are you ready for s347

8/12/2019 Are You Ready for S347

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S347 Metals and Life

Are you ready for S347?

Contents

1 Introduction 1

2 Suggested prior study 2

3 Key concepts for S347 2

3.1Atomic structure and the Periodic Table 2

3.2 Molecules and chemical bonding 3

3.3 Chemical reactions 3

3.4 The solid state crystal lattices 3

3.5 Chemical implications of the second law of thermodynamics 43.6 Spectroscopy 4

3.7 Biological concepts 5

3.8 Suggested further reading 5

4 Mathematical skill s 5

5 Other skills 5

5.1 Basic study skills 5

5.2 Writing skills 6

6 Self-assessment questions 67Answers to the self-assessment quest ions 11

1 IntroductionIf you are intending to study S347, you will want to make sure that you have the

necessary background knowledge and skills to be able to enjoy the course fully

and to give yourself the best possible chance of completing it successfully.

Please read through these notes carefully, and work through the self-assessment

questions given in Section 6. You will find this a useful exercise, even if you

have already studied other Open University Science courses and have completed

the recommended prior courses for S347. Working through these questions will

serve as a reminder of some of the facts, skills and conceptual knowledge, which

is assumed you will bring with you from earlier courses. If you are coming to

S347 without having studied the recommended prior courses, then it is

essential that you establish whether or not your background and experience

give you a sound basis on which to tackle the work. If you find that you have

difficulty in answering a significant number of the self-assessment questions,

you are advised not to study S347 without first completing the study of the

recommended prerequisite courses.

We advise you to seek further help and advice on the recommended prior courses

Copyright 2009 The Open University WEB 01420 5

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from a Science Staff Tutor at your Regional/National Centre, especially if, after

working through these notes, you are not sure about whether or not S347 is the

right course for you.

At the end of Section 3 we have suggested sources of further reading that would

help you to fill any gaps in your knowledge, or revise any areas of weakness.

2 Suggested prior studyThe main recommended prior course is the Open University Level 2 chemistry

course S205 The Molecular World. (Alternatively, the former courses

S247Inorganic Chemistry: Concepts and Case Studiesand S246 Organic

Chemistryare acceptable.) If you have studied S205, it is likely that you have

also studied a Level 1 Science Course (S104 or S103). Many concepts from the

biology and chemistry parts of that course are developed further in S347.

3 Key concepts for S347S205 includes a study of the chemistry of the typical (main-Group) elements.

S347 continues a study of the chemistry of the elements by looking at the varied

and versatile chemistry of the transition elements, in particular the crucial role

they play in living systems. Some of the same concepts are used, and many new

ones are introduced. Topics covered in S347 include the chemistry of the

transition metals, focusing on the synthesis, stereochemistry, oxidation states,

bonding, magnetic properties and thermodynamics of both simple compounds

and of complexes. The course looks in particular at the essential role that metals

play in life, looking closely at how organisms acquire metals and how they are

transported and stored. This is extended to include some of the key biological

processes that include metals, as well as topics in medicinal coordination

chemistry, and techniques required to study transition metals and biologicallyimportant metal coordination complexes.

The key concepts given below have been introduced in S205 or in S104 (or

S103): a basic understanding of these concepts is essential for S347. Many of

them are briefly described again and developed further in the S347 course

materials. Self-assessment questions amplifying and illustrating a selection of this

list of topics are given in Section 6. You should be able to establish the level of

understanding needed to study this Course by working through them. The

answers to the questions are given in Section 7.

3.1 Atomic structure and the Periodic Table

The electronic configuration of the elements.

Atomic orbitals: s, p, d and f.

Electron spin.

The Periodic Table: the names of the elements.

Trends in bonding, covalent and ionic radii, electronegativity and ionisation

energies.

Radioactivity, half-life, chain-reaction, - and -radiation.

(SAQ 1)

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3.2 Molecules and chemical bonding

Covalent bonding by sharing electron pairs: single, double and triple bonds.

Ionic bonding by electron transfer.

Lewis structures and resonance structures of molecules and ions.

Multicentre bonding for example diborane, B2H6.

Coordinate (dative) bonding and coordination compounds.

