South Pasadena • AP Chemistry
Name
8 ▪ Atomic Theory
Period
UNIT TEST
Date
–
PRACTICE
Part 1 – Multiple Choice
You should allocate 25 minutes to finish this portion of the test. No calculator should be used. A periodic table
and data table will be provided. Select the answer that best responds to each question.
1. Sodium exists in only one naturally occurring
isotope: Na-23. However, chlorine exists in two
naturally occurring isotopes: Cl-35 and Cl-37.
Which of the following can be the number of
neutrons in one Na-Cl compound?
(A) 30
(B) 58
(C) 58.5
(D) 60
4. Which of the following requires a photon of light
with the shortest wavelength to break the bond?
(A) The C–O single bond in methanol, CH3OH.
(B) The C−
...O bond in the carbonate ion, CO32–,
which has a bond order of 1.3.
(C) The C=O double bond in formaldehyde,
CH2O.
(D) The C≡O triple bond in carbon monoxide,
CO.
2. Fluorine is one of nineteen elements that are said
to be mononuclidic, meaning it has only one stable
naturally occurring isotope. What are the numbers
of protons, neutrons, and electrons in the most
common ion of naturally-occurring F?
(A) 9 protons, 10 neutrons, 10 electrons
(B) 9 protons, 10 neutrons, 9 electrons
(C) 9 protons, 9 neutrons, 10 electrons
(D) 9 protons, 9 neutrons, 9 electrons
5. When ultraviolet light of a particular wavelength
shines upon a piece of zinc metal, several
electrons are ejected, and their speeds measured.
Electron A
5.0 × 105 m/s
Electron B
8.0 × 105 m/s
Which electron has a higher binding energy?
(A) Electron A
(B) Electron B
(C) They have the same binding energy
(D) Their binding energies cannot be compared
with the data given
3. The mass spectrometry plot for the element boron,
B, is shown above. The plot provides evidence for
which of the following?
(A) The B-11 isotope has a greater nuclear charge
than the B-10 isotope.
(B) The B-11 isotope has more electrons than the
B-10 isotope.
(C) The B-11 isotope is denser than the B-10
isotope.
(D) The B-11 isotope is more abundant than the
B-10 isotope.
6. Ultraviolet light of the following wavelengths
were detected from the hydrogen emission
spectrum:
122 nm, 103 nm, 97 nm, 95 nm
There were no ultraviolet light observed with
longer wavelengths. The 122 nm line corresponds
to electrons that drop between which energy
levels?
(A) n = 2 to n = 1
(B) n = 3 to n = 1
(C) n = 3 to n = 2
(D) n = 4 to n = 2
7. Which of the following statements about the 3p
orbital in the aluminum atom is incorrect?
(A) A 3p orbital has two nodes.
(B) Electrons in the 3p orbital are always
further away from the nucleus than
electrons in the 3s orbital.
(C) The 3p orbital is higher energy than the 3s
orbital.
(D) There are three degenerate 3p orbitals in this
subshell.
ψ
radial distance
8. The wave function, ψ, for which atomic orbital is
shown above?
(A) 2s
(B) 2p
(C) 3s
(D) 3p
11. The PES for lithium, Li, is shown above. Which of
the following correctly identifies and explains the
peak on the right?
(A) The peak to the right represents electrons in
the 1s orbital because electrons in lower
energy shells have shorter peaks.
(B) The peak to the right represents electrons in
the 1s orbital because electrons in lower
energy levels are closer to and more attracted
to the nucleus.
(C) The right peak represents electrons in the 2s
orbital because electrons in higher energy
shells have shorter peaks.
(D) The right peak represents electrons in the
2s orbital because electrons in higher
energy shells are further away from and
less attracted to the nucleus.
9. How many electrons in the ground state of Te
have ℓ = 1?
(A) 4
(B) 6
(C) 22
(D) 52
10. Which of the following could be the set of
quantum numbers that describes an electron with
the highest energy in the ground state electron
configuration of Rh2+?
(A) n = 3, ℓ = 2, mℓ = 0, ms = +½
(B) n = 4, ℓ = 2, mℓ = –1, ms = +½
(C) n = 5, ℓ = 0, mℓ = 0, ms = +½
(D) n = 5, ℓ = 2, mℓ = 1, ms = +½
12. The spectra for nitrogen (solid) and oxygen
(dotted) shown above provide evidence for which
of the following?
(A) The first ionization energy of nitrogen is
greater than that of oxygen.
(B) The 2p electron of nitrogen is further from the
nucleus than that of oxygen.
(C) The nuclear charge of nitrogen is greater than
that of oxygen.
(D) The ionic radius of nitrogen is greater than
that of oxygen.
13. The electron affinity of carbon (–122 kJ mol–1) is
more negative than boron (–27 kJ mol–1). This can
primarily be explained by:
(A) Carbon has a smaller atomic radius than
boron, so it is easier to add electrons.
(B) Carbon has more core electrons than boron
does, so the outer electrons experience greater
shielding.
(C) Carbon has more protons than boron, so it
has greater effective nuclear charge.
(D) Carbon has more valence electrons, so it has
greater electron repulsions than boron.
15. Which element would most likely have the
following ionization energies?
First Ionization Energy
738 kJ mol–1
Second Ionization Energy 1451 kJ mol–1
Third Ionization Energy
7733 kJ mol–1
Fourth Ionization Energy 10,540 kJ mol–1
(A) Na
(B) Mg
(C) Al
(D) Si
14. Which of the following quantities is larger for Mg
than Al?
I. Atomic Radius
II. Ionic Radius
III. Ionization Energy
(A) I only
(B) III only
(C) II and III only
(D) I, II and III
Part 2 – Free Response
You should allocate 30 minutes to finish this portion of the test. You may use a scientific calculator. A periodic
table and data table will be provided. Respond to each part of the questions completely. Be sure to show your
work clearly for questions that involve calculations.
