Modern atomic theory received a “shot in the arm” when it was recognized that the individual atom has light absorption and emission spectra occurring at narrow lines of the spectrum at specific wavelengths, as opposed to the broad bands typical of the polyatomic molecules and compounds. Since the line spectrum of each element is characteristic of that element, atomic spectroscopy can be used for precise elementary analysis of many types of chemically simple and complex materials. These studies make use of the wave character of light, as well as light’s particle character.
A light beam can be viewed as a form of energy emitted from a source. A light beam can be imagined in the form of a sine wave, as shown in Fig. 8-1. The distance between two waves, usually measured from the peak of the waves, is the wavelength, given the symbol lambda, λ. The frequency is a statement of the number of waves passing a point in space per second; it is given the symbol nu, ν. The hertz is commonly used as the unit for frequency; 1 Hz = 1s–1.
Fig. 8-1
The product of the wavelength and the frequency is equal to the velocity of light, usually designated by c:
c = λv
The velocity of light in a vacuum is the same for all wavelengths, 2.998 × 108 m/s. The formula above tells us that there is a simple inverse relationship of wavelength and frequency, λ = c/ν or ν = c/λ. The velocity of light in the atmosphere is a little slower than in a vacuum; however, the drop in velocity is less than 0.1 percent of that in a vacuum, making 2.998 × 108 m/s acceptable for most applications due to the low percent error introduced. Another term is often used, the wave number, 
, defined as 1/λ (or as v/c). The most common unit for wave number is cm–1.
The energy of light is emitted, absorbed, or converted to other forms of energy in individual units referred to as quanta (singular: quantum). The unit of light energy is often referred to as the particle of light called the photon.
The energy of a photon is proportional to the frequency:
ε = hν = (6.626 × 10–34 J · S)ν
Planck’s constant, h, is the universal proportionality constant.
Chemists’ studies are performed using radiation whose wavelengths vary from 0.1 nm (x-rays) to several centimeters (microwaves). Note that visible light is in the range of 400 to 700 nm.
A major step toward the understanding of atomic structure was Bohr’s explanation of hydrogen’s spectrum. His postulates were
1. The electron of the hydrogen atom revolves around the nucleus in stable circular orbits.
2. In each stable orbit the electrostatic attraction between the negatively charged electron and the positively charged nucleus provides the centripetal force (pulling the electron toward the nucleus) needed for the circular motion of the electron. The energy of the atom is the sum of the potential energy of the electrostatic interaction between the nucleus and the electron and the kinetic energy of the motion of the electron.
3. Only certain stable orbits are allowed—those for which the angular momentum of the electron is an integer, n, times the constant h/2π. (The h is Planck’s constant introduced above.)
4. An electron can move from one stable orbit to another only by absorbing or releasing an amount of energy exactly equal to the difference between the energies of the two orbits. Energy is absorbed when going to a higher level (out from the nucleus) and released (not necessarily the same wavelength) when the electron falls back to the original orbit. If this energy is absorbed or released as light, a single photon of absorbed or emitted light must account for the required energy difference as
hν = |ΔE|
where ΔE is the difference between the energies of the final and initial orbits.
Bohr satisfactorily accounted for the observed series of spectral lines for hydrogen with this theory. The predicted orbit energies in the simplest form of the theory are given by
where m and e are the mass and charge of the electron, ε0 is permittivity of free space, and Z is the atomic number of the nucleus (1 for hydrogen). Note that the energy is negative with respect to the state in which the electron and nucleus are widely separated, which is the zero energy state. In SI units,
The predicted allowed wave numbers observed in the spectrum are
where n1 < n2 in absorption and n1 > n2 in emission. The quantity R, the Rydberg constant, is defined purely in terms of universal constants and has the value of 109,737 cm–1:
Note: The observed value for 1H is 0.06 percent lower because of the reduced mass effect. In accordance with a refinement of the simple Bohr theory, R increases with increasing nuclidic mass toward a maximum of 109,737 cm–1. The predicted radii of the orbits are
The predicted value for hydrogen’s first Bohr orbit (n = 1) is a0 = 5.29 × 10–11 m = 0.529 Å.
A note of caution: The Bohr theory, even when improved and amplified, applies only to hydrogen and hydrogen-like species, such as He+ and Li+. The theory explains neither the spectra of atoms containing even as few as two electrons, nor the existence and stability of chemical compounds. The next advance in the understanding of atoms requires an understanding of the wave nature of matter.
De Broglie proposed that not only does light have the dual properties of waves and particles, but also particles of matter have properties of waves. The wavelength of those particle waves is given by
where m and v are the mass and velocity of the particle. Planck’s constant, h, is so small that the wavelengths are in an observable range only for particles of atomic or subatomic mass.
