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© Nelson Australia Pty Ltd 2007 Student Activity Manual solutions and information, Chapter 4- 1 -

Activity 4.1 Patterns in the atomic volumesof the elements

Information and advice

This is an activity in which students process and interpret second-hand data, which not only

provides training in many important skills but also enables them to understand and appreciate some

of the rigorous thinking behind the development of the Periodic Table.

Suggested solutions to pre-lab activity: Focus questions

1 a The rapid movement of the electrons of the atom, and their maximum distance from the

nucleus define the radius of the atom and hence its volume.

b They should have the same size since they have the same number of protons in their

nucleus attracting the electrons towards the nucleus and the same electron

configuration of the uncharged atom.

2 a 7.9 g cm –3

b 52 cm3

c Factors include the masses of the atoms, the volume of the particles present (atoms or

molecules) and how closely they pack together in the solid.

d When making a particular-sized object, it affects the weight of the object and hence how

much fuel is needed to transport it and how much energy is required to lift it in place.

3 a 3

3

4 r V π=

b 0.03 cm3

c 2.1 10 –22 cm3

d 2.70 10 –20 cm3

e 130 times greater

f The uncharged sodium atom has two more occupied shells of electrons than the

uncharged hydrogen atom.

4 This law stated that if the elements were listed in order of increasing atomic weight, about

every 8th element would have similar properties. It was proposed by John Newlands.

5 Student responses.

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Suggested solutions to calculations from second-hand data

1 Table 4.1 Properties of the elements

Element Symbol Atomic weight

(relativeatomic mass)

Density

(g cm –3)

Relative

atomic volume

(cm3)*

Hydrogen H 1.0 0.07 (l) 1

Lithium Li 6.9 0.53 13

Beryllium Be 9.0 1.85 4.9

Boron B 10.8 2.34 4.62

Carbon C 12.0 2.26 (graphite) 5.31

Nitrogen N 14.0 0.81 (l) 17Oxygen O 16.0 1.15 (O2, l) 13.9

Fluorine F 19.0 1.51 (l) 12.6

Sodium Na 23.0 0.97 24

Magnesium Mg 24.3 1.74 14.0

Aluminium Al 27.0 2.70 10.0

Silicon Si 28.1 2.33 12.1

Phosphorus P 31.0 1.82 (white) 17.0

Sulfur S 32.1 2.07 (rhombic) 15.5

Chlorine Cl 35.5 1.56 (l) 22.8

Potassium K 39.1 0.86 45

Calcium Ca 40.1 1.55 25.9

Scandium Sc 45.0 3.0 15

Titanium Ti 47.9 4.50 10.6

Vanadium V 50.9 6.1 8.3

Chromium Cr 52.0 7.19 7.23

Manganese Mn 54.9 7.43 7.39

Iron Fe 55.8 7.86 7.10

Nickel Ni 58.7 8.90 6.60

Cobalt Co 58.9 8.90 6.62

Copper Cu 63.5 8.96 7.09

Zinc Zn 65.4 7.14 9.16

Gallium Ga 69.7 5.91 11.8

Germanium Ge 72.6 5.32 13.6

Arsenic As 74.9 5.72 (grey) 13.1

Selenium Se 79.0 4.80 (grey) 16.5

Bromine Br 79.9 3.12 25.6

Rubidium Rb 85.5 1.53 55.9

Strontium Sr 87.6 2.6 34

Yttrium Y 88.9 4.47 19.9Zirconium Zr 91.2 6.49 14.1

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Element Symbol Atomic weight

(relative

atomic mass)

Density

(g cm –3)

Relative

atomic volume

(cm3

)*Niobium Nb 92.9 8.57 10.8

Molybdenum Mo 95.9 10.2 9.40

Ruthenium Ru 101.1 12.4 8.15

Rhodium Rh 102.9 12.4 8.30

Palladium Pd 106.4 12.0 8.87

Silver Ag 107.9 10.5 10.3

Cadmium Cd 112.4 8.65 13.0

Indium In 114.8 7.31 15.7

Tin Sn 118.7 7.30 (white) 16.3

Antimony Sb 121.8 6.68 18.2

Iodine I 126.9 4.94 25.7

Tellurium Te 127.6 6.24 20.4

Caesium Cs 132.9 1.87 71.1

Barium Ba 137.3 3.50 39.2

* Note: According to conventional practice, the number of significant figures in the answer has been determined

by the number of significant figures in the experimental data (in this case, density).

