# Chemical Principles/Electronic Structure and Atomic Properties

 “ I have been asked sometimes how one can be sure that elsewhere in the universe there may not be further elements) other than those in the periodic system. I have tried to answer by saying that it is like asking how one knows that elsewhere in the universe there may not be another whole number between 4 and 5. Unfortunately) some persons think that is a good question) too. ”

George Wald, 1964

## Introduction

We now know the wave functions and energy levels for a hydrogen atom. With this information and the aufbau (or buildup) process, we can go on to determine the electronic structures for atoms of all the elements. These structures lead directly to the periodic table of Figures 7-3 and 7-4. As we shall see, the structures explain the stability of eight-electron shells in noble gases and the trends in ionization energies and electron affinities of the elements.

## Buildup of Many-Electron Atoms

Although we cannot solve the Schrödinger equation exactly for many electron atoms, we can show that no radical new features are expected as the atomic number increases. There are the same quantum states, the same four quantum numbers (n, l, m, and s), and virtually the same electronic probability functions or electron-density clouds. The energies of the quantum levels are not identical for all elements; rather, they vary in a regular fashion from one element to the next.

In studying the electronic structure of a many-electron atom, we shall assume the existence of a nucleus and the required number of electrons. We shall assume that the possible electronic orbitals are hydrogenlike, if not identical to the hydrogen orbitals. Then we shall build the atom by adding electrons one at a time, placing each new electron in the lowest-energy orbital available. In this way we shall build a model of an atom in its ground state, or the state of the lowest electronic energy. Wolfgang Pauli (1900 - 1958) first suggested this treatment of many-electron atoms, and called it the aufbau, or buildup, process.

The aufbau process involves three principles:

1. No two electrons in the same atom can be in the same quantum state. This principle is known as the Pauli exclusion principle. It means that no two electrons can have the same n, l, m, and s values. Therefore, one atomic orbital, described by n, l, and m, can hold a maximum of two electrons: one of spin $+ \textstyle\frac{1}{2}$ and one of spin $- \textstyle\frac{1}{2}$. We can represent an atomic orbital by a circle and an electron by an arrow:
File:Chemical Principles Equation 9.1.png

When two electrons occupy one orbital with spins $+ \textstyle\frac{1}{2}$ and one of spin $- \textstyle\frac{1}{2}$, we say that their spins are paired. A paired spin is represented as follows:

File:Chemical Principles Equation 9.2.png
2. Orbitals are filled with electrons in order of increasing energies. The s orbital can hold a maximum of 2 electrons. The three p orbitals can hold a total of 6 electrons, the five d orbitals can hold 10, and the seven f orbitals can hold 14. We must decide on the order of increasing energies of the levels before we can begin the buildup process. For atoms with more than one electron, in the absence of an external electric or magnetic field, energy depends on n and l (the size and shape quantum numbers) and not on m, the orbital-orientation quantum number.
3. When electrons are added to orbitals of the same energy (such as the five 3d orbitals), one electron will enter each of the available orbitals before a second electron enters any one orbital. This follows Hund's rule, which states that in orbitals of identical energy electrons remain unpaired if possible. This behavior is understandable in terms of electron-electron repulsion. Two electrons, one in px orbital and one in a py orbital, remain farther apart than two electrons paired in the same px orbital (Figure 8-22). A consequence of Hund's rule is that a half-filled set of orbitals (each orbital containing a single electron) is a particularly stable arrangement. The sixth electron in a set of five d orbitals is forced to pair with another electron in a previously occupied orbital. The mutual repulsion of negatively charged electrons means that less energy is required to remove this sixth electron than to remove one of the five in a set of five half-filled d orbitals. Similarly, the fourth electron in a set of three p orbitals is held less tightly than the third.