An experimental confirmation of the de Broglie relation for a beam of uniformly energetic electrons was followed by a theoretical development of quantum mechanics, also referred to as wave mechanics. Not only was Bohr’s successful prediction of the stable energy levels of the hydrogen atoms reproduced, but the concept can be applied to many-electron atoms and to many-atom molecules. The Bohr postulates are replaced by the Schrödinger equation, which must be solved by the methods of partial differential equations. The equation has mathematical similarities to descriptions of physics waves, and the arbitrary introduction of integers in the Bohr theory receives justification in quantum mechanics in the requirement that wave-like solutions of the equation must be continuous, finite, and single-valued. The steady-state solutions of the wave equation corresponds to states of fixed energy, as in the Bohr theory, and in many cases, including the hydrogen atom, of fixed angular momentum. Many other descriptive properties, like the location of an electron, are not definitely fixed, but are represented by probabilities of distribution over a range of numerical values. In the hydrogen atom, the electron is not confined to a two-dimensional orbit, but is represented by a wave that extends over three-dimensional space. The wave amplitude indicates the varying probability of finding the electron at different locations with respect to the nucleus.
Orbitals
A solution to the Schrödinger equation for an electron must satisfy three quantum conditions corresponding to the three dimensions of space. Each quantum condition introduces an integer, called a quantum number, into the solution. A separate solution, describing a probability distribution of finding the electron at various locations, exists for each allowed set of three quantum numbers. Such a solution is called an orbital; it is similar to a hypothetical time-exposure photograph of an electron taken over a time interval large enough so that each region of space is represented by the weighted probability of finding the electron in that region. These three quantum numbers are usually designated as follows:
1. n, principal quantum number: This number almost exclusively determines the energy of the orbital in one-electron systems, and is still the principal determinant of energy in many-electron systems.
2. l, angular momentum quantum number: This number defines the shape of the orbital and, together with n, defines the average distance of the electron from the nucleus.
3. ml, magnetic quantum number: This number determines the orientation of the orbital in space.
In addition to the three dimensions in space describing the relative positions of the electron and the nucleus, there is a fourth dimension internal to the electron itself, related to the spin of the electron around an internal axis and characterized by the magnetic moment associated with this spin. The quantum number associated with electron spin is usually designated as ms.
Each of the four quantum numbers may have only certain values, which are:
(a) n may be any positive integer. Electrons having a given n-value are said to be in the same shell (orbit). Shells are designated by capital letters as follows:
(b) The value of l varies from 0 to n – 1. The values of l are designated by letters as follows:
An orbital is usually referred to by its principal quantum number (n) followed by the letter corresponding to its l-value. Examples are 2s, 2p, 4d, and 5f.
(c) ml may have any integral value from –l to +l. This rule gives the number of orbitals, 2l + 1, associated with a given (n, l) combination. There are three p orbitals corresponding to ml = 1, 0, and –1. However, it is usually more convenient in chemistry to use a new set of three orbitals oriented along the x, y, and z axes to display the shapes and directions of these orbitals. Further, there are 5 d orbitals and 7 f orbitals having different shapes and orientations in space.
(d) The values of ms may either be + 
or – .
The probabilities of finding an electron at various distances, r, from the hydrogen nucleus are given in Fig. 8-2 for several (n, l) combinations. The shapes of the s, p, and d orbitals are shown in Fig. 8-3. An inspection of Fig. 8-2 shows a detail which emerges as an exact mathematical consequence of the theory: when l = n – 1, the most probable distance of the electron from the nucleus is exactly equal to the radius of the Bohr orbit, n2a0. In general, the average distance of the electron from the nucleus increases with increasing n. This same figure shows another interesting feature for all allowed l values, except the maximum for a given n; namely, the existence of minima of zero probability, corresponding to spherical nodal surfaces around the nucleus, at which the electron will not be found. Figure 8-3 shows that s orbitals take the shape of a sphere with the nucleus in the center. Each p orbital is concentrated along the + and – portions of one of the Cartesian axes (x, y, or z) with a nodal plane of zero probability perpendicular to that axis at the nucleus. Of the five allowed d orbitals for a given n, only four have the same shape—a three dimensional four-leaf clover—with the highest probability along the x and y axes in the case of dx2–y2 and in between the axes in the case of dxy, dxz, and dyz. Each of these orbitals has two perpendicular nodal planes through the nucleus. The fifth d orbital, dz2, has maxima in the +z and –z directions and a secondary concentration in the (x, y) plane. This orbital has two conical nodal surfaces through the nucleus, one projecting above the (x, y) plane and one below. This separates the +z and –z lobes from the doughnut-shaped ring of probability concentrated in that plane.
Fig. 8-2
Fig. 8-3 Angular dependence of orbitals.
The Pauli exclusion principle states that no two electrons in an atom may have the same set of all four quantum numbers. This principle places the following limits on the number of electrons for the various (n, l) orbital designations:
Because the Schrödinger equation cannot be solved exactly for polyelectron atoms, it has become the practice to approximate the electron configuration by assigning electrons to hydrogen-like orbitals. These orbitals are designated by the same labels as for hydrogen’s orbitals and have the same spatial characteristics described in the previous section, “Orbitals.”
