explanation for these unusual electron configurations is that the d sub-level is particularly
stable when it is half-full (5 electrons) or completely full (10 electrons). Since the 4s and 3d orbitals are close in energy, this added stabilization is enough to
change the location of one valence electron.
TA B L E 9.4: Atomic Radii for 3d Metals
Element
Scandium (Sc)
Titanium (Ti)
Vanadium (V)
Chromium (Cr)
Manganese (Mn)
Iron (Fe)
Cobalt (Co)
Nickel (Ni)
Copper (Cu)
Zinc (Zn)
Atomic Radii (pm)
164
147
135
129
137
126
125
125
128
137
Table 9.4 lists the atomic radii for the first row of the transition metals. It can be seen from this table that the period trend in atomic radii is not followed
as closely by the transition metals. Since we are adding electrons to the 3d orbitals, we are actually adding to the core electrons and not to the valence
orbitals. Although the nuclear charge is going up, the shielding is also increasing with each added electron. Because of this, there is less atomic contraction
throughout the transition metals.
The graph of the number of protons versus the atomic radii for the 3d metals is shown below. Compared to the same graph for the elements in period 2,
the graph for transition metals shows the trend for atomic radii is not as straightforward.
Manganese vs Iron
Nickel vs Copper vs Zinc
Atomic Radii Trend - Summary
• Atomic size is the distance from the nucleus to the valence shell.
• Atomic size is difficult to measure because it has no definite boundary.
• Atomic radius is a more definite and measurable way of defining atomic size. It is half the distance from the center of one atom to the center of another
atom in a homonuclear diatomic molecule.
• There are three factors that help in the prediction of the trends in the periodic table: 1) number of protons in the nucleus, 2) number of energy levels,
16
and 3) the shielding effect.
• Atomic radius increases from top to bottom within a group. This is caused by electron shielding.
• Atomic radius decreases from left to right within a period. This is caused by the increase in the number of protons and electrons across a period. One
proton has a greater effect than one electron; thus, a lot of electrons will get pulled towards the nucleus, resulting in a smaller radius.
• This trend is not as systematic for the transition metals because other factors come into play.
Group and Period Trends in Ionic Size - Summary
• In chemical reactions, atoms can gain or lose electrons. This results in the formation of an ion. An ion is basically an atom with a positive or negative
charge.
• Atoms of metallic elements tend to form positive ions (cations) by losing one or more electrons.
• Atoms of nonmetallic elements tend to form negative ions (anions) by gaining one or more electrons.
• Cations are smaller than the atoms from which they are formed.
• Anions are larger than the atoms from which they were formed.
• Ionic size increases from top to bottom down a group of elements in the periodic table.
• From left to right across a period, the ionic size decreases as long as you are comparing all metals or all nonmetals. Between the metals and nonmetals,
the ionic size increases as you switch from cations to anions.
Figure 6. Periodic Table showing Atomic Radius Trend
(5) Periodic Trends for Melting Point
Melting points are the amount of energy required to break a bond(s) to change the solid phase of a substance to a liquid. Generally, the stronger the bond
between the atoms of an element, the higher the energy requirement in breaking that bond. Since temperature is directly proportional to energy, a high
bond dissociation energy correlates to a high temperature. Melting points are varied and don't generally form a distinguishable trend across the periodic
table. However, certain conclusions can be drawn from the following graph.
• Metals generally possess a high melting point.
• Most non-metals possess low melting points.
• The non-metal carbon possesses the highest boiling point of all the elements. The semi-metal boron also possesses a high melting point.
Figure 7. Chart of Melting Points of Various Elements
(6) Periodic Trends for Metallic Character
The metallic character of an element can be defined as how readily an atom can lose an electron. As you move from right to left across a period,
metallic character increases because the attraction between valence electron and the nucleus is weaker, thus enabling an easier loss of electrons.
Metallic character increases as you move down a group because the atomic size is increasing. When the atomic size increases, the outer shells are
farther away. The principle quantum number increases and average electron density moves farther from nucleus. The electrons of the valence shell
have less of an attraction to the nucleus and, as a result, can lose electrons more readily, causing an increase in metallic character.
•
Metallic characteristics decrease from left to right across a period. This is caused by the decrease in radius (above it is stated that Zeff causes
this) of the atom which allows the outer electrons to ionize more readily.
17
•
Metallic characteristics increase down a group. Electron shielding causes the atomic radius to increase thus the outer electrons ionizes more
readily than electrons in smaller atoms.
•
Metallic character relates to the ability to lose electrons, and nonmetallic character relates to the ability to gain electrons.
•
Another easier way to remember the trend of metallic character is that as you move from left and down towards the bottom-left corner of the
periodic table, metallic character increases because you are heading towards Groups 1 and 2, or the Alkali and Alkaline metal groups. Likewise, if
you move up and to the right to the upper-right corner of the periodic table, metallic character decreases because you are passing by to the right
side of the staircase, which indicate the nonmetals. These include the Group 8, the noble gases, and other common gases such as oxygen and
nitrogen.
o In other words:
o Move left across period and down the group: increase metallic character (heading towards alkali and alkaline metals)
o Move right across period and up the group: decrease metallic character (heading towards nonmetals like noble gases)
Figure 8. Periodic Table of Metallic Character Trend