1. Electrons in Atoms and Periodic Properties
Chapter 7
Electrons in Atoms and
Periodic Properties

2. Light is electro/magnetic (EM) radiation

3. Terms
Wavelength () is the distance from crest to crest or trough to trough on a wave.
The frequency () of a wave expresses the number times a wave passes a given point in some unit of time.
Amplitude of a wave is the height of the crest or depth of the trough with respect to the center line of the wave.

4. Visible light is a small portion of the entire spectrum of EM radiation
Increasing Energy and Frequency
(decreasing wavelength)

5. EM Spectrum

6. EM Spectrum and Classification

7. Electromagnetic Radiation
The mathematical relationship between wavelength and frequency for EM radiation is:
c = l۰n
c = 2.998E8 m/s (speed of light in a vacuum)
l = wavelength (in meters) (Greek letter, lambda)
n = frequency (in Hertz or s-1) (Greek letter, nu)

8. Electromagnetic Radiation
Using the mathematical relationship between wavelength and frequency:
c = ln
Calculate the wavelength associated with Montana Tech’s student radio station, KMSM-FM which broadcasts at a frequency of MHz.

9. KMSM-FM broadcasts at a frequency of
107.1 MHz

10. What is the frequency of light, in hertz, if it has a wavelength of 1
What is the frequency of light, in hertz, if it has a wavelength of 1.05 x 10-7 m and is traveling in vacuum?  What portion of the electromagnetic spectrum does this “light” belong to?
Electromagnetic radiation slows down as it travels through matter.  What fraction and percentage of “c” is the velocity of 475 nm light, if its frequency is 6.00 x 1014 Hz?

11. Behavior of Waves
Waves refract or bend when they pass from one medium to another with different densities.
Diffraction is the bending of electromagnetic radiation as it passes around the edge of an object or through narrow openings.
Interference is the interaction of waves that results in either reinforcing their amplitudes or canceling them out.

12. Refraction
Refraction – the change in direction of a beam of Electromagnetic Radiation (light) as it passes from one medium into another.

13. Diffraction and Interference

14. Laser light is monochromatic (one wavelength) and spatially and temporally coherent

15. He-Ne Laser Spectrum
632.8 nm
Wavelength (nm)

16. Missing lines of light (dark lines) are found in solar spectra
Missing lines of light (dark lines) are found in solar spectra. In distant stars these lines are shifted towards longer wavelengths (red shift). These red shifts are caused by the “Doppler effect”.
What other places in your regular life to do find examples of the Doppler effect?

17. Fraunhofer Lines (dark) in the Solar Spectrum
Shifts to lower energies (red-shift) of these lines suggested to Hubble et al. that more distance galaxies were moving away more rapidly. This would be the expected result assuming the universe began with a Big Bang

18. Redshift calculations
Using wavelength = v/c
Using frequency = v/c
’ represents the longer wavelength and ’ represents the lower frequency

19. Try this example problem:
A spectral line for atomic hydrogen (H) is known to occur at 485 nm. Studying the stars in a distant galaxy, it is noted that the spectral line now appears at 558 nm ( a shift to longer wavelength). At what percentage of “c” is the galaxy moving away from earth? What is the velocity of the galaxy relative to earth?

20. Atomic Emission Spectra…the light from atoms

22. Types of Spectra
Atomic emission spectra consist of bright lines on a dark background.
Atomic absorption spectra consist of characteristic series of dark lines produced when free gaseous atoms are illuminated by external sources of radiation.

23. Emission versus Absorption Spectra

24. Absorption Spectra
How do the spectra change in going from H to Ne? Why?

25. Black Body Radiation and the end of classical physics…the UV catastrophe.

26. Quantum Theory
From work on BB radiation, Max Planck proposed that light can have both wavelike and particle-like properties.
A quantum is the smallest discrete quantity of a particular form of energy.
Particles of radiant energy are known as Photon.
Quantum theory is based on the idea that energy is absorbed and emitted in discrete quanta…at least in small (nm) spaces.

