In a gaseous atom, the potential energy of an electron can be estimated from its ionization energy.. The first ionization energy of an element is defined as the minimum amount of energy
Trang 1GENERAL CHEMISTRY TOPICAL:
Electronic Structure and the Periodic Table Test 1
Time: 22 Minutes*
Number of Questions: 17
* The timing restrictions for the science topical tests are optional If you are using this test for the sole purpose of content reinforcement, you may want to disregard the time limit
Trang 2DIRECTIONS: Most of the questions in the following test
are organized into groups, with a descriptive passage preceding each group of questions Study the passage, then select the single best answer to each question in the group Some of the questions are not based on a descriptive passage; you must also select the best answer
to these questions If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain Indicate your selection by blackening the corresponding circle on your answer sheet A periodic table is provided below for your use with the questions
PERIODIC TABLE OF THE ELEMENTS
1
H
1.0
2
He
4.0 3
Li
6.9
4
Be
9.0
5
B
10.8
6
C
12.0
7
N
14.0
8
O
16.0
9
F
19.0
10
Ne
20.2 11
Na
23.0
12
Mg
24.3
13
Al
27.0
14
Si
28.1
15
P
31.0
16
S
32.1
17
Cl
35.5
18
Ar
39.9 19
K
39.1
20
Ca
40.1
21
Sc
45.0
22
Ti
47.9
23
V
50.9
24
Cr
52.0
25
Mn
54.9
26
Fe
55.8
27
Co
58.9
28
Ni
58.7
29
Cu
63.5
30
Zn
65.4
31
Ga
69.7
32
Ge
72.6
33
As
74.9
34
Se
79.0
35
Br
79.9
36
Kr
83.8 37
Rb
85.5
38
Sr
87.6
39
Y
88.9
40
Zr
91.2
41
Nb
92.9
42
Mo
95.9
43
Tc
(98)
44
Ru
101.1
45
Rh
102.9
46
Pd
106.4
47
Ag
107.9
48
Cd
112.4
49
In
114.8
50
Sn
118.7
51
Sb
121.8
52
Te
127.6
53
I
126.9
54
Xe
131.3 55
Cs
132.9
56
Ba
137.3
57
La *
138.9
72
Hf
178.5
73
Ta
180.9
74
W
183.9
75
Re
186.2
76
Os
190.2
77
Ir
192.2
78
Pt
195.1
79
Au
197.0
80
Hg
200.6
81
Tl
204.4
82
Pb
207.2
83
Bi
209.0
84
Po
(209)
85
At
(210)
86
Rn
(222) 87
Fr
(223)
88
Ra
226.0
89
Ac †
227.0
104
Rf
(261)
105
Ha
(262)
106
Unh
(263)
107
Uns
(262)
108
Uno
(265)
109
Une
(267)
*
58
Ce
140.1
59
Pr
140.9
60
Nd
144.2
61
Pm
(145)
62
Sm
150.4
63
Eu
152.0
64
Gd
157.3
65
Tb
158.9
66
Dy
162.5
67
Ho
164.9
68
Er
167.3
69
Tm
168.9
70
Yb
173.0
71
Lu
175.0
†
90
Th
232.0
91
Pa
(231)
92
U
238.0
93
Np
(237)
94
Pu
(244)
95
Am
(243)
96
Cm
(247)
97
Bk
(247)
98
Cf
(251)
99
Es
(252)
100
Fm
(257)
101
Md
(258)
102
No
(259)
103
Lr
(260)
GO ON TO THE NEXT PAGE.
