1. Chapter 6 Electronic Structure of Atoms
Lecture Presentation
Chapter 6 Electronic Structure of Atoms
James F. Kirby
Quinnipiac University
Hamden, CT

2. Electronic Structure
This chapter is all about electronic structure—the arrangement and energy of electrons.
It may seem odd to start by talking about waves. However, extremely small particles have properties that can only be explained in this manner!

3. Waves
To understand the electronic structure of atoms, one must understand the nature of electromagnetic radiation.
The distance between corresponding points on adjacent waves is the wavelength ().

4. Waves
The number of waves passing a given point per unit of time is the frequency ().
For waves traveling at the same velocity, the longer the wavelength, the smaller the frequency.
If the time associated with the lines to the left is one second, then the frequencies would be 2 s–1 and 4 s–1, respectively.

5. Electromagnetic Radiation
All electromagnetic radiation travels at the same velocity: The speed of light (c) is 3.00  108 m/s.
c = 

6. Sample Exercise 6.1 Concepts of Wavelength and Frequency
Two electromagnetic waves are represented in the margin. (a) Which wave has the higher frequency? (b) If one wave represents visible light and the other represents infrared radiation, which wave is which?
Solution
(a) Wave 1 has a longer wavelength (greater distance between peaks). The longer the wavelength, the lower the frequency (ν = c ⁄ λ). Thus, Wave 1 has the lower frequency, and Wave 2 has the higher frequency.
(b) The electromagnetic spectrum (Figure 6.4) indicates that infrared radiation has a longer wavelength than visible light. Thus, Wave 1 would be the infrared radiation.

7. Sample Exercise 6.2 Calculating Frequency from Wavelength
The yellow light given off by a sodium vapor lamp used for public lighting has a wavelength of 589 nm. What is the frequency of this radiation?
Solution
Analyze We are given the wavelength, λ, of the radiation and asked to calculate its frequency, ν.
Plan The relationship between the wavelength and the frequency is given by Equation 6.1. We can solve for ν and use the values of λ and c to obtain a numerical answer. (The speed of light, c, is 3.00 × 108 m ⁄ s to three significant figures.)
Solve Solving Equation 6.1 for frequency gives (ν = c ⁄ λ). When we insert the values for c and λ, we note that the units of length in these two quantities are different. We can convert the Wavelength from nanometers to meters, so the units cancel:
Check The high frequency is reasonable because of the short wavelength. The units are proper because frequency has units of “per second,” or s–1.

8. The Nature of Energy
The wave nature of light does not explain how an object can glow when its temperature increases.

9. The Nature of Energy—Quanta
Max Planck explained it by assuming that energy comes in packets called quanta (singular: quantum).

10. The Photoelectric Effect
Einstein used quanta to explain the photoelectric effect.
Each metal has a different energy at which it ejects electrons. At lower energy, electrons are not emitted.
He concluded that energy is proportional to frequency:
E = h
where h is Planck’s constant,  10−34 J∙s.

11. Sample Exercise 6.3 Energy of a Photon
Calculate the energy of one photon of yellow light that has a wavelength of 589 nm.
Solution
Analyze Our task is to calculate the energy, E, of a photon, given λ = 589 nm.
Plan We can use Equation 6.1 to convert the wavelength to frequency:
ν = c ⁄ λ
We can then use Equation 6.3 to calculate energy:
E = hν
Solve The frequency, ν, is calculated from the given wavelength, as shown in Sample Exercise 6.2:
ν = c ⁄ λ = 5.09 × 1014 s–1
The value of Planck constant, h, is given both in the text and in the table of physical constants on the inside back cover of the text, and so we can easily calculate E:
E = (6.626 × 10–34 J-s)(5.09 × 1014 s–1) = 3.37 × 10–19 J
Comment If one photon of radiant energy supplies 3.37 × 10–19 J, we calculate that one mole of these photons will supply:
(6.02 × 1023 photons ⁄ mol)(3.37 × 10–19 J ⁄ photon) = 2.03 × 105 J ⁄ mol

12. Atomic Emissions
Another mystery in the early twentieth century involved the emission spectra observed from energy emitted by atoms and molecules.

13. Continuous vs. Line Spectra
For atoms and molecules, one does not observe a continuous spectrum (the “rainbow”), as one gets from a white light source.
Only a line spectrum of discrete wavelengths is observed. Each element has a unique line spectrum.

14. The Hydrogen Spectrum
Johann Balmer (1885) discovered a simple formula relating the four lines to integers.
Johannes Rydberg advanced this formula.
Neils Bohr explained why this mathematical relationship works.

15. The Bohr Model
Niels Bohr adopted Planck’s assumption and explained these phenomena in this way:
Electrons in an atom can only occupy certain orbits (corresponding to certain energies).

