NAT ION AL Q UAL IF ICAT ION S C UR R IC UL UM SU PP ORT


Acknowledgements
This document is produced by Learning and Teaching Scotland as part of the National
Qualifications support programme for Chemistry. Grateful thanks are due to the Education
Division of the Royal Society of Chemistry, Scottish Region, for comment on the drafts.
The support of the Higher Still Development Unit and the editorial advice of Douglas
Buchanan are also acknowledged.
All information given by Goodfellow Cambridge Limited (formerly Goodfellow Technology
Limited) is supplied as a guide only. Alt hough every effort has been made to ensure that the information is correct, no warranty is given as to its completeness or accuracy.
First published 2001
Electronic version 2002
© Learning and Teaching Scotland 2001
This publication may be reproduced in whole or in part for educational purposes by educational establishments in Scotland provided that no profit accrues at any stage.
ISBN 1 85955 912 3

Introduction
Section 1: Electronic structure
Section 2: Chemical bonding
Section 3: Some chemistry of the Periodic Table
17
25
1
3
C H E M I S T R Y i i i

Note on safety
Teachers/lecturers are advised that it is their responsibility to take notice of employers’ regulations with regard to the safe practices to be followed. Where necessary, prior to using chemicals, the relevant advice may be consulted in the Hazardous Chemicals Manual (SSERC), either in printed form or in CD -ROM version. i v C H E M I S T R Y

In these notes we have tried to provide some amplification of what is contained in the Advanced Higher Chemistry course booklet Unit 1:
Electronic Structure and the Periodic Table by Archie Gibb, Arthur Sandison and Andrew Watson, produced by Learning and Teaching Scotland (2000), without straying far from the subject matter, nor introducing any new concepts. Most of what is provided is a series of experiments that may be carried out either by staff or, if there is time, by the class. These experiments, which illustrate the concepts being developed in the unit, are numbered and titled in bold italic s and all the experimental detail is provided.
The rest of the notes try to give examples of the relevance of the various different aspects of the unit to everyday life. We hope they may be of interest. It is not necessarily proposed that all staff shou ld cover all of the material in these support notes, but it is hoped that some of the experiments and descriptions may appeal to all teachers and lecturers.
Colour versions of some of the pictures that do not display well in monochrome, together with links to other sites, are available at http://chemistry.st-andrews.ac.uk/ah
The order follows the order of the course booklet and all the page references here are to the course booklet cited above.
C H E M I S T R Y 1

2 C H E M I S T R Y

E L E C T R O N I C S T R U C T U R E
Li ght emi ssi on an d a b sorpti on ( p p 4 – 22, a l so rel evant to p 62)
Absorpti o n
St udent s and ot hers ar e of t en conf used a bout e mi ssi on an d absorpt i on of l i ght , si nce ei t her c an be r esponsi bl e for c ol our. T he experi me nt descr i bed i n t he cours e boo kl et i s w or t h doi ng i f you can get h ol d of a pr i sm, possi bl y f r o m t he ph ysi cs depar t ment . Set up a proj ect or, sl i t and pr i s m as descri bed i n Experi ment 1, bel o w. T he l i ght t hat shi n es on a card hel d on t he ot her si de of t he pri s m wi l l l oo k l i ke a r ai nb ow.
If you use a neon l i ght or l i ght -e mi t t i ng di od es, onl y so me of t he col our s wi l l be seen .
Experi m ent 1: Dem onst rat i ng l i ght absorpt i on usi ng col oured sol ut i ons
Use a pri s m t o prod uc e a spect r u m and t hen pl ace col oured sol ut i o ns bet ween t he pri s m and a sheet of car d.
Pr oj ect i ng a spect r u m i sn’t qui t e as eas y as p ut t i ng a pri s m bet ween a l i ght bul b and a s heet of card – t he l i ght i s t oo di sper se. You need so me wa y of proj ect i ng a nar row bea m of l i ght t hrou gh t he pri s m – use a sl i de proj ect or cont ai ni ng a 35 m m sl i de bl acked out apart fro m a
0.25 mm wi de ver t i cal sl i t down t he mi ddl e. T he i ma ge of t he sl i t c an be projected through a 50 mm 60° prism onto a sheet of white card to pr oduce a br i ght s pect ru m.
Fi gure 1
6 0 ° p ri s m
5 0 m m g l a s s t a n k
S l i d e p r o j e c t o r
For colour, see http://chemistry.st -andrews.ac.uk/ah
C H E M I S T R Y 3