The filling of the orbitals recognizes that electrons will first occupy those orbitals having the lowest energy. The build-up (Aufbau, German) principle describes the way electrons assort into the orbitals. Both n and l are accounted for by the process; however screening occurs, which reduces the nuclear charge in addition to the reduction due to distance. Essentially, an inner electron cloud screens the positive nuclear charge so that the electrons in successive orbitals are attracted less than predicted by the inverse square law (electromagnetic phenomena drop in intensity as the inverse of the square of the distance changes). Due to these factors and others which influence the process, the order for filling of the orbits can be taken as 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s, etc., which is difficult to memorize. Figure 8-4 displays a layout that is easy to set up and remember. Note that the rows are the orbitals predicted to fill and the filling order is indicated by a set of diagonal arrows. The Aufbau principle predictions are for the configurations in their ground (lowest energy) state. An exception to the sequence exists and is covered by Problem 8.10.
Fig. 8-4
Note: All electron configurations in this book are in the ground state unless otherwise indicated.
The position of an element in the Periodic Table of the Elements (inside front cover) is a clue to the configuration of the outermost electron shell. Let n be the number of the period in which the element is found on the table and the superscript on s state the number of electrons in the s orbital. The group IA and IIA (1 and 2) electron configurations are ns1 and ns2. The other A groups, IIIA to VIIIA (13 to 18), have the configurations ns2np1 to ns2np6. These two sets of elements are the main group elements and are often referred to as the s-block and p-block elements. The exception to the pattern is helium (1s2), which, although in group VIIIA, does not have sufficient electrons to conform to the group’s configuration.
The transition metals in periods 4 and 5 have the outer configurations of ns2(n – 1)d1 to ns2(n – 1)d10, but there are some anomalies and the filling of the d shell is not uniform; refer to your text. The transition metals in periods 6 and 7 fill the (n – 2)f orbitals before the (n – 1)d’s are completed. There are many anomalies in the course of filling the f shell. The lanthanides, period 6, and the actinides, period 7, are displayed separately below the body of the table and are handled by the guidelines outlined.
The outermost electrons, often called the valence electrons, are primarily responsible for the chemical properties of the elements. It follows that the elements in a specific group will show similar characteristic oxidation numbers (charges, also called valences) and display a trend in characteristics. Even though electron configurations were not known when the earliest periodic tables were formulated, the elements were placed by similarity of characteristics.
When considering the filled orbitals, it is important to note that electrons are in pairs as shown below. However, the individual orbitals fill one electron at a time until the orbital is completed. The electrons in unfilled orbitals can be represented by upward and downward arrows that indicate the opposite spins of the electrons in a pair. Figure 8-5 shows the placement of the p electrons in order represented by superscripts and the filled orbital with the electrons represented by the arrows commonly sketched; the d and f orbitals are handled similarly. Arrows are placed upward in the order of filling (1→3), then downward (4→6) to form pairs.
Fig. 8-5
The inner electrons, those below the valence electrons, are arranged as is the noble gas element (Group VIIIA) before the element under consideration. If we were to consider titanium (Ti, Z = 22), the electron configuration could be expressed as [Ar]3d24s2. Note that, although the Aufbau principle indicates the 4s filling before the 3d, it is common practice to present the electron configuration in numerical order with respect to n, rather than the filling order.
The electron cloud around an atom makes the concept of atomic size somewhat imprecise. Even so, it is useful to refer to an atomic size or an atomic radius. Operationally, one can divide the experimentally determined distance between the centers of two chemically bonded atoms to arrive at the two atomic radii. If the bonding is covalent (see Chapter 9), the radius is called a covalent radius; if the bonding is ionic, the radius is an ionic radius. The radius for a nonbonded situation may also be defined in terms of the distance of closest nonbonding approach and is called a van der Waals radius. These concepts of size are illustrated in Fig. 8-6. The following generalizations are observed regarding atomic size:
Fig. 8-6 (a) Portion of NaCl crystal showing the ionic radii (i). (b) Two Cl2 molecules in contact in liquid chlorine showing the covalent radii (c) and the van der Waals radii (v).
1. Within a given group of the Periodic Table, the radius increases with increasing atomic number. This fact is related to the increased n of the outermost shell.
2. Within a given period of the Periodic Table, the covalent radii generally decrease with increasing atomic number. This is related to the facts that (i) the size of an atom depends on the average distance of its outermost electron(s), (ii) there is no change in n of the outermost electron(s) within a given period, and (iii) there is increasing nuclear charge with increasing atomic number.
3. Ionic radii of cations (positive ions) are fairly small compared to the covalent radii for the corresponding atoms since all the outermost electrons (highest n) are usually removed. The radii of anions (negative ions) are only slightly larger than the van der Waals radii for the corresponding atoms since the extra electron(s) have the same n. However, the covalent radii of these atoms are appreciably less since they are bonded to their neighbors by electron sharing.