### Relative Energies of Atomic Orbitals

File:Chemical Principles Fig 9.1.png
Figure 9-1 Radial distribution fnctions for electrons in the 3s, 3p, and 3d atomic orbitals of hydrogen. These curves are obtained by spinning the orbital in all directions around the nucleus to smear out all details that depend on direction away from the nucleus, and then by measuring the smeared electron probabil ity as a function of distance from the nucleus. The 35 orbital, which is already spherically symmetrical without the smearing operation. has a most probable radius at 13 atomic units and two minor peaks close to the nucleus. The 3p orbital has a maximum density near r = 12 atomic units, one spherical node at r = 6 atomic units and a density peak close to the nucleus. The 3d orbital has only one density peak, which occurs very close to the Bohr orbit radius of 9 atomic units. The shapes of the three orbitals before the spherical smearing process are to the right of each curve. An electron in the hydrogen atom with n = 2 will be in the neighborhood of r = 4 atomic units. The scale of distances changes in many electron atoms, but relative distances in different orbitals in the atom are the same as in H. An electron in a 3s orbital is more stable than one in a 3p or 3d orbital because it has a greater probability of being inside the orbital of n = 2 electrons, in which it experiences a greater attraction from the nucleus. The 3p orbital is similarly more stable than the 3d.

The 3s, 3p, and 3d orbitals in the hydrogen atom have the same energy but differ in the closeness of approach of the electron to the nucleus (Figure 9-1).

The energy of an electron in an orbital depends on the attraction exerted on it by the positively charged nucleus. Electrons with low principal quatum numbers will lie close to the nucleus and will screen some of this electrostatic attraction from electrons with higher principal quantum numbers. In the Li+ ion, the effective nuclear charge beyond 1 or 2 atomic units from the nucleus is not the true nuclear charge of +3, but a net charge of +1 produced by the nucleus plus the two 1s electrons. Similarly, the lone n = 3 electron in sodium experiences a net nuclear charge of approximately +1 rather than the full nuclear charge of +11.

If the net charge from the nucleus and the filled inner orbitals were concentrated at a point at the nucleus, then the energies of 3s, 3p, and 3d orbitals would be the same. But the screening electrons extend over an appreciable volume of space. The net attraction that an electron with a principal quantum number of 3 experiences depends on how close it comes to the nucleus, and whether it penetrates the lower screening electron clouds. As in Sommerfeld's elliptical-orbit model, the s orbital comes closer to the nucleus and is somewhat more stable than the p, and the p is more stable than the d. This is the reason for the variation of the l energy levels in the lithium energy-level diagram in Figure 8-13.

For a given value of the principal quantum number, n, the order of increasing energy is s, p, d, f, g, . . . . It is less easy to decide whether and when the high l-value orbitals of one n overtake the low-l orbitals of the next: for example, whether a 4f orbital has a higher energy than a 5s, or a 3d a higher energy than a 4s. The question was originally settled empirically by choosing the order of overlap that accounted for the observed structure of the periodic table. The energies have since been calculated theoretically, and (fortunately for quantum mechanics) they agree with the observed order of levels. The sequence of energy levels is shown in Figure 9-2.

### Orbital Configurations and First Ionization Energies

File:Chemical Principles Fig 9.2.png
Figure 9-2 Idealized diagram of the energy levels of the hydrogenlike atomic orbitals during the buildup of many-electron atoms. On each level are written the symbols of those elements that are completed with the addition of electrons on that level. Note the nearly equal energies of 4s and 3d, of 5s and 4d, of 6s, 4f, and 5d, and finally of 7s, 5f, and 6d. The near equivalence of energies is reflected in some irregularity in the order of filling levels in the transition metals and inner transition metals. Elements with such irregularities are circled. For example, after the 6s and 7s orbitals fill in lanthanum and actinium, the next electron goes into a d rather than an f orbital. See Figure 9-3 for details.
File:Chemical Principles Fig 9.3.png
Figure 9-3 "Superlong" form of the periodic table. Column head indicates last electron to be added in the Pauli buildup process. Those elements whose electronic structures in the ground state differ from this simple buildup model are in color. In Gd, Cm, Cr, Mo, Cu, Ag, and Au, this difference arises from the extra stability of half-filled (f7, d5) or completely filled (d10) shells. Other deviations arise from the extremely small energy differences between d and s, or d and f levels. These deviations are less important to us now than the overall patterns of buildup and the way they account for the structure of the periodic table. He is placed over Be in Group IIA since the second electron is added to complete the s orbital in each of these elements. In the usual periodic table (inside front cover), He is placed over Ne, Ar, and the other noble gases to indicate that the entire valence shell is filled in these elements.

