1. Chapter 6 Electronic Structure of Atoms
Lecture Presentation
Chapter 6 Electronic Structure of Atoms
(With chapter 7)
6 out of 60
FRQ almost every year
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. 6.1 The Wave Nature of Light
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 ().
1nm = 10-9m 1A=10-10m

4. Waves
Amplitude – height from origin to crest,
The number of waves passing a given point per unit of time is the frequency () or nu.1Hz=1s-1
1MHz= 1x106Hz
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 s–1, respectively.
The speed of light (c) is 3.00  108 m/s c = 

5. If a wave (a) has a wavelength of 2. 0 m and a frequency of 1
If a wave (a) has a wavelength of 2.0 m and a frequency of 1.5 x 108 cycles/s, what are the wavelength and a frequency of a wave (b)?

6. Our bodies are penetrated by X rays but not by visible light
Our bodies are penetrated by X rays but not by visible light. Is this because X rays travel faster than visible light?
Yes: X-rays travel at at faster speeds than visible light.
No: both X-rays and visible light travel at the speed of light, c.
Answer: b

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

8. Speed of Light Problem
1) What is the wavelength of a microwave having a frequency of 3.44 x 109 Hz?
2)A helium-neon laser emits light with a wavelength of 633 nm. What is the frequency of this light?
8.72 x x 10 14

9. Is the wavelength of a microwave longer or shorter than the wavelength of visible light? By how many orders of magnitude do the waves differ in wavelength?

10. 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.

11. Sample Exercise 6.1 Concepts of Wavelength and Frequency
Continued
Practice Exercise 1
A source of electromagnetic radiation produces infrared light. Which of the following could be the wavelength of the light? (a) 3.0 nm (b) 4.7 cm (c) 66.8 m (d) 34.5 μm (e) 16.5 Å
Practice Exercise 2
If one of the waves in the margin represents blue light and the other red light, which wave is which?

12. 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.

13. Sample Exercise 6.2 Calculating Frequency from Wavelength
Continued
Practice Exercise 1
Consider the following three statements: (i) For any electromagnetic radiation, the product of the wavelength and the frequency is a constant. (ii) If a source of light has a wavelength of 3.0 Å, its frequency is 1.0 × 1018 Hz. (iii) The speed of ultraviolet light is greater than the speed of microwave radiation. Which of these three statements is or are true? (a) Only one statement is true. (b) Statements (i) and (ii) are true. (c) Statements (i) and (iii) are true. (d) Statements (ii) and (iii) are true. (e) All three statements are true.
Practice Exercise 2
(a) A laser used in orthopedic spine surgery produces radiation with a wavelength of 2.10 μm. Calculate the frequency of this radiation. (b) An FM radio station broadcasts electromagnetic radiation at a frequency of MHz (megahertz; 1 MHz = 106 s–1). Calculate the wavelength of this radiation. The speed of light is × 108 m ⁄ s to four significant figures.

14. 6.2 Quantized Energy and Photons
Some phenomena can not be explained using a wave model of light.
1. Blackbody radiation: emission of light from hot objects.
2. The photoelectric effect: emission of electrons from metal surfaces on which light shines.
3. Emission spectra: emission of light from electronically excited gas atoms.

15. Particle Nature of Light
The wave nature of light does not explain how an object can glow when its temperature increases(blackbody radiation).
Q: Which area in the photograph corresponds to the highest T?

16. The Nature of Energy—Quanta
Max Planck explained it by assuming that energy comes in packets called quanta (singular: quantum).
Quantum is the smallest amount of energy that can be emitted or absorbed as electromagnetic radiation.

17. Consider the notes that can be played on a piano
Consider the notes that can be played on a piano. In what way is a piano an example of a quantized system?
Multiple notes can be played at once.
Only certain notes exist with none in between.
Answer: b

18. 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.

19. What is the source of the energy that causes electrons to be emitted from the source?

20. In Figure 6.7, will the kinetic energy of an emitted electron equal the energy of the photon that causes its emission?
Yes No
Can’t tell from information given
Answer: b

26. 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

27. Sample Exercise 6.3 Energy of a Photon
Continued
Practice Exercise 1
Which of the following expressions correctly gives the energy of a mole of photons with wavelength λ?
Practice Exercise 2
(a) A laser emits light that has a frequency of 4.69 × 1014 s–1. What is the energy of one photon of this radiation? (b) If the laser emits a pulse containing 5.0 × 1017 photons of this radiation, what is the total energy of that pulse? (c) If the laser emits 1.3 × 10–2 J of energy during a pulse, how many photons are emitted?

