Chapter 8: Periodic Properties of the Elements

Periodic Trends in the Size of Atoms and Effective Nuclear Charge

In previous chapters, we saw that the volume of an atom is taken up primarily by its electrons (Chapter 2) occupying quantum-mechanical orbitals (Chapter 7). We also saw that these orbitals do not have definite boundaries, but represent only statistical probability distributions for where the electrons are found. So how do we define the size of an atom? One way to define atomic radii is to consider the distance between nonbonding atoms in molecules or atoms that are in direct contact. For example, krypton can be frozen into a solid in which the krypton atoms are touching each other but are not bonded together. The distance between the centers of adjacent krypton atoms—which can be determined from the solid's density—is twice the radius of a krypton atom. An atomic radius determined in this way is the nonbonding atomic radius or the van der Waals radius. The van der Waals radius represents the radius of an atom when it is not bonded to another atom. The bonding atomic radius or covalent radius, another way to indicate the size of an atom, is defined differently for nonmetals and metals, as follows: Nonmetals: one-half the distance between two of the atoms bonded together Metals: one-half the distance between two of the atoms next to each other in a crystal of the metal For example, the distance between Br atoms in Br2 is 228 pm; therefore, the Br covalent radius is one-half of 228 pm or 114 pm. Covalent radii can be similarly assigned to all elements in the periodic table that form chemical bonds or form metallic crystals. A more general term, the atomic radius, refers to a set of average bonding radii determined from measurements on a large number of elements and compounds. The atomic radius represents the radius of an atom when it is bonded to another atom and is always smaller than the van der Waals radius. The approximate bond length of any two covalently bonded atoms is simply the sum of their atomic radii. For example, the approximate bond length for ICl is iodine's atomic radius (133 pm) plus chlorine's atomic radius (99 pm), for a bond length of 232 pm. (The actual experimentally measured bond length in ICl is 232.07 pm.) FigurE 8.9▲ shows the atomic radius plotted as a function of atomic number for the first 57 elements in the periodic table. Notice the periodic trend in the radii. Atomic radii peak with each alkali metal. FigurE 8.10▼ is a relief map of atomic radii for most of the elements in the periodic table. The general trends in the atomic radii of main-group elements, which are the same as trends observed in van der Waals radii, are stated as follows: 1. As you move down a column (or group) in the periodic table, atomic radius increases. 2. As you move to the right across a row (or period) in the periodic table, atomic radius decreases. We can understand the observed trend in radii as we move down a column in light of the trends in the sizes of atomic orbitals. The atomic radius is largely determined by the valence electrons, the electrons farthest from the nucleus. As we move down a column in the periodic table, the highest principal quantum number (n) of the valence electrons increases. Consequently, the valence electrons occupy larger orbitals, resulting in larger atoms. The observed trend in atomic radii as we move to the right across a row, however, is a bit more complex. To understand this trend, we revisit some concepts from Section 8.3, including effective nuclear charge and shielding.

Electron Configurations for Multielectron Atoms

Now that we know the energy ordering of orbitals in multielectron atoms, we can build ground state electron configurations for the rest of the elements. Since we know that electrons occupy the lowest energy orbitals available when the atom is in its ground state, and that only two electrons (with opposing spins) are allowed in each orbital, we can systematically build up the electron configurations for the elements. The pattern of orbital filling that reflects what we have just learned is known as the aufbau principle (the German word aufbau means "build up"). For lithium, which has three electrons, the electron configuration and orbital diagram are: Notice that the 2p electrons occupy the p orbitals (of equal energy) singly, rather than pairing in one orbital. This way of filling orbitals is known as Hund's rule, which states that when filling degenerate orbitals, electrons fill them singly first, with parallel spins. Hund's rule is a result of an atom's tendency to find the lowest energy state possible. When two electrons occupy separate orbitals of equal energy, the repulsive interaction between them is lower than when they occupy the same orbital because the electrons are spread out over a larger region of space. Summarizing Orbital Filling: ▶ Electrons occupy orbitals so as to minimize the energy of the atom; therefore, lower energy orbitals fill before higher energy orbitals. Orbitals fill in the following order: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s. ▶ Orbitals can hold no more than two electrons each. When two electrons occupy the same orbital, their spins are opposite. This is another way of expressing the Pauli exclusion principle (no two electrons in one atom can have the same four quantum numbers). ▶ When orbitals of identical energy are available, electrons first occupy these orbitals singly with parallel spins rather than in pairs. Once the orbitals of equal energy are half-full, the electrons start to pair (Hund's rule). Consider the electron configurations and orbital diagrams for elements with atomic numbers 3-10. Notice that, as a result of Hund's rule, the p orbitals fill with single electrons before the electrons pair The electron configuration of neon represents the complete filling of the n = 2 principal level. When writing electron configurations for elements beyond neon, or beyond any other noble gas, we can abbreviate the electron configuration of the previous noble gas—sometimes called the inner electron configuration—with the symbol for the noble gas in square brackets. For example, the electron configuration of sodium is: Na 1s 2 2s 2 2p6 3s 1 We can write this configuration more succinctly by using [Ne] to represent the inner electrons: Na 3Ne4 3s 1 [Ne] represents 1s 2 2s 2 2p6 , the electron configuration for neon. To write an electron configuration for an element, first find its atomic number from the periodic table—this number equals the number of electrons. Then use the order of filling to distribute the electrons in the appropriate orbitals. Remember that each orbital can hold a maximum of two electrons. Consequently, The s sublevel has only one orbital and can therefore hold only 2 electrons. The p sublevel has three orbitals and can therefore hold 6 electrons. The d sublevel has five orbitals and can therefore hold 10 electrons. The f sublevel has seven orbitals and can therefore hold 14 electrons. ExamPLE 8.1 Electron Configurations Write electron configurations for each element. (a) Mg (b) P (c) Br (d) Al SOLUTION (a) Mg Magnesium has 12 electrons. Distribute two of these into the 1s orbital, two into the 2s orbital, six into the 2p orbitals, and two into the 3s orbital. (b) P Phosphorus has 15 electrons. Distribute two of these into the 1s orbital, two into the 2s orbital, six into the 2p orbitals, two into the 3s orbital, and three into the 3p orbitals. (c) Br Bromine has 35 electrons. Distribute two of these into the 1s orbital, two into the 2s orbital, six into the 2p orbitals, two into the 3s orbital, six into the 3p orbitals, two into the 4s orbital, ten into the 3d orbitals, and five into the 4p orbitals. Al Aluminum has 13 electrons. Distribute two of these into the 1s orbital, two into the 2s orbital, six into the 2p orbitals, two into the 3s orbital, and one into the 3p orbital. ExamPLE 8.2 writing Orbital Diagrams Write an orbital diagram for sulfur and determine the number of unpaired electrons. SOLUTION Since sulfur is atomic number 16, it has 16 electrons and the electron configuration is 1s2 2s2 2p6 3s2 3p4. Draw a box for each orbital, putting the lowest energy orbital (1s) on the far left and proceeding to orbitals of higher energy to the right Distribute the 16 electrons into the boxes representing the orbitals, allowing a maximum of two electrons per orbital and remembering Hund's rule. You can see from the diagram that sulfur has two unpaired electrons.