4. Within a given group, ionic radii increase with increasing atomic number. Within a given period, the radii of cations decreases with increasing atomic number, as do the radii of anions. Note that the radius of the last cation and the first anion in a given period do not conform to the trend. The reason is that an ion becomes a cation by electron(s) loss, but an anion is the result of the gain of an electron(s).
The Bohr formula for the hydrogen atom energy levels predicts that higher energy levels get closer and closer together and tend to approach a limit of zero as n → ∞. At this limit, the atom has been ionized. The minimum energy required to ionize an isolated gaseous atom may be determined by spectroscopic, thermochemical, or electrical means. In the electrical method, a measurement is made of the accelerating potential that will impart to a projectile electron (not the electron within the atom) an amount of kinetic energy just sufficient to dislodge the bound electron from its atom. This means that the ionization energy can be measured directly in electrical terms. The electron volt, ev, is the energy imparted to one electron accelerated by a potential difference of 1 V (Note:1V = 1J/C). The value of an electron volt is expressed as follows:
1eV = (e–’s charge) × (potential difference) = (1.6022 × 10–19 C) × (1 J/C) = 1.6022 × 10–19 J
Ionization energies, I.E., have been measured for all the atoms. They are all positive, corresponding to an endothermic process. Some interesting trends have been observed.
1. Within a given group of the periodic table, the first ionization energy decreases with increasing atomic number. This is related to the increase in atomic radius and the decreasing attraction of the nucleus for the increasingly distant outermost electron. It should be mentioned that this trend is not uniformly noted for the transition metals.
2. Within a given period, the trend is an increase in ionization energy with increasing atomic number. However, atoms just beginning a new subshell or the second half of a subshell may have a smaller ionization energy than the previous atom.
3. The ionization of each succeeding stage of ionization is always greater than for the preceding state. As an example, the second ionization energy of magnesium is considerably greater than the first (about twice as great) because the Mg2+ has twice the electron static attraction for the removed electron as does Mg1+. However, the second ionization energy of sodium is many times the first because the second electron must be removed from the n = 2 shell rather than the n = 3 shell (further from the nucleus).
Some free atoms can capture an extra electron to form a stable gaseous anion, particularly with elements having almost-completed p subshells (Groups VIA and VIIA, especially). For example,
For most elements, the first electron gained is exothermic (energy given off). Chlorine’s electron affinity (E.A.) is –349 kJ per mole of electrons gained. The higher the first electron affinity (chlorine’s is high), the more likely the element is to form an anion. The electron affinities of the elements display loose trends to increase in a period with the atomic number.
The magnetic properties of matter depend on the properties of the individual atoms. We have noted that electron spin has a magnetic moment associated with it. Two electrons occupying the same orbital (same n, l, and ml) have their magnetic moments canceled out because the two values of ms are of equal magnitude and have opposite spin angular momentum (equal but opposed magnet moment). This means that atoms, ions, or compounds in which at least one orbital is singly occupied then have a net magnetic moment. Such a substance is said to be paramagnetic and is attracted into a magnetic field. The magnitude of the magnetic moment (and the number of unpaired electron spins) can be determined experimentally by measuring the force of attraction of an externally imposed magnetic field for the substance. Substances without unpaired electron spins do not have magnetic moments and are repelled by a magnetic field; they are referred to as being diamagnetic. The repulsion of diamagnetic substances is much smaller in magnitude than the attraction of paramagnetic substances.
Magnetic measurement is an important experimental tool for determining the electron assignment into orbitals for atoms, ions, and compounds. Several rules have been developed for the assignment of electrons within a subshell.
1. Electrons within a subshell for which l > 0 tend to avoid pairing within the same orbital. This rule is Hund’s rule and reflects the relatively greater electrostatic repulsion between two electrons in the same orbital as compared with occupancy of two orbitals having differing values for ml.
2. Electrons in singly occupied orbitals tend to have their spins in the same direction so as to maximize the net magnetic moment.
3. Looking at the electron configuration allows for a prediction whether or not an atom (or ion) is paramagnetic. Note that the prediction only applies to the free individual atom. Any firm conclusion based on the prediction is risky when applied to collections of atoms (or ions). For example, the aluminum atom has one unpaired electron, but a piece of aluminum is diamagnetic.
8.1. Determine the frequencies of electromagnetic radiation of the following wavelengths: (a) 0.10 nm; (b) 5000 Å; (c) 4.4 μm; (d)89m; (e) 562nm.
(a)
(b)
(c)
(d)
(e)
8.2. (a) What change in molar energy (J/mol) would be associated with an atomic transition giving rise to radiation at 1 Hz? (b) For any given photon, what is the numerical relationship between its wavelength in nanometers and its energy in electron volts?