28. Do you think that the formation of a rainbow is more a demonstration of the wave-like or particle-like behavior of light?
Wave-like because the colors represent continuous wavelengths.
Particle-like because the colors represent specific energies.
Answer: a

29. 6.3 Atomic Emissions
Another mystery in the early twentieth century involved the emission spectra observed from energy emitted by atoms and molecules.
Bright line spectrum which is basically when u pass energy through an atom or molecules. They stars to glow

30. monochromatic

32. 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.

33. Bohr’s Model
Bohr noted that line spectra of certain elements and assumed electrons were confined to specific energy states. These were called orbits.
Bohr’s line spectra video

35. 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).

36. The Bohr Model
Electrons in permitted orbits have specific, “allowed” energies; these energies will not be radiated from the atom.
3. 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

37. With reference to Figure 6
With reference to Figure 6.6, in what way is the Bohr model for the H atom more like steps than a ramp?
If the climber slips, he or she can fall to the base of the staircase.
Only certain stairs are available.
Two stairs can be occupied simultaneously.
Answer: b

38. 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.

39. Limitations of the Bohr Model

40. What is the significance of the minus sign in front of ΔE in the following equation?
The photon energy is absorbed by the electron transition.
The photon energy is released by the electron transition.
Answer: b

41. What is the relationship between 1/λ and ΔE for a transition of the electron from a lower value of n to a higher one?
1/λ = ΔE/hc
1/λ = –ΔE/hc
1/λ = hc/ΔE
1/λ = hc/–ΔE
Answer: a

42. Sample Exercise 6.4 Electronic Transitions in the Hydrogen Atom
Using Figure 6.12, predict which of these electronic transitions produces the spectral line having the longest wavelength: n = 2 to n = 1, n = 3 to n = 2, or n = 4 to n = 3.
Solution
The wavelength increases as frequency decreases (λ = c ⁄ ν). Hence, the longest wavelength will be associated with the lowest frequency. According to Planck’s equation, E = hv, the lowest frequency is associated with the lowest energy. In Figure 6.12 the shortest vertical line represents the smallest energy change. Thus, the n = 4 to n = 3 transition produces the longest wavelength (lowest frequency) line.

43. Sample Exercise 6.4 Electronic Transitions in the Hydrogen Atom
Continued
Practice Exercise 1
In the top part of Figure 6.11, the four lines in the H atom spectrum are due to transitions from a level for which ni > 2 to the nf = 2 level. What is the value of ni for the blue-green line in the spectrum? (a) 3 (b) 4 (c) 5 (d) 6 (e) 7
Practice Exercise 2
For each of the following transitions, give the sign of ΔE and indicate whether a photon is emitted or absorbed: (a) n = 3 to n = 1; (b) n = 2 to n = 4.

44. 6.4 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.

45. 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:

46. 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.

47. Sample Exercise 6.5 Matter Waves
Continued
Practice Exercise 1
Consider the following three moving objects: (i) a golf ball with a mass of 45.9 g moving at a speed of 50.0 m ⁄ s, (ii) An electron moving at a speed of 3.50 × 105 m ⁄ s, (iii) A neutron moving at a speed of 2.3 × 102 m ⁄ s. List the three objects in order from shortest to longest de Broglie wavelength.
(a) i < iii < ii (b) ii < iii < i (c) iii < ii < i (d) i < ii < iii (e) iii < i < ii
Practice Exercise 2
Calculate the velocity of a neutron whose de Broglie wavelength is 505 pm. The mass of a neutron is given in the table inside the back cover of the text.

48. A baseball pitcher throws a fastball that moves at 95 miles per hour
A baseball pitcher throws a fastball that moves at 95 miles per hour. Does that moving baseball generate matter waves? If so, can we observe them?
Yes, but they are not observable because the matter wavelength is too small for our eyes to detect.
Yes and they are observable because matter waves of macroscopic objects are in the visible region of light.
Matter waves are not produced for macroscopic objects and thus none are observable.
Answer: a

49. 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
In many cases, our uncertainty of the whereabouts of an electron is grater than the size of the atom itself!

50. What is the principal reason we must consider the uncertainty principle when discussing electrons and other subatomic particles but not when discussing our macroscopic world?
The charge of subatomic particles compared to zero charge for macroscopic particles
The small mass and size of subatomic particles compared to the mass and size of macroscopic particles
The small volume occupied by subatomic particles compared to the volume occupied by macroscopic particles
The slower speeds of subatomic particles compared to the speeds of macroscopic particles
Answer: b

51. 6.5 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.