27. Quantum Theory
Something that is quantized has values that are restricted to whole-number multiples of a specific base value.
The energy of a quantum of radiation is:
E = h where h is Planck’s constant
h = x J•s
Or E = hc/

28. Particle Nature of Light
Each packet of electromagnetic radiation energy is called a quantum.
Einstein called the packets photons. A mole of photons is called an Einstein.

29. What is the wavelength of a photon, in vacuum, with an energy of
1.25 x J? 
What portion of the electromagnetic spectrum does this photon belong to?
E = hn = hc/l
l = hc/E = (6.626E-34 J s)(2.998E8 m/s)/(1.25E-20 J)
l = 1.59E-5 m

30. EM Spectrum

31. Planck and Einstein (1929) in happier days.
Plank was among the few that recognized the significance of Special Relativity. But did not accept that the “quantum” was a real phenomenon.

32. A new scientific truth does not triumph by convincing its opponents and making them see the light, but rather because its opponents eventually die, and a new generation grows up that is familiar with it.
Max Planck

33. Photoelectric Effect
The photoelectric effect is the release of electrons from a metal as a result of electromagnetic radiation.
The photoelectric effect can be explained if the electromagnetic radiation is treat as being composed of tiny particles (wave packets) called photons.

34. Only electrons with sufficient energy will displace electrons
Only electrons with sufficient energy will displace electrons. This energy (or threshold frequency) is known as the Work Function (F)

35. F = hn0 = Ebound electron
The work function depends on the type of metal. If the element surface is irradiated with light of frequency greater than the threshold, the excess energy appears in the kinetic energy of the electron.
KEelectron = hn – hn0 = hv - F

36. Work Function for various elements (eV)

37. Problem
When light of frequency 1.30E15 s-1 shines on the surface of cesium metal, electrons are ejected with a maximum kinetic energy of
5.2E-19J.
Calculate the wavelength of this light.
b) Calculate the work function for cesium.
c) Calculate the longest wavelength of light that will displace electrons.

42. The Hydrogen Spectrum
Johannes Rydberg revised Balmer’s equation to describe the complete hydrogen spectrum.
N1 is a whole number that remains fixed for a series of calculations in which n2 is also a whole number with values of n1+1, n1+2,… for successive line in the spectrum.

43. Problem
What is the wavelength of the line in the visible spectrum corresponding to n1 = 2 and n2 = 4?
1/l = 1.097E-2 nm-1(1/22 – 1/42)
= 1.097E-2 nm-1(3/16)
= 2.057E-3 nm-1
l = 486 nm

44. The Bohr Model for Electrons in the Hydrogen Atom
The electron in a hydrogen atom occupies a discrete energy level and may exist only in the available energy levels.
The electron may move between energy levels by either absorbing or emitting specific amounts of energy.
Each energy level is designated by a specific value for n, called the principal quantum number.

45. The Bohr model of the hydrogen atom places electrons in concentric orbits with certain “allowed” orbital energies for the electrons in the field of the nucleus (~Ze).
Z is the atomic number, and e is the fundamental charge (1.6E-19 C)

46. Balmer Series of The Hydrogen Emission Spectrum

47. Energy of Electronic Transitions
Neil Bohr derived the following formula for the energy levels of hydrogen-like orbitals
En = - Z2e4m
8e0n2h2
Z is the atomic number, is the vacuum electric permittivity
m and e is the mass and charge of the electron.

48. Hydrogen Spectrum
An energy level is an allowed state that an electron can occupy in an atom.
Movements of electrons between energy levels are called electronic transitions.