(a) If each of NA atoms gives off one 1-Hz photon,
ΔE = NA(hv) = (6.022 × 1023 mol–1)(6.626 × 10–34 J · s)(1 s–1) = 3.990 × 10–10 J · mol–1
Since ΔE and v are proportional, we may treat the ratio
as a “conversion factor” between Hz and J · mol–1, provided we understand just what this “conversion” means. Then, for 1-MHz (106 Hz) radiation,
ΔE = (106 Hz)(3.990 × 10–10 J · mol–1 · Hz–1) = 3.990 × 10–4 J · mol–1
(b) First, let us find the frequency equivalent of 1 eV from the Planck equation, then find the wavelength from the frequency.
Because of the inverse proportionality between wavelength and energy, the relationship may be written as
8.3. In the photoelectric effect, an absorbed quantum of light results in the ejection of an electron from the absorber. The kinetic energy of the ejected electron is equal to the energy of the absorbed photon minus the energy of the longest-wavelength photon that causes the effect. Calculate the kinetic energy of a photoelectron (eV) produced in cesium by 400-nm light. The critical (maximum) wavelength for the photoelectric effect in cesium is 660 nm.
Using the result of Problem 8.2(b),
8.4. It has been found that gaseous iodine molecules dissociate into separate atoms upon absorption of light at wavelengths less than 499.5 nm. If each quantum is absorbed by one molecule of I2, what is the minimum input, in kJ/mol, needed to dissociate I2 by this photochemical process?
8.5. A beam of electrons accelerated through 4.64 V in a tube containing mercury vapor was partly absorbed by the vapor. As a result of absorption, electronic changes occurred within a mercury atom and light was emitted. If the full energy of a single incident electron was converted into light, what was the wave number of the emitted light?
Using the result of Problem 8.2(b),
8.6. An electron diffraction experiment was performed with a beam of electrons accelerated by a potential difference of 10 keV. What was the wavelength of the electron beam in nm?
We can use the de Broglie equation; take the mass of an electron as 0.922 × 10–30kg. The velocity of the electron is found by equating its kinetic energy, mv2, to the 10 keV loss of electric potential energy.
and 
Now, the de Broglie equation gives us
The results calculated above are somewhat in error because of the law of relativity, which becomes more and more relevant as the velocity approaches the speed of light. For instance, for a potential difference of 300 kV, the velocity calculated as above would exceed c, an invalid result since no particle can have a velocity greater than the speed of light.
8.7. The Rydberg constant for deuterium (2H or 2D) is 109,707 cm–1. Calculate (a) the shortest wavelength in the absorption spectrum of deuterium, (b) the ionization energy of deuterium, and (c) the radii of the first three Bohr shells (orbits).
(a) The shortest-wavelength transition would correspond to the highest frequency and to the highest energy. The transition would be from the lowest energy state (the ground state), for which n = 1, to the highest, for which n = ∞.
(b) The transition calculated in (a) is indeed the ionization of the atom in its ground state. From the result of Problem 8.2(b),
This value is slightly greater than the value for 1H.
(c) From the equation (Z = 1),
The radii are 1, 4, and 9 times a0, or 0.529, 2.116, and 4.76 Å. The reduced-mass correction, involving an adjustment of 3 parts in 104, is not significant and a0 for the first Bohr shell of 1H is a good substitution.
8.8. (a) Neglecting reduced-mass effects, what optical transition in the He+ spectrum would have the same wavelength as the first Lyman transition of hydrogen (n = 2 to n = 1)? (b) What is the second ionization energy of He? (c) What is the radius of the first Bohr orbit for He+?
(a) He+ has only one electron. It is classified as a hydrogen-like species with Z = 2, and the Bohr equations may be applied. From the equation
the first Lyman transition for hydrogen would be given by
The assumption regarding reduced-mass effects is equivalent to considering R for He+ the same as for 1H. The Z2-term can just be compensated by increasing n1 and n2 by a factor of 2 each.
The transition in question is then the transition from n = 4 to n = 2.
(b) The second ionization energy for He is the same as the first ionization energy for He+, and the Bohr equations may be applied to the ground state of He+ for which Z = 2 and n = 1.
The result is 4 times the result for deuterium in Problem 9.7. Since is proportional to energy, the ionization energy will likewise be 4 times that for deuterium.
(c) 
8.9. (a) Write the electron configurations for the ground states of N, Ar, Fe, Fe2+, and Pr3+. (b) How many unpaired electrons would there be in each of these isolated particles?
(a) The atomic number of N is 7. The first shell (orbit) will contain its maximum of 2 electrons, the next 5 electrons will be in the second shell, 2 filling the lower-energy s subshell (orbital) and the remaining 3 electrons filling the p subshell. A common notation is
The second notation forms shows only those electrons beyond those in the noble gas (Group VIIIA or 18) found before the element in question and identifies that element in square brackets.
The atomic number of Ar is 18, which results in the filling of shells 1 through 3 (K, L, and M).
1s22s22p63s23p6
The atomic number of Fe is 26. Beyond the electron configuration of argon, the order of filling is 4s, then 3d, until 26 electrons are assigned (8 beyond argon).
