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Transcript
Worked solutions to student book questions
Chapter 2 A particle view of matter
Q1.
Dalton and Thomson each proposed a model of an atom.
a What experimental evidence did Thomson have that was not available to Dalton?
b As a result of this experimental evidence, how did Thomson’s model of an atom
differ from that of Dalton’s?
A1.
a
b
Thomson had evidence that atoms contained positively charged matter and
negatively charged particles.
Thomson proposed an atomic model in which the negatively charged particles
were embedded in a sphere of positively charged matter. In this model, the
number of negatively charged particles would distinguish the atoms of one
element from those of all other elements. Dalton described the atom as an
indivisible particle that was the basic unit of an element. The mass of an atom
distinguished one element from all others.
Q2.
Scientists often use models to help them to understand something that is either too
small to see or too large to imagine. Give two examples of scientific models you have
used, or perhaps made, in the past.
A2.
Solar system, water cycle, plant structure, volcanoes, plant and animal cells
Q3.
What does the word nucleus mean?
A3.
Nucleus: the central part of a system around which other parts are arranged or
grouped.
Q4.
Describe the experimental evidence that led to Rutherford’s model of a nuclear atom.
A4.
Rutherford ‘fired’ a stream of alpha particles at a piece of gold foil. When most of the
alpha particles passed through the foil in straight lines or with minor deflection,
Rutherford concluded that the gold atoms were largely empty space. The observation
that a small fraction of the positively charged alpha particles were deflected through
large angles led him to propose that the atoms had a very small region of positive
charge (i.e. the nucleus). The positive alpha particles were repelled by the positively
charged nuclei.
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Worked solutions to student book questions
Chapter 2 A particle view of matter
Q5.
Define the following terms: atomic number, mass number, isotope.
A5.
Atomic number is the number of protons in the nucleus of an atom.
Mass number equals the sum of the number of protons plus the number of neutrons in
the nucleus of an atom.
Isotopes are atoms with the same atomic number (i.e. same element) but different
mass numbers (i.e. different numbers of neutrons).
Q6.
For the cadmium atom containing the nucleus
a
b
c
112
Cd,
48
state:
the number of protons
the number of electrons
the number of neutrons present in the atom.
A6.
The number of neutrons can be calculated by subtracting the atomic number from the
mass number. The number of electrons is always equal to the atomic number for a
neutral atom. (An ion may have more or fewer electrons than protons.)
a 48
b 48
c 64
Q7.
Use the isotopic symbol convention shown in Question 6 to describe the following
atoms:
a a carbon atom that has 6 protons, 6 neutrons and 6 electrons
b a carbon atom that has 6 protons, 7 neutrons and 6 electrons
c a carbon atom that has 6 protons, 8 neutrons and 6 electrons
d an aluminium atom that has 13 protons, 14 neutrons and 13 electrons
A7.
a
12
6
C
b
13
6
C
c
14
6
C
d
27
13
Al
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Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)
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Worked solutions to student book questions
Chapter 2 A particle view of matter
Q8.
Represent each of the following atoms using the isotopic symbol convention you used
in Question 7. (You may need to refer to the list of elements in Appendix 4, page
389.)
a an atom that has 8 protons, 8 neutrons and 8 electrons
b an atom that has 16 protons, 18 neutrons and 16 electrons
c an atom that has 56 protons, 74 neutrons and 56 electrons
d an atom that has 56 protons, 82 neutrons and 56 electrons
A8.
a
16
8
O
b
34
16
S
c
130
56
Ba
d
138
56
Ba
Q9.
Complete the following table by filling in the missing detail about each ion. The first
one has been completed as an example.
Atomic number
Mass number
Number of
Formula of ion
electrons
13
27
10
12
24
10
12
25
10
16
34
18
7
15
19
40
27
13
Al 3+
40
20
Ca 2 +
15
7
N 3−
18
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Worked solutions to student book questions
Chapter 2 A particle view of matter
A9.