49. Mathematically in the Bohr model, the energy of each orbital is:
En = (- 2.18E-18J) (1/n2)
Where n= 1, 2, 3,…∞
The constant in the equation equals: Rhc
Where R = E7 m-1 (Rydberg constant)
Note that the orbital energies are mathematically all negative (< 0) in energy ( corresponding to bound electronic states)

50. The Bohr model of the hydrogen atom places electrons in concentric orbits with certain “allowed” orbital energies for the electrons in the field of the nucleus (~Ze).
Note pattern of orbital spacings…

52. Problem
What is the minimum wavelength of light that can ionize a hydrogen atom…in the gas phase? (Big Hint…assume nf = ∞).

54. Electronic States
The lowest energy level (n) available to an electron in an atom is its ground state.
An excited state of an electron in an atom (or molecule) is any energy state above the ground state.

55. Particle or Waves?
If electromagnetic radiation behaves as a particle, de Broglie reasoned, why couldn’t a particle in motion, such as an electron, behave as a wave?
de Broglie’s Equation
 = h/mu (m in kg and u in m/s)

56. De Broglie Wave Equation

57. Wavelengths of matter
= h/mv = h/r
Calculate the wavelength of an electron (mass = 9.109E-28 g) in an electron microscope moving at 3.62E6 m/s.

58. De Broglie Wave Equation

59. Wave Equations
Wave equation for a standing wave:
L = nl/2
The wave equation for electrons is called the Schrödinger Equation
ĤY = EY
Where Y(psi) is a wave function, and Ĥ is the Hamiltonian operator:
Ĥ = -iħ (t)

60. Wave equations describe a “quantized” electron (sec 3.5 & 3.6)
Mathematical equations known as wave equations are use to describe probabilities of finding electrons around the nucleus.
The wave equations for electrons yield three quantum numbers (n, l, ml) that define the energy, shape, and orientation for the electron orbitals.
A fourth quantum number (ms) gives the quantized (relativistic) spin of an electron in an orbital.

61. l (ang. momentum) positive less than n 0, 1,..(n-1)
Quantum Numbers (QN)
QN Restrictions Range
n (principal) positive integers 1, 2, …, 
l (ang. momentum) positive less than n 0, 1,..(n-1)
ml (magnetic) integers between –l and l
s (spin) half-integers –½, ½ ; –½, ½

62. Possible sets of quantum number for n = 1, 2, 3
As n increases the possible number of orbitals increases by n2. Only two electrons can occupy a single orbital, thus there are 2n2 electrons per n-shell

63. Quantum Numbers

64. Shell and Orbital Names
Value of n 1 2 3 5
Shell Name* K L M N O
Value of l 1 2 3 4
Orbital Name(s) s p d f g
*Primarily used in X-ray spectroscopy

65. Practice n=3, l=2 n=3, l=0 n=4, l=3 n=5, l=2, ml =1
What orbitals and how many electrons are identified by the following combinations of quantum numbers?
n=3, l=2
n=3, l=0
n=4, l=3
n=5, l=2, ml =1

66. Problem
What are the letter designations of all the subshells in the n = 5 energy level or shell? What is total number of orbitals in the n = 5 shell?

68. Problem
For the following sets of quantum numbers, determine which describe actual orbits, and list why others are non-existent.
n l ml s
(a)
(b) ½
(c) ½
(d) ½

69. Shape and Sizes of Orbitals
Psi squared, 2, defines the probability of an electron in some region of space around the nucleus .
A radial distribution plot is a graphical representation of the probability of finding an electron in a thin spherical layer near the nucleus of an atom.

70. The shape of an atomic orbital is determine by “l” the angular momentum quantum number
node + -
l=1; p-orbital
l=2; d-orbital
l=0; s-orbital

71. Probability Electron Density for 1s (l=0) Orbital

72. rmp r90 Electron density in the 1s orbital of the hydrogen atom
The probability of the electron density is a function of the square of the wave-function (Y2)

73. Probability Density of s-Orbitals

74. The 2p (l=1) Orbitals

75. The Five 3d (l=2) Orbitals

76. Assigning Quantum Number for Electrons
Pauli’s exclusion principle - no two electrons in an atom may have the same set of four quantum numbers.
An orbital can only hold two electrons and they must have opposite spins.