The iron(II) ion, Fe2+, contains two fewer electrons than the iron atom. Although 4s is of lower energy than 3d for atomic numbers 19 and 20 (K and Ca), this order reverses for higher nuclear charges. In general, the electrons most easily lost from any atom are those with the largest principal quantum number.
[Ar]3d6
Pr3+ has 56 electrons, 3 fewer than the Pr atom (atomic number 59). The order of filling of those elements in the period following Xe is 6s2, then one 5d electron, then the whole 4f subshell, then the rest of the 5d subshell followed by the 6p subshell. There are frequent replacements of the first 5d assignment with an additional 4f, or one of the 6s with an additional 5d, but these irregularities are of no consequence in the assignment of electrons in Pr3+. The 3 electron removed from the neutral atom to form the ion follow the general rule of removal first from the outermost shell (highest n) and then from the next-to-outermost (next highest n); in this case, the 6s electrons are removed first and no 5d electrons would remain even if there were one in the neutral atom.
(b) Complete subshells have no unpaired electrons; it is necessary to examine only the electrons past a noble gas core.
For N, the 2p is the only incomplete subshell. Following Hund’s rule, the three electrons in this subshell will singly occupy the three available subshells. There will be three unpaired electrons.
Argon has no incomplete subshells and, therefore, no unpaired electrons.
Fe has six 3d electrons in the only unfilled subshell. The maximum unpairing occurs with double occupancy of one of the available d subshells and single occupancy of the remaining four. There will be 4 unpaired electrons.
Fe2+ also has 4 unpaired electrons for the same reason as Fe.
Pr3+ has 2 unpaired electrons that are in two of the seven 4f subshells.
8.10. Nickel’s electron configuration is [Ar]3d84s2. How can we account for the fact that the configuration of the next element, Cu, is [Ar]3d104s1?
In the hypothetical procedure of making up the electronic complement of Cu by adding one electron to the configuration of the providing element, Ni, we might have expected only a ninth 3d electron. For atomic number 19, the 3d subshell is decidedly of higher energy than is 4s; then potassium has a single 4s electron and no 3d electron. After the 3d subshell begins to fill with Sc, atomic number 21 (4s filled), the addition of each succeeding 3d electron is accompanied by a lowering of the average energy of the 3d level. This is because each succeeding element has an increased nuclear charge which is only partially screened from a test 3d electron by the additional electron in the same subshell. The energy of the 3d subshell decreases gradually as the subshell undergoes filling and drops below the level of the 4s toward the end of the transition series.
Another factor is that the configuration 3d104s1 has a spherically symmetrical distribution of electron density, a stabilizing arrangement characteristic of all filled or half-filled subshells. On the other hand, the configuration 3d94s2 has a “hole” (a missing electron) in the 3d subshell, destroying the symmetry and any extra stabilization.
8.11. The ionization energies of Li and K are 5.4 and 4.3 eV. What can we predict for the I.E. of Na?
The first I.E. of Na should be intermediate between that of Li and K; it can be determined by a simple average of the two values. That average is 4.9 eV, which is reasonably close to the 5.1 eV observed.
8.12. The first ionization energies of Li, Be, and C are 5.4, 9.3, and 11.3 eV. What can be predicted for the ionization energies of B and N?
There is a general increasing trend in I.E. with increasing atomic number in a given period. This is true of the values we have here, but there is a larger increase between Li and Be than between Be and C. The filling of the 2s subshell gives Be a greater stability than would be suggested by a smooth progression across the periodic table. The next element, B, would have an I.E. that represents a balancing of two oppositely directed factors: an increase for Be because of the increased Z (nuclear charge increase) and a decrease because a new subshell is beginning to be filled in the case of B (between Be and C). We could guess that the I.E. for B would be less than that for Be—that is the case. The observed I.E. for B is 8.3 eV.
The increase in going from Z = 5 to Z = 6 is 3.0 eV. We could expect the increase in going to N (Z = 7) to be similar, bringing the I.E. of N to about 14.3 eV. Because of the extra stability of the half-filled p subshell, the I.E. is even greater. A value of 14.5 eV is observed.
8.13. In the ionic compound KF, the K+ and F– ions are found to have practically identical radii, about 0.134 nm. What can be predicted about the relative covalent radii of K and F?
The covalent radius of K should be much greater than 0.134 nm and that of F much smaller, since atomic cations are smaller than their parent atoms, while atomic ions are larger than their parents. The observed covalent radii of K and F are 0.20 and 0.06 nm.
8.14. The single covalent radius of P is 0.11 nm. What is the prediction for the single covalent radius of Cl?
P and Cl are members of the same period. Cl should have a smaller radius in keeping with the usual trend across the period. The experimental value is 0.10 nm.
8.15. The first I.E. for Li is 5.4 eV and the electron affinity of Cl is 3.61 eV. Compute ΔH (kJ/mol) for the reaction carried out at such low pressures that the resulting ions do not combine with each other.