E L E C T R O N I C S T R U C T U R E
A 50 mm cubic glass tank (available from Philip Harris Education) containing solutions of various transition metal ions in turn can be placed between the prism and the screen (Figure 1). The vessel that is used to contain the solutions must be flat sided – a test-tube is unsuitable since it acts as a lens and disperses the light. Note that the concentrations of the solutions used must be adjusted to suit the path length and the intensity of the light source.
Students could try several different solutions (transition met al salts are good) and try to determine the approximate wavelength at which they absorb. This part of the course could also be linked into the section on ultra -violet/visible spectroscopy of the transition metals (see pp 64 –66).
Table 1
Ion
Titanium(III)
Cobalt(II)
Nickel(II)
Copper(II)
[Cu(NH
3
)
4
] 2+
Approx. concentration
0.08 mol l –1
0.05 mol l –1
0.4 mol l –1
0.4 mol l –1
0.02 mol l –1
Region absorbed
Green
Most of the green
Red and violet
Red
Red to green
Note: In all cases, the colour seen is what is transmitted, so it is the complementary colour to that which is absorbed.
To make a Ti 3+ solution, take some titanium granules (0.1 g). Add concentrated hydrochloric acid (3 cm 3 ) and heat over a bunsen burner (fume cupboard). After a few minutes of boiling the solution will turn purple.
Once a reasonable colour has developed, cool and decant the liquid slowly into an equal volume of water (do not add water to the concentrated acid solution). The solution predominantly contains purple [Ti(H
2
O)
6
] 3+ , which absorbs at c.
500 nm or in the green region of the spectrum.
A couple of more interesting (dynamic) experiments are described below.
The absorbance of the green region of the spectrum by permanganate ions can be demonstrated by filling the glass tan k with water then sprinkling in a few crystals of KMnO
4
. As the solution is stirred gently, the middle of the spectrum gradually disappears.
A similar demonstration can be done for the [Fe(SCN)
6
] 3– complex. The tank is filled with 0.05 mol l –1 thiocyanate solution and stirred while a solution containing 0.1 mol l –1 iron(III) ammonium sulphate and
0.2 mol l –1 sulphuric acid is added dropwise. The violet to green r egi on of t he spect ru m i s gra dua l l y absorbe d.
4 C H E M I S T R Y

E L E C T R O N I C S T R U C T U R E
Now for something mildly spectacular! The ta nk can be filled with a solution containing 0.1 mol l –1 ammonium metavanadate and 1 mol l –1 sulphuric acid
(1.4 g NH
4
VO
3
+ 120 cm 3 1 mol l –1 H
2
SO
4
(aq)). This yellow solution contains dioxovanadium(V) ions and absorbs the violet to blue region of the spect rum, transmitting red to green. 1 g of sodium metabisulphite is then added and dissolved by stirring. The colour changes to sky blue as dioxovanadium(V) is reduced to oxovanadium(IV), which absorbs the red to yellow region and transmits violet to green. The green region of the spectrum is not absorbed by either species. The exciting part is seeing the shift as the violet end of the spectrum appears while the red end of the spectrum disappears.
Emission
Experiment 2: Flame tests to show light emission
The easiest way to demonstrate colour from emission is to carry out flame tests. The classical way to do this is to use a platinum wire fused into the end of a piece of glass. The wire is first cleaned by dipping it in HCl(aq) (dilute,
5 mol l –1 , is sufficiently concentrated), usually in a watchglass, then heating it in a hot bunsen burner. This process is repeated until no colour is observed in the flame. Once the wire is clean it is dipped again into the HCl(aq) and then into a small pile of solid crystals of the metal salt being tested. It is then put back in the flame and the colour of the flame noted. This colour is characteristic of the metal ion present and it would be interesting for students to try compounds of various different elements – chlorides are best – and to look at the different colours. It is worth noting that only elements with relatively low ionisation potentials give coloured flames. The students should be able to get a qualitative feel for ionisation energies (discussed on p
18) by carrying out flame tests on a variety of different compounds.
If you do not have any platinum wire, a cheaper alternative (although with careful handling, the platinum wires should last a very long time, so only the initial outlay is significant) is to dissolve metal chlorides in methanol in a small beaker or petrie dish and set fire to the vapour. Elements with low ionisation energies, such as the alkali metals, will give a characteristically coloured flame.
These flame tests can also be used as a first indication of the nature of the positive ion in salts. Why not let students do a little detective work?
You could give them a few salts and ask them to identify the positive ion present. You could either give them a table of colours or some salts for which they are told the positive ion and they could then do a comparison.
This could then lead on to a discussion of why some
C H E M I S T R Y 5