Effective Nuclear Charge

The trend in atomic radii as we move to the right across a row in the periodic table is determined by the inward pull of the nucleus on the electrons in the outermost principal energy level (highest n value). According to Coulomb's law, the attraction between a nucleus and an electron increases with increasing magnitude of nuclear charge. For example, compare the H atom to the He+ ion. H 1s 1 He+ 1s 1 It takes 1312 kJ>mol of energy to remove the 1s electron from hydrogen, but 5251 kJ>mol of energy to remove it from He+. Why? Although each electron is in a 1s orbital, the electron in the helium ion is attracted to the nucleus with a 2+ charge, while the electron in the hydrogen atom is attracted to the nucleus by only a 1+ charge. Therefore, the electron in the helium ion is held more tightly, making it more difficult to remove and making the helium ion smaller than the hydrogen atom. As we saw in Section 8.3, any one electron in a multielectron atom experiences both the positive charge of the nucleus (which is attractive) and the negative charges of the other electrons (which are repulsive). Consider again the outermost electron in the lithium atom: Li 1s 2 2s 1 As shown in FigurE 8.11▼, even though the 2s orbital penetrates into the 1s orbital to some degree, the majority of the 2s orbital is outside of the 1s orbital. Therefore, the electron in the 2s orbital is partially screened or shielded from the 3+ charge of the nucleus by the 2- charge of the 1s (or core) electrons, reducing the net charge experienced by the 2s electron. As we have seen, we can define the average or net charge experienced by an electron as the effective nuclear charge. The effective nuclear charge experienced by a particular electron in an atom is the actual nuclear charge (Z) minus the charge shielded by other electrons (S). For lithium, we can estimate that the two core electrons shield the valence electron from the nuclear charge with high efficiency (S is nearly 2). The effective nuclear charge experienced by lithium's valence electron is therefore slightly greater than 1+. Now consider the valence electrons in beryllium (Be), with atomic number 4. Its electron configuration is: Be 1s 2 2s 2 To estimate the effective nuclear charge experienced by the 2s electrons, we must distinguish between two different types of shielding: (1) the shielding of the outermost electrons by the core electrons and (2) the shielding of the outermost electrons by each other. The key to understanding the trend in atomic radii is understanding the difference between these two types of shielding. In general: Core electrons efficiently shield electrons in the outermost principal energy level from nuclear charge, but outermost electrons do not efficiently shield one another from nuclear charge For example, the two outermost electrons in beryllium experience the 4+ charge of the nucleus through the shield of the two 1s core electrons without shielding each other from that charge very much. We can estimate that the shielding (S) experienced by any one of the outermost electrons due to the core electrons is nearly 2, but that the shielding due to the other outermost electron is nearly zero. The effective nuclear charge experienced by beryllium's outermost electrons is therefore slightly greater than 2+. Notice that the effective nuclear charge experienced by beryllium's outermost electrons is greater than that experienced by lithium's outermost electron. Consequently, beryllium's outermost electrons are held more tightly than lithium's, resulting in a smaller atomic radius for beryllium. The effective nuclear charge experienced by an atom's outermost electrons continues to become more positive as we move to the right across the rest of the second row in the periodic table, resulting in successively smaller atomic radii. The same trend is generally observed in all main-group elements Summarizing Atomic Radii for Main-Group Elements: ▶ As you move down a column in the periodic table, the principal quantum number (n) of the electrons in the outermost principal energy level increases, resulting in larger orbitals and therefore larger atomic radii. ▶ As you move to the right across a row in the periodic table, the effective nuclear charge (Zeff) experienced by the electrons in the outermost principal energy level increases, resulting in a stronger attraction between the outermost electrons and the nucleus and therefore smaller atomic radii. EFFECTIVE NuCLEAR CHARGE Which electrons experience the greatest effective nuclear charge? (a) The valence electrons in Mg (b) The valence electrons in Al (c) The valence electrons in S

Trends in Second and Successive Ionization Energies

Notice the trends in the first, second, and third ionization energies of sodium (group 1A) and magnesium (group 2A), as shown in the margin. For sodium, there is a huge jump between the first and second ionization energies. For magnesium, the ionization energy roughly doubles from the first to the second, but then a huge jump occurs between the second and third ionization energies. What is the reason for these jumps? We can understand these trends by examining the electron configurations of sodium and magnesium: Na 3Ne4 3s 1 Mg 3Ne4 3s 2 The first ionization of sodium involves removing the valence electron in the 3s orbital. Recall that these valence electrons are held more loosely than the core electrons and that the resulting ion has a noble gas configuration, which is particularly stable. Consequently, the first ionization energy is fairly low. Conversely, the second ionization of sodium involves removing a core electron from an ion with a noble gas configuration. This requires a tremendous amount of energy, making the value of IE2 very high. As with sodium, the first ionization of magnesium involves removing a valence electron in the 3s orbital. However, this requires a bit more energy than the corresponding ionization of sodium because of the trends in Zeff that we discussed earlier (Zeff increases as we move to the right across a row). The second ionization of magnesium also involves removing an outer electron in the 3s orbital, but this time from an ion with a 1+ charge (instead of from a neutral atom). This requires roughly twice the energy as removing the electron from the neutral atom. The third ionization of magnesium is analogous to the second ionization of sodium—it requires removing a core electron from an ion with a noble gas configuration. This requires a tremendous amount of energy, making the value of IE3 very high. As Table 8.1 illustrates, similar trends exist for the successive ionization energies of many elements. Ionization energy increases fairly uniformly with each successive removal of an outermost electron, but then it takes a large jump with the removal of the first core electron.

Energy Level Diagrams

Prior to 1922, atomic emission was used to qualitatively identify elements, but was too imprecise for quantitative analysis. Lester Strock developed the use of internal standards (see our page on internal standards as well) to compensate for the sample-to-sample and time-dependent variations in signal generation from a fixed quantity of analyte. Up to the end of the Second World War, the common atomic emission sources were flames, arcs, and sparks (the latter the result of applying a sufficiently high electrical potential between two conducting electrodes to cause ionization of the gas between the electrodes). In the next three decades, there was a burst of creativity, leading to the use of low pressure sources (hollow cathode lamps, glow discharges, electrodeless discharge lamps), atmospheric pressure plasmas (inductively coupled plasma, DC plasma), and laser-initiated plasmas (laser microprobe, laser-induced breakdown) for sampling and excitation of materials. While the details and experimental parameters vary among these sources, the essentials of turning bulk materials into individual atoms, putting energy into the atoms to excite them, and then observing light emitted by those atoms as they drop to lower energies is universal. Atoms can only exist in certain discrete energy levels. Typically, at low energies, the levels are far apart. At higher energies, they are closer together. At sufficiently high energy, an electron is removed and the atom ionized. The ion has widely spaced energy levels, that in turn get closer and closer together until another electron can be removed. This process can continue until all the electrons are ionized and the nucleus is bare. The only common fully-ionized species is H ion, a bare proton. A visual way to understand that light emission and absorption occur at the same wavelengths is to sketch a Grotrian diagram (named for Walter Grotrian, a German astronomer from the first half of the 20th century). An atom (or molecule) can go from a lower to a higher energy state by absorbing light or by colliding with another atom or ion with sufficient energy. By convention, emission occurs from an atom that was excited to its upper state by collision or absorption of ambient light. Fluorescence occurs from an atom that was excited to its upper state by light specfically supplied for that purpose.