We shall build up the electronic structures of the atoms in the periodic table by adding electrons to the hydrogenlike orbitals in order of increasing energy, and by increasing the nuclear charge by one at each step. During this process we shall pay particular attention to the relationship between the orbital electronic configurations of atoms and their first ionization energies. The first ionization energy (IE1) of an atom is the energy required to remove one electron:

atom(g) + energy(IE1) positive ion(g) + e-

Numerical IE1 values are given in Table 9-1.* Use the periodic table in Figure 9-3 as an aid as you follow this building process.

A hydrogen (H) atom has only one electron, which in the ground state must go in the 1s orbital. So we write 1s1 (the superscript represents the number of electrons in the orbital), and illustrate the electronic configuration as follows:

*It is common to refer to IE1 simply as IE: This is done in Table 9-1.

File:Chemical Principles Table 9.1.png

File:Chemical Principles Equation 9.3.png

In helium (He), the second electron also can be in the 1s orbital if its spin is paired with that of the first electron. In spite of electron - electron repulsion , this electron is more stable in the 1s orbital than in the higher-energy 2s orbital: File:Chemical Principles Equation 9.4.png

Because of the electron - electron repulsion, the first ionization energy of He is less than we might might have expected for an atom with a nuclear charge of +2. A simple calculation illustrates this point. If electron - electron repulsion were not important, each electron would feel the full force of the +2 nuclear charge, and the first ionization energy could be calculated from the one-electron formula:

IE1 = -E1 = $\textstyle\frac{Z^2k}{n^2} = \frac{(2)^2(1312 kJ mole^{-1})}{(1)^2}$
= 5248 kJ mole-1

However, the experimental value of IE1 for He is much less, 2372 kJ mole-1. Although the strong attraction of a 1s electron to the +2 He nucleus is partially counterbalanced by the electron - electron repulsion, the IE1 is still very large, showing how tightly each electron is bound in He.

Lithium (Li) begins the next period in the periodic table. Two electrons fill the 1s orbital; the third electron in Li must, by the Pauli exclusion principle, occupy the next lowest-energy orbital, namely, the 2s: File:Chemical Principles Equation 9.5.png

The fourth electron in beryllium (Be) fills the 2s orbital, and the fifth electron in boron (B) must occupy one of the higher-energy 2p orbitals: File:Chemical Principles Equation 9.6.png

For B, the first ionization energy is less than that of Be because its outermost electron is in a less-stable (higher-energy) orbital. In carbon (C), two of the three 2p orbitals contain an electron. As Hund's rule predicts, in nitrogen (N) the three p electrons are found in all three 2p orbitals, instead of two being paired in one: File:Chemical Principles Equation 9.7.png

The fourth 2p electron in an oxygen (O) atom is held less tightly than the first three because of the electron - electron repulsion with the other electron in one of the 2p orbitals. The first ionization energy of O is accordingly low.

The general trend across this period is for each new electron to be held more tightly because of the increased charge on the nucleus. Because the others 2s and 2p electrons are approximately the same distance from the nucleus, they do not shield the new electron from the steadily increasing charge. This increased charge overcomes the electron repulsion as the fifth 2p electron is added in fluorine (F). Therefore, the fifth electron is held very tightly in F, and the first ionization energy increases again. The most stable configuration results when the sixth 2p electron is added to complete the n = 2 shell with the noble gas neon (Ne): File:Chemical Principles Equation 9.8.png

The complete n = 1 shell of two electrons is often given the symbol K, and the complete n = 2 shell of eight electrons is given the symbol l. A briefer representation of the Ne atom then is

Ne: KL

For all but a few atoms, writing the complete orbital electronic structure is a tedious procedure. It is also unnecessary because only the outer electrons are important in chemical reactions. We call the chemically important or outer electrons the valence electrons. The valence electrons of an atom are the electrons in the s and p orbitals beyond the closed-shell configurations. For example, in Li the two 1s electrons in He, they are chemically unreactive. Thus we say that the valence electronic structure of Li is 2s. Similarly, the valence electronic structure of Be is 2s2; of B, 2s22p1; of C, 2s22p2; of N, 2s22p3; of O, 2s22p4; and of F, 2s22p5.

The buildup of the third period of the periodic table proceeds exactly as that of the preceding period did. Each new electron is bound more firmly because of the increasing nuclear charge, except for the fluctuations at aluminum (Al) and sulfur (S) produced by the filling of 3s in magnesium (Mg) and the half-filling of 3p in phosphorus (P): File:Chemical Principles Equation 9.9.png

The outermost electron for each element in this period is bound less firmly than the outermost electron in the corresponding element of the previous period because the n = 3 electrons are farther from the nucleus. Therefore, the first ionization energies for the n = 3 elements are smaller than for the corresponding n = 2 elements. With the completion of the 3s and 3p orbitals, we have again reached a particularly stable electronic configuration with the noble gas argon (Ar).