E L E C T R O N I C S T R U C T U R E elements give flame tests and some others do not, then to what the origin of the emission is and the relative ease of ionising the elements.
Table 2
P osi ti ve i on
Li +
Na +
K +
Ca 2 +
Sr 2 +
Ba 2 +
Cu 2 +
F l ame col our
Cr i mson
Yel l ow
Li l ac
Oran ge -r ed
Scarl et
Appl e gr een
E mer al d gr een ( so met i mes wi t h a bl ue cent re )
If you want t o use t he met hanol met hod, pr oc eed as f ol l ows .
Mi x t he met al chl or i de s (100 –200 mg) wi t h met hanol (1 c m 3 ) i n separ at e 10 c m 3 bea ke rs cont ai ned i n a met al t ray. T he met han ol c an t hen be i gni t ed usi n g a burni n g wooden spl i nt . T he resul t s are l i st ed i n T abl e 3.
Tabl e 3
Sal t
CuCl
2
.2H
2
O
Li Cl
Comm ents
Sol ubl e
Sol ubl e
F l ame col our
E meral d green
Cri mson
NaCl
K Cl
Spari n gl y sol ubl e Yel l ow
Spari n gl y sol ubl e None
CaCl
Sr Cl
2
2
. 6H
.6 H
2
2
O
O
Sol ubl e
Sol ubl e
None. Occasi onal fl ash of redt owards t he en d.
Red i ni t i al l y, but fade s hal f wa y t hrou gh.
BaCl
2
. 2H
2
O Spari n gl y sol ubl e Appl e gr een i ni t i al l y, but soon f ades.
If wat ch gl asses ar e us ed i nst ead of bea ker s , onl y copper, l i t hi u m a nd st r ont i um col our t he f l a mes, b ut t he y do i t i mpressi vel y.
6 C H E M I S T R Y

E L E C T R O N I C S T R U C T U R E
Experiment 3: Light emission from ‘King Alfred’s cakes’
Atomic emission in another guise is what gives fireworks their spectacular colours and one can make ‘King Alfred’s cakes’ by mixing copper(II) chloride crystals into molten wax and pouring some of the mixture into paper cake cups. Once solid, they can be put in a fire and give a very pleasant blue colour to the flames. With rather fewer open fires available now, this may not be appropriate for everyone.
Experiment 4: Analysis of mixtures using flame tests
Some teachers will remember that in order to see the lila c flame from potassium salts, they used to look through deep blue ‘cobalt’ glass. This is no longer necessary. This may be because some years ago it was difficult to obtain potassium salts entirely free from sodium and the extremely bright yellow flame from the sodium masked the pale lilac from the potassium.
Being blue the cobalt glass absorbed yellow light. Nowadays, the salts are purer so this is not a problem. However, a nice test for the brightest students might be to give them a mixture of sodium and potassium chlorides and ask them to identify the positive ions present. They should immediately see the yellow of sodium and then you could discuss with them how they might see other colours. This should lead to a good discussion of absorption and emission. Once they have decided that they have to absorb the yellow light, you could then discuss what colour filters will do this and then, when they have identified the appropriate colour as blue, you could produce the cobalt glass and let them work out what other positive ion is present. Of course, you could use a positive ion other than potassium – calcium is good or strontium, but this has such an intense colour that it will compete even with sodium.
Various colour combinations and different filter s could be used to identify different positive ions present in mixtures.
C H E M I S T R Y 7

E L E C T R O N I C S T R U C T U R E
An absorption spectrum of cobalt glass is shown in Figure 2.
Figure 2 u.v. / violet / blue / green / yellow / orange / red / IR
Visible absorption spectrum of cobalt glass
For colour, see http://chemistry.st-andrews.ac.uk/ah
Experiment 5: The d lines of sodium through a spectroscope
If you look very carefully, using a spectroscope, at the emission line of sodium, it is actually two lines that are very close toge ther (589.0 and 589.6 nm). It may be that your physics department has a spectroscope that you can use. With the slits very narrow, you can point it at a bunsen flame into which you have put some sodium chloride (using the platinum wire method). If you have a sodium discharge lamp, this is better since it is brighter and more stable than a flame. You should be able to resolve the two lines. This is an important experiment because it was this experiment that first showed that the electron has a spin. The lines are much too close together to be caused by the electron falling from different levels, but the two electronic spins give the excited el ect r ons sl i ght l y di ff er ent ener gi es, s o t wo l i nes are see n.
8 C H E M I