78. Aufbau Principle
As protons are added one by one to the nucleus to build up the elements, electrons are similarly added to these hydrogen-like orbitals.

79. Orbital Energy Levels for Hydrogen-Like Atoms3p 3d 2s 2p 1s

80. Many Electron Atoms
The presence of more than one electron in an atom effect the relative orbital energies which only depend on n in the one-electron orbital (hydrogen-like atom).
In the many electron atom, orbital energies depend on both n and l (n+l rule).
In general, for a given value of n, the lower the value of l, the lower in energy the orbital subshell (4s < 3d

81. The effective nuclear charge (Zeff) felt by the 2s electron in lithium is less than the effective nuclear charge in hydrogen or helium.
The presence of electron density in the 1s orbital “screens” the outer electron in the 2s orbital from the nuclear charge. This results a lowering in the outer shell. orbital energy

82. Orbital Energies in Multi-electron Atoms
Note the spacing of the orbitals and that the ordering of the energies depends on n + l.

83. The 2s orbital is lower in energy than the 2p orbital because there is some 2s density closer to the nucleus

84. Terminology
Orbitals that have the exact same energy level are said to be degenerate (e.g 2px and 2py).
Core electrons are those in the filled, inner shells in an atom and are not involved in chemical reactions.
Valence electrons are those in the outermost shell of an atom and have the most influence on the atom’s chemical reactivity.

85. Electron Configuration
1s s p
H: 1s1
He: 1s2
Li: 1s22s1
Be: 1s22s2
B: 1s22s22p1

86. Orbital diagrams and electron configurations describe how the electrons fill orbitals
Hund’s rules state that the lowest energy electron configuration will have a maximum in unpaired electrons. Note the filling of the 2p orbitals in C, O and N.

87. Electron Configuration
1s s p
C: 1s22s22p2 or
C: 1s22s22p2
Hund’s Rule tells us which configuration is correct.

88. Hund’s Rule
The lowest energy configuration for an atom is the one having the maximum number of unpaired electrons allowed by the Pauli principle in a particular set of degenerate orbitals.

89. Electron Configuration
1s s p
C: 1s22s22p2
N: 1s22s22p3
O: 1s22s22p4
F: 1s22s22p5
Ne: 1s22s22p6

90. Electron Configurations of the Fourth Period
K 1s22s22p63s23p64s or [Ar]4s1 4s
Ca 1s22s22p63s23p64s or [Ar]4s2
Sc 1s22s22p63s23p64s23d or [Ar]4s23d1
Ti 1s22s22p63s23p64s23d or [Ar]4s23d2
V 1s22s22p63s23p64s23d or [Ar]4s23d3 3d
Cr 1s22s22p63s23p64s13d or [Ar]4s13d5
Mn 1s22s22p63s23p64s23d or [Ar]4s23d5 •
Cu 1s22s22p63s23p64s13d or [Ar]4s13d10

91. Anomalies in Configurations
Chromium and Copper do not follow the pattern of the other elements.
You should remember these two families, because other elements in these families exhibit the same types of configurations
You can use the Periodic Table to guide you in writing electron configurations.

92. The energies of orbitals in multi-electron atoms are different than for hydrogen due to electron-electron interactions (repulsion and exchange).

93. The organization of the periodic table is based on the chemical properties of the elements which is determined by their electron configurations. n
Zeff

94. Mendeleev’s Periodic Table organized elements according to their chemical combining properties.

95. Electron Configurations of Ions
Start with the configuration for the neutral atom, then add or remove electrons from the valence shells to make the desired ion.
Atoms or ions that are isoelectronic with each other have identical numbers and configurations of electrons.

