AP Chemistry
Quarter 1
PRACTICE Exam 1 Solutions
The first 4 questions refer to the following atoms. The same answer may be used more than once.
(A) Ba
(B) B
(C) Br
(D) Be
(E) Bi
1. Write your exam form on the “Subject” line of the scantron sheet. Then answer the questions. Which two
of these have the same number of valence electrons?
Ba and Be are both in group 2. They have 2 valence electrons.
2. Which one has the most unpaired electrons in the ground state?
Remember from Hund’s rule that electrons will fill empty orbitals before occupied ones. The 2 valence
electrons in Ba and Be fill an s-subshell, so have 0 unpaired electrons. B has 1 unpaired electron in the 2psubshell, Br has 1 unpaired electron in the 4p-subshell, Bi has 3 unpaired electrons in the 6p-subshell.
Therefore the answer is Bi.
3. Which one is least electronegative?
Electronegativity is the ability of an atom to attract a bonded electron pair and is a function of Zeff and electron
shielding. Of the atoms above, Ba has the lowest Zeff (which affects the force on the valence electrons) and has
a high degree of electron shielding, this gives it the lowest electronegativity value.
4. Which one’s valence energy level is n=4?
Valence energy level can be found by using the period number on the periodic table. Br is in the 4th period and
is the answer.
5. Which of the following puts the particles in order from least to greatest mass?
(A) proton, electron, alpha particle, 2H
(B) alpha particle, electron, proton, 2H
(C) 2H, electron, proton, alpha particle
(D) electron, proton, 2H, alpha particle
(E) electron, 2H, alpha particle, proton
6. Which of the following correctly gives a possible set of the four quantum numbers (n, l, ml, ms) for the
highest-energy electron in a ground-state atom of tin (Sn)?
(A) 5, 1, 1, +½
(D) 5, 0, 0, +½
(B) 5, 2, 0, +½
(E) 5, 2, 1, −½
(C) 5, 1, 2, −½
The highest-energy electron is in the 5p-subshell. This means the principal energy level, n, is 5 and the sublevel, l, is 1. This leaves us with A + C as options. The values for ml can only be between –l to l, so C
cannot be the answer.
7. Which of the factors below contribute(s) significantly to the decrease in electronegativity from top to
bottom in the periodic table?
I. The increased mass of the nucleus down a group
II. The increased charge of the nucleus down a group
III. The increased size of the valence orbitals down a group
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III
While the increased charge does affect the electronegativity, the Zeff in a group stays constant, so it has no
effect in this case. If the increased nuclear charge were a factor, it would tend to increase the
electronegativity, not decrease it.
The 4 questions that follow refer to these electron configurations.
A = 1s22s22p63s2
B = [Ar]4s23d104p6 C = [Kr]5s24d66s1
D = [Xe]6s1 E = [Ne]3s2
8. Which two are equivalent to each other? A + E
9. Which one represents an excited state? C, look for configurations that don’t follow the normal progression
10. Which one represents a noble gas? B
11. Which of these has/have no unpaired electrons (there may be one or more than one)? A, B, E
12. Consider Rutherford’s famous experiment, in which shot alpha particles, which are positively charged, at a
thin gold foil. A small percentage of the alpha particles bounced off to the side or bounced back. The fact
that only a small percentage of the particles were deflected demonstrated which of the following about the
nucleus of an atom?
I. The nucleus has a relatively large mass.
II. The nucleus has a small volume.
III. The nucleus contains neutral particles.
A) I only
B) II only
C) III only
D) I and II only
E) I, II, and III
13. Which electron configuration shown below does NOT violate the Pauli exclusion principle?
A) 1s22s22p63s23p64s23d104p65s1
B) 1s22s22p63s23p64s23d104p65s24d12
C) 1s22s22p63s23p64s23d104p65s24d125p2
D) 1s22s22p63s23p64s23d104p65s24d125p5
E) 1s22s22p63s23p64s23d104p65s24d125p66s24f14
Answers B – E all have 12 electrons in the 4d-subshell. The maximum possible is 10.
14. Which of the following is/are true about atoms of a pair of isotopes?
I. They have the same number of protons.
II. They have the same number of neutrons.
III. They have different masses.
A) I only
B) II only
C) III only
D) I and III only
E) I, II, and III
15. Potassium consists of only two naturally-occurring isotopes, 39K and 41K. Based on the information on the
periodic table, the relative abundance of 39K is roughly
A) 5%
B) 25%
C) 50%
D) 75%
E) 95%
The atomic weight is 39.1, so most of it has to be 39K. Since 0.1 (the increase from 39) is 5% of the way
between 39 and 41, 95% has to be 39K.
16. If 87.5% of a sample of a particular radioisotope decays in 3 days, what is the half life of the isotope
A) 12 hours B) 18 hours
C) 24 hours
D) 30 hours
E) 36 hours
First, figure out the number of half-lives result in 87.5% of a sample decaying (or think about it as 12.5%
remaining). The progression is 50-25-12.5, so 3 half-lives have occurred. Divide the amount of time (3 days),
by the number of half-lives (3) and you get 24 hours.
17. A student is attempting to explain why the atomic radius of chlorine (Cl) is smaller than the atomic radius of
bromine (Br). Which of the following factors contribute(s) significantly the observed difference in size?
I. Bromine has more protons than chlorine.
II. Chlorine needs one more electron to reach a noble gas configuration.
III. Bromine’s valence energy level is 4, while chlorine’s is 3.
A) I only
B) II only
C) III only
D) I and III only
E) I, II, and III
18. Very high temperatures are generally required for nuclear fusion to occur. Which of the following best
explains this phenomenon?
A) A nucleus is unlikely to break apart at low temperatures.
B) High temperatures are needed in order to produce the neutrons needed for fusion.
C) Two nuclei repel each other strongly, so they need to be moving quickly in order to collide.
D) Fusion releases energy, and processes that release energy generally require high temperatures.
E) Fusion requires atoms to be radioactive, which only occurs at high temperatures.
19. Which of the following represents the ground-state electron configuration for the Mn3+ ion?
A) 1s22s22p63s23p63d4
B) 1s22s22p63s23p63d54s2
C) 1s22s22p63s23p63d24s2
D) 1s22s22p63s23p63d84s2
E) 1s22s22p63s23p63d34s1
Remember, transition elements lose their s-subshell electrons first.
20. Fill in the missing isotope in the nuclear equation below.
A)
107
Mo
B) 108Mo
C) 109Mo
D) 107Nb
E) 109Nb
Open Response Section
You may use your periodic table, formula sheet, and calculator.
Please write all answers in the space provided.
21. The half-life of 11C is roughly 20 minutes. The half-life of 10C is roughly 20 seconds. If 12 µg of 11C is
injected into a patient for a medical procedure, roughly what mass of 11C remains 1.5 hours later? A
ballpark estimate is acceptable.
First, determine how many half-lives have passed. 1.5 hrs = 90 minutes. 90 minutes/20 minutes per half-life =
4.5 half-lives. Now you can make a chart
Mass remaining 12 µg
6.0 µg
3.0 µg
1.5 µg
0.75 µg
# of half-lives
0
1
2
3
4
So make an estimate somewhere in between 0.75 µg and 0.375 µg.
0.38 µg
5
As an alternative you could use the equation A = Ao(1/2)(t/h) where A = the desired value, Ao = 12 µg, t = 90
min and h = 20 min. This leaves us with the answer of 0.53 µg.
22. Hund’s rule states that each orbital in a sublevel will be singly occupied before any of them becomes doubly
occupied. Explain this rule based on fundamental principles.
Electrons have the same charge. So two electrons will repel each other. If forced to be in the same orbital,
they will have higher PE than they will if allowed to be further apart. Therefore, they choose to be further
apart.
23. For each part of this question, two situations are depicted or described. Circle which situation (on the left or
on the right) involves higher potential energy. Then explain your answer.
a. An electron in a 1s orbital or an electron in a 2s orbital in the same atom
An electron in the 2s orbital is further away from the nucleus. Since the electron is attracted to the nucleus
(positive charge), it has greater PE when it is further away.
b. Three charges arranged like this
or
like this
The arrangement on the right is higher in PE because the negative charge is further away from the left-hand
positive charge. Note that the positive charges are equally close in both pictures, so the interaction between
these two has no effect on PE.
c. A 55Cr nucleus
a 55Mn nucleus
or
In this question, think about the stability of these atoms. 55Cr is further away from the average atomic mass of
chromium than 55Mn is from the average atomic mass of manganese, meaning it is less stable. Less stable
atoms have more potential to release energy in a nuclear decay, therefore have more potential energy. This
question was a bit of a stretch, since we did not focus on this aspect of nuclear stability this year.
24. For each of the following subshells, give the n and l quantum numbers, and state how many electrons could
occupy that subshell.
subshell
n
l
Max # of e-
1
3
5
1s
3p
5f
0
1
3
2
6
14
25. Two oppositely charged ions are a distance of 20 nm apart (measured from center to center) as shown.
A) If the two ions were 10 nm apart would the force on them be weaker, equal strength, or stronger?
Weaker
Equal
Stronger
When charges move closer together, the force is increased.
B) If the two ions were 10 nm apart would their potential energy be less, equal, or greater?
Less
Equal
Greater
Like charges prefer to be closer together. Since we are decreasing the distance between the charges, when
we let them go, we could get less work out of the system, therefore there is less potential energy.
C) If the ions shown above (O2- and Mg2+) were replaced by a Ca2+ ion and a Ba2+ ion at the same distance,
would the force on them be weaker, equal strength, or stronger?
Weaker
Equal
Stronger
The magnitude of the charges did not change, only the direction.
D) If the ions shown above (O2- and Mg2+) were replaced by a Na+ ion and a S2- ion at the same distance,
would the force on them be weaker, equal strength, or stronger?
Weaker
Equal
Stronger
The system replaced a +2 ion with a +1 ion, decreasing the magnitude of the charges involved. This
weakens the force.
26. Give the ground state electron configuration for each of the following atoms or ions. You may use
appropriate abbreviations.
A) F [He]2s22p5
B) Co2+ [Ar]3d7
C) Cl- [Ne]3s23p6
D) Which of the above has the most unpaired electrons? _Co2+__ How many does it have? __2__
27. Each of the nuclei listed below is unstable. Write a balanced nuclear equation for each decay.
A) 241Am undergoes alpha decay
241
B)
Am  4α + 237Np
11
C undergoes positron emission
C  01β + 11Β
Note: The β could be symbolized by “e” as well.
11
C)
23
Ne undergoes β− decay
Ne  0-1β + 23Na
Note: The β could be symbolized by “e” as well.
23
28. The visible spectrum of light consists of wavelengths from roughly 400 nm – 700 nm.
(6 pt)
A) Which wavelength (400 nm or 700 nm) corresponds to the red end of the spectrum? Explain.
The 700 nm wavelength corresponds to the red end of the spectrum. A way to remember this is ultraviolet light
has more energy than infrared, so in the visible red has the lowest energy (which corresponds to the longest
wavelength). After all, you don’t worry about IR rays from the sun giving you skin cancer – you worry (or
should worry, at least) about the UV.
B) Calculate the amount of energy carried by a photon with a wavelength of 520 nm.
Combine the equations C=λ*f and e=h*f to get e=h*c/λ.
𝐸=
ℎ𝑐 (6.626 × 10−34 𝐽 ∙ 𝑠)(3 × 108 𝑚/𝑠)
=
= 3.82 × 10−19 𝐽
10−9 𝑚
λ
(520 𝑛𝑚) �
�
1 𝑛𝑚