16. The Bohr Model
Electrons in permitted orbits have specific, “allowed” energies; these energies will not be radiated from the atom.
Energy is only absorbed or emitted in such a way as to move an electron from one “allowed” energy state to another; the energy is defined by
E = h

17. The Bohr Model
The energy absorbed or emitted from the process of electron promotion or demotion can be calculated by the equation
E = −hcRH
( ) 1
nf2
ni2 –
where RH is the Rydberg constant,  107 m−1, and ni and nf are the initial and final energy levels of the electron.

18. Limitations of the Bohr Model
It only works for hydrogen!
Classical physics would result in an electron falling into the positively charged nucleus. Bohr simply assumed it would not!
Circular motion is not wave-like in nature.

19. Important Ideas from the Bohr Model
Points that are incorporated into the current atomic model include the following:
Electrons exist only in certain discrete
energy levels.
2) Energy is involved in the transition of
an electron from one level to another.

20. The Wave Nature of Matter
Louis de Broglie theorized that if light can have material properties, matter should exhibit wave properties.
He demonstrated that the relationship between mass and wavelength was
 = h mv
The wave nature of light is used to produce this electron micrograph.

21. Sample Exercise 6.5 Matter Waves
What is the wavelength of an electron moving with a speed of 5.97 × 106 m ⁄ s? The mass of the electron is 9.11 × 10–31 kg.
Solution
Analyze We are given the mass, m, and velocity, v, of the electron, and we must calculate its de Broglie wavelength, λ.
Plan The wavelength of a moving particle is given by Equation 6.8, so λ is calculated by inserting the known quantities h, m, and v. In doing so, however, we must pay attention to units.
Solve Using the value of Planck constant:
h = × 10–34 J-s
we have the following:

22. Sample Exercise 6.5 Matter Waves
Continued
Comment By comparing this value with the wavelengths of electromagnetic radiation shown in Figure 6.4, we see that the wavelength of this electron is about the same as that of X rays.

23. The Uncertainty Principle
Heisenberg showed that the more precisely the momentum of a particle is known, the less precisely is its position is known:
(x) (mv)  h 4

24. Quantum Mechanics
Erwin Schrödinger developed a mathematical treatment into which both the wave and particle nature of matter could be incorporated.
This is known as quantum mechanics.

25. Quantum Mechanics
The solution of Schrödinger’s wave equation is designated with a lowercase Greek psi ().
The square of the wave equation, 2, gives the electron density, or probability of where an electron is likely to be at any given time.

26. Quantum Numbers
Solving the wave equation gives a set of wave functions, or orbitals, and their corresponding energies.
Each orbital describes a spatial distribution of electron density.
An orbital is described by a set of three quantum numbers.

27. Principal Quantum Number (n)
The principal quantum number, n, describes the energy level on which the orbital resides.
The values of n are integers ≥ 1.
These correspond to the values in the Bohr model.

28. Angular Momentum Quantum Number (l)
This quantum number defines the shape of the orbital.
Allowed values of l are integers ranging from 0 to n − 1.
We use letter designations to communicate the different values of l and, therefore, the shapes and types of orbitals.

29. Angular Momentum Quantum Number (l)

30. Magnetic Quantum Number (ml)
The magnetic quantum number describes the three-dimensional orientation of the orbital.
Allowed values of ml are integers ranging from −l to l:
−l ≤ ml ≤ l
Therefore, on any given energy level, there can be up to 1 s orbital, 3 p orbitals, 5 d orbitals, 7 f orbitals, and so forth.

31. Magnetic Quantum Number (ml)
Orbitals with the same value of n form an electron shell.
Different orbital types within a shell are subshells.

32. Sample Exercise 6.6 Subshells of the Hydrogen Atom
(a) Without referring to Table 6.2, predict the number of subshells in the fourth shell, that is, for n = 4. (b) Give the label for each of these subshells. (c) How many orbitals are in each of these subshells?
Solution
Analyze and Plan We are given the value of the principal quantum number, n. We need to determine the allowed values of l and ml for this given value of n and then count the number of orbitals in each subshell.
Solve There are four subshells in the fourth shell, corresponding to the four possible values of l (0, 1, 2, and 3).
These subshells are labeled 4s, 4p, 4d, and 4f. The number given in the designation of a subshell is the principal quantum number, n; the letter designates the value of the angular momentum quantum number, l: for l = 0, s; for l = 1, p; for l = 2, d; for l = 3, f.
There is one 4s orbital (when l = 0, there is only one possible value of ml: 0). There are three 4p orbitals (when l = 1, there are three possible values of ml: 1, 0, −1). There are five 4d orbitals (when l = 2, there are five allowed values of ml: 2, 1, 0, −1, −2). There are seven 4f orbitals (when l = 3, there are seven permitted values of ml: 3, 2, 1, 0, −1, −2, −3).
Practice Exercise 1
An orbital has n = 4 and ml = –1. What are the possible values of l for this orbital?
(a) 0, 1, 2, 3 (b) −3, −2, −1, 0, 1, 2, 3 (c) 1, 2, 3 (d) −3, −2 (e) 1, 2, 3, 4