96. Write the electron configurations for the following:
a) C
b) S
d) Ti
e) Ti4+

97. How many unpaired electrons are in the followingSc
Ag+
Cd2+

99. Sizes of Atoms and Ions

100. Orbital Penetration and Effective Nuclear Charge
Orbital penetration occurs when an electron in an outer orbital has some probability of being close to the nucleus
Penetration ability follows this order: s > p > d > f.
Effective nuclear charge (Zeff) is the attractive force toward the nucleus experienced by an electron in an atom.

101. Penetration Ability of s Orbitals

102. Radii of Atoms and Ions

103. X(g) ---> X+(g) + e-(g)
Ionization Energy
The quantity of energy required to remove 1 mole of electrons from 1 mole of the gaseous atom or ion.
X(g) ---> X+(g) + e-(g)

104. Ionization Energy Trends

105. Ionization Energies

106. Periodic Trends First ionization energy:
increases from left to right across a period;
decreases going down a group.

107. Successive Ionization Energies (kJ/mol)

108. The Uncertainty Principle
Quantum mechanics allows us to predict the probabilities of where we can find an electron.
We cannot map out on the path an electron travels.
The Heisenberg’s uncertainty principle says that you cannot determine the position and momentum of an electron at the same time.

109. Heisenberg’s Uncertainty Principle (the quantum mechanical rules of the road)
Example…
The position and momentum of an electron in an atom can not be simultaneously determined.
Dx*Dp ≥ h/4p
From the ChemQuizzes CD-ROM by David Laws and Alexander Pines © 2003 W. W. Norton & Company Inc. All rights reserved.

110. ChemTour: Electromagnetic Radiation
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This ChemTour explores the relationship of frequency, wavelength, and energy using animations, interactive graphs, and equations. The quantitative exercises include graph reading and calculations using Planck’s constant and the speed of light.

111. ChemTour: Light Diffraction
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This animation recreates Thomas Young’s double-slit experiment and demonstrates how constructive and destructive interference occur.

112. ChemTour: Doppler Effect
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A boat moving with or against the direction of wave movement demonstrates the motion-induced shifts in wavelengths and frequency that are examples of the Doppler effect.

113. ChemTour: Light Emission and Absorption
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This ChemTour examines the emission and absorption spectra for sodium and hydrogen and relates them to energy level transitions.

114. ChemTour: Bohr Model of the Atom
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This ChemTour explores the idea that energies of electrons surrounding atomic nuclei are quantized. In Practice Exercises, students learn to calculate the energies of specific states of hydrogen, and the energies involved in electronic transitions.

115. ChemTour: de Broglie Wavelength
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In this ChemTour, students learn to apply the de Broglie equation to calculate the wavelength of moving objects ranging from baseballs to electrons. Includes Practice Exercises.

116. ChemTour: Quantum Numbers
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In this ChemTour, students explore the rules for designating quantum numbers. Includes Practice Exercises.

117. ChemTour: Electron Configuration
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This ChemTour explains how electrons are distributed within atomic orbitals. Students learn how to determine an element’s electron configuration and learn how to complete an orbital box diagram. Includes practice exercises.

118. Suppose two photons combine in a crystal to form a single photon of green light or "green photon."
Which of the following could be the colors of the two combining photons?
A) Green & green B) Blue & yellow C) Infrared & infrared
© 2008 W. W. Norton & Company Inc. All rights reserved.
Combining Two Photons

119. Please consider the following arguments for each answer and vote again:
A green photon can only be produced by the combination of two other green photons of the same wavelength.
The color green is the result of combining the colors blue and yellow, just as a green photon will result from the combination of blue and yellow photons.
Only two infrared photons have the proper total energy needed to form a green photon.
Answer: B
Combining Two Photons

120. Absorption and Fluorescence of Light
An electron in the ground state absorbs a single photon of light and then relaxes back to the ground state by emitting an infrared photon (1200 nm) followed by an orange photon (600 nm).
What is the wavelength of the absorbed photon?
A) 400 nm B) 600 nm C) nm
© 2008 W. W. Norton & Company Inc. All rights reserved.
Absorption and Fluorescence of Light