Trang 3Passage I (Questions 1–6)
In addition to kinetic energy, electrons also have
potential energy due to their positions relative to the
nucleus, other electrons, and other nuclei In a gaseous
atom, the potential energy of an electron can be estimated
from its ionization energy The first ionization energy of
an element is defined as the minimum amount of energy
required to remove an electron from the gaseous element in
its ground state First ionization energies range from less
than 400 kJ/mole for cesium to nearly 2400 kJ/mole for
helium Table 1 gives the first ionization energies of
selected elements
Table 1
H He Li Be B C N F Na
1312 2372 520 899 800 1086 1402 1680 496 Photoionization is a method by which first ionization
energies can be conveniently measured A gaseous sample
of an element is irradiated and the energy of absorbed
photons is determined by the following equation:
I = hν
Equation 1
where I is the ionization energy, ν is the frequency of the
light absorbed, and h is Planck’s constant, 6.63 × 10–34 J •
s
1 Sodium has a first ionization energy of 496 kJ/mol.
What is the frequency of light absorbed per photon in the photoionization of a gas phase sodium sample?
A 1.24 × 1012 Hz
B 1.24 × 1015 Hz
C 7.48 × 1026 Hz
D 7.48 × 1038 Hz
2 The first ionization energy of aluminum is 577 kJ/mol
while that of magnesium is 738 kJ/mol This difference can most likely be attributed to:
A the higher energy of the 3p electron of aluminum
as opposed to the lower lying 3s electrons of
magnesium
B the larger nuclear charge of magnesium compared
to that of aluminum
C the increased internuclear attraction between
aluminum atoms
D the electron-electron repulsion of aluminum 3p
electrons
3 The second ionization energy is defined as the
minimum amount of energy needed to remove an electron from a gaseous +1 cation in its ground state Which of the following must be true for any element?
A Second ionization energies are always higher than
first ionization energies
B Second ionization energies may be higher or lower
than first ionization energies
C Second ionization energies are usually lower than
first ionization energies, but exceptions exist in the inner transition metals
D Second ionization energies are always lower than
first ionization energies due to the destabilization
of the element brought about by ionization
4 Which of the following elements has the smallest
atomic radius?
A Potassium
B Calcium
C Cesium
D Barium
GO ON TO THE NEXT PAGE.
Trang 45 When cesium is photoionized, an electron is removed
from which of the following atomic orbitals?
A 4p
B 5p
C 5d
D 6s
6 Why must the photoionization experiment in the
passage be carried out on atoms in the gas phase?
A Gas phase atoms are the only ones capable of
absorbing visible light
B The refractive index of most solids and liquids is
greater than one and thus obscures the true
frequency of absorption
C The proximity of neighboring atoms in solids and
liquids increases the potential energy of an
electron, making ionization impossible
D Factors such as lattice energy and van der Waals
forces in the solid and liquid states alter the
environment of the electron, thereby distorting the
frequency of absorption
GO ON TO THE NEXT PAGE.
Trang 5Passage II (Questions 7–12)
The Aufbau, or Building-up, Principle states that the
electronic configuration of an atom or molecule is the
result of the placement of electrons in orbitals of increasing
energy until the correct number of electrons has been
accommodated This principle takes into consideration the
attraction of the electron to the nucleus and inter-electron
repulsion The electronic configuration of ions can be
determined by subtracting electrons from the highest energy
orbital for cations or by adding electrons according to the
Building-up Principle
There has been, however, experimental evidence that
the Aufbau principle is an oversimplification of electronic
structure The electronic configurations of transition
elements and their ions often do not correspond to those
predicted by the Aufbau Principle These inconsistencies
are due to electrons filling orbitals in such a way as to give
the atom the lowest possible energy The total energy of an
atom can be described as:
E = F – G
Equation 1
where E is the total energy, F is the sum of the orbital
energies of the atom, and G is the electron repulsion
energy The F term increases with increasing values of the
quantum numbers n and l, but the orbitals that minimize F
may not necessarily minimize E As a result, the lowest
total energy of some elements is achieved by placing
electrons in orbitals with a higher energy (i.e., a higher
principal quantum number) even though there are still
unfixed subshells in the lower energy orbital This is
especially true for the transition elements, where the s
orbital of a higher electron shell fills before the d orbital
with a lower n value.