33. Sample Exercise 6.6 Subshells of the Hydrogen Atom
Continued
Practice Exercise 2
(a) What is the designation for the subshell with n = 5 and l = 1? (b) How many orbitals are in this subshell? (c) Indicate the values of ml for each of these orbitals.

34. s Orbitals The value of l for s orbitals is 0.
They are spherical in shape.
The radius of the sphere increases with the value of n.

35. s Orbitals For an ns orbital, the number of peaks is n.
For an ns orbital, the number of nodes (where there is zero probability of finding an electron) is n – 1.
As n increases, the electron density is more spread out and there is a greater probability of finding an electron further from the nucleus.

36. p Orbitals The value of l for p orbitals is 1.
They have two lobes with a node between them.

37. d Orbitals The value of l for a d orbital is 2.
Four of the five d orbitals have four lobes; the other resembles a p orbital with a doughnut around the center.

38. f Orbitals Very complicated shapes (not shown in text)
Seven equivalent orbitals in a sublevel
l = 3

39. Energies of Orbitals—Hydrogen
For a one-electron hydrogen atom, orbitals on the same energy level have the same energy.
Chemists call them degenerate orbitals.

40. Energies of Orbitals— Many-electron Atoms
As the number of electrons increases, so does the repulsion between them.
Therefore, in atoms with more than one electron, not all orbitals on the same energy level are degenerate.
Orbital sets in the same sublevel are still degenerate.
Energy levels start to overlap in energy (e.g., 4s is lower in energy than 3d.)

41. Spin Quantum Number, ms
In the 1920s, it was discovered that two electrons in the same orbital do not have exactly the same energy.
The “spin” of an electron describes its magnetic field, which affects its energy.
This led to the spin quantum number, ms.
The spin quantum number has only two allowed values, +½ and –½.

42. Pauli Exclusion Principle
No two electrons in the same atom can have exactly the same energy.
Therefore, no two electrons in the same atom can have identical sets of quantum numbers.
This means that every electron in an atom must differ by at least one of the four quantum number values: n, l, ml, and ms.

43. Electron Configurations
The way electrons are distributed in an atom is called its electron configuration.
The most stable organization is the lowest possible energy, called the ground state.
Each component consists of
a number denoting the energy level;
4p5

44. Electron Configurations
The way electrons are distributed in an atom is called its electron configuration.
The most stable organization is the lowest possible energy, called the ground state.
Each component consists of
a number denoting the energy level;
a letter denoting the type of orbital;
4p5

45. Electron Configurations
The way electrons are distributed in an atom is called its electron configuration.
The most stable organization is the lowest possible energy, called the ground state.
Each component consists of
a number denoting the energy level;
a letter denoting the type of orbital;
a superscript denoting the number of electrons in those orbitals.
4p5

46. Orbital Diagrams Each box in the diagram represents one orbital.
Half-arrows represent the electrons.
The direction of the arrow represents the relative spin of the electron.

47. Hund’s Rule
“For degenerate orbitals, the lowest energy is attained when the number of electrons with the same spin is maximized.”
This means that, for a set of orbitals in the same sublevel, there must be one electron in each orbital before pairing and the electrons have the same spin, as much as possible.

48. Sample Exercise 6.7 Orbital Diagrams and Electron Configurations
Draw the orbital diagram for the electron configuration of oxygen, atomic number 8. How many unpaired electrons does an oxygen atom possess?
Solution
Analyze and Plan Because oxygen has an atomic number of 8, each oxygen atom has eight electrons. Figure 6.25 shows the ordering of orbitals. The electrons (represented as half arrows) are placed in the orbitals (represented as boxes) beginning with the lowest-energy orbital, the 1s. Each orbital can hold a maximum of two electrons (the Pauli exclusion principle). Because the 2p orbitals are degenerate, we place one electron in each of these orbitals (spin-up) before pairing any electrons (Hund’s rule).
Solve Two electrons each go into the 1s and 2s orbitals with their spins paired. This leaves four electrons for the three degenerate 2p orbitals. Following Hund’s rule, we put one electron into each 2p orbital until all three orbitals have one electron each. The fourth electron is then paired up with one of the three electrons already in a 2p orbital, so that the orbital diagram is
The corresponding electron configuration is written 1s22s22p4. The atom has two unpaired electrons.