121. Absorption and Fluorescence of Light
Please consider the following arguments for each answer and vote again:
The wavelength is inversely proportional to the energy, so for energy to be conserved the absorbed photon must have a wavelength of 400 nm.
The wavelength of the absorbed photon is the difference of the wavelength of the two emitted photons, which is 600 nm.  
For the energy to be conserved, the sum of the wavelengths must be conserved. So the wavelength of the absorbed photon is 1800 nm.
Answer: A
Absorption and Fluorescence of Light

122. Two-Slit Diffraction and Interferometry
The diagram to the left depicts the interference pattern that results from the constructive and destructive interference of light waves that are diffracted as they pass through two slits. If the pattern is the result of green light passing through two slits, which of the following patterns would be the result of blue light passing through the same two slits?
A) B) C)
© 2008 W. W. Norton & Company Inc. All rights reserved.
Two-Slit Diffraction and Interferometry

123. Two-Slit Diffraction and Interferometry
Please consider the following arguments for each answer and vote again:
The wavelength of blue light is shorter than that of green light, so constructive and destructive interference occurs at smaller intervals.
The interference pattern is dependent only on the width of and distance between the two slits. Therefore, the interference pattern should not change.
Blue light is higher in energy than green light and therefore would be less affected by the two slits.
Answer: A
Two-Slit Diffraction and Interferometry

124. Photoelectric Effect: Red and Yellow Light
When a photon of red light hits metal X, an electron is ejected. Will an electron be ejected if a photon of yellow light hits metal X?
© 2008 W. W. Norton & Company Inc. All rights reserved.
A) Yes B) No C) Can't tell
Photoelectric Effect: Red and Yellow Light

125. Photoelectric effect: Red and Yellow Light
Please consider the following arguments for each answer and vote again:
Photons of yellow light possess more energy than photons of red light, so a yellow photon also must eject an electron.
Each metal has a specific wavelength of light that will cause electrons to be ejected. If red light has the correct wavelength, yellow cannot.
Whether a yellow photon will eject an electron from the metal will depend on how tightly the electron is bound to the metal.
Answer: A
Photoelectric effect: Red and Yellow Light

126. Photoelectric Effect: Blue and Green Light
When a photon of blue light hits metal X, an electron is ejected. Will an electron be ejected if a photon of green light hits metal X?
© 2008 W. W. Norton & Company Inc. All rights reserved.
A) Yes B) No C) Can't tell
Photoelectric Effect: Blue and Green Light

127. Photoelectric effect: Blue and Green Light
Please consider the following arguments for each answer and vote again:
So long as enough photons of light hit the metal, an electron will always be ejected, regardless of the wavelength of the light.
The energy of a blue photon is higher than the energy of a green photon so an electron removed with blue light will not be removed with green light.
Whether a green photon will eject an electron from the metal will depend on how tightly the electron is bound to the metal.
Answer: C
Photoelectric effect: Blue and Green Light

128. Photoelectric Effect: Kinetic Energies of Electron
A 300-nm photon can eject an electron from a metal surface with a certain kinetic energy. What photon wavelength would be required to eject an electron from the same metal surface with twice the kinetic energy?
A) 150 nm B) 200 nm C) 600 nm
© 2008 W. W. Norton & Company Inc. All rights reserved.
Photoelectric Effect: Kinetic Energies of Electron

129. Photoelectric Effect: Kinetic Energies of Electrons
Please consider the following arguments for each answer and vote again:
To eject an electron with twice the kinetic energy, twice the energy must be provided by the photon, so the photon wavelength must be halved.
A photon with a wavelength of 200 nm will overcome the work function and provide twice the kinetic energy.
To double the kinetic energy of the ejected electron, the wavelength of the impacting photon also must be doubled.
Answer: B
Photoelectric Effect: Kinetic Energies of Electrons