7 Which of the following best explains why chromium
exhibits a valence electron configuration of 3d54s1 instead of the predicted 3d44s2?
A The 4s orbital has a higher energy and therefore
fills last
B After the first 3d electron is added to an atom, each
successive electron will go into the 3d subshell
until it is filled
C Since the 3d subshell is half-filled, the total
energy of the atom is minimized
D Chromium exhibits multiple oxidation states and
so must have many unpaired electrons
8 Based on information in the passage, which of the
following statements is true?
A By nature of its principal quantum number alone,
the 3d orbital has lower energy and fills before the 4s orbital.
B By nature of its principal quantum number alone,
the 3d orbital has lower energy but fills after the 4s orbital because of electron repulsion energies.
C By nature of its principal quantum number alone,
the 4s orbital has lower energy and fills before the 3d orbital.
D By nature of its principal quantum number alone,
the 4s orbital has lower energy but fills after the 3d orbital because of electron repulsion energies.
9 What are the ground-state electronic configurations of
the Fe2+ and Fe3+ ions respectively?
A [Ar]3d6 and [Ar]3d5
B [Ar]3d6 and [Ar]3d44s1
C [Ar]3d54s1 and[Ar]3d44s1
D [Ar]3d44s2 and [Ar]3d34s2
1 0 What is the electronic configuration of indium
(Z = 49)?
A 1s22s22p63s23p64s23dl04p65s24d105p1
B 1s22s22p63s23p64s23dl04p65s14d105p2
C 1s22s22p63s23p64s23dl04p65s14d95p3
D 1s22s22p63s23p64s23dl04p65s24d85p3
GO ON TO THE NEXT PAGE.
Trang 61 1 What is the maximum number of electrons shell
number 4 can hold?
A 32
B 18
C 8
D 2
1 2 Which set of elements has electrons added to the 4f
subshell as the atomic number increases?
A Metalloids
B Actinides
C Alkali metals
D Lanthanides
GO ON TO THE NEXT PAGE.
Trang 7Questions 13 through 17 are NOT
based on a descriptive passage
1 3 All halogens have similar reactivity because:
A they have the same number of protons.
B they have the same number of electrons.
C they have similar outer shell electron
configurations
D they have valence electrons with the same
quantum numbers
1 4 Electron affinity is defined as:
A the change in energy when a gaseous atom in its
ground state gains an electron
B the pull an atom has on the electrons in a
chemical bond
C the energy required to remove a valence electron
from a neutral gaseous atom in its ground state
D the energy difference between an electron in its
ground and excited states
1 5 The modern periodic table is ordered on the basis of
A atomic mass.
B atomic radius.
C atomic charge.
D atomic number.
1 6 Two electrons with the same n, l, and ml values:
A must be in different atoms.
B are in different orbitals of the same subshell.
C are in the same orbital of the same subshell with
opposite spins
D are indistinguishable from each other.
1 7 Which of the following would be different in a
ground-state and an excited-ground-state neon atom?