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2

Figure 4.1

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3

Figure 4.2

Suggested solutions to discussion questions

1 Student response. (If you examine the peaks of the graphs of atomic volume against atomic

weight, Na is the 8th element counting from Li, and K is the 8th element counting from Na, but

the other peaks do not occur at the 8th element. If you examine the troughs on this graph, Al

is the 8th element after B, but these are the only ones that appear to obey the rule. Hence we

can conclude that for the lighter elements only, there is some evidence to support the law.)

2 Student response. A possible answer is shown next, based on positions of peaks in graph of

atomic volume against atomic weight and troughs of graph of density against atomic weight.

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Table 4.2 Sorting the elements

H

Li Be B C

N O F

Na Mg Al Si P S Cl

K Ca Sc Ti V Cr Mn Fe Ni Co Cu Zn Ga As Se Br

Rb Sr Y Zr Nb Mo Ru Rh Pd Ag Cd In Sn Sb I Te

Cs Ba

3 Student response. A possible answer is shown next, based on fact that B and Al show similar

patterns, Li, Na, K, Rb and Cs show similar patterns but not the same as H, and removing N

from pattern since it is not a solid and therefore its density and hence calculated atomic

volume are not compared fairly against those of most of the other elements. We placeelements that display peaks or troughs or very similar values in same vertical columns.

Table 4.2 Sorting the elements

H

Li Be B C N O F

Na Mg Al Si P S Cl

K Ca, Sc,

Ti, V, Cr,

Mn, Fe

Ni, Co,

Cu

Ga As Se Br

Rb Sr, Y, Zr,

Nb, Mo

Ru, Rh,

Pd

Ag Cd In, Sn Sb, I, Te

Cs Ba

4 a 24 times bigger

b This is much smaller than the theoretical answer of 130.

c The assumption is that the atoms occupy the entire volume of the sample, i.e. there is

no empty space between the atoms, which would decrease the density of the sample.

Since atoms are spherical, which means there must be some unoccupied space at least

when they pack together, and given the fact many elements do not exist as separate

atoms, this assumption leads to quite significant error.

5 The data have been obtained from the work of someone else or a group of other people andnot directly by the person now using the data.

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Activity 4.2 Patterns in the valencies of the elements

Information and advice

This is another activity in which students process and interpret second-hand data, which not only

provides training in many important skills but also enables them to understand and appreciate some

of the rigorous thinking behind the development of the Periodic Table.

Suggested solutions to pre-lab activity: Focus questions

1 An element is one of the possible ingredients from which all substances are made and cannot

be further broken down chemically. Its atoms have a unique atomic number.

A compound is a substance composed of two or more elements that are chemically combinedtogether in definite proportions and can be broken down into its elements by heating it or

passing an electric current through it.

2 a In this compound, there are 3 atoms of hydrogen for each atom of nitrogen.

b In this compound, for each 2 atoms of boron there are 3 atoms of oxygen.

3 4

4 a 1

b Cl2O

c 2

d ZnCl2

5 Al2O3

Suggested solutions to calculations from second-hand data

1 Table 4.4

Element Symbol Atomicweight


Formula of hydride

Formula of

principal

oxide(s)

Valency

(valencies)

of element *

Hydrogen H 1.0 Not applicable H2O 1

Lithium Li 6.9 LiH Li2O 1Beryllium Be 9.0 – BeO 2

Boron B 10.8 Simplest is

B2H6 *

B2O3 3

Carbon C 12.0 Simplest is

CH4 *

CO, CO2 2, 4

Nitrogen N 14.0 NH3 N2O, NO,

NO2, N2O3,

N2O4, N2O5

1, 2, 3, 4, 5

Oxygen O 16.0 H2O not applicable 2

Fluorine F 19.0 HF F2O 1

Sodium Na 23.0 NaH Na2O, Na2O2 1, 2

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Element Symbol Atomic

weight


Formula of

hydride

Formula of

principaloxide(s)