130. De Broglie Wavelengths of H20 Molecules
Suppose a hydrogen molecule (1H2) is traveling at 800 m/s and a deuterium molecule (2H2) is traveling at 400 m/s. What can be said of the de Broglie wavelengths of the two molecules?
© 2008 W. W. Norton & Company Inc. All rights reserved.
A) λH > lD B) λH < lD C) λH = lD
De Broglie Wavelengths of H20 Molecules

131. De Broglie Wavelengths of H2O Molecules
Please consider the following arguments for each answer and vote again:
The kinetic energy of the deuterium molecule is twice that of the hydrogen molecule. Therefore, the deuterium molecule will have a shorter de Broglie wavelength.
Because the speed of the hydrogen molecule is greater than the speed of the deuterium molecule, the de Broglie wavelength of the hydrogen molecule will be shorter.
The hydrogen molecule and the deuterium molecule have the same momentum and therefore will have the same de Broglie wavelength.
Answer: C
De Broglie Wavelengths of H2O Molecules

132. Laser Cooling of Sodium Atoms
One method for decreasing the temperature of atoms, known as laser cooling, involves bombarding an atom with photons of light, decreasing its overall momentum and thus its kinetic energy (just like one could slow a fast-moving car by colliding it with another car).
A sodium atom at a temperature of 60 K has a de Broglie wavelength of 66 pm (6.6x10-11 m). Approximately how many photons of red light (at λ = 660 nm) would it take to stop a sodium atom at 60 K?
A) ~1 B) ~102 C) ~104
© 2008 W. W. Norton & Company Inc. All rights reserved.
Laser Cooling of Sodium Atoms

133. Laser Cooling of Sodium Atoms
Please consider the following arguments for each answer and vote again:
A photon travels ~105 times faster than a sodium atom. Therefore, only one photon is required.
The kinetic energy of a sodium atom is ~100 times less than the kinetic energy of a red photon.
The de Broglie wavelength of a sodium atom at 60 K is ~104 times shorter than the wavelength of a red photon, so it will take 104 photons to stop a single sodium atom.
Answer: C
Laser Cooling of Sodium Atoms

134. Transmission of Light through a Color Filter
What color will a yellow object appear when it is seen through a filter with the absorption spectrum shown to the left?
© 2008 W. W. Norton & Company Inc. All rights reserved.
A) Yellow B) Blue C) Black
Transmission of Light through a Color Filter

135. Transmission of Light through a Color Filter
Please consider the following arguments for each answer and vote again:
The filter absorbs no yellow light, so the object will appear yellow.
Blue light is absorbed by the filter, so an object seen through the filter will appear blue.
No yellow light is absorbed by the filter, so the object will appear black.
Answer: A
Transmission of Light through a Color Filter

136. Photon emission from a system possessing the energy level diagram to the left would produce which of the following spectra? A) B) C)
© 2008 W. W. Norton & Company Inc. All rights reserved.
Emission Spectra

137. Consider the following arguments for each answer and vote again:
The photon wavelength depends only on the energy of the lowest state, so only 1 wavelength is possible.
There are 2 possible transitions—one from each of the 2 upper levels. Thus, 2 wavelengths of light are emitted.
The 3 energy levels lead to 2 high-energy transitions and 1 low-energy transition. Therefore, 3 different photon wavelengths are possible.
Answer: C
Emission Spectra

138. Emission from which of the following energy level diagrams would produce the spectrum shown to the left? A) B) C)
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Energy Levels

139. Consider the following arguments for each answer and vote again:
The arrangement of the energy levels reflects the arrangement of the lines in the emission spectrum.
This energy level diagram allows only 1 low-energy transition, consistent with the emission spectrum.
Only this energy level diagram allows 3 high-energy transitions and 1 low-energy transition.
Answer: C
Energy Levels