49. Sample Exercise 6.8 Electron Configurations for a Group
What is the characteristic valence electron configuration of the group 7A elements, the halogens?
Solution
Analyze and Plan We first locate the halogens in the periodic table, write the electron configurations for the first two elements, and then determine the general similarity between the configurations.
Solve The first member of the halogen group is fluorine (F, element 9). Moving backward from F, we find that the noble-gas core is [He]. Moving from He to the element of next higher atomic number brings us to Li, element 3. Because Li is in the second period of the s block, we add electrons to the 2s subshell. Moving across this block gives 2s2. Continuing to move to the right, we enter the p block. Counting the squares to F gives 2p5. Thus, the condensed electron configuration for fluorine is
F: [He]2s22p5
The electron configuration for chlorine, the second halogen, is
Cl: [Ne]3s23p5
From these two examples, we see that the characteristic valence electron configuration of a halogen is ns2np5, where n ranges from 2 in the case of fluorine to 6 in the case of astatine.

50. Condensed Electron Configurations
Elements in the same group of the periodic table have the same number of electrons in the outer most shell. These are the valence electrons.
The filled inner shell electrons are called core electrons. These include completely filled d or f sublevels.
We write a shortened version of an electron configuration using brackets around a noble gas symbol and listing only valence electrons.

51. Periodic Table We fill orbitals in increasing order of energy.
Different blocks on the periodic table correspond to different types of orbitals: s = blue, p = pink (s and p are representative elements); d = orange (transition elements); f = tan (lanthanides and actinides, or inner transition elements)

52. Sample Exercise 6.9 Electron Configurations from the Periodic Table
(a) Based on its position in the periodic table, write the condensed electron configuration for bismuth, element 83. (b) How many unpaired electrons does a bismuth atom have?
Solution
(a) Our first step is to write the noble-gas core. We do this by locating bismuth, element 83, in the periodic table. We then move backward to the nearest noble gas, which is Xe, element 54. Thus, the noble-gas core is [Xe].
Next, we trace the path in order of increasing atomic numbers from Xe to Bi. Moving from Xe to Cs, element 55, we find ourselves in period 6 of the s block. Knowing the block and the period identifies the subshell in which we begin placing outer electrons, 6s. As we move through the s block, we add two electrons: 6s2.
As we move beyond the s block, from element 56 to element 57, the curved arrow below the periodic table reminds us that we are entering the f block. The first row of the f block corresponds to the 4f subshell. As we move across this block, we add 14 electrons: 4f 14.
With element 71, we move into the third row of the d block. Because the first row of the d block is 3d, the second row is 4d and the third row is 5d. Thus, as we move through the ten elements of the d block, from element 71 to element 80, we fill the 5d subshell with ten electrons: 5d10.

53. Sample Exercise 6.9 Electron Configurations from the Periodic Table
Continued
Moving from element 80 to element 81 puts us into the p block in the 6p subshell. (Remember that the principal quantum number in the p block is the same as that in the s block.) Moving across to Bi requires three electrons: 6p3. The path we have taken is

54. Sample Exercise 6.9 Electron Configurations from the Periodic Table
Continued
Putting the parts together, we obtain the condensed electron configuration: [Xe]6s24f 145d106p3. This configuration can also be written with the subshells arranged in order of increasing principal quantum number: [Xe]4f 145d106s26p3.
Finally, we check our result to see if the number of electrons equals the atomic number of Bi, 83: Because Xe has 54 electrons (its atomic number), we have = 83. (If we had 14 electrons too few, we would realize that we have missed the f block.)
(b) We see from the condensed electron configuration that the only partially occupied subshell is 6p. The orbital diagram representation for this subshell is
In accordance with Hund’s rule, the three 6p electrons occupy the three 6p orbitals singly, with their spins parallel. Thus, there are three unpaired electrons in the bismuth atom.
Practice Exercise 1
A certain atom has an [noble gas]5s24d105p4 electron configuration. Which element is it?
(a) Cd (b) Te (c) Sm (d) Hg (e) More information is needed
Practice Exercise 2
Use the periodic table to write the condensed electron configuration for (a) Co (element 27), (b) In (element 49).

55. Some Anomalies
Some irregularities occur when there are enough electrons to half-fill s and d orbitals on a given row.

56. Chromium as an Anomaly
For instance, the electron configuration for chromium is
[Ar] 4s1 3d5
rather than the expected
[Ar] 4s2 3d4.
This occurs because the 4s and 3d orbitals are very close in energy.
These anomalies occur in f-block atoms with f and d orbitals, as well.