Ligands and simple metal complexes monodentate, bidentate, tridentate,

tetradentate and hexadentate ligands.

The shapes of molecules valence-shell electron-pair repulsion theory.

Centre of symmetry.

Chirality.

Molecular orbitals: bonding and antibonding, and .

The relationship between the overlap of atomic orbitals and the separation of

the bonding and antibonding orbitals.

Molecular-orbital energy-level diagrams of diatomic molecules and simple

polyatomic molecules (e.g. H2O).

The properties of metals and of non metals.

The nomenclature of simple organic molecules.

(SAQs 27)

3.3 Chemical reactions

Chemical equations.

Molar quantities.

Oxidationreduction (redox) equations.

Standard electrode potentials.

Oxidation states (oxidation numbers).

Equilibrium and equilibrium constant.

Acids and bases pH, strong and weak, and their reactions.

(SAQs 79)

3.4 The sol id state crystal latticesClose-packing of spheres (cubic close-packing and hexagonal close-packing).

Octahedral and tetrahedral holes in close-packed lattices.

The body-centred cubic lattice and the structures of metals.

Unit cells.

Ionic solids the sodium chloride, caesium chloride, fluorite and corundum

lattices.

Ionic radius.

Defects in crystals Schottky and Frenkel defects.

(SAQ 10)

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3.5 Chemical impl ications of the second law ofthermodynamics

The second law of thermodynamics: for a natural (a spontaneous) process in an

isolated system, .total > 0S

The definition of G(the free energy change or the Gibbs function):

= G H T S

K

The criterion for a spontaneous reaction (at constant temperature and

pressure).

< 0G

Standard molar enthalpies ( ), entropies ( ) and free energies ( ).mH

mS

mG

The standard molar enthalpy change for a chemical reaction:

m f f(reaction) = (products) (reactants)H H H

The standard molar change in entropy for a chemical reaction ( ):mS

m m m(reaction) =

(products)

(reactants)S S S

The standard molar free energy change for a chemical reaction:

m f f(reaction) = (products) (reactants)G G G

The relationship of the standard molar free energy change ( ) to the

equilibrium constant (K):

mG

m = 2.303 logG RT

The relationship of the standard molar free energy change ( ) to the standard

redox potential

mG

:E

m = G nFE

(F = 96 485 Cmol1)

(SAQs 11 13)

3.6 Spectroscopy

The electromagnetic spectrum.

Wavelength (), frequency () and wavenumber ().

The relationshipsE= h(whereEis the energy of a photon), = c, and

= 1/.

The rotational and vibrational spectra of simple molecules.

Infrared and Raman spectra.

Molecular structure the effect of symmetry on group frequencies.

Nuclear magnetic resonance spectroscopy.

Mass spectrometry.

(SAQs 14 18)

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3.7 Biological concepts

The three fundamental attributes of living things; metabolism, growth and

reproduction.

The fundamental similarities between all organisms at the cellular and

biochemical level.

The differences between eukaryotic and prokaryotic cells.

The biochemical building blocks of living things and the relationship

between structure of a variety of biological molecules and the functions they

carry out in the cell/organism.

The importance to the cell/organism of metabolic processes including

photosynthesis, glucose oxidation and protein synthesis.

To recognise that enzymes are proteins that catalyse biochemical reactions,

and all organisms rely on enzymic reactions for their survival.

The basic structure of nucleic acids.

(SAQs 19 25)

3.8 Suggested further reading

S205 The Molecular World(The Open University). This is the major

recommended prerequisite for this Course.

If you have not studied S205, you should contact your Regional Enquiry Service,

and they will be able to tell you where you can see reference copies of the Books.

There are also many good general inorganic textbooks, for example:

P. Atkins, T. Overton, J. Rourke, M. Weller, Armstrong,Inorganic Chemistry,

(2006), Fourth Ed, Oxford University Press.

C. Housecroft and A. G. Sharpe,Inorganic Chemistry, (2007), Prentice Hall.

4 Mathematical skillsOnly basic mathematical skills are needed, such as the ability to carry out simple

calculations, plot graphs and obtain information from them, and to calculate

chemical formulae from experimental analytical data.