140. The diagram to the left shows the spacing of the first five energy levels for a hydrogen atom. Which of the following transitions in He+ has the same wavelength as the 4→2 transition in H?
© 2008 W. W. Norton & Company Inc. All rights reserved.
A) 4→2 B) 8→4 C) 16→8
Transition in H and He+

141. Consider the following arguments for each answer and vote again:
He+ has the same electron configuration as H; therefore, the energy level diagram will be the same.
The atomic number of He+ is twice that of H. Therefore, to produce the same energy splitting, the energy levels must be twice that of H.
The energy of the electron is proportional to Z2, which is 4 for He+. Therefore, the two levels, 4 and 2, must be increased by a factor of 4 to 16 and 8, respectively.
Answer: B
Transition in H and He+

142. Electron Configurations
Periodic Table
Which atom or ion can have the electron configuration 1s22s22p1?
A) Li B) Be- C) B+
© 2008 W. W. Norton & Company Inc. All rights reserved.
Electron Configurations

143. Electron Configurations
Consider the following arguments for each answer and vote again:
The answer must be lithium because it is the first element in row 2 to possess only one unpaired electron.
Beryllium in its ground state has the electron configuration 1s22s2, so Be- in its ground state will have the configuration 1s22s22p1.
In its ground state, boron has the electron configuration 1s22s22p1, so B+ must also have this configuration.
Answer: B
Electron Configurations

144. A) H(1s1) B) He(1s13p1) C) He+(4p1)
Which of the following has the lowest ionization energy?
A) H(1s1) B) He(1s13p1) C) He+(4p1)
© 2008 W. W. Norton & Company Inc. All rights reserved.
Ionization Energies

145. Consider the following arguments for each answer and vote again:
Hydrogen has a lower nuclear charge than helium, so it always has a lower ionization energy than any helium atom or ion.
He(1s13p1) has almost the same ionization energy as H(3p1), which has a lower ionization energy than either H(1s1) or He+(4p1).
Because the electron in He+(4p1) is in the fourth shell, the ionization energy of He+(4p1) is the lowest.
Answer: B
Ionization Energies

146. Ionization Energies of He(1s2)
How does the ionization energy of He(1s2) compare to the ionization energies of H(1s1) and He+(1s1)?
A) Higher B) Lower C) In-between
© 2008 W. W. Norton & Company Inc. All rights reserved.
Ionization Energies of He(1s2)

147. Ionization Energies of He(1s2)
Consider the following arguments for each answer and vote again:
It is harder to remove an electron from a doubly occupied orbital than from a singly occupied orbital.
Each electron offsets the charge of one of the protons, giving an effective nuclear charge of zero.
Each electron partially shields the other, leading to an effective nuclear charge that is between 1 and 2.
Answer: C
Ionization Energies of He(1s2)

148. Which of the following atoms (or ions) has the smallest radius?
A) K+ B) Ar C) Cl-
© 2008 W. W. Norton & Company Inc. All rights reserved.
Atomic and Ionic Radii

149. Consider the following arguments for each answer and vote again:
K+ has the highest nuclear charge and so has the smallest atomic radius.
Because it is a noble gas, Ar has the smallest atomic radius.
Cl- has the nucleus with the lowest mass, so it has the smallest atomic radius.
Answer: A
Atomic and Ionic Radii

150. Electron Affinity of Halogen Atoms
Suppose an electron is transferred from a potassium atom to an unknown halogen atom. For which of the following halogen atoms would this process require the least amount of energy?
A) Cl B) Br C) I
© 2008 W. W. Norton & Company Inc. All rights reserved.
Electron Affinity of Halogen Atoms

151. Electron Affinity of Halogen Atoms
Consider the following arguments for each answer and vote again:
Chlorine has the greatest affinity for electrons and so would release the most energy when an electron is added.
Electron donation is most favorable energetically when it occurs between atoms on the same row of the periodic table.
Because of its massive nuclear charge and large electron cloud, an iodine atom can most easily accept an additional electron.
Answer: A
Electron Affinity of Halogen Atoms