52. 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.
The “electron cloud” is another way of expressing the probability density.

53. Quantum Mechanics

54. What is the difference between stating “The electron is located at a particular point in space” and “There is a high probability that the electron is located at a particular point in space”?
In the first statement, knowing the position of the electron enables us to know its momentum whereas in the second statement, the momentum is already known.
Both statements suggest that the momentum and position are known.
The first statement presents a known position whereas the second statement says we don’t know exactly the position.
Answer: c

55. 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.
There are 3 quantum numbers:

56. Quantum Numbers
NOTE: While we no longer need to know the symbols and numbers associated with these quantum numbers for the AP Test, we still need to understand the concepts related to them to understand and explain the electronic structure of atoms.

57. 1. Principal Quantum Number (n) (Energy Level)
The principal quantum number, n, describes the energy level on which the orbital resides.
Describes the distance from the nucleus and general energy.
The values of n are integers ≥1.n=1,2,3,4, ...
These correspond to the values in the Bohr model. The higher the energy level the larger the atom and farther the e- is from the nucleus.

58. 2. Angular Momentum Quantum Number (l) Sublevel –orbital shape (s,p,d,f)
This quantum number defines the shape of the orbital. These shapes are represented by the symbols. s,p,d, or f.
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.

59. Orbital: a three-dimensional region around the nucleus in which an electron moves and is found 90% of the time.
-Each orbital can hold up to two electrons
The total number of orbitals on a level=n2

60. Angular Momentum Quantum Number (l)

61. 3. Magnetic Quantum Number (ml) Orientation- which orbital on the sublevel
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.

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

63. Other Terms for the AP Test
Orbitals with the same value of n form an electron shell.
Different orbital types within a shell are subshells.

64. What is the difference between an orbit in the Bohr model of the hydrogen atom and an orbital in the quantum mechanical model?
An orbital is composed of some integral number of orbits.
An orbit is a well-defined circular path around the nucleus while an orbital is a wave function that gives the probability of finding the electron at any point in space.
An orbit is a well-defined circular path around the nucleus while an orbital is the object (electron) that is moving around the nucleus.
There is no difference between the definitions of the terms “orbit” and “orbital.” They simply were proposed by different scientists.
Answer: b

65. In Figure 6.18, why is the energy difference between the n = 1 and n = 2 levels much greater than the energy difference between the n = 2 and n = 3 levels?
Because –1/(2)2 and –1/(1)2 are of much greater difference than –1/(3)2 and –1/(2)2 in Bohr’s equation that describes the hydrogen atom.
Because the Rydberg in Bohr’s equation constant is higher for n = 1 to n = 2 transitions.
Because the energy difference between the 1st and 2nd energy level of hydrogen is smaller than the energy difference between the 2nd and 3rd energy level.
Because the electron in the hydrogen atom never occupies energy levels of higher than n = 1.
Answer: a

66. 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

67. Sample Exercise 6.6 Subshells of the Hydrogen Atom
Continued
Practice Exercise 2
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.

68. 6.6 Representations of Orbitals 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.

69. Comparing the radial probability distributions for the 1s, 2s, and 3s orbitals reveals three trends:
1. For an ns orbital, the number of peaks is n.
2. For an ns orbital, the number of nodes (where there is zero probability of finding an electron) is n – 1.
3. As n increases, the electron density is more spread out and there is a greater probability of finding an electron further from the nucleus.

70. How many maxima would you expect to find in the radial probability function for the 4s orbital of the H atom? How many nodes would you expect in this function?

71. p Orbitals
The value of l for p orbitals is 1.2nd energy level and above.
They have two lobes with a node between them.
p sublevels can hold up to 6 electrons.

72. d Orbitals
The value of l for a d orbital is 2.3rd energy level and above.
Four of the five d orbitals have four lobes; the other resembles a p orbital with a doughnut around the center.

73. f Orbitals Very complicated shapes (not shown in text)
Seven equivalent orbitals in a sublevel
l = 3. 4th energy level and above.

74. 6.7 Many-Electron Atoms 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 (same energy).

75. In a many-electron atom, can we predict unambiguously whether the 4s orbital is lower or higher in energy than the 3d orbitals?
Yes, the figure shows that 4s is lower than 3d.
Yes, the n quantum number 4 is higher in energy than the quantum number 3.
No, because this will be different depending on the identity of the atom.
Answer: c

76. 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.)