Useful reading: A. Northedge, J Thomas, A Lane and A Peasgood, (1997), The

Sciences Good Study Guide, The Open University.

This contains a maths reference section.

5 Other skills

5.1 Basic study skil ls

To organise time for study, learning to pace study, effective reading: to extract

relevant information and data from scientific texts, diagrams and accounts.

Useful reading:The Sciences Good Study Guide.

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5.2 Writing skills

To write coherently and structure arguments, and to present a scientific account

with appropriate diagrams.

Useful reading:The Sciences Good Study Guide.

6 Self-assessment questions

SAQ 1

This question refers to the Periodic Table.

(a) Give the electronic configurations of the elements in Group 14 (IV): C,

Si, Ge, Sn and Pb.

(b) Are the elements in Group 14 (IV) metals or non-metals?

(c) In which Group do the atoms have the largest covalent radius of any in

their Period?

(d) In which Period do the atoms have the largest covalent radius in any of

their Group?

(e) In which Group do the atoms have the largest electronegativity in their

Period?

(f) In which Period do the atoms have the highest electronegativity in their

Period?

(g) Which element(s) in the 2nd period have/has a first ionisation energy

lower than the element that precedes it/them?

SAQ 2

Draw Lewis structures and structural formulae for the following species: note if

resonance structures are needed.

(a) Silicon tetrafluoride, SiF4;

(b) nitrogen trifluoride, NiF3;

(c) sulfur hexafluoride, SF6;

(d) the sulfite ion, SO32;

(e) diborane, B2H6.

SAQ 3

Draw structural formulae and explain the structures of:

(a) The tetraammino copper(II) complex ion ([Cu(NH3)4]2+);

(b) The bis-ethylenediamine copper(II) complex ion

[Cu (ethylenediamine)2]2+; ethylenediamine is NH2CH2CH2NH2.

SAQ 4

The molecular orbital energy-level diagram for the dinitrogen molecule, N2(derived from the N 2s and 2p levels only) is shown in Figure 1.

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Figure 1 The molecular orbital energy-level diagram for dinitrogen, N2.

Account for the appearance of the molecular-orbital diagram in terms of the

overlap of the 2s and 2p energy levels.

What is the bond order of the N2molecule? Why is it acceptable to consider only

the 2s and 2p energy levels?

Carbon monoxide (CO) is isoelectronic with dinitrogen. Explain how their

molecular-orbital energy-level diagrams will differ.

SAQ 5

(a) What are the important differences between the type of bonding present

in isolated molecules and that present in solids?

(b) What type of bonding is predominant in:

(i) sodium chloride,

(ii) diamond,

(iii) lithium metal?

SAQ 6

State some of the properties of metals and non-metals.

SAQ 7

(a) Calculate the oxidation numbers of the metal atoms in the species:

Cr2O72, VO2+, CrO2and CrO42, and of Cl in OCl.

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(b) Write balanced equations for the reactions:

(i) between dichromate ions (Cr2O72

) and vanadium(III) (V3+

) ions

in acidic aqueous solution, when VO2+ions and chromium(III) ions are

produced;

(ii) between sodium hypochlorite (NaOCl) and CrO2ions in basic

aqueous solution (OH), when CrO42and chloride ions are produced.

(c) Describe the redox changes involved for the reactions in part (b).

SAQ 8

Potassium permanganate in acid solution oxidises thallium(I) to thallium(III) if

excess fluoride ions are present. 20.00 cm3of a solution of 0.025 molar

(mol dm3) KMnO4reacts with 10 cm3of 0.1 molar thallium(I) solution.

Calculate the number of moles of thallium(I) that react with one mole of

permanganate, and hence the oxidation number of the manganese in the product.

Write a balanced equation for the reaction (in the products, the manganese is asix-coordinate fluoride complex and the oxygen appears in water).

SAQ 9

Acetic acid is a weak acid with a dissociation constant (Ka) in aqueous solution of

1.8 105mol dm3. What is the concentration of H+ions in a 0.01 molar

solution? What is the pH of this solution?

Assume that [H+]


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SAQ 12

Use the thermodynamic data given below to determine whether:

(a) sodium chloride will dissolve in water at room temperature (25 C), and

(b) diamond will react with oxygen at 100 C.