Something unusual happens in the fourth period. The 4s orbital penetrates closer to the nucleus than does the 3d orbital, and at this point in the buildup process the 4s has slightly lower energy than the 3d. Hence, the one and two electrons that are added to form potassium (K) and calcium (Ca) go into the 4s orbital before the 3d orbital is filled in the elements scandium (Sc) through zinc (Zn). If we assume a constant inner electronic configuration of KL 3s23p6, the valence electronic configurations for the 4s and 3d elements are

 K 3d04s1 Mn 3d54s2 Ca 3d04s2 Fe 3d64s2 Sc 3d14s2 Co 3d74s2 Ti 3d24s2 Ni 3d84s2 V 3d34s2 Cu 3d104s1 Cr 3d54s1 Zn 3d104s2

There are two anomalies in this order of filling. The half-filled (d5) and filled (d10) levels are particularly stable, therefore the chromium (Cr) and copper (Cu) atoms have only one 4s electron each.

 The valence electronic configuration for the ground state of chromium is 3d54s1. Predict the configuration of the first (i.e., lowest-energy) excited state of chromium. Solution The aufbau process predicts 3d44s2 for the ground state, but the extra stability of a half-filled level makes the 3d54s1 configuration slightly lower in energy than 3d44s2. The latter configuration thus becomes the first excited state. For those elements whose ground-state configurations differ from those predicted by the aufbau process, the predicted configuration is that of an excited state usually only slightly higher in energy than the ground state.

Although the 4s orbital penetrates closer to the nucleus than the 3d and therefore has a lower energy, the majority of the probability density of the 4s orbital is farther from the nucleus than in the 3d. An electron in a 4s orbital is simultaneously farther from the nucleus, on the average, than a 3d electron and more stable because of the small but not negligible probability that it will be very close to the nucleus. In chemical bonding, the energies of electrons in such a closely spaced levels in atoms are not as significant as distances of the electrons from the nucleus. Therefore, the 4s electrons have more of an effect on chemical properties than the relatively buried 3d electrons. With the exception of Cr and Cu, all the elements from Ca through Zn have the same outer electronic structure: two 4s electrons. The chemical properties of this series of elements will vary less rapidly than those in a series in which s or p electrons are being added. This is the reason for the relatively unchanging properties of the transition metals.

After the 3d orbitals are filled, the 4p orbitals fill, in a straightforward manner, to form the representative elements from gallium (Ga), 3d104s24p1, to the noble gas krypton (Kr), 3d104s24p6. The first ionization energy, which had risen with increasing nuclear charge in the transition metals, plummets at Ga when the next electron is placed in the less stable 4p orbital.

The fifth period repeats the same pattern: first the filling of the 5s orbitals, then an interruption while the buried 4d orbitals are filled in another series of transition metals, and finally the filling of the 5p orbitals, ending with the noble gas xenon (Xe), 4d105s25p6. The common feature of all noble gases is the outermost electronic arrangement s2p6. This is the origin of the stable eight-electron shells that we mentioned in Chapter 7. The late filling of the d orbitals (and f orbitals) produces the observed lengths of the periods of the periodic table: first 2, then 8, then only 8 instead of 18 for n = 3, then only 18 instead of 32 for n = 4.

According to the energy diagram in Figure 9-2, the 6s orbital is more stable than the 5d, although the difference is small and there are exceptions. The idealized filling pattern is for the 6s orbital to fill in cesium (Cs) and barium (Ba), followed by the deeply buried 4f orbitals in the 14 inner transition elements lanthanum (La) through ytterbium (Yb). There are minor deviations from this pattern, as shown in Figure 9-3. The most important of these deviations is that the first electron after Ba goes into the 5d orbital in La and not into the 4f. Lanthanum is more properly a transition metal than an inner transition metal. It is more relevant to understand the idealized filling pattern, however, than to worry about the individual exceptions to it.