77. 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.

78. 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.

79. 6.8 Electron Configurations

80. 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.

82. 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.

84. 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

85. 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

86. 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

89. 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.

90. 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.

91. Sample Exercise 6.7 Orbital Diagrams and Electron Configurations
Continued
Practice Exercise 1
How many of the elements in the second row of the periodic table (Li through Ne) will have at least one unpaired electron in their electron configurations? (a) 3 (b) 4 (c) 5 (d) 6 (e) 7
Practice Exercise 2
(a) Write the electron configuration for silicon, element 14, in its ground state. (b) How many unpaired electrons does a ground-state silicon atom possess?

92. 6.9 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)

95. Some Anomalies
Some irregularities occur when there are enough electrons to half-fill s and d orbitals on a given row.
Stability is about minimizing repulsions and lower potential energy.

96. Chromium as an Anomaly
For instance, the electron configuration for chromium is
[Ar] 4s1 3d5
rather than the expected
[Ar] 4s2 3d4.
These anomalies occur in f-block atoms with f and d orbitals, as well.

103. Are the atoms weakly attracted to a magnetic field?
Diamagnetic – electrons are all paired so the substance is NOT attracted to a magnetic field. Zn
Paramagnetic – unpaired electrons cause the substance to attracted to a magnetic field. Cr

104. Is Carbon paramagnetic?
1s2 2s22p2
Yes! It has 2 unpaired electrons.

105. 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.

106. Sample Exercise 6.8 Electron Configurations for a Group
Continued
Practice Exercise 1
A certain atom has an ns2np6 electron configuration in its outermost occupied shell. Which of the following elements could it be? (a) Be (b) Si (c) I (d) Kr (e) Rb
Practice Exercise 2
Which family of elements is characterized by an ns2np2 electron configuration in the outermost occupied shell?

107. 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.

108. 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

109. 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).

110. Sample Integrative Exercise Putting Concepts Together
Boron, atomic number 5, occurs naturally as two isotopes, 10B and 11B, with natural abundances of 19.9% and 80.1%, respectively. (a) In what ways do the two isotopes differ from each other? Does the electronic configuration of 10B differ from that of 11B? (b) Draw the orbital diagram for an atom of 11B. Which electrons are the valence electrons? (c) Indicate three major ways in which the 1s electrons in boron differ from its 2s electrons. (d) Elemental boron reacts with fluorine to form BF3, a gas. Write a balanced chemical equation for the reaction of solid boron with fluorine gas. (e) ΔHf° for BF3(g) is − kJ ⁄ mol. Calculate the standard enthalpy change in the reaction of boron with fluorine. (f) Will the mass percentage of F be the same in 10BF3 and 11BF3? If not, why is that the case?
Solution
(a) The two isotopes of boron differ in the number of neutrons in the nucleus (Sections 2.3 and 2.4) Each of the isotopes contains five protons, but 10B contains five neutrons, whereas 11B contains six neutrons. The two isotopes of boron have identical electron configurations, 1s22s22p1, because each has five electrons.
(b) The complete orbital diagram is 1s 2s 2p
The valence electrons are the ones in the outermost occupied shell, the 2s2 and 2p1 electrons. The 1s2 electrons constitute the core electrons, which we represent as [He] when we write the condensed electron configuration, [He]2s22p1.

111. Sample Integrative Exercise Putting Concepts Together
Continued
(c) The 1s and 2s orbitals are both spherical, but they differ in three important respects: First, the 1s orbital is lower in energy than the 2s orbital. Second, the average distance of the 2s electrons from the nucleus is greater than that of the 1s electrons, so the 1s orbital is smaller than the 2s. Third, the 2s orbital has one node, whereas the 1s orbital has no nodes (Figure 6.19).

112. Sample Integrative Exercise Putting Concepts Together
Continued
(d) The balanced chemical equation is
2 B(s) + 3 F2(g) → 2 BF3(g)
(e) ΔH ° = 2(−1135.6) − [0 + 0] = − kJ. The reaction is strongly exothermic.
(f) As we saw in Equation 3.10 (Section 3.3), the mass percentage of an element in a substance depends on the formula weight of the substance. The formula weights of 10BF3 and 11BF3 are different because of the difference in the masses of the two isotopes (the isotope masses of 10B and 11B are and amu, respectively). The denominators in Equation 3.10 would therefore be different for the two isotopes, whereas the numerators would remain the same.