NaCl(s) Na+(aq) Cl(aq) C(diamond) O2(g) CO2(g)

1f / kJmolH

411.2 240.1 167.2 1.9 0 393.5

1/ J K molS 72.1 59.0 56.5 2.4 205.1 213.7

SAQ 13

For the reaction of aluminium metal with an acid,

Al(s) + 3H+(aq) = Al3+(aq) + 32

H2(g) = 485 kJ molmG 1

(a) Give this equation in the usual form for a standard redox potential and

calculate the value of E (Al3+|Al).

(b) Does the value you have obtained suggest that aluminium is a powerful

or weak oxidising or reducing agent?

SAQ 14

Convert the frequency 1 1013Hz (Hz = 1 s1

) into its equivalent in (a) kilojoules

per mole, (b) metres and (c) wavenumbers (cm1).

(Planck constant, h = 6.626 1034J s;

Avogadro constant,NA= 6.022 1023mol1; speed of light, c= 3.0 108m s1.)

SAQ 15

Two of the absorptions in spectra of the molecule OCS are at 1.216 1010Hz

and 2.580 1013Hz.

(a) What wavelengths and wavenumbers correspond to these frequencies?

(b) In what regions of the electromagnetic spectrum do these frequencies lie?

(c) What type of molecular energy change is likely to be responsible for each

of these absorptions?

SAQ 16

Which of the following molecules will have a vibration that is active in (a) the

infrared spectrum, (b) the Raman spectrum?

N2, HF, HCl, Br2and NO

SAQ 17

The 13C n.m.r. spectrum in Figure 2 was obtained from a compound of formula

C7H14O. Values of between 0 and 55 p.p.m. correlate to aliphatic carbons, andvalues in the range 175225 p.p.m. correlate to aldehyde or ketone groups. The

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labels d, t and q on the spectrum refer to doublets, triplets and quartets produced

by coupling to the hydrogen atoms attached to each particular carbon atom.

Identify the structure of the unknown compound.

Figure 2 The 13C n.m.r. spectrum of an unknown compound.

SAQ 18

What features distinguish a prokaryotic cell from a eukaryotic cell?

SAQ 19

Certain molecules found in cells often have hydrophilic and hydrophobic regions

within their structure. Comment on how such a molecule would interact with

water.

SAQ 20

Proteins are formed from the polymerisation of what kind of molecule?

SAQ 21

By means of a diagram, describe the generalised structure of an amino acid

SAQ 22

What is a peptide bond and how is it formed? Use a diagram in your answer.

SAQ 23

Which of the following statements about metabolism are true?

(a) The conversion of ADP + Pito ATP is coupled to energy-requiring

reactions in the cell.

(b) ATP is a carrier of chemical energy.

(c) All the ATP produced in the cell by catabolic pathways is used in

biosynthetic pathways.

(d) Energy derived from the oxidation of glucose is packaged as ATP and

used within minutes.

SAQ 24

State two reasons why photosynthesis is a vital process for life on Earth.

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SAQ 25

What is meant by the active site of an enzyme?

7 Answers to the self-assessmentquestions

SAQ 1 Answer

(a)

C: 1s22s22p2

Si: 1s22s22p63s23p2

Ge: 1s22s22p63s23p63d104s24p2

Sn: 1s22s22p63s23p63d104s24p64d105s25p2

Pb: 1s22s22p63s23p63d104s24p64d104f145s25p65d106s26p2

(b) Tin and lead are metals; carbon and silicon are non-metals. Germanium

shows both metallic and non-metallic properties.

(c) Group 1 (I) (Li, Na, K, Rb and Cs). Across a Period, the contribution to

the covalent radius due to the increasing nuclear charge dominates over the

increase in size caused by adding outer electrons.

(d) 6th Period (Cs to Rn). The increase in size as a result of adding the outer

electrons predominates (NoteCovalent radii are not known for many elements of

the 7th Period, and are likely to be reduced by the lanthanide contraction.)

(e) Group 17 (VII) (F, Cl, Br, I and At). Electronegativity is a measure of the

power of an atom to attract electrons to itself when entering into chemical

combination. This is most affected by the increasing nuclear charge across the

Period.