The chemical properties of the inner transition metals cesium (ce) to lutetium (Lu) vary even less than the properties of the transition metals, because successive electrons are in the deeply buried 4f orbitals. After the 4f orbitals are filled, the balance of the third transition-metal series, hafnium (Hf) to mercury (Hg), occurs with the filling of the 5d orbitals. The representative elements thallium (Tl) through radon (Rn) are formed as the 6p orbitals fill.

The seventh and last period begins in the same way. First the 7s orbital fills, then the inner transition metals from actinium (Ac) to nobelium (No) - with the irregularities shown in Figure 9-3 - and finally the beginning of a fourth transition-metal series with lawrencium (Lr). There are more deviations from this simple f-first, d-next filling pattern in the actinides than in the lanthanides (Figure 9-3), and consequently the first few actinide elements show a greater diversity of chemical properties than do the lanthanides.

In summary, the idealized sequence of filling of orbitals across a period is as follows:

1. For period n, the ns orbital is filled first with two electrons. These elements are the alkali metals (Group IA) and the alkaline earths (Group IIA) and are classed with the representative elements.
2. The very deeply buried (n - 2) f orbitals are filled next. They exist only for (n - 2) greater than 3, or for Periods 6 and 7. These elements, which have virtually identical outer electronic structure and therefore virtually identical chemical properties, are the inner transition metals.
3. The less deeply buried (n -1)d orbitals are then filled if they exist. They exist only for (n -1) greater than 2, or for Period 4 and greater. These elements are similar to one another, but not as similar as the inner transition metals. They are called the transition metals (B groups).
4. Finally, the three np orbitals are filled to form the remaining representative elements (Groups IIIA-VIIA) and to conclude in each period with the outermost s2p6 configuration of the noble gases.

We can now explain many of the facts that we presented in Chapter 7. The structure of the periodic table, with its groups and periods, can be seen to be a consequence of the order of energy levels (Figure 9-2). Elements in the same group have similar chemical properties because they have the same outer electronic structure in the s and p orbitals. The outer valence electrons that are so important in chemistry are these s and p electrons. The closed, inert shell of the noble gases is the completely filled s2p6 configuration. We can understand the mechanism of formation of the transition metals and the inner transition metals in terms of the filling of inner d and f orbitals. We can see the reasons for general trends across a period or down a group, and for local fluctuations within a period.

### Electron Affinities

Another atomic property that depends strongly on the orbital electronic configuration is the electron affinity (EA), which is the energy change that accompanies the addition of an electron to a gaseous atom to form a negative ion:

Atom (g) + e- negative ion (g)

If energy is released when an atom adds an electron to form a negative ion, the EA has a positive value. If energy is required, the EA is negative. (Values for the known atomic electron affinities are given in Table 9-1.)

Within a period, the halogens have the highest electron affinities because, after the effect of screening electrons in lower quantum levels has been accounted for, the net nuclear charge is greater for a halogen than for any other element in the period. The noble gases have negative electron affinities because the new electron must be added to the next higher principal quantum level in each atom. Not only would the added electron be farther from the nucleus than the other electrons, it also would receive the full screening effect from all the others.

Lithium and sodium have moderate electron affinities; beryllium has a negative electron affinity, and magnesium has a near zero electron affinity. In Be and Mg the valence s orbital is full and the added electron must go into a higher-energy p orbital. Nitrogen and phosphorus have low electron affinities because an added electron must pair with an electron in one of the half-filled p orbitals.

Example 2
Write the ground-state orbital electronic configurations of Cl+, Cl, and Cl-. What are the valence electronic configurations of these species?
Solution

The atomic number of Cl is 17. Therefore the positive ion Cl+ has 16 electrons, Cl has 17, and Cl- has 18. The ground-state orbital electronic configurations are as follows:

 Cl+: 1s22s22p63s23p4 or KL 3s23p4 Cl: 1s22s22p63s23p5 or KL 3s23p5 Cl-: 1s22s22p63s23p6 or KL 3s23p6

The valence electronic configurations are 3s23p4 for Cl+, 3s23p5 for Cl, and 3s23p4 for Cl-.