Valency

(valencies)of element *

Magnesium Mg 24.3 MgH2 MgO 2

Aluminium Al 27.0 AlH3 Al2O3 3

Silicon Si 28.1 Simplest is

SiH4 *

SiO2 4

Phosphorus P 31.0 PH3 PO2, P4O6,

P2O5

2, 3, 4, 5

Sulfur S 32.1 H2S SO2, SO3 2, 4, 6

Chlorine Cl 35.5 HCl Cl2O, ClO2,

Cl2O

7

1, 4, 7

Potassium K 39.1 KH K2O, K2O2,

KO2

1, 2, 4

Calcium Ca 40.1 CaH2 CaO 2

Scandium Sc 45.0 – Sc2O3 3

Titanium Ti 47.9 – TiO2 4

Vanadium V 50.9 – VO, V2O3,

VO2, V2O5

2, 3, 4, 5

Chromium Cr 52.0 – Cr 2O3, CrO2,

CrO3

3, 4, 6

Manganese Mn 54.9 – MnO, Mn2O3,

MnO2, Mn2O7

2, 3, 4, 7

Iron Fe 55.8 – FeO, Fe2O3 2, 3

Nickel Ni 58.7 – NiO 2

Cobalt Co 58.9 – CoO, Co2O3 2, 3

Copper Cu 63.5 – Cu2O, CuO 1, 2

Zinc Zn 65.4 – ZnO 2

Gallium Ga 69.7 GaH, Ga2H6 Ga2O, Ga2O3 1, 3

Germanium Ge 72.6 GeH4, Ge2H6 GeO, GeO2 2, 4

Arsenic As 74.9 AsH3 As2O3, As2O5 3, 5

Selenium Se 79.0 H2Se SeO2, SeO3 2, 4, 6Bromine Br 79.9 HBr BrO2 1, 4

Rubidium Rb 85.5 RbH Rb2O, Rb2O2,

RbO2

1, 2, 4

Strontium Sr 87.6 SrH2 SrO 2

Yttrium Y 88.9 – Y2O3 3

Zirconium Zr 91.2 – ZrO2 4

Niobium Nb 92.9 – – –

Molybdenum Mo 95.9 – MoO2, MoO3 4, 6

Ruthenium Ru 101.1 – – –

Rhodium Rh 102.9 – – –

Palladium Pd 106.4 – – –

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Element Symbol Atomic

weight


Formula of

hydride

Formula of

principaloxide(s)

Valency

(valencies)of element *

Silver Ag 107.9 – Ag2O 1

Cadmium Cd 112.4 – CdO 2

Indium In 114.8 InH In2O3 1, 3

Tin Sn 118.7 SnH4 SnO, SnO2 2, 4

Antimony Sb 121.8 SbH3 Sb2O3, Sb2O4,

Sb2O5

3, 4, 5

Iodine I 126.9 HI I2O4, I2O5 1, 4, 5

Tellurium Te 127.6 – TeO2, TeO3 4, 6

Caesium Cs 132.9 CsH Cs2O, CsO2 1, 4Barium Ba 137.3 BaH2 BaO, BaO2 2, 4

*Note that these have been based on the assumption from the data in the exercise that oxygen only exhibits a

valency of 2. The bolded number is the valency of the element in its hydrogen compound. Where a hydrogen

compound does not exist, the lowest valency of the element in its oxygen compound is bolded instead.

2

Figure 4.3

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Suggested solutions to discussion questions

1 Student response. One possible answer is shown below, based primarily on the formulas of hydrides and then on formulas of oxides. The order of I and Te has been switched so that they

fall into columns that better suit the formulas of their compounds. The italicised elements in

row 5 may be placed in a separate column of their own.

H

Li Be B C N O F

Na Mg Al Si P S Cl

K Ca (Sc, Ti, V, Cr, Mn,

Fe, Ni, Co, Cu, Zn)

Ga Ge As Se Br

Rb Sr (Y, Zr, Nb, Mo, Ru,

Rh, Pd , Cd, Ag)

In Sn Sb Te I

Cs Ba

2 Student response. (Students should put H in a different group to the other elements in the

above column (Li etc), given that its density and atomic volume do not fall into the same

patterns as those of Li etc.)

3 Possible answers may include: The data for density jumps about and is difficult to compare for

substances that are not solid at room temperature. There are no clear-cut rules, as there is

still variation within each column. The data may be inaccurate, which may affect the patterns

observed. The data on formulas of compounds did not show a clear pattern unless the graph

focused on one of the values of valency for each element—where possible, deduced from itshydride. More data than this are needed to confirm positions, particularly for elements where

no hydride or oxide was known. A spiral table would better show the continual increase in

atomic weight and the corresponding trends in properties.

4 a The early periodic table was based on experimental data only, including atomic weights

of elements that were deduced from relative weights of substances reacting. The

elements were listed in order of increasing atomic weight and placed into the same

vertical column as those with similar properties. In the modern Periodic Table the

elements are listed in order of increasing atomic number (which was deduced from

experimental data), and they are still primarily placed into columns that are based on

properties, although electron configurations also are taken into account.b The basis of organisation has changed because it was found that there were elements

that needed to be placed out of order of atomic weight, if they were to be placed in the

same column as elements with similar properties. Listing in order of increasing atomic

number overcomes this problem.

The repeating patterns in properties of the elements are due to repeating patterns in

their electron configurations, since their properties are primarily determined by their

electron configurations.

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