Sublevel Energy Splitting in Multielectron Atoms

A major difference in the (approximate) solutions to the Schrödinger equation for multi electron atoms compared to the solutions for the hydrogen atom is the energy ordering of the orbitals. In the hydrogen atom, the energy of an orbital depends only on n, the princi- pal quantum number. For example, the 3s, 3p, and 3d orbitals (which are all empty for hydrogen in its lowest energy state) all have the same energy—they are degenerate. By contrast, the orbitals within a principal level of a multielectron atom are not degenerate— their energy depends on the value of l. We say that the energies of the sublevels are split. In general, the lower the value of l within a principal level, the lower the energy of the corresponding orbital. Thus, for a given value of n: E(s orbital) 6 E(p orbital) 6 E(d orbital) 6 E( f orbital) In order to understand sublevel energy splitting, we must examine three key concepts associated with the energy of an electron in the vicinity of a nucleus: (1) Coulomb's law, which describes the interactions between charged particles; (2) shielding, which describes how one electron can shield another electron from the full charge of the nucleus; and (3) penetration, which describes how one atomic orbital can overlap spatially with another, thus penetrating into a region that is close to the nucleus (and therefore less shielded from nuclear charge). Coulomb's Law The attractions and repulsions between charged particles, first introduced in Section 2.3, are described by Coulomb's law, which states that the potential energy (E) of two charged particles depends on their charges (q1 and q2) and their separation (r). In this equation, e0 is a constant (e0 = 8.85 * 10-12 C2>J # m). The potential energy is positive for charges of the same sign (plus * plus, or minus * minus) and negative for charges of opposite sign (plus * minus, or minus * plus). The magnitude of the potential energy depends inversely on the separation between the charged particles. We can draw three important conclusions from Coulomb's law: - For like charges, the potential energy (E) is positive and decreases as the particles get farther apart (as r increases). Since systems tend toward lower potential energy, like charges repel each other (in much the same way that the like poles of two magnets repel each other). - For opposite charges, the potential energy is negative and becomes more negative as the particles get closer together (as r decreases). Therefore opposite charges (like opposite poles on a magnet) attract each other. - The magnitude of the interaction between charged particles increases as the charges of the particles increase. Consequently, an electron with a charge of 1- is more strongly attracted to a nucleus with a charge of 2+ than it would be to a nucleus with a charge of 1+. According to Coulomb's law, what happens to the potential energy of two oppositely charged particles as they get closer together? (a) Their potential energy decreases. (b) Their potential energy increases. (c) Their potential energy does not change.

Orbital blocks in the Periodic Table

A pattern similar to what we just saw for the first 18 elements exists for the entire periodic table, as shown in FigurE 8.7▼. Note that, because of the filling order of orbitals, the periodic table can be divided into blocks representing the filling of particular sublevels. The first two columns on the left side of the periodic table comprise the s block, with outer electron configurations of ns1 (the alkali metals) and ns2 (the alkaline earth metals). The six columns on the right side of the periodic table comprise the p block, with outer electron configurations of ns2 np1 , ns2 np2 , ns2 np3 , ns2 np4 , ns2 np5 (halogens), and ns2 np6 (noble gases). The transition elements comprise the d block, and the lanthanides and actinides (also called the inner transition elements) comprise the f block. (For compactness, the f block is normally printed below the d block instead of being embedded within it.) You can see that the number of columns in a block corresponds to the maximum number of electrons that can occupy the particular sublevel of that block. The s block has 2 columns (corresponding to one s orbital holding a maximum of two electrons); the p block has 6 columns (corresponding to three p orbitals with two electrons each); the d block has 10 columns (corresponding to five d orbitals with two electrons each); and the f block has 14 columns (corresponding to seven f orbitals with two electrons each). Notice also that, except for helium, the number of valence electrons for any maingroup element is equal to its lettered group number. For example, we can tell that chlorine has 7 valence electrons because it is in group number 7A. Lastly, note that, for main-group elements, the row number in the periodic table is equal to the number (or n value) of the highest principal level. For example, chlorine is in row 3, and its highest principal level is the n = 3 level Summarizing Periodic Table Organization: ▶ The periodic table is divisible into four blocks corresponding to the filling of the four quantum sublevels (s, p, d, and f ). ▶ The group number of a main-group element is equal to the number of valence electrons for that element. ▶ The row number of a main-group element is equal to the highest principal quantum number of that element.

Metallic Character

As we discussed in Chapter 2, metals are good conductors of heat and electricity; they can be pounded into flat sheets (malleability); they can be drawn into wires (ductility); they are often shiny; and they tend to lose electrons in chemical reactions. Nonmetals, by contrast, have more varied physical properties; some are solids at room temperature, others are gases, but in general nonmetals tend to be poor conductors of heat and electricity, and they all tend to gain electrons in chemical reactions. As we move to the right across a row in the periodic table, ionization energy increases and electron affinity becomes more negative, which means that elements on the left side of the periodic table are more likely to lose electrons than elements on the right side of the periodic table, which are more likely to gain them. The other properties associated with metals follow the same general trend (even though we do not quantify them here). Consequently, as shown in FigurE 8.17▶ As we move to the right across a period (or row) in the periodic table, metallic character decreases. As we move down a column in the periodic table, ionization energy decreases, making electrons more likely to be lost in chemical reactions. Consequently, As we move down a column (or family) in the periodic table, metallic character increases. These trends, based on the quantum-mechanical model, explain the distribution of metals and nonmetals that we discussed in Chapter 2. Metals are found on the left side and toward the center of the periodic table and nonmetals on the upper right side. The change in chemical behavior from metallic to nonmetallic can be seen most clearly as we proceed to the right across period 3, or down along group 5A, of the periodic table, as can be seen in FigurE 8.18▶. ExamPLE 8.9 Metallic Character On the basis of periodic trends, choose the more metallic element from each pair (if possible) (a) Sn or Te (b) P or Sb (c) Ge or In (d) S or Br SOLUTION (a) Sn or Te Sn is more metallic than Te because, as you trace the path between Sn and Te on the periodic table, you move to the right within the same period. Metallic character decreases as you go to the right. (b) P or Sb Sb is more metallic than P because, as you trace the path between P and Sb on the periodic table, you move down a column. Metallic character increases as you go down a column. (c) Ge or In In is more metallic than Ge because, as you trace the path between Ge and In on the periodic table, you move down a column (metallic character increases) and then to the left across a row (metallic character increases). These effects add together for an overall increase. (d) S or Br Based on periodic trends alone, you cannot tell which is more metallic because as you trace the path between S and Br, you go to the right across a row (metallic character decreases) and then down a column (metallic character increases). These effects tend to oppose each other, and it is not easy to tell which will predominate.

