ch2 - ChemistryVCE

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Worked solutions to textbook 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, which 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 Thomson had evidence that atoms contained positively charged matter and negatively charged particles. b 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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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

112

48

Cd, state: a the number of protons b the number of electrons c 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

2

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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 d

130

56

Ba

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 electrons Formula of ion

13 27 10

27

13

Al

3

12 24 10

12

16

25

34

10

18

7

19

15

40 18

40

20

Ca

2

15

7

N

3

3

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A9.

Atomic number Mass number

13 27

12

12

16

20

7

19

24

25

34

40

15

40

Number of electrons

10

10

10

18

18

10

18

Formula of ion

27

13

Al

3

24

12

Mg

2

25

Mg

12

2

34

16

S

2

40

20

Ca

2

15

N

3

7

40

19

K

Q10.

Using Table 2.3 on page 26, 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 2 b 3 c 1 d 4 e 2 f 8

4

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Q11.

Using Table 2.3 on page 26, 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 fluorine and bromine b silicon c oxygen and sulfur d boron and aluminium

Q12.

Write the electronic configuration of: a Be b sulfur c Ar d magnesium e Ne

A12. a 2,2 b 2,8,6 c 2,8,8 d 2,8,2 e 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 helium, He b fluorine, F c aluminium, Al d nitrogen, N e chlorine, Cl

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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 number) boron (5) carbon (6) lithium (3) fluorine (9) chlorine (17)

Electronic configuration

(using the shell model)

2,3

Electronic configuration

(using the subshell model)

1s

2

2s

2

2p

1 sodium (11) aluminium (13) neon (10) argon (18) potassium (19) calcium (20) scandium (21) nickel (28) bromine (35)

A14.

Element (atomic number) boron (5) carbon (6) lithium (3) fluorine (9) chlorine (17) sodium (11) aluminium (13) neon (10) argon (18) potassium (19) calcium (20) scandium (21) nickel (28) bromine (35)

Electronic configuration (using the shell model)

2,3

2,4

2,1

2,7

2,8,7

2,8,1

2,8,3

2,8

2,8,8

2,8,8,1

2,8,8,2

2,8,9,2

2,8,16,2

2,8,18,7

Electronic configuration

(using the subshell model)

1s 2 2s 2 2p 1

1s

2

2s

2

2p

2

1s

2

2s

1

1s

2

2s

2

2p

5

1s

2

2s

2

2p

6

3s

2

3p

5

1s

2

2s

2

2p

6

3s

1

1s

2

2s

2

2p

6

2s

2

2p

1

1s 2 2s 2 2p 6

1s

2

2s

2

2p

6

3s

2

3p

6

1s 2 2s 2 2p 6 3s 2 3p 6 4s 1

1s

2

2s

2

2p

6

3s

2

3p

6

4s

2

1s

2

2s

2

2p

6

3s

2

3p

6

3d

1

4s

2

1s

2

2s

2

2p

6

3s

2

3p

6

3d

8

4s

2

1s

2

2s

2

2p

6

3s

2

3p

6

3s

2

3d

10

4s

2

4p

5

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 textbook questions

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 × 10 23

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

27 g of aluminium contains 6.02 × 10 23

atoms so 2.7 g of aluminium contains x atoms

By ratio, 2.7/27 = x /6.02 × 10 23 x = 6.0 × 10

22 i.e. 2.7 g of aluminium contains 6.20 × 10 22

aluminium atoms b Applying the same method as in part a :

1/27 = x

/6.02 × 10 23 so x = 2.2 × 10

22 i.e. 1.00 g of aluminium contains 2.2 × 10 22

atoms c 0.16 g of aluminium contains 6.02 × 10 22 i.e. 3.6 × 10 21

atoms

× 0.16/27 atoms d

4.8 kg = 4.8 × 10 3

g

4.8 × 10 3 g of aluminium contains 6.02 × 10 23 × 4.8 × 10 3 /27 aluminium atoms i.e. 1.1 × 10 26

atoms

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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

235

92

U. Give its atomic number and mass 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

He b

19

9

F c d

23

11

Na

40

20

Ca

A23.

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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 × 10

13

).

Therefore, the radius of the atom must also be increased by a factor of 5 × 10 13

.

This means the radius of the scaled atom is 10

–9

× 5 × 10 13 cm = 5 × 10 4 cm or 500 m.

Q28. a List the chief assumptions of the Bohr theory for the behaviour of electrons in atoms. b In what respects did the theory prove inadequate? c What modifications to the theory were introduced by quantum mechanics?

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A28. a 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. b 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. c 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 1s 2 b 1s

2

2s

2

2p

2 c 1s

2

2s

2

2p

5 d 1s

2

2s

2

2p

6

3s

2

3p

1 e 1s

2

2s

2

2p

6

3s

2

3p

6 f 1s

2

2s

2

2p

6

3s

2

3p

6

3d

8

4s

2 g 1s

2

2s

2

2p

6

3s

2

3p

6

3d

10

4s

2

4p

5

Q30. a State whether atoms with the following electronic configurations are in the ground state or an excited state. i 1s 2 2s 2 2p 1 ii 1s

2

2s

2

3s

2 iii 1s

2

2s

2

2p

6

3s

1

3p

1 iv 1s

2

2s

2

2p

6

3s

2

3p

6

4s

1 v 1s

2

2s

2

2p

6

3s

2

3p

6

3d

2

4s

2 b Identify the elements that could have the electron arrangements given in part a .

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A30. a i ground state ii excited state iii excited state iv ground state v ground state b i boron ii carbon iii magnesium iv potassium v titanium

Q31.

Write electronic configurations for each of the following species in their lowest energy states: a

16

8

O b

32

16

S

2– c d

37

17

Cl

25

12

Mg

2+

A31. a 1s

2

2s

2

2p

4 b 1s 2 2s 2 2p 6 3s 2 3p 6 c 1s

2

2s

2

2p

6

3s

2

3p

6 d 1s

2

2s

2

2p

6

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 s- and 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 1s 2 2s 2 2p 5 .

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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 textbook questions 13 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 i Thomson’s ii

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. b Ockham may well have preferred the model represented by Figure 2.25c, as it more simply accounts for the presence of isotopes.

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

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. b 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. 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.

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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 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. b 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. c 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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