Li(g) + Cl(g) → Li+(g) + Cl–(g)
The overall reaction may be expressed as two partial reactions:
where e– stands for an electron. Although ΔH for each of the above partial reactions differs slightly from ΔE (by the term pΔV), the ΔH of the overall reaction is the sum of the two ΔE’s (the overall volume change is zero). Since the values of I.E. and E.A. are given on a per-atom basis, the factor 6.02 × 1023 atoms/mol is required to obtain ΔH on the conventional per-mol basis.
8.16. What is the wavelength in meters of the radiation from (a) a low-range TV station broadcasting at a frequency of 55 MHz, (b) an AM radio station at 610 kHz, and (c) a microwave oven operating at 14.6 GHz?
Ans. (a) 5.5 m; (b) 492 m; (c) 0.0205 m
8.17. The critical wavelength for producing the photoelectric effect in tungsten is 260 nm. (a) What is the energy of a quantum at this wavelength in joules and in electron volts? (b) What wavelength would be necessary to produce photoelectrons from tungsten having twice the kinetic energy of those produced at 220 nm?
Ans. (a)7.65 × 10–19J = 4.77 eV; (b) 191 nm
8.18. In a measurement of the quantum efficiency of photosynthesis in green plants, it was found that 8 quanta of red light at 685 nm were needed to evolve one molecule of O2. The average energy storage in the photosynthetic process is 469 kJ per mole of O2 evolved. What is the energy conversion efficiency?
Ans. 33.5%
8.19. O2 undergoes photochemical dissociation into one normal oxygen atom and one oxygen atom 1.967 eV more energetic than normal. The dissociation of O2 into two normal oxygen atoms is known to require 498 kJ/mol O2. What is the maximum wavelength effective for O2’s photochemical dissociation?
Ans. 174 nm
8.20. Acriflavine is a dye that, when dissolved in water, has its maximum light absorption at 453 nm, and its maximum fluorescence emission at 508 nm. The number of fluorescence quanta averages 53% of the number of quanta absorbed. Using the wavelengths of maximum absorption and emission, what percentage of absorbed energy is emitted as fluorescence?
Ans. 47%
8.21. The prominent yellow line in the spectrum of a sodium vapor lamp has a wavelength of 590 nm. What minimum accelerating potential will excite this line in an electron tube containing Na vapor?
Ans. 2.10 V
8.22. Show by substitution in the formula given in the text (“Interaction of Light with Matter”) that ao, the radius of the first Bohr orbit for hydrogen, is 5.29 × 10–11 m.
8.23. A sample of an unknown compound is exposed to light of wavelength 1080 Å. Nitrogen is found to be given off during the irradiation, hinting that there may have been N≡N bonds in the sample. Calculate the amount of energy (bond energy) necessary to break a mole of these bonds.
Ans. 1.1 × 103kJ/mol
8.24. What accelerating potential is needed to produce an electron beam with an effective wavelength of 0.0256 nm?
Ans. 2.30 kV
8.25. What is the wavelength of a beam of neutrons (1.67 × 10–24 g) that has a velocity of 2.50 × 105 cm/s?
Ans. 1.59 × 10–10m or 1.59 Å
8.26. In studies of electron spin resonance (esr) the energy differences between spin states are very small, of the order of 1 × 10–4 eV, compared with about 3 eV in visible spectroscopy. What wavelength of radiation is required for esr, and which types in Problem 8.16 does it resemble?
Ans. 0.012 m, microwave region
8.27. What accelerating potential must be imparted to a proton beam to give it an effective wavelength of 0.0050 nm?
Ans. 33 V
8.28. All atoms with odd Z values must have at least one unpaired electron. Can an even Z atom have unpaired electrons? If so, give examples from the first three periods.
Ans. Yes; C, O, Si, and S
8.29. Which atoms in the first transition period (Z = 21 – 30) are diamagnetic? Give their configurations.
Ans. Only Zn; [Ar]3d10 4s2
8.30. The configuration of Cr differs from that derived by the Aufbau procedure. Deduce the actual configuration and explain the anomaly.
Ans. Cr; [Ar]3d5 4s1. Aufbau would give 3d4 4s2, so an s shifts to d to gain the stability of the half-filled subshell. Note that the actual configuration has complete spherical symmetry.
8.31. The configurations of negative ions follow the Aufbau rule. Write the electron configurations for H–, N3–, F–, and S2–.
Ans. H– is 1s2 or [He]; N3– is [Ne]; F– is [Ne]; S2– is [Ar]
8.32. The Rydberg constant for Li2+ is 109,729 cm–1. (a) What is the long-wavelength limit in the absorption spectrum of Li2+ at ordinary temperatures (all the ions in ground state)? (b) What would be the shortest wavelength in the emission line spectrum of Li2+ within the visible (400 to 750 nm) region? (c) What would be the subshell radius of the ground state of Li2+? (d)Li2+ has what ionization energy?