Ions: Electron Configurations, Magnetic Properties, Ionic Radii, and Ionization Energy

As we have seen, ions are simply atoms (or groups of atoms) that have lost or gained electrons. In this section, we examine periodic trends in ionic electron configurations, magnetic properties, ionic radii, and ionization energies. Electron Configurations and Magnetic Properties of Ions We can deduce the electron configuration of a main-group monoatomic ion from the electron configuration of the neutral atom and the charge of the ion. For anions, we add the number of electrons indicated by the magnitude of the charge of the anion. For example, the electron configuration of fluorine (F) is 1s 2 2s 2 2p5 and that of the fluoride ion (F -) is 1s 2 2s 2 2p6 . We determine the electron configuration of cations by subtracting the number of electrons indicated by the magnitude of the charge. For example, the electron configuration of lithium (Li) is 1s 2 2s 1 and that of the lithium ion (Li+) is 1s 2 2s 0 (or simply 1s 2 ). For main-group cations, we remove the required number of electrons in the reverse order of filling. However, an important exception occurs for transition metal cations. When writing the electron configuration of a transition metal cation, we remove the electrons in the highest n-value orbitals first, even if this does not correspond to the order of filling. For example, the electron configuration of vanadium is: V 3Ar4 4s 2 3d3 The V2+ ion, however, has the following electron configuration: V2+ 3Ar4 4s 0 3d3 In other words, for transition metal cations, the order in which electrons are removed upon ionization is not the reverse of the filling order. During filling, we normally fill the 4s orbital before the 3d orbital. When a fourth-period transition metal ionizes, however, it normally loses its 4s electrons before its 3d electrons. Why this odd behavior? The full answer to this question is beyond the scope of this text, but the following two factors contribute to this behavior. - As discussed previously, the ns and (n-1)d orbitals are extremely close in energy and, depending on the exact configuration, can vary in relative energy ordering. - As the (n-1)d orbitals begin to fill in the first transition series, the increasing nuclear charge stabilizes the (n-1)d orbitals relative to the ns orbitals. This happens because the (n-1)d orbitals are not outermost (or highest n) orbitals and are therefore not effectively shielded from the increasing nuclear charge by the ns orbitals. The experimental observation is that an ns0 (n-1)dx configuration is lower in energy than an ns2 (n-1)dx-2 configuration for transition metal ions. Therefore, when writing electron configurations for transition metals, we remove the ns electrons before the (n-1)d electrons. The magnetic properties of transition metal ions support these assignments. An unpaired electron generates a magnetic field due to its spin. Consequently, if an atom or ion contains unpaired electrons, it is attracted by an external magnetic field, and we say that the atom or ion is paramagnetic. An atom or ion in which all electrons are paired is not attracted to an external magnetic field—it is in fact slightly repelled—and we say that the atom or ion is diamagnetic. The zinc atom, for example, is diamagnetic. The magnetic properties of the zinc ion provide confirmation that the 4s electrons are indeed lost before 3d electrons in the ionization of zinc. If zinc lost two 3d electrons upon ionization, then the Zn2+ ion would become paramagnetic (because the two electrons would come out of two different filled d orbitals, leaving each of them with one unpaired electron). However, the zinc ion, like the zinc atom, is diamagnetic because the 4s electrons are lost instead. Similar observations in other transition metals confirm that the ns electrons are lost before the (n-1)d electrons upon ionization. ExamPLE 8.6 Electron Configurations and Magnetic Properties for Ions Write the electron configuration and orbital diagram for each ion and determine whether the ion is diamagnetic or paramagnetic. (a) Al3+ (b) S2- (c) Fe3+ SOLUTION (a) Al3+ Begin by writing the electron configuration of the neutral atom. Since this ion has a 3+ charge, remove three electrons to write the electron configuration of the ion. Write the orbital diagram by drawing half-arrows to represent each electron in boxes representing the orbitals. Because there are no unpaired electrons, Al3+ is diamagnetic. (b) S2- Begin by writing the electron configuration of the neutral atom. Since this ion has a 2- charge, add two electrons to write the electron configuration of the ion. Write the orbital diagram by drawing half-arrows to represent each electron in boxes representing the orbitals. Because there are no unpaired electrons, S2- is diamagnetic. (c) Fe3+ Begin by writing the electron configuration of the neutral atom. Since this ion has a 3+ charge, remove three electrons to write the electron configuration of the ion. Since it is a transition metal, remove the electrons from the 4s orbital before removing electrons from the 3d orbitals. Write the orbital diagram by drawing half-arrows to represent each electron in boxes representing the orbitals. Because there are unpaired electrons, Fe3+ is paramagnetic.

Quantum-mechanical theory describes the behavior of electrons in atoms. Since chemical bonding involves the transfer or sharing of electrons, quantum-mechanical theory helps us understand and describe chemical behavior. As we saw in Chapter 7, electrons in atoms exist within orbitals. An electron configuration for an atom shows the particular orbitals that electrons occupy for that atom. For example, consider the ground state—or lowest energy state—electron configuration for a hydrogen atom: The electron configuration indicates that hydrogen's one electron is in the 1s orbital. Electrons generally occupy the lowest energy orbitals available. Since the 1s orbital is the lowest energy orbital in hydrogen (see Section 7.5), the electron occupies that orbital. The solutions to the Schrödinger equation (the atomic orbitals and their energies) described in Chapter 7 are for the hydrogen atom. What do the atomic orbitals of other atoms look like? What are their relative energies? Unfortunately, the Schrödinger equation for multielectron atoms is so complicated— because of new terms introduced into the equation by the interactions of the electrons with one another—that it cannot be solved exactly. However, approximate solutions indicate that the orbitals in multielectron atoms are hydrogen-like—they are similar to the s, p, d, and f orbitals that we discussed in Chapter 7. In order to see how the electrons in multielectron atoms occupy these hydrogen-like orbitals, we need to examine two additional concepts: the effects of electron spin, a fundamental property of all electrons (first discussed in Section 7.5) that affects the number of electrons allowed in one orbital; and sublevel energy splitting, which applies to multielectron atoms and determines the order of orbital filling within a level.