(f) The 2nd Period (Li, Be, B, C, N, O, F). The small size predominates over

the increase in the nuclear charge on moving down a Group.

(g) Boron and oxygen. The first ionisation energy of boron (801 kJ mol1) is

less than that of beryllium (899 kJ mol1) because the electron being removed in

beryllium is from the filled s orbital (Be Be++ e

is 1s22s21s22s2). The 2p

orbital is less likely to be found close to the nucleus than the 2s, and therefore the

electron can be removed more easily.

The first ionisation energy of oxygen (1 314 kJ mol1) is less than that of nitrogen

(1 402 kJ mol1) because the electron being removed from nitrogen is from a

half-filled p orbital (p3p2), whereas that being removed from oxygen is from a

p orbital with two electrons (p4p3). The electron being removed from oxygen

is paired with another electron and is more easily removed than an electron in

nitrogen where no electrons are paired, but all three have parallel spins.

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SAQ 2 Answer

F Si

F

F

F F N

F

F

F

F

S

F

F

F

FO S

O

OH

HB

H

HB

H

H

2

(a) (b) (c) (d) (e)

Si

F

F

F

F N

F

F

FS

F

F

F

FF

FB

H

H

H

H

B

H

H

For SO32there are three resonance hybrids:

O S

O

O O S

O

O O S

O

O

Notice the two different depictions of the structure of SO32 a Lewis structure

and three structural formulae for the resonance hybrids.

In B2H6each bridging atom is bonded to the two boron atoms by a three-centre

two-electron bond.

SAQ 3 Answer

Cu NH3

NH3

NH3

H3N

H2C

H2CN

NH2

H2

Cu

H2

H2

CH2

CH2N

N

2+ 2+

(a) (b)

In both (a) and (b) the N\Cu bond is a coordinate bond. Ammonia is a

monodentate ligand, whereas ethylenediamine is a bidentate ligand.

SAQ 4 Answer

With reference to the molecular orbital energy-level diagram for dinitrogen

shown in Figure 1:

the N 2s orbitals overlap to give 2sgand 2sumolecular orbitals;

the N 2pzorbitals overlap to give 2pgand 2pumolecular orbitals;

the N 2pxand N 2pyorbitals overlap to give two degenerate pairs of

molecular orbitals: 2puand 2pg

the 2sgand 2su(bonding and antibonding) orbitals and the 2puand

2pu(bonding) orbitals are filled, and the 2pgand 2puorbitals are

empty.

The bond order is (8 2)/2 = 6/2 = 3; and so N2has a triple bond.

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We need consider only the 2s and 2p energy levels, because the 1s levels are

much lower in energy and are not involved in bonding.

Carbon monoxide differs from dinitrogen in that the 2s and 2p atomic orbitals on

carbon and oxygen differ in energy. The atomic ionisation energies are

C(2s) 26.6 1019J, C(2p) 18.0 1019J, O(2s) 45.6 1019J and

O(2p) 21.8 1019J. These are to be compared with N(2s) 32.6 1019J and

N(2p) 23.3 1019J. The molecular orbital diagram for CO is shown in Figure 3.

Figure 3 Orbital energy-level diagram for CO

The most significant effect is that the O 2p orbital falls between that of the C 2s

and the C 2p. Because of this, the O 2p orbital of orientation interacts not only

with the C 2p orbital but also with the C 2s orbital. Hence the 2p bonding

molecular orbital is formed from not just the O 2p and C 2p orbitals, but also

from the C 2s; similarly the 2suantibonding molecular orbital is not just formedfrom C 2s and O 2s orbitals, but also from the O 2p orbitals. This results in the

energy of the 2pbonding molecular orbital being higher than in dinitrogen, and

the energy of the 2suantibonding molecular orbital being lower than in

dinitrogen. In CO the 2sbonding molecular orbital (5) is higher in energy than

the 2pmolecular orbitals (1). The net result is that the O 2p atomic orbitals

contribute more to the bonding molecular orbitals than do the C 2s and C 2p

atomic orbitals. Also in the N2molecule, three of the six electrons come from

each N atom. In CO four electrons are from the O atom and two from the C atom.

Note also that N2has a centre of symmetry, whereas CO does not. That is why

the N2molecular orbitals have subscript labels g or u, but those of CO do not.