Atomic number
Mass number
Number of
electrons
Formula of ion
13
27
10
27
13
Al 3+
12
24
10
24
12
Mg 2 +
12
25
10
25
12
Mg 2 +
16
34
18
34
16
S 2−
20
40
18
40
20
Ca 2 +
7
15
10
15
7
N 3−
19
40
18
40
19
K+
Q10.
Using Table 2.3 on page 26 of the student book, give the number of valence electrons
in atoms of each of the following elements:
a magnesium
b boron
c K
d carbon
e Be
f Ar
A10.
a
b
c
d
e
f
2
3
1
4
2
8
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Worked solutions to student book questions
Chapter 2 A particle view of matter
Q11.
Using Table 2.3 on page 26 of the student book, name the element(s) that has (have):
a the same number of valence electrons as chlorine
b the same number of valence electrons as C
c one more valence electron than P
d two fewer valence electrons than nitrogen.
A11.
a
b
c
d
fluorine and bromine
silicon
oxygen and sulfur
boron and aluminium
Q12.
Write the electronic configuration of:
a Be
b sulfur
c Ar
d magnesium
e Ne
A12.
a
b
c
d
e
2,2
2,8,6
2,8,8
2,8,2
2,8
Q13.
Write the name and symbol of the element with the electronic configuration:
a 2
b 2,7
c 2,8,3
d 2,5
e 2,8,7
A13.
a
b
c
d
e
helium, He
fluorine, F
aluminium, Al
nitrogen, N
chlorine, Cl
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Worked solutions to student book questions
Chapter 2 A particle view of matter
Q14.
Complete the following table by using the examples given to write the electronic
configuration of each of the atoms in its electronic ground state.
Element (atomic
Electronic configuration Electronic configuration
number)
(using the shell model)
(using the subshell
model)
Boron (5)
2,3
1s22s22p1
Lithium (3)
Chlorine (17)
Sodium (11)
Neon (10)
Potassium (19)
Scandium (21)
Iron (26)
Bromine (35)
A14.
Element (atomic
number)
Boron (5)
Lithium (3)
Chlorine (17)
Sodium (11)
Neon (10)
Potassium (19)
Scandium (21)
Iron (26)
Bromine (35)
Electronic
configuration (using the
shell model)
2,3
2,1
2,8,7
2,8,1
2,8
2,8,8,1
2,8,9,2
2,8,16
2,8,18,7
Electronic configuration
(using the subshell model)
1s22s22p1
1s22s1
1s22s22p63s23p5
1s22s22p63s1
1s22s22p6
1s22s22p63s23p64s1
1s22s22p63s23p63d14s2
1s22s22p63s23p63d64s2
1s22s22p63s23p63s23d104s24p5
Q15.
In terms of energy levels, what is the essential difference between the shell model and
the subshell model of the atom?
A15.
The subshell is a refinement of the shell model. The shell model proposed that all
electrons in the one shell were of equal energy. Evidence from emission spectra
indicated that there were different electronic energy levels (called subshells) within a
shell.
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Worked solutions to student book questions
Chapter 2 A particle view of matter
Chapter review
Q16.
Thomson concluded that all atoms of an element had the same mass. Given
Thomson’s plum pudding model, in what ways would an atom of one element have
differed from atoms of other elements?
A16.
Thomson’s plum pudding model could have accounted for atoms of different mass if
one assumed the mass of the positive matter in atoms was different for each element.
(The total mass of the electrons would also vary from one element to the next but
electrons contribute very little to the mass of atoms.)
Q17.
The neutron was not discovered until more than 30 years after the discovery of the
proton and the electron. Why was the neutron more difficult to detect?
A17.
Most of the instruments used for investigating the structure of the atom are based on
the use or measurement of electric charge. Since the neutron is an uncharged particle,
it was not detected by these instruments.
Q18.