 The first ionization energy of p, 1063 kJ mole-1, is greater than that of S, 1000 kJ mole-1. Explain this difference in terms of the valence orbital electronic configurations of P and S atoms. Solution The valence orbital electronic configuration of P is 3s23p3; of S, 3s23p4. The 3p shell is exactly half-filled in a P atom, whereas there is one extra electron in S that is forced to pair with another 3p electron: Chemical Principles Equation 9.10.png As a result of the added electron-electron repulsion of the paired 3p electrons in atomic s, the normal trend of increasing first ionization energies with increasing atomic number in a given period is reversed, with the IE1 of P being grater than the IE1 of S. This effect illustrates the special stability associated with a half-filled p shell. After the half-filled p shell is disrupted (p3 to p4), the electron-electron repulsions associated with the addition of the fifth and sixth p electrons in Cl and Ar are not large enough to override the attractive effect of the increasing positive nuclear charge. Thus the ionization energies of S, Cl, and Ar increase in the usual order (S < Cl < Ar).

 In each of the following orbital electronic configurations, does the configuration represent a ground state, an excited state, or a forbidden state (that is, a configuration that cannot exist): (a) 1s22s22p24s1; (b) 1s12s22p1; (c) 1s22s22p6; (d) 1s22s22p53s3 ? Solution (a) The ground-state configuration for an atom or ion with 7 electrons is 1s22s22p3. If it is an atom, then it must be N (atomic number 7). The configuration 1s22s22p24s1 represents a state in which a 2p electron has been excited to a 4s orbital. Therefore the configuration represents an excited state. (b) The configuration 1s22s22p1 represents an excited state (for a 4-electron atom or ion, 1s22s2 is the ground state; if it is an atom, it is Be). (c) The configuration 1s22s22p6 represents a ground state (F-, Ne, Na+, Mg2+). (d) The ground-state configuration for an atom or ion with 12 electrons is 1s22s22p63s2 (magnesium atom). In the configuration 1s22s22p53s3, 3 electrons are placed in the 3s orbital, which can take only 2 (one with s = $+\textstyle\frac{1}{2}$ and one with s = $-\textstyle\frac{1}{2}$). The configuration with three 3s electrons violates the Pauli principle and cannot exist. It represents a forbidden state.

 The electron affinity of Si, 138 kJ mole-1, is much larger than that of P, 75 kJ mole-1. Explain why this is so in terms of valence-orbital electronic configurations. Solution (The valence-orbital configuration of Si is 3s23p2; of P, 3s23p3. Adding one electron to Si to give Si

## Types of Bonding

File:Chemical Principles Fig 9.4.png
Figure 9-4 The two hydrogen 1s orbitals overlap to form an electron-pair covalent bond in H2

A covalent bond forms between combining atoms that have electrons of similar, or equal, valence-orbital energies. For example, two atoms of hydrogen are joined by a covalent bond in the H2 molecule. The energy required to separate two bonded atoms is known as the bond energy. For H2, the bond energy (corresponding to the process H2 H + H) is 432 kJ mole-1.

The two electrons in H2 are shared equally by the two hydrogen 1s orbitals. This, in effect, gives each hydrogen atom a stable, closed-shell (helium-type configuration. An orbital representation of the covalent electron-pair bond in H2 is shown in Figure 9-4.

An ionic bond is formed between atoms with very different ionization energies and electron affinities. This situation allows one atom in a two-atom pair to transfer one or more valence electrons to its partner. An atom of Na is so different from an atom of Cl, for example, that it is not possible for atoms in NaCl to share their electrons equally. The Na atom has a relatively low IE1 of 498 kJ mole-1 and a small EA of 117 kJ mole-1. Therefore, it will readily form Na+ in the presence of an atom with a high EA. The chlorine atom has an EA of 356 kJ mole-1 and an IE1 of 1255 kJ mole-1. Rather than lose an electron, a Cl atom has a strong tendency to gain one. The result is that in diatomic NaCl an ionic bond is formed, Na+Cl-, in which the 3s valence electron in Na is transferred to the one vacancy in the 3p orbitals of Cl.

The separation of the nuclei of two atoms that are bonded together (such as H2 or Na+Cl-) is called the bond distance. In the hydrogen molecule, H2, the bond distance is 0.74 Å. Each hydrogen atom in H2 may be assigned an atomic radius of 0.37 Å. The average radii of atoms o some representative elements shown in the periodic-table arrangement of Figure 9-5 were determined from experimentally observed bond distances in many molecules. The atomic radius in most cases is compared with the size of the appropriate closed-shell positive or negative ion.