SAQ 5 Answer

(a) In an isolated molecule the electrons are in well-separated discrete energy

levels. In a solid the levels can be regarded as energy bands. The relative energies

of bands containing electrons and empty bands, and their full or partial

occupation by electrons, determine the type of solid metal, non-metal,

semiconductor, ionic or covalent.

(b) (i) Sodium chloride: ionic bonding; (ii) diamond: covalent bonding; (iii)

lithium metal; metallic bonding.

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SAQ 6 Answer

Metals are the elements at the left-hand side of the Periodic Table and non-metals

are those on the right-hand side. The most important property of a metal atom is

that it will lose electrons to form a cation (positive ion) in aqueous solution, and

that of a non-metal atom that it will gain electrons to form an anion (negative ion)

in aqueous solution. This property is reflected in the standard redox potentials (a

measure of the ease of electron gain), which are negative for metals (in general)and positive for non-metals. The bonding in metallic solids is delocalised metallic

bonding. The bonding in non-metallic elements is covalent. Solid metals conduct

heat and electricity; they are ductile and malleable. Solid non-metals are

insulators.

SAQ 7 Answer

(a) Chromium +6 in Cr2O72; vanadium +4 in VO2+;

chromium +3 in CrO2; chromium +6 in CrO4

2; chlorine +1 in ClO.

(b) (i) Cr2O72(aq) + 2H+(aq) +6V3+(aq)+= 2Cr3+(aq) + 6VO2+(aq) + H2O(l)

(ii) 3ClO(aq) + 2CrO2(aq) + 2OH(aq) = 3Cl(aq) + 2CrO4

2(aq) +

H2O(l)

(c) (i) Chromium is reduced from an oxidation state of +6 to +3; vanadium is

oxidised from oxidation state +3 to +4.

(ii) Chlorine is reduced from an oxidation state of +1 to 1; chromium is

oxidised from oxidation state +3 to +6.

SAQ 8 Answer

The number of moles of MnO4is3

3 20 dm0.025 mol dm1 000

.

3 3(since 1dm = 1 000 cm )

The number of moles of Tl+is3

3 10 dm0.1 mol dm = 1 101 000

3

Therefore, the ratio of Tl+: MnO4

= 1 103

: 5 104

= 0.001 : 0.000 5 = 2 : 1.

The thallium changes oxidation state by +2. If two moles of thallium(I) are

needed to reduce 1 mole of MnO4, then the oxidation state of the manganese

must change by 2 2 = 4. In MnO4the manganese is in oxidation state +7,

therefore in the product it is + (7 4) = +3. The reaction is represented by the

equation

MnO4+ 2Tl+(aq) + 6F(aq) + 8H+(aq) = MnF6

3(aq) + 2Tl3+(aq) + 4H2O(l)

SAQ 9 Answer

The acid dissociation constant for acetic acid (HAc) is given by the expression

+ [H (aq)][Ac (aq)]=

[HAc]

aK

The concentrations of H+and Ac

are equal (sayx). Therefore

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1.8 105mol dm3

=2

30.01 mol dm

x

x

Asxis very small, (0.01 x) 0.01. Therefore

x2= 1.8 105mol dm3

0.01 mol dm3

= 1.8 107mol2dm6

Hence,x= 4.24 104mol dm3. The pH of the solution is given by

pH = log (4.24 104) = 3.37

SAQ 10 Answer

(a) In a face-centred cubic (cubic close-packed, ccp) lattice there is an atom

at each of the eight corners of a cube and an atom in each of the six faces.

An atom at the corner of a cube is shared by eight cubes (eight cubes meet at a

corner); therefore each unit cell contains 18

of an atom at each corner. As there

are eight corners, the total number of atoms at corners is 18

8 1 = .

An atom at the centre of a face of a cube is shared by two cubes (two cubes meetat a face); therefore each unit cell contains 1

2of an atom in each face. As there

are six faces, the total number of atoms in faces is 12

6 3 = .

Therefore the total number of atoms in the unit cell is 1 + 3 = 4.