27 g of aluminium contains approximately 6.02 × 1023 aluminium atoms. Calculate
the number of aluminium atoms in the following masses of aluminium:
a 2.70 g
b 1.00 g
c 0.16 g
d 4.8 kg
A18.
a
b
c
27 g of aluminium contains 6.02 × 1023 atoms
so 2.7 g of aluminium contains x atoms
2.7
x
By ratio,
=
27 6.02 ×1023
x = 6.0 × 1022
i.e. 2.7 g of aluminium contains 6.20 × 1022 aluminium atoms
Applying the same method as in part a:
1
x
=
27 6.02 ×1023
so x = 2.2 × 1022
i.e. 1.00 g of aluminium contains 2.2 × 1022 atoms
6.02 × 1022 × 0.16
0.16 g of aluminium contains
atoms
27
i.e. 3.6 × 1021 atoms
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Worked solutions to student book questions
Chapter 2 A particle view of matter
d
4.8 kg = 4.8 × 103 g
4.8 × 103 g of aluminium contains
i.e. 1.1 × 1026 atoms
6.02 × 1023 × 4.8 ×103
aluminium atoms
27
Q19.
How are protons, electrons and neutrons arranged in an atom?
A19.
The protons and neutrons form the nucleus. The electrons are grouped in shells and
occupy the space around the nucleus.
Q20.
Compare the mass and charge of protons, electrons and neutrons.
A20.
The mass of a proton is approximately equal to the mass of a neutron and is about
1840 times the mass of an electron. The proton and electron have equal but opposite
charges and the neutron has no charge.
Q21.
An atom of uranium can be represented by the symbol
and mass number.
235
92
U. Give its atomic number
A21.
atomic number is 92; mass number is 235
Q22.
What is the maximum number of electrons in the second shell of an atom?
A22.
8
Q23.
Make a sketch representing each of the following atoms, showing electrons in their
major cells.
a
4
2
b
19
9F
c
23
11 Na
d
40
20
He
Ca
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Worked solutions to student book questions
Chapter 2 A particle view of matter
A23.
Q24.
Two atoms both have 20 neutrons in their nucleus. The first also has 19 protons and
the other has 20 protons. Are they isotopes? Why or why not?
A24.
No. Isotopes have the same number of protons in their nuclei.
Q25.
Explain why the number of electrons in an atom equals the number of protons.
A25.
Atoms are electrically neutral. The positive charge on one proton balances the
negative charge on one electron. Therefore, for electrical neutrality, there must be
equal numbers of protons and electrons.
Q26.
Using the element bromine as an example, explain why elements are best identified
by their atomic number and not by their mass number.
A26.
Most elements have more than one isotope, so they will have more than one mass
number. All bromine atoms have 35 protons in their nuclei. No other type of atom has
35 protons in its nucleus (i.e. no other atom has an atomic number of 35). Isotopes of
bromine, however, differ in their mass numbers, so mass number is not fixed for an
element (except for those elements such as sodium, which have only one naturally
occurring isotope). In addition, an isotope of one element may have the same mass
number as an isotope of another element.
Q27.
The nucleus of an atom has a radius of the order of 10–13 cm. The radius of the atom
itself is of the order of 10–9 cm. If the nucleus could be scaled up to the size of an
orange (radius 5 cm), what would be the radius of the atom at that same scale?
A27.
The radius of the nucleus and the atom are increased by the same factor.
We are told the radius of the atom increases from 10–13 cm to 5 cm, that is, by a factor
of 5/10–13
(i.e. by 5 × 1013).
Therefore, the radius of the atom must also be increased by a factor of 5 × 1013.
This means the radius of the scaled atom is 10–9 × 5 × 1013 cm = 5 × 104 cm or 500 m.
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Worked solutions to student book questions
Chapter 2 A particle view of matter
Q28.
a
b
c
List the chief assumptions of the Bohr theory for the behaviour of electrons in
atoms.
In what respects did the theory prove inadequate?
What modifications to the theory were introduced by quantum mechanics?
A28.
a
b
c
The main assumptions of the Bohr theory were that electrons in atoms circled the
nucleus without loss of energy, electrons moved only in certain fixed orbits of
particular energies, and an electron’s orbit depended on its energy.