Ans. (a) 13.5 nm; (b) 415.4 nm (n = 8 → n = 5); (c) 0.176 Å; (d) 122.4 eV
8.33. Magnesium and aluminum form an alloy used in the automobile, aerospace, and other industries requiring a light-weight, strong material for the construction of parts. (a) What is the complete electron configuration of each metal? (b) Manganese, nickel, and titanium are also used in the same industries. Provide the electron configurations of these metals. (c) Display the complete electron configuration for thallium, a very toxic metal that must be handled with care.
Ans.
8.34. The first ionization energy of Li is found experimentally to be 5.363 eV. If the electron in the second shell (n = 2) is assumed to move in a central field of an effective nuclear charge, Zeff, consisting of the nucleus and the other electrons, by how many units of charge is the nucleus shielded by the other electrons? Assume that the ionization energy can be calculated from Bohr theory.
Ans. Zeff = 1.26; since the nuclear charge is 3+, the effective shielding by the two 1s electrons is 1.74 charge units.
8.35. What are the electron configurations of Ni2+, Re3+, and Ho3+? How many unpaired electron spins are in each of these ions?
Ans. Ni2+ is [Ar]3d8 with 2 unpaired spins; Re3+ is [Xe]4f145d4 with 4 unpaired spins; Ho3+ is [Xe]4f10 with 4 unpaired spins.
8.36. Which shell (letter and n value) would be the first to have a g subshell? Note that two-thirds of the elements have electrons in that shell and higher. Why, then, are there no g electrons?
Ans. O, n = 5. In accord with their high l value (4), g subshells are too high in energy to receive any electrons via the Aufbau principle in known elements in their ground states.
8.37. Predict for the atomic number of the noble gas beyond Rn, if such an element had sufficient stability to be prepared or observed. Assume no g subshells are occupied in the preceding elements.
Ans. 118
8.38. All the lanthanides form stable compounds containing the 3+ cation. Of the few other ionic forms known, Ce forms the most stable 4+ series of ionic compounds and Eu the most stable 2+ series. Account for these unusual ionic forms in terms of their electronic configurations.
Ans. Ce4+ has the stable electron configuration of the noble gas, Xe. Eu2+, with 61 electrons, could have the configuration [Xe]4f7, with the added stability of a half-filled 4f subshell.
8.39. For the gaseous reaction K + F → K+ + F–, ΔH was calculated to be 91 kJ under conditions where the cations and anions were prevented by electrostatic separation from combining. K’s ionization energy is 4.34 eV. What is F’s electron affinity?
Ans. 3.40 eV
8.40. The ionic radii of S2– andTe2– are 1.84 and 2.21 Å. What would you predict for the ionic radius of Se2– and for P3–?
Ans. Since Se falls in between S and Te, it is expected that an intermediate value is correct; observed value is 1.98 Å. Since P is just to the left of S, a slightly larger value is expected; observed value is 2.12 Å.
8.41. Van der Waals radii for S and Cl are 1.85 and 1.80 Å, respectively. What would you predict for the van der Waals radius of Ar?
Ans. Moving toward the right in the same period, the trend is toward smaller size. The observed value is 1.54 Å.
8.42. The first ionization energy of C is 11.2 eV. Would you expect the first ionization energy of Si to be greater or less than this amount?
Ans. Less, because Si is larger; the observed value is 8.1 eV.
8.43. The first ionization energies of Al, Si, and S are 6.0, 8.1, and 10.3 eV. What would be your prediction for the first ionization energy of P?
Ans. A value of about 9.2 eV for P would be halfway between the values for Si and S, but because of the stability of the half-filled subshell, a significantly high energy might be required to remove an electron from P, perhaps higher than for S; observed value is 10.9 eV.
8.44. Several experimenters have attempted to synthesize super-heavy elements by bombarding atoms of the actinide series with heavy ions. While waiting for confirmation and acceptance of the results, some investigators in the early 1970s referred to elements 104 and 105 as eka-hafnium and eka-tantalum. Why were these names chosen?
Ans. Mendeleev had used the prefix eka- (Sanskrit word for first) to name elements whose existence he predicted, applying the prefix to a known element in the same periodic group as the predicted element. His eka-boron, eka-aluminum, and eka-silicon were later discovered, confirmed, and named scandium, gallium, and germanium. Elements 104 and 105 were predicted to have electronic structures analogous to Hf and Ta.
8.45. Consider the trends within the Periodic Table of the Elements. Suppose we knew (previous experience) that the temperature at which manganese melts (melting point) is 1246°C, and that of rhenium is 3186°C. What is the predicted melting point of the synthetic element technetium? Note that these elements are in group VIIB.
Ans. The general rule is that an element will have a physical characteristic between the elements directly above and below (or to the right and left) of that element. Of course, taking an average to the two elements’ characteristics and rigidly declaring it to be the characteristic of the element in question is very risky. However, applying the concept, the melting point for technetium should be about 2216°C. The observed is 2157°C, which gives us a prediction reasonably close to the actual.