Electron Configurations: How Electrons Occupy Orbitals

Electron Affinities and Metallic Character

Electron affinity and metallic character also exhibit periodic trends. Electron affinity is a measure of how easily an atom accepts an additional electron. Since chemical bonding involves the transfer or sharing of electrons, electron affinity is crucial to chemical bonding. Metallic character is important because of the high proportion of metals in the periodic table and the crucial role metals play in our lives. Of the roughly 110 elements, 87 are metals. In this section of the chapter, we examine each of these periodic properties—electron affinity and metallic character—individually. Electron Affinity The electron affinity (EA) of an atom or ion is the energy change associated with the gaining of an electron by the atom in the gaseous state. Electron affinity is usually—though not always—negative because an atom or ion usually releases energy when it gains an electron. (The process is analogous to an exothermic reaction, which releases heat and therefore has a negative ∆H.) In other words, the coulombic attraction between the nucleus of an atom and the incoming electron usually results in the release of energy as the electron is gained. For example, the electron affinity of chlorine can be represented with the equation: Cl(g) + 1 e- S Cl-(g) EA = -349 kJ>mol as regular as trends in other properties we have examined. For example, we might expect electron affinities to become relatively more positive (so that the addition of an electron is less exothermic) as we move down a column because the electron is entering orbitals that are successively farther from the nucleus. This trend applies to the group 1A metals but does not hold for the other columns in the periodic table. There is a more regular trend in electron affinity, however, as we move to the right across a row. Based on the periodic properties we have examined so far, should we expect more energy to be released when an electron is gained by Na or Cl? We know that Na has an outer electron configuration of 3s 1 and Cl has an outer electron configuration of 3s 2 3p5 . Since adding an electron to chlorine gives it a noble gas configuration and adding an electron to sodium does not, and since the outermost electrons in chlorine experience a higher Zeff than the outermost electrons in sodium, we would expect chlorine to have a more negative electron affinity—the process should be more exothermic for chlorine. This is in fact the case. For main-group elements, electron affinity generally becomes more negative (more exothermic) as we move to the right across a row in the periodic table. The halogens (group 7A) therefore have the most negative electron affinities. However, exceptions do occur. For example, notice that nitrogen and the other group 5A elements do not follow the general trend. These elements have ns2 np3 outer electron configurations. When an electron is added to this configuration, it must pair with another electron in an already occupied p orbital. The repulsion between two electrons occupying the same orbital causes the electron affinity to be more positive for these elements than for elements in the previous column. Summarizing Electron Affinity for Main-Group Elements: ▶ Most groups of the periodic table do not exhibit any definite trend in electron affinity. Among the group 1A metals, however, electron affinity becomes more positive as we move down the column (adding an electron becomes less exothermic). ▶ Electron affinity generally becomes more negative (adding an electron becomes more exothermic) as we move to the right across a period (or row) in the periodic table.

Electronegativity Trends

Electronegativity can be understood as a chemical property describing an atom's ability to attract and bind with electrons. Because electronegativity is a qualitative property, there is no standardized method for calculating electronegativity. However, the most common scale for quantifying electronegativity is the Pauling scale (Table A2), named after the chemist Linus Pauling. The numbers assigned by the Pauling scale are dimensionless due to the qualitative nature of electronegativity. Electronegativity values for each element can be found on certain periodic tables. An example is provided below. Electronegativity measures an atom's tendency to attract and form bonds with electrons. This property exists due to the electronic configuration of atoms. Most atoms follow the octet rule (having the valence, or outer, shell comprise of 8 electrons). Because 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 with the energy required to lose electrons. As a result, the elements on the left side of the periodic table generally lose electrons when 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. The nature of electronegativity is effectively described thus: the more inclined an atom is to gain electrons, the more likely that atom will pull electrons toward itself. From left to right across a period of elements, electronegativity increases. If the valence shell of an atom is less than half full, it requires less energy to lose an electron than to gain one. Conversely, if the valence shell is more than half full, it is easier to pull an electron into the valence shell than to donate one. From top to bottom down a group, electronegativity decreases. This is because 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 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, although they have electronegativity values, there is little variance among them across the period and up and down a group. This is because their metallic properties affect their ability to attract electrons as easily as the other elements.

Shielding

For multielectron atoms, any one electron experiences both the positive charge of the nucleus (which is attractive) and the negative charges of the other electrons (which are repulsive). Within an atom or ion, we can think of the repulsion of one electron by other electrons as screening or shielding that electron from the full effects of the nuclear charge. Consider a lithium ion (Li+). Since the lithium ion contains two electrons, itselectron configuration is identical to that of helium: Li+ 1s2 Now imagine bringing a third electron toward the lithium ion. When the third electron is far from the nucleus, it experiences the 3+ charge of the nucleus through the screen FigurE 8.2 Shielding and Penetration (a) An electron far from the nucleus is partly shielded by the electrons in the 1s orbital, reducing the effective net nuclear charge that it experiences. (b) An electron that penetrates the electron cloud of the 1s orbital experiences more of the nuclear charge. or shield of the 2- charge of the two 1s electrons, as illustrated in FigurE 8.2(a)▲. We can think of the third electron as experiencing an effective nuclear charge (Zeff) of approximately 1+ (3+ from the nucleus and 2- from the electrons, for a net charge of 1+). The inner electrons in effect shield the outer electron from the full nuclear charge. Penetration Now imagine allowing this third electron to come closer to the nucleus. As the electron penetrates the electron cloud of the 1s electrons, it begins to experience the 3+ charge of the nucleus more fully because it is less shielded by the intervening electrons. If the electron could somehow get closer to the nucleus than the 1s electrons, it would experience the full 3+ charge, as shown in FigurE 8.2(b)▲. In other words, as the outer electron penetrates into the region occupied by the inner electrons, it experiences a greater nuclear charge and therefore (according to Coulomb's law) a lower energy.

Exceptions to Trends in First Ionization Energy

If we carefully examine Figure 8.15, we can see some exceptions to the trends in first ionization energies. For example, boron has a smaller first ionization energy than beryllium, even though it lies to the right of beryllium in the same row. This exception is caused by the move from the s block to the p block. Recall from Section 8.3 that the 2p orbital penetrates into the nuclear region less than the 2s orbital. Consequently, the 1s electrons shield the electron in the 2p orbital from nuclear charge more than they shield the electrons in the 2s orbital. The result, as we saw in Section 8.3, is that the 2p orbitals are higher in energy, and therefore the electron is easier to remove (it has a lower ionization energy). Similar exceptions occur for aluminum and gallium, both directly below boron in group 3A. Another exception occurs between nitrogen and oxygen. Although oxygen is to the right of nitrogen in the same row, it has a lower ionization energy. This exception is caused by the repulsion between electrons when they must occupy the same orbital. Examine the electron configurations and orbital diagrams of nitrogen and oxygen. Nitrogen has three electrons in three p orbitals, while oxygen has four. In nitrogen, the 2p orbitals are half-filled (which makes the configuration particularly stable). Oxygen's fourth electron must pair with another electron, making it easier to remove. Similar exceptions occur for S and Se, directly below oxygen in group 6A.