The theory was inadequate because it did not explain why electrons moved only
in circular orbits. Also, calculations, based on the model, of the energy of lines in
emission spectra of atoms with more than one electron agreed poorly with
measured values of the energies.
According to quantum mechanics, electrons have wave-like behaviour. By
applying equations that describe the behaviour of waves to the electron, many
new ideas emerged, including that of the existence of subshells and orbitals.
Q29.
Write electronic configurations, using subshell notation, for atoms in the ground state
of the following elements. The atomic number of each element is shown in brackets:
a helium (2)
b carbon (6)
c fluorine (9)
d aluminium (13)
e argon (18)
f nickel (28)
g bromine (35)
A29.
a
b
c
d
e
f
g
1s2
1s22s22p2
1s22s22p5
1s22s22p63s23p1
1s22s22p63s23p6
1s22s22p63s23p63d84s2
1s22s22p63s23p63d104s24p5
Q30.
a
b
State whether atoms with the following electronic configurations are in the
ground state or an excited state.
i 1s22s22p1
ii 1s22s23s2
iii 1s22s22p63s13p1
iv 1s22s22p63s23p64s1
v 1s22s22p63s23p63d24s2
Identify the elements that could have the electron arrangements given in part a.
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Worked solutions to student book questions
Chapter 2 A particle view of matter
A30.
a
b
ground state
excited state
excited state
ground state
ground state
boron
carbon
magnesium
potassium
titanium
i
ii
iii
iv
v
i
ii
iii
iv
v
Q31.
Write electronic configurations for each of the following species in their lowest
energy states:
a
16
8O
b
32
16
S2–
c
37
17
Cl–
d
25
12
Mg2+
A31.
a
b
c
d
1s22s22p4
1s22s22p63s23p6
1s22s22p63s23p6
1s22s22p6
Q32.
Using the fluorine atom as an example, explain the difference between the terms shell,
subshell and orbital.
A32.
A fluorine atom contains nine electrons. The electrons are arranged in energy levels
called shells; two electrons are in the first shell and seven electrons are in the second
shell, which has higher energy. The electron arrangement in the shells can be written
as 2,7.
Shells are regarded as being made up of energy levels called subshells. The first shell
contains an s-type subshell, which is labelled ‘1s’. The second shell contains both sand p-type subshells, labelled ‘2s’ and ‘2p’, respectively.
Within subshells, electrons occupy regions of space known as orbitals. An orbital can
hold up to two electrons. Subshells of an s-type contain one orbital, whereas p-type
subshells contain three orbitals. The electron arrangement in the subshells of a
fluorine atom can be represented as 1s22s22p5.
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Worked solutions to student book questions
Chapter 2 A particle view of matter
Q33.
Today’s atomic model describes an atom as consisting of rapidly moving electrons at
a relatively large distance from a very small central nucleus. What is there between
those electrons and the nucleus?
A33.
Nothing.
Q34.
The alchemists spent a great deal of time trying to make precious substances, such as
gold, from base (cheap) metals. Explain, in terms of modern atomic theory, why they
were unsuccessful.
A34.
Elements differ in the number of protons in their atomic nuclei. The early alchemists
did not have the means to generate the amount of energy required to change atomic
nuclei. Rather, their chemical reactions only involved the rearrangement of the atoms’
electrons.
Q35.
The Englishman William of Ockham (1280– c. 1349) had some interesting thoughts
on the development of explanations and theories. He is quoted as saying ‘Pluralitas
non est ponenda sine necessitate’, which can be translated as ‘Entities should not be
multiplied unnecessarily’. Today, we would probably use the expression ‘keep it
simple’. A more subtle interpretation of Ockham’s Razor would suggest the
following:
• If two competing theories have the very same predictions, then we adopt the
simpler of the two theories.