(b) There is a tetrahedral hole between each corner atom and the atoms in the

centres of the three adjacent faces. As there are eight corner atoms, there are eight

such tetrahedral holes and they are all within the unit cell. Therefore the number

of tetrahedral holes is eight.

There is an octahedral hole at the centre of a face-centred cube, where the six

atoms in the six faces meet. There is also an octahedral hole in the centre of eachof the twelve edges of the cube. This is where two atoms at corners and two in

adjacent faces meet two other atoms in faces of adjacent unit cells. These holes

are shared by four unit cells (four cubes meet at an edge), so the total number of

holes in edges is 14

12 3 = . The total number of octahedral holes in the unit

cell is therefore 1 + 3 = 4.

(c) (i) NaCl is a cubic close-packed (ccp) array of Clions with Na+ions in

all the octahedral holes. It would also be true to describe it as a ccp array of Na +

ions with Clions in all the octahedral holes, although the Clions are much

larger than the Na+ions. The convention is to describe the array in terms of the

largest ion (here Cl).

(ii) CaF2is a ccp array of Ca2+ions, with Fions in all of the tetrahedral holes.

Note that here it is the cations (Ca2+) forming the close-packed array.

(iii) -Alumina (Al2O3) is a hexagonal close-parked (hcp) array of O2ions, with

Al3+ions in two-thirds of the octahedral holes.

SAQ 11 Answer

(a) For a reaction to be thermodynamically feasible,

(= ) should be negative.mG

mH

mT S

Assuming that and are independent of T, at 100 C is given bymH

mS

mG

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mG = [131.3 (373 133.9)/1 000] kJ mol1= 81.35 kJ mol1

This is positive, so the reaction will not take place. (Note the factor of 1 000; this

is needed because is given in J KmS 1mol1.)

(b) for the reverse reaction will be 81.35 kJ molmG 1. This is negative,

so the reaction is thermodynamically possible, but will only occur in practice if it

is also kinetically favourable.

(c) The reaction becomes thermodynamically possible when = 0, that

is, when = . Hence this occurs at a temperature

mG

mH

mS

m

m

=H

TS

Hence1

1 1

131.3 1 000 J mol= = 980.6 K

133.9 J mol KT

SAQ 12 Answer

(a) (i) The reaction is NaCl(s) = Na+(aq) + Cl(aq). can be calculated

using the equation

mG

mG = mH

mT S

again assuming and to be independent of T.mH

mS

Now:

m f f

1

(reaction) = (products) (reactants)

= [240.1 167.2 (411.2)] kJ mol = +3.9 kJ mol

H H H

1

and m m m1 1


= [59.0 + 56.5 72.1] kJ mol = +43.4 JK mol

S S S

1

1m

1

At 25C (= 298 K), = = [3.9 (298 43.4)/1 000] kJ mol

= 9.0 kJ mol

G H T S

mG is negative, so the reaction is thermodynamically favourable at

25 C (= 298 K)

(b) The reaction is C(diamond)(s) + O2(g) = CO2(g)

m f f

1


= [393.5 1.9 0] kJ mol = 395.4 kJ mol

H H H

1

and m m m1 1 1


= [213.7 2.4 205.1] J K mol = 6.2 J K mol

S S S

1

1

1

At 100 C (=373 K), = = [395.4 (373 6.2)/1 000] kJ mol

= 397.7 kJ mol

G H T S

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This is a large negative value, so the reaction is thermodynamically very

favourable. In fact, it does not occur, which means that diamond is kinetically

stable in oxygen at 100 C.

SAQ 13 Answer

Standard redox potentials are reduction potentials by definition, so the equation

must be transposed from right to left. For this reaction = 485 kJ molmG

1.

To obtain the equation in the usual form, you must recognise that

H2(g) = H++ e (where e = an electron). Therefore the equation becomes:

Al3+(aq) + 3H+(aq) + 3e = Al(s) + 3H+(aq).

Eliminating the H+from both sides of the equation gives Al3+(aq) + 3e = Al(s).

Now = nFmG E. Therefore

485 103J mol

1= (3 96 485 C mol

1) E ; that is, E = 1.68 V

(1 V = 1 J C1)

(b) This is a high negative potential (cf. E (Zn2+| Zn) = 0.76 V and

E (Mg2+| Mg) = 2.36 V), so aluminium is expected to be a powerful reducing

agent.