Atomic Radii and the Transition Elements

In Figure 8.10, we can see that as we move down the first two rows of a column within the transition metals, the elements follow the same general trend in atomic radii as the main-group elements (the radii get larger). However, with the exception of the first couple of elements in each transition series, the atomic radii of the transition elements do not follow the same trend as the main-group elements as we move to the right across a row. Instead of decreasing in size, the radii of transition elements stay roughly constant across each row. Why? The difference is that, across a row of transition elements, the number of electrons in the outermost principal energy level (highest n value) is nearly constant (recall from Section 8.3, for example, that the 4s orbital fills before the 3d). As another proton is added to the nucleus with each successive element, another electron is added as well, but the electron goes into an nhighest-1 d orbital. The number of outermost electrons stays constant, and they experience a roughly constant effective nuclear charge, keeping the radius approximately constant. ExamPLE 8.5 Atomic Size On the basis of periodic trends, choose the larger atom from each pair (if possible). Explain your choices. (a) N or F (b) C or Ge (c) N or Al (d) Al or Ge SOLUTION (a) N atoms are larger than F atoms because, as you trace the path between N and F on the periodic table, you move to the right within the same period (row). As you move to the right across a row, the effective nuclear charge experienced by the outermost electrons increases, resulting in a smaller radius. (b) Ge atoms are larger than C atoms because, as you trace the path between C and Ge on the periodic table, you move down a column. Atomic size increases as you move down a column because the outermost electrons occupy orbitals with a higher principal quantum number that are therefore larger, resulting in a larger atom. (c) Al atoms are larger than N atoms because, as you trace the path between N and Al on the periodic table, you move down a column (atomic size increases) and then to the left across a row (atomic size increases). These effects add together for an overall increase (d) Based on periodic trends alone, you cannot tell which atom is larger, because as you trace the path between Al and Ge you go to the right across a row (atomic size decreases) and then down a column (atomic size increases). These effects tend to oppose each other, and it is not straightforward to tell which one dominates.

Ionization Energy Trends

Ionization energy is the energy required to remove an electron from a neutral atom in its gaseous phase. Conceptually, ionization energy is the opposite of electronegativity. The lower this energy is, the more readily the atom becomes a cation. Therefore, the higher this energy is, the more unlikely it is 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.

Electron Configurations, Valence Electrons, and the Periodic Table

Recall that Mendeleev arranged the periodic table so that elements with similar chemical properties lie in the same column. We can begin to make the connection between an element's properties and its electron configuration by superimposing the electron configurations of the first 18 elements onto a partial periodic table, as shown in FigurE 8.6▼. As we move to the right across a row, the orbitals fill in the correct order. With each subsequent row, the highest principal quantum number increases by one. Notice that as we move down a column, the number of electrons in the outermost principal energy level (highest n value) remains the same. The key connection between the macroscopic world (an element's chemical properties) and the microscopic world (an atom's electronic structure) lies in these outermost electrons. An atom's valence electrons are those that are important in chemical bonding. For main-group elements, the valence electrons are those in the outermost principal energy level (Figure 8.6). For transition elements, we also count the outermost d electrons among the valence electrons (even though they are not in an outermost principal energy level). The chemical properties of an element depend on its valence electrons, which are important in bonding because they are held most loosely (and are therefore the easiest to lose or share). We can now see why the elements in a column of the periodic table have similar chemical properties: They have the same number of valence electrons. Valence electrons are distinguished from all of the other electrons in an atom, which are called core electrons. Core electrons are those in complete principal energy levels and those in complete d and f sublevels. For example, silicon, with the electron configuration 1s 2 2s 2 2p6 3s 2 3p2 , has 4 valence electrons (those in the n = 3 principal level) and 10 core electrons. ExamPLE 8.3 Valence Electrons and Core Electrons Write the electron configuration for Ge and identify the valence electrons and the core electrons SOLUTION Write the electron configuration for Ge by determining the total number of electrons from germanium's atomic number (32) and distributing them into the appropriate orbitals as described in Example 8.1 Since germanium is a main-group element, its valence electrons are those in the outermost principal energy level. For germanium, the n = 1, 2, and 3 principal levels are complete (or full) and the n = 4 principal level is outermost. Consequently, the n = 4 electrons are valence electrons, and the rest are core electrons. Note: In this book, electron configurations are always written with the orbitals in the order of filling. However, writing electron configurations in order of increasing principal quantum number is also common. The electron configuration of germanium written in order of increasing principal quantum number is: Ge1s22s22p63s23p63d104s24p2

The Transition and Inner Transition Elements

The electron configurations of the transition elements (d block) and inner transition elements ( f block) exhibit trends that differ somewhat from those of the main-group elements. As we move to the right across a row in the d block, the d orbitals fill as shown here: However, the principal quantum number of the d orbitals that fill across each row in the transition series is equal to the row number minus one. In the fourth row, the 3d orbitals fill, and in the fifth row, the 4d orbitals fill, and so on. This happens because, as we discussed in Section 8.3, the 4s orbital is generally lower in energy than the 3d orbital (because it more efficiently penetrates into the region occupied by the core electrons). The result is that the 4s orbital fills before the 3d orbital, even though its principal quantum number (n = 4) is higher. Keep in mind, however, that the 4s and the 3d orbitals are extremely close to each other in energy, so that their relative energy ordering depends on the exact species under consideration, which causes some irregular behavior in the transition metals. For example, in the first transition series of the d block, the outer configuration is 4s 2 3dx with two exceptions: Cr is 4s 1 3d5 and Cu is 4s 1 3d10. This behavior is related to the closely spaced 3d and 4s energy levels and the additional stability associated with a half-filled or completely filled sublevel. Actual electron configurations are always determined experimentally (through spectroscopy) and do not always conform to the general patterns, as is the case for Cr and Cu. Nonetheless, the patterns we have described allow us to accurately predict electron configurations for most of the elements in the periodic table. As we move across the f block (the inner transition series), the f orbitals fill. However, the principal quantum number of the f orbitals that fill across each row is the row number minus two. (In the sixth row, the 4f orbitals fill, and in the seventh row, the 5f orbitals fill.) In addition, within the inner transition series, the close energy spacing of the 5d and 4f orbitals sometimes causes an electron to enter a 5d orbital instead of the expected 4f orbital. For example, the electron configuration of gadolinium is 3Xe4 6s24f 75d1 (instead of the expected 3Xe4 6s24f 8).