(From AP Chemistry 1999 #2)
16. Answer the following questions regarding light and its interactions with molecules, atoms, and ions.
(a) The longest wavelength of light with enough energy to break the Cl–Cl bond in Cl2(g) is 495 nm.
i. Calculate the frequency, in s–1, of the light.
c 3.0 × 108 m/s109 nm
ν= =
= 6.06 × 1014 Hz
λ
495 nm  1 m 
ii. Calculate the energy, in J, of a photon of the light.
E = h·ν = (6.626 × 10−34 J·s)(6.06 × 1014 Hz) = 4.02 × 10−19 J
iii. Calculate the minimum energy, in kJ moL–1, of the Cl–Cl bond.
4.02 × 10−19 J 1 kJ 6.02 × 1023 bonds
1 bond 1000 J
1 mol
 = 242 kJ/mol
(b) A certain line in the spectrum of atomic hydrogen is associated with the electronic transition in the H
−2.178 × 10−18
atom from the sixth energy level (n = 6) to the second energy level (n = 2). En =
J
n2
i. Indicate whether the H atom emits energy or whether it absorbs energy during the transition. Justify
your answer.
−2.178 × 10−18
−2.178 × 10−18
E2 – E6 =
J−
J = −4.84 × 10−19 J
2
2
62
Energy is emitted because ∆E < 0.
When electrons drop from a higher energy to lower energy level, energy is emitted.
ii. Calculate the wavelength, in nm, of the radiation associated with the spectral line.
h·c (6.626 × 10−34 J·s)(3.0 × 108 m/s) 109 nm
λ=
=
E
(4.84 × 10−19 J)
 1 m  = 411 nm
iii. Account for the observation that the amount of energy associated with the same electronic transition
(n = 6 to n = 2) in the He+ ion is greater than that associated with the corresponding transition in the H
atom.
He+ has 2 protons, so the electron shells are more attracted to the nucleus than those in H
resulting in greater electron transition energies.
(From AP Chemistry 2000 #7)
17. Answer the following questions about the element selenium, Se (atomic number 34).
(a) Samples of natural selenium contain six stable isotopes. In terms of atomic structure, explain what these
isotopes have in common, and how they differ.
Isotopes have the same number of protons, but different number of neutrons.
(b) Write the complete electron configuration (e.g., 1s2 2s2… etc.) for a selenium atom in the ground state.
Indicate the number of unpaired electrons in the ground-state atom, and explain your reasoning.
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4
(c) In terms of atomic structure, explain why the first ionization energy of selenium is
i. less than that of bromine (atomic number 35), and
Se and Br are both in the 4p subshell, but while Se has 34 protons, Br has 35 protons. Outer
electrons are less attracted to a nucleus with fewer protons (lower Zeff), resulting in a lower IE.
Because Se has fewer protons than Br, it has a lower IE than Br.
ii. greater than that of tellurium (atomic number 52).
Outer electron of Se is in the 4p subshell, but that of Te is in the 5p subshell. Electrons in lower
subshells are more attracted to the nucleus, and require more energy to remove. Because outer
electron of Se is in a lower subshell than that of Te, Se has a higher IE than Te.
South Pasadena • AP Chemistry
Name
8 ▪ Atomic Theory
Period
UNIT TEST
Date
BLUEPRINT
Part 1: Multiple Choice
 Format: 15 questions, four answer choices: (A)-(D)
 Expected time: 25 minutes
 Allowed resources: Periodic Table, Equations and Constants. No calculators.
Q Lesson
Topic
Objective
1 8.0
2 8.0
3 8.1
Atomic Structure
Isotopes
Mass Spectroscopy
4 8.1
Light
5 8.1
Photoelectric Effect
6 8.1
7 8.1
Hydrogen Spectrum
Atomic Orbitals
Explain how the quantum numbers for a wave function determines the electron
distribution of atomic orbitals
Atomic Orbitals
Explain how the quantum numbers for a wave function determines the electron
distribution of atomic orbitals
Electron
Predict the ground state electron configuration of an element or ion based on its
Configuration
position on the periodic table.
Electron
Predict properties of an element based on its valence shell electron configuration.
Configuration
PES
Explain an element’s electron configuration using its photoelectron
spectrograph.
PES
Explain an element’s electron configuration using its photoelectron
spectrograph.
Periodic Trends –
Use Coulomb’s Law to explain how attractions between the nucleus and valence
Elements
electrons change between elements on the Periodic Table or ions.
Periodic Trends –
Compare properties (i.e. size, ionization energy, electronegativity, ion charge,
Ions
reactivity) between two elements or ions by comparing attractions between the
nucleus and valence electrons.
Periodic Trends –
Compare properties (i.e. size, ionization energy, electronegativity, ion charge,
Successive IE
reactivity) between two elements or ions by comparing attractions between the
nucleus and valence electrons.
8 8.2
9 8.2
10 8.2
11 8.2
12 8.3
13 8.3
14 8.3
15 8.3
Describe an atom with isotopic notation.
Describe an atom with isotopic notation.
Calculate the atomic mass of an element using isotopic abundance information
from mass spectroscopy data.
Compare and calculate the wavelength, frequency, and energy of particular
electromagnetic radiation.
Explain how the photoelectric effect demonstrated the particle nature of light,
and the diffraction of electrons demonstrated the wave nature of matter.
Part 2: Free Response




Format:
o 1 long question (5-8 parts)
o 1 short questions (2-4 parts)
Expected time: 30 minutes
Allowed resources: Periodic Table, Equations and Constants, and scientific calculators.
Topics: Any