A The number of neutrons
B The number of electrons
C The atomic weight
D The electronic configuration
END OF TEST
Trang 8ANSWER KEY:
Trang 9ELECTRONIC STRUCTURE AND THE PERIODIC TABLE TEST 1 EXPLANATIONS
Passage I (Questions 1–6)
1 B
In order to answer this question correctly you need to understand and use Equation 1 The question-stem states that the ionization energy of sodium is 496 kJ/mole; in other words, 496 kilojoules of energy is required to remove one mole of
electrons from one mole of sodium atoms Well, all you have to do is plug the energy into Equation 1 and divide by Planck's constant to get the frequency of the photon, right? Well, essentially, yes, but remember that 496 kJ is the amount of energy for one mole of these photons, not for an individual photon So if one mole of photons has an energy of 496 kJ, one photon has
496 kJ divided by 6.02 ∞ 1023, or 8.24 ∞ 10–22, kilojoules of energy So, converting that to joules to match units with Planck's constant and plugging into Equation 1, we get that the frequency of the photons is 8.24 ∞ 10–19 joules divided by 6.62
∞ 10–34 joule seconds, or 1.24 ∞ 1015 hertz, answer choice B If you failed to recognize that the energy you were given was per mole, not per photon, you would have gotten answer choice D If you had forgotten that Planck's constant is in joule seconds and the energy was in kilojoules so there had to be a conversion, you would have gotten choice A Choice C represents a combination of calculation errors, including moving the decimal point the wrong way when converting kilojoules to joules
2 A
The first thing you may have noticed about this question is that in describing the first ionization energies of
magnesium (atomic number 12) and aluminum (atomic number 13), the trend in the periodic table that shows increasing first ionization energies across a period doesn't apply Due to the increased nuclear pull, first ionization energies generally increase across a period, but here, the first ionization energy has decreased First of all, remember that the trends in the periodic table are not absolute Second, the question is essentially asking us to explain why these two elements don't follow the trend Why is the first ionization energy of aluminum lower than the first ionization energy of magnesium? To answer that, we have to look
at the electronic structure of these two atoms Magnesium is in the second column of the periodic table, indicating that its two
valence electrons are in the 3s subshell Aluminum, in column 13, has one of its three valence electrons in the 3p subshell and the other two in the 3s Aluminum's lone "p" electron has a higher energy than magnesium's "s" electrons and is, therefore,
held less tightly by the nucleus Because it is held less tightly, it requires less energy to remove it A similar situation exists between beryllium and boron Choice B is wrong because aluminum, with one more proton than magnesium, has a greater nuclear charge Choice C is wrong because the two positively charged nuclei of neighboring atoms repel each other; they don't
attract as indicated here D is also incorrect because there is only one "p" electron in aluminum; no electron-electron repulsions
exist
3 A
As the question states, and is stated in the question, the second ionization energy is the energy required to remove an electron from the +1 cation; in essence, the energy needed to remove the second most energetic electron from an atom after the first is removed All atoms have as many ionization energies as they do electrons, but we hardly ever deal with anything higher than the sixth; after all, when was the last time you had to deal with a +20 cation? Anyway, which of the statements is correct? Choice A says that the second ionization energy of an atom is always greater than the first ionization energy of the same atom That seems right After all, the second electron to be removed must have had a lower energy than the first to begin with, so it would take more energy to remove it Further, it requires more energy to remove an electron from a positively charged ion So choice A looks good, but we should check on the other three Choice B says that the second ionization energies can be lower or higher than the first The only way that the second ionization energy can be lower than the first is if the second electron to be removed has a greater energy than the first However, if the second electron had more energy, it would have been the first electron removed, so choice B is wrong Choice C is wrong because, as we've just discussed, second ionization energies are not lower than first ionization energies Choice D is wrong for the same reason, but also because ionization does not always destabilize a species Sodium tends to be found not as neutral Na, but as the Na+ cation because the cation is more stable than the neutral atom So choice A is the best answer
4 B
This question is only indirectly related to the passage The distance from the nucleus is a factor in ionization energy
In general, the further an electron is from the nucleus, the greater its overall energy is and the easier it is to strip from the atom
It should also be apparent to you that the further away from the nucleus the outermost electron is, the larger the atomic radius
is The trend in atomic radius in the periodic table is just the opposite of that for ionization energy, that is, it generally
decreases as you move across a period and increases as you move down a column The radius decreases from left to right across
a period because the effective nuclear charge is increasing: the greater the effective nuclear charge, the stronger the electrons are