Trends in First Ionization Energy

The first ionization energies of the elements through Xe are shown in FigurE 8.14▶. Notice the periodic trend in first ionization energy, peaking at each noble gas. Based on what we have learned about electron configurations and effective nuclear charge, how can we account for the observed trend? As we have seen, the principal quantum number, n, increases as we move down a column. Within a given sublevel, orbitals with higher principal quantum numbers are larger than orbitals with smaller principal quantum numbers. Consequently, electrons in the outermost principal level are farther away from the positively charged nucleus—and are therefore held less tightly—as we move down a column. This results in a lower first ionization energy as we move down a column, as we can see in FigurE 8.15▼. What about the trend as we move to the right across a row? For example, would it take more energy to remove an electron from Na or from Cl, two elements on either end of the third row in the periodic table? We know that Na has an outer electron configuration of 3s 1 and Cl has an outer electron configuration of 3s 2 3p5 . As we discussed previously, the outermost electrons in chlorine experience a higher effective nuclear charge than the outermost electrons in sodium (which is why chlorine has a smaller atomic radius than sodium). Consequently, we would expect chlorine to have a higher ionization energy than sodium, which is in fact the case. A similar argument can be made for other main-group elements so that first ionization energy generally increases as we move to the right across a row in the periodic table, as illustrated in Figure 8.15. Summarizing Trends in Ionization Energy for Main-Group Elements: ▶ First ionization energy generally decreases as we move down a column (or group) in the periodic table because electrons in the outermost principal level are farther away from the positively charged nucleus and are therefore held less tightly. ▶ First ionization energy generally increases as we move to the right across a period (or row) in the periodic table because electrons in the outermost principal energy level generally experience a greater effective nuclear charge (Zeff). ExamPLE 8.8 First Ionization Energy On the basis of periodic trends, determine the element with the higher first ionization energy in each pair (if possible). (a) Al or S (b) As or Sb (c) N or Si (d) O or Cl SOLUTION (a) Al or S S has a higher first ionization energy than Al because, as you trace the path between Al and S on the periodic table, you move to the right within the same row. Ionization energy increases as you go to the right because of increasing effective nuclear charge. (b) As or Sb As has a higher first ionization energy than Sb because, as you trace the path between As and Sb on the periodic table, you move down a column. Ionization energy decreases as you go down a column because of the increasing size of orbitals with increasing n. (c) N or Si N has a higher first ionization energy than Si because, as you trace the path between N and Si on the periodic table, you move down a column (ionization energy decreases) and then to the left across a row (ionization energy decreases). These effects sum together for an overall decrease. (d) O or Cl Based on periodic trends alone, it is impossible to tell which has a higher first ionization energy because, as you trace the path between O and Cl, you go to the right across a row (ionization energy increases) and then down a column (ionization energy decreases). These effects tend to oppose each other, and it is not obvious which will dominate.

Ionization Energy

The ionization energy (IE) of an atom or ion is the energy required to remove an electron from the atom or ion in the gaseous state (see Figure 7.19). Ionization energy is always positive because removing an electron always takes energy. (The process is similar to an endothermic reaction, which absorbs heat and therefore has a positive ∆H.) The energy required to remove the first electron is the first ionization energy (IE1). We represent the first ionization energy of sodium with the following equation: Na(g) S Na+(g) + 1 e- IE1 = 496 kJ>mol The energy required to remove the second electron is the second ionization energy (IE2), the energy required to remove the third electron is the third ionization energy (IE3), and so on. We represent the second ionization energy of sodium as follows: Notice that the second ionization energy is not the energy required to remove two electrons from sodium (that quantity would be the sum of IE1 and IE2), but rather the energy required to remove the second of two electrons from Na+. In this section of the chapter, we consider trends in IE1 and IE2 separately.

writing an Electron Configuration for an Element from Its Position in the Periodic Table

The organization of the periodic table allows us to write the electron configuration for any element based on its position in the periodic table. For example, suppose we want to write an electron configuration for Cl. The inner electron configuration of Cl is that of the noble gas that precedes it in the periodic table, Ne. So we can represent the inner electron configuration with [Ne]. The outer electron configuration—the configuration of the electrons beyond the previous noble gas—is obtained by tracing the elements between Ne and Cl and assigning electrons to the appropriate orbitals, as shown here. Remember that the highest n value is given by the row number (3 for chlorine). So, we begin with [Ne], then add in the two 3s electrons as we trace across the s block, followed by five 3p electrons as we trace across the p block to Cl, which is in the fifth column of the p block. The electron configuration is Cl 3Ne4 3s 2 3p5 Notice that Cl is in column 7A and therefore has 7 valence electrons and an outer electron configuration of ns2 np5. ExamPLE 8.4 writing Electron Configurations from the Periodic Table Use the periodic table to write the electron configuration for selenium (Se) SOLUTION The atomic number of Se is 34. The noble gas that precedes Se in the periodic table is argon, so the inner electron configuration is 3Ar 4 . Obtain the outer electron configuration by tracing the elements between Ar and Se and assigning electrons to the appropriate orbitals. Begin with 3Ar 4 . Because Se is in row 4, add two 4s electrons as you trace across the s block (n = row number). Next, add ten 3d electrons as you trace across the d block (n = row number -1). Lastly, add four 4p electrons as you trace across the p block to Se, which is in the fourth column of the p block (n = row number).

The Explanatory Power of the Quantum-Mechanical Model

We can now see how the quantum-mechanical model accounts for the chemical properties of the elements, such as the inertness of helium or the reactivity of hydrogen, and (more generally) how it accounts for the periodic law. The chemical properties of elements are largely determined by the number of valence electrons they contain. Their properties are periodic because the number of valence electrons is periodic. Since elements within a column in the periodic table have the same number of valence electrons, they also have similar chemical properties. The noble gases, for example, all have eight valence electrons, except for helium, which has two. Although we do not cover the quantitative (or numerical) aspects of the quantum-mechanical model in this book, calculations of the overall energy of atoms with eight valence electrons (or two for helium) show that they are particularly stable. In other words, when a quantum level is completely full, the overall energy of the electrons that occupy that level is particularly low. Those electrons cannot lower their energy by reacting with other atoms or molecules, so the corresponding atom is unreactive or inert. Consequently, the noble gases are the most chemically stable or relatively unreactive family in the periodic table. Elements with electron configurations close to those of the noble gases are the most reactive because they can attain noble gas electron configurations by losing or gaining a small number of electrons. For example, alkali metals (group 1A) are the most reactive metals because their outer electron configuration (ns1 ) is one electron beyond a noble gas configuration. They react to lose the ns1 electron, obtaining a noble gas configuration. This explains why—as we saw in Chapter 2—the group 1A metals tend to form 1+ cations. Similarly, alkaline earth metals, with an outer electron configuration of ns2 , also tend to be reactive metals, losing their ns2 electrons to form 2+ cations. This does not mean that forming an ion with a noble gas configuration is in itself energetically favorable. In fact, forming cations always requires energy. But when the cation formed has a noble gas configuration, the energy cost of forming the cation is often less than the energy payback that occurs when that cation forms ionic bonds with anions, as we shall see in Chapter 9. On the right side of the periodic table, halogens are among the most reactive nonmetals because of their ns2 np5 electron configurations. They are only one electron short of a noble gas configuration and tend to react to gain that one electron, forming 1- ions. FigurE 8.8▲, first introduced in Chapter 2, shows the elements that form predictable ions. Notice how the charges of these ions reflect their electron configurations. In their reactions, these elements form ions with noble gas electron configurations.