SAQ 14 Answer

(a) The energy of one photon is given by the equationE= hv, therefore the

energy per mole is:

34 3 1 23 1 1A

1

= = (6.626 10 J s) (1 10 s ) (6.022 10 mol ) = 3 990.18 J mol

= 3.990 kJ mol

E hnN

(b) 8 1 13 1 5= / = (3 10 m s )/(1 10 s ) = 3 10 mc v

(c) 5 3= 1/= 1/(3 10 m) = 1/(3 10 cm) = 333.33 cm 1

SAQ 15 Answer

(a) = c/vand = 1/.

For 1.216 1010Hz:

= (3 108m s1)/(1.216 1010103s1) = 2.467 102m,

And = 1/(2.467 cm) = 0.4054 cm1.

For 2.580 1013Hz: = (3 108m s1)/(2.580 1013Hz) = 1.163 105m

And = 1/(1.163 103cm) = 859.8 cm1.

(b) The absorption at 1.216 1010Hz is in the microwave region, and that

at 2.580 1013Hz is in the infrared region.

(c) The microwave transition is likely to be caused by a rotational change,

and the infrared transition will be caused by a vibrational change in the molecule.

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SAQ 16 Answer

(a) Only those molecules in which the dipole moment changes during the

vibration are infrared active: these are the heteronuclear molecules HF, HCl and

NO.

(b) Molecules that change their polarizability during vibration are active in

the Raman spectrum. All diatomic molecules do this, so the vibrations of all the

molecules are Raman active.

SAQ 17 Answer

There are seven peaks in the n.m.r. spectrum shown in Figure 1, so there must be

seven non-equivalent carbon atoms present in the molecule. A peak at = 202

p.p.m. is indicative of an aldehyde or a ketone group. This peak is labelled as a

doublet. Therefore the carbon atom producing this peak must have one attached

hydrogen atom: that is, the functional group present is an aldehyde. The

remaining six peaks all fall in the region occupied by aliphatic carbon atoms. Five

of the peaks are labelled as triplets and one is labelled as a quartet. Therefore

there must be five CH2groups present and one CH3group. The only way theseseven groups can be put together to yield a molecule with satisfactory valencies is

heptanal CH3CH2CH2CH2CH2CH2CHO. All the CH2groups are non equivalent

due to symmetry.

SAQ 18 Answer

The eukaryotic cell has a nucleus, and membrane bound organelles such as

mitochondria, and chloroplasts. The prokaryotic cell lacks these, i.e. no clearly

defined structures are visible under the microscope.

SAQ 19 Answer

The hydrophilic groups would interact with water, but the hydrophobic groups

would interact with one another i.e. avoid water as far as possible, this determines

the molecules three dimensional structure in water.

SAQ 20 Answer

Proteins are formed by polymerization of amino acids (a chain of amino acids is

called a polypeptide).

SAQ 21 Answer

N C C

R

H

O

HH

HThe generalised structure of an amino acid is shown below, where R is the

variable side chain. The Ccarbon is linked to an amino group, a carboxylic acid

group and a side chain.

SAQ 22 Answer

peptidebond

A peptide bond is the bond formed between two amino acid residues in a

polypeptide. It is formed by the elimination of a water molecule between the

amino group of one amino acid and the carboxyl group of another. The atoms

within the box comprise the peptide group. N CH C N C

R O

H

H O

HH

peptide g roup

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SAQ 23 Answer

(a) False. The conversion of ADP + P ito ADP is coupled to energy-releasing

reactions.

(b) True. ATP is a carrier of chemical energy in that it transfers energybetween energy-releasing and energy-requiring reactions.

(c) False. ATP is used in many processes other than biosynthesis, such asmuscle contraction.

(d) True. ATP is a short-lived energy store, typically being converted toADP within a minute.

SAQ 24 Answer

Photosynthesis is essential for life on Earth because:

it converts solar energy into chemical energy in sugars, which can beused to fuel all the energy-requiring processes of life

it releases oxygen, which is required for respiration.

SAQ 25 Answer

The active site of an enzyme is the site at which a substrate binds and undergoes

chemical modification.

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