• However, if we have two competing theories that give different predictions, then
we need to experiment to identify the most logical theory.
a i Suppose William of Ockham had been alive at the end of the 20th century;
do you think he would have supported Dalton’s atomic theory or that of
Thomson?
ii On what basis might he have made his decision?
iii Since Thomson’s plum pudding model, successive models of the atom have
become more complex. Explain whether or not you think this is against
William of Ockham’s preference to ‘keep it simple’.
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Worked solutions to student book questions
Chapter 2 A particle view of matter
b
Before the discovery of the neutron by Chadwick in 1930, scientists had predicted
the existence of a neutral nuclear particle. Suppose Figure 2.25 represents three
possible models of a sodium atom at this time. Which one do you think William
of Ockham would have preferred and why?
Figure 2.25
A35.
a
b
Thomson’s
Thomson’s theory better explains the evidence available at the end of the
19th century of the existence of subatomic particles than does Dalton’s
model of an indivisible atom.
iii Ockham favoured the simplest theory or model that was consistent with
available information. The increasing complexity of atomic models reflects
the increasing amount of experimental information about subatomic particles
and their arrangement within atoms.
Ockham may well have preferred the model represented by Figure 2.25c, as it
more simply accounts for the presence of isotopes.
i
ii
Q36.
New models for the atom have evolved as scientists become aware of inconsistencies
between current models and experimental data. Outline the problems with the existing
model of the atom that led to the modifications suggested by the following scientists:
a Rutherford
b Bohr
c Schrödinger
A36.
a
b
Until Rutherford’s work, the plum pudding model of the atom was widely
accepted. However, his discovery that a beam of alpha particles directed at thin
gold foil causes a few particles to deflect through high angles led to the
development of a new atomic model.
Although Rutherford’s atomic model accounted for a number of atomic
properties, it was not able to account for the characteristic emission spectra that
each element produces. The model was also in conflict with the principles of
classical physics, which suggested that electrons moving in circular orbits should
continuously lose energy and spiral into the nucleus.
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Worked solutions to student book questions
Chapter 2 A particle view of matter
c
The Bohr model of the atom did not adequately explain why electrons adopted
some energy levels but not others. In addition, calculated frequencies for lines in
the emission spectra of atoms with more than one electron gave poor agreement
with measured values.
Q37.
It is approximately 200 years since John Dalton proposed that matter was composed
of indivisible particles called atoms. Most of the evidence for the existence of
subatomic particles was not found until the early part of the 20th century.
a Ernest Rutherford and his co-workers fired alpha particles at a thin sheet of gold
foil and measured the deflection produced in the path of the alpha particles.
Briefly explain the results of this experiment and the concepts about atomic
structure that came from this.
b In 1913, Niels Bohr proposed that electrons circled the nucleus in fixed orbits, but
this idea was modified in the 1920s in light of knowledge of quantum mechanics.
Outline the current model of electron behaviour by using the terms shell, subshell
and orbital.
c After learning about the work of chemists such as Marie Curie in isolating new
elements, a student asked the teacher, ‘Do you think there are elements lighter
than uranium that have not been discovered yet?’ Describe how you think the
teacher would respond.
A37.
a
b
c
Most of the alpha particles passed through the gold foil as if there was nothing
there, showing that the atom is mainly empty space. Occasionally alpha particles
were deflected. From the frequency and angles of deflection, Rutherford was able
to deduce that most of the matter (mass) of the atom was concentrated in a very
small nucleus at the centre of the atom.
Electrons possess particular amounts of energy. The energies of the electrons of
an atom are grouped into energy levels called shells. Within each shell, the
energy levels subdivide into subshells. When an electron has the energy of a
particular subshell it is most likely to be found in a region of space around the
nucleus called an orbital.
Work begun by Henry Moseley allowed atomic numbers to be assigned to the
elements, beginning with hydrogen as 1. The atomic number is equal to the
number of protons in the nucleus of an atom of an element. Since there are no
atomic numbers between 1 and 92 (uranium) for which elements have not been
discovered, there are no elements lighter than uranium still to be discovered.
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