Electron Spin and the Pauli Exclusion Principle

We can represent the electron configuration of hydrogen (1s1) in a slightly different way with an orbital diagram, which gives similar information, but symbolizes each electron as an arrow in a box that represents the orbital. The orbital diagram for a hydrogen atom is In an orbital diagram, the direction of the arrow (pointing up or pointing down) represents the orientation of the electron's spin. Recall from Section 7.5 that the orientation of the electron's spin is quantized, with only two possibilities—spin up (ms = +/2) and spin down (ms = -1/2). In an orbital diagram, ms = +1/2 is represented with a half-arrow pointing up (↑), and ms = -1/2 is represented with a half-arrow pointing down (↓). In a collection of hydrogen atoms, the electrons in about half of the atoms are spin up, and the other half is spin down. Since no other electrons are present within the atom, we conventionally represent the hydrogen atom electron configuration with its one electron as spin up. Helium is the first element on the periodic table that contains two electrons. Its two electrons occupy the 1s orbital. He 1s² How do the spins of the two electrons in helium align relative to each other? The answer to this question is described by the Pauli exclusion principle, formulated by Wolfgang Pauli (1900-1958) in 1925: Pauli exclusion principle: No two electrons in an atom can have the same four quantum numbers. Since two electrons occupying the same orbital have three identical quantum numbers (n, l, and ml ), they must have different spin quantum numbers. As there are only two possible spin quantum numbers (+1/2 and -1/2), the Pauli exclusion principle implies that each orbital can have a maximum of only two electrons, with opposing spins. By applying the exclusion principle, we can write an electron configuration and orbital diagram for helium as follows:

Electron Spatial Distributions and Sublevel Splitting

We now have examined the concepts we need to understand the energy splitting of the sublevels within a principal level. The splitting is a result of the spatial distributions of electrons within a sublevel. Recall from Section 7.6 that the radial distribution function for an atomic orbital shows the total probability of finding the electron within a thin spherical shell at a distance r from the nucleus. FigurE 8.3▶ shows the radial distribution functions of the 2s and 2p orbitals superimposed on one another (the radial distribution function of the 1s orbital is also shown). Notice that, in general, an electron in a 2p orbital has a greater probability of being found closer to the nucleus than an electron in a 2s orbital. We might initially expect, therefore, that the 2p orbital would be lower in energy. However, exactly the opposite is the case—the 2s orbital is actually lower in energy, but only when the 1s orbital is occupied. (When the 1s orbital is empty, the 2s and 2p orbitals are degenerate.) Why? The reason is the bump near r = 0 (near the nucleus) for the 2s orbital. This bump represents a significant probability of the electron being found very close to the nucleus. Even more importantly, this area of the probability penetrates into the 1s orbital—it gets into the region where shielding by the 1s electrons is less effective. In contrast, most of the probability in the radial distribution function of the 2p orbital lies outside the radial distribution function of the 1s orbital. Consequently, almost all of the 2p orbital is shielded from nuclear charge by the 1s orbital. The end result is that the 2s orbital—since it experiences more of the nuclear charge due to its greater penetration—is lower in energy than the 2p orbital. The results are similar when we compare the 3s, 3p, and 3d orbitals. The s orbitals penetrate more fully than the p orbitals, which in turn penetrate more fully than the d orbitals, as shown in FigurE 8.4▲. FigurE 8.5▼ shows the energy ordering of a number of orbitals in multielectron atoms. Notice these points about the diagram in Figure 8.5: - Because of penetration, the sublevels of each principal level are not degenerate for multielectron atoms. - In the fourth and fifth principal levels, the effects of penetration become so significant that the 4s orbital lies lower in energy than the 3d orbitals and the 5s orbital lies lower in energy than the 4d orbitals. - The energy separations between one set of orbitals and the next become smaller for 4s orbitals and beyond, so that the relative energy ordering of these orbitals can actually vary among elements. These variations result in anomalies in the electron configurations of the transition metals and their ions (as we shall see later in this chapter). PENETRATION AND SHIELDING Which statement is true? (a) An orbital that penetrates into the region occupied by core electrons is more shielded from nuclear charge than an orbital that does not penetrate and will therefore have higher energy. (b) An orbital that penetrates into the region occupied by core electrons is less shielded from nuclear charge than an orbital that does not penetrate and will therefore have higher energy. (c) An orbital that penetrates into the region occupied by core electrons is less shielded from nuclear charge than an orbital that does not penetrate and will therefore have lower energy. (d) An orbital that penetrates into the region occupied by core electrons is more shielded from nuclear charge than an orbital that does not penetrate and will therefore have lower energy

Ionic Radii

What happens to the radius of an atom when it becomes a cation? An anion? Consider, for example, the difference between the Na atom and the Na+ ion. Their electron configurations are: Na 3Ne4 3s 1 Na+ 3Ne4 The sodium atom has an outer 3s electron and a neon core. Since the 3s electron is the outermost electron, and since it is shielded from the nuclear charge by the core electrons, it contributes greatly to the size of the sodium atom. The sodium cation, having lost the outermost 3s electron, has only the neon core and carries a charge of 1+. Without the 3s electron, the sodium cation (ionic radius = 95 pm) becomes much smaller than the sodium atom (covalent radius = 186 pm). The trend is the same with all cations and their atoms, as shown in FigurE 8.12▼. In general, Cations are much smaller than their corresponding atoms. What about anions? Consider, for example, the difference between Cl and Cl-. Their electron configurations are as follows: Cl 3Ne43s 2 3p5 Cl- 3Ne43s 2 3p6 The chlorine anion has one additional outermost electron, but no additional proton to increase the nuclear charge. The extra electron increases the repulsions among the outermost electrons. As a result, the chloride anion is larger than the chlorine atom. The trend is the same with all anions and their atoms, as shown in FigurE 8.13▼. In general, Anions are much larger than their corresponding atoms. We see a characteristic trend in ionic size in the radii of an isoelectronic series of ions— ions with the same number of electrons. For example, consider the following ions and their radii: S2- (184 pm) Cl- (181 pm) K+ (133 pm) Ca2+(99 pm) 18 electrons 18 electrons 18 electrons 18 electrons 16 protons 17 protons 19 protons 20 protons Each of these ions has 18 electrons in exactly the same orbitals, but the radii of the ions get successively smaller. Why? Because this set of nuclei has a progressively greater number of protons. The S2- ion has 16 protons, and therefore a charge of 16+ pulling on 18 electrons. The Ca2+ ion, however, has 20 protons, and therefore a charge of 20+ pulling on the same 18 electrons. The result is a much smaller radius for the calcium cation. For a given number of electrons, a greater nuclear charge results in a smaller atom or ion. EXAMPLE 8.7 Ion Size Choose the larger atom or ion from each pair. (a) S or S2- (b) Ca or Ca2+ (c) Br- or Kr SOLUTION (a) The S2- ion is larger than an S atom because anions are larger than the atoms from which they are formed. (b) A Ca atom is larger than Ca2+ because cations are smaller than the atoms from which they are formed. (c) A Br- ion is larger than a Kr atom because, although they are isoelectronic, Br- has one fewer proton than Kr, resulting in a lesser pull on the electrons and therefore a larger radius. FOR PRACTICE 8.7 Choose the larger atom or ion from each pair. (a) K or K+ (b) F or F- (c) Ca2+ or Cl FOR MORE PRACTICE 8.7 Which is larger, Ar or Cl- . IONS, ISOTOPES, AND ATOMIC SIZE In the previous sections, we have seen how the number of electrons and the number of protons affect the size of an atom or ion. However, we have not considered how the number of neutrons affects the size of an atom. Why not? Would you expect isotopes—for example, C-12 and C-13—to have different atomic radii?