You will notice (Figure 9-5) that the atomic radii become smaller across a given row (or period) of the periodic table. This shrinkage occurs because in any given period s and p orbitals acquire additional electrons, which are not able to shield each other effectively from the increasing positive nuclear charge. Thus an increase in the effective nuclear charge, thereby decreasing the effective atomic radius. This is why a Be atom, for example, is smaller than a Li atom.

From H to Li there is a large increase in effective atomic radius; the third electron in a Li atom is in an orbital that has a much larger effective radius than the H 1s orbital has. According to the Pauli principle, the third electron in Li must be in an orbital with a larger principal quantum number, namely the 2s orbital. Seven more electrons can be added to the 2s and 2p orbitals, which have approximately the same radii. However, these electrons do not effectively shield each other from the positive nuclear charge as it increases, and the result is an increase in the effective nuclear charge and a corresponding decrease in radii in the series Li (Z = 3) through Ne (Z = 10). After Ne, additional electrons cannot be accommodated by the n = 2 level. Thus an eleventh electron must go into the n = 3 level, specifically, into the 3s orbital. Since the effective radii increase from the n = 1 to n = 2 to n = 3 valence orbitals, the effective size of an atom also increases with increasing atomic number within each group in the periodic table.

File:Chemical Principles Fig 9.5.png
Figure 9-5 Relative atomic radii of some elements compared with the radii of the appropriate closed-shell ions. Radii are in angstroms. Solid spheres represent atoms and dashed circles represent ions. Notice that positive ions are smaller than their neutral atoms and negative ions are larger. Why is this so?
 Predict the order of decreasing atomic radii for S, Cl, and Ar. Predict the order for Ca2+, Cl-, Ar, K+, and S2-. Solution The order S > Cl > Ar is correct for the radii of these atoms, because the nuclear charge increases by one unit from S to Cl and by one unit from Cl to Ar. The valence electrons are attracted more strongly to the nuclei with higher positive charges in any given period, so the atomic radii decrease correspondingly. For isoelectronic atomic and ionic species (those having the same number of electrons), radii decrease as the nuclear charge (atomic number) increases, again because of increasing electron-nucleus attraction. Thus the correct order for the isoelectronic species is S2- > Cl- > Ar > K+ > Ca2+.

## Electronegativity

Most bonds in molecules fall somewhere between the extremes of the covalent and ionic types. The bond in the HF molecule, for example, is neither purely covalent nor purely ionic. Just how unequal is the sharing of electrons in HF? And which atom in HF is able to attract the greater share of the bonding electrons, H or F? To anser the second question, and to provide a qualitative guide to the first, Linus Pauling (b. 1901) defined a quantity called electronegativity, χ (the Greek letter chi), in 1932: two years later R. S. Mulliken (b. 1896) showed that electronegativity could be related to the average of the electron affinity and the ionization energy of an atom.

Pauling obtained electronegativity values by comparing the energy of a bond between unlike atoms, AB, with the average energies of the A2 and B2 bonds. If HF formed a covalent bond as in H2 and F2, then we would expect the bond energy in HF to be close to the average (say, the arithmetic mean or the geometric mean) of the bond energies in H2 and F2. However, in molecules such as HF, the bonds are stronger than predicted from such averages. The bond energy of HF is 565 kJ mole-1, whereas the bond energies of H2 and F2 are 432 and 139 kJ mole-1, respectively. The geometric mean of the last two values is (139 $\times$ 432)1/2 = 245 kJ mole-1, which is much less than the observed bond energy of HF. This "extra" bond energy (designated Δ) in an AB molecule is assumed to be a consequence of the partial ionic character of the bond due to the electronegativity difference between atoms A and B. The electronegativity difference between A and B may be defined as

      χA - χB = 0.102Δ1/2                             (9-1)


in which χA and χB are the electronegativities of atoms A and B, and Δ is the extra bond energy in kilojoules per mole. The extra bond energy is calculated from the equation

Δ = DEAB - [(DEA2)(DEB2)]1/2

in which DE is the particular bond dissociation energy.

In equation 9-1 the square root of Δ is used because it gives a more consisten set of atomic electronegativity values. Since only differences are obtained from equation 9-1, one atom must be assigned a specific electronegativity value, and then the values for the other atoms can be calculated easily. In a widely adopted electronegativity scale, the most electrnegative atom, F, is assigned a value of 3.98. (Electronegativity values based on this assignment are given in Table 9-1.)