Trang 10attracted to the nucleus, the smaller the radius The radii increase in going down a group because in the higher periods, the value of n increases: the larger n is, the larger the orbitals are, and the further away the electrons are from the nucleus
Applying this knowledge to the question, we can eliminate choices C and D pretty quickly since they are in the sixth period and the other choices are in the fourth We know that the atomic radii of potassium and calcium will be much smaller than those of cesium and barium Since calcium is to the right of potassium in the same period, we can say that choice B, calcium, has the smallest radius
5 D
This answer shouldn't come as a great surprise to you By now, you should have a handle on the fact that ionization,
be it photoionization or some other method for removing an electron from an atom, takes an electron from the valence shell of
an atom That means that this question essentially asks you what the valence shell of cesium is Cesium is element 55 It is the first element in the sixth period of the periodic table Here's where some knowledge of how the periodic table is set up
comes in handy In the period table, the first two columns are the s-block elements Their valence electrons are in the s subshell of whatever energy level characterizes their outermost shell The next ten columns are the d-block elements, also called the transition elements The next six columns are the p-block elements and the 28 elements below in the lanthanide and
actinide series, the inner transition metals, are the f-block elements Notice that the number of columns in a block is the same
as the maximum number of electrons the subshell that block is named after can hold Now, don't think that the elements in any one block can only use the electrons in the subshell the block is named for as valence electrons Metals in the fourth period
transition elements often use both 4s and 3d electrons in reactions and elements in the p-block often utilize the electrons in the s
subshell too However, the blocks are named as such because each electron added in these blocks goes into the subshell the
block is named for; the block name identifies the valence subshell Anyway, let's answer the question Since cesium is an s-block element, its valence electrons are in the s subshell, so the answer must be choice D Further, since cesium is in the sixth
period, it must have valence electrons in the sixth shell Again, only choice D reflects electrons in the sixth shell So by both the shell and subshell identity, the only possible answer to this question is D
6 D
The best way to answer this question is by eliminating each incorrect choice In doing this, it forces you to read each answer choice in turn before deciding which one is the best So here we're looking for the best reason that photoionization, for the sake of determining the first ionization energy, should take place in the gas phase Choice A says that it is because only gas phase atoms can absorb visible light Well, this is untrue If solids or liquids were unable to absorb at least some visible light, there would be no colored solids or liquids So choice A must be wrong B says that the atoms must be in the gas phase because the refractive index is greater than one, so the frequency of the absorbed light is obscured This is untrue, the frequency
of absorption can be measured; what effect the photon has on the material is another story entirely choice B is incorrect Choice C essentially says that solids and liquids cannot be ionized and so electrons cannot be removed or shifted in solids or liquids Not true! Think about electricity How could solid metal wires conduct electricity if their electrons couldn't be moved around? How could redox reactions of solids take place if electrons couldn't be gained or lost? That leaves choice D as the right answer It says that other energy factors on the electrons between the atoms of solids or liquids distorts the light frequency that
an electron can absorb by altering the energy of the electron That makes sense After all, when bonds are formed, energy is released To remove the bonding electrons, energy must first be added to break the bond and then to release the electron from the nuclear pull When atoms are associated through some sort of intermolecular force like hydrogen bonds, energy must be added
to break the intermolecular association and then to release the electron So only for elements in the gas phase, where the distances between atoms is so great that there is very little interatomic interactions, will photoionization give a measurement for the ionization energy that is not affected by other energy factors
Passage II (Questions 7–12)
7 C
The answer to this question comes from the passage Electrons fill the orbitals around the nucleus of an atom in such
a way as to give the lowest possible atomic energy That was right in the passage This knowledge, along with the fact that half-filled and fully filled subshells have increased stability, gives us the answer In the case of chromium, a lower energy is
achieved when one of the 4s electrons is transferred to the 3d subshell, making it half-filled Choice C is the correct answer.
Other anomalies include molybdenum, which behaves similarly to chromium; copper, silver, and gold, all of which promote an
"s" electron to give a fully filled "d" subshell All right, choice A is wrong because there would be no 4s electrons in this atom
if the 4s orbital filled last Even with five 3d electrons, the subshell has room for 5 more Choice B is wrong because it implies that the 4s electrons would have remained where they were while the 3d orbitals filled That doesn't seem to be the case here since a 4s electron has moved into the 3d subshell Finally, choice D is more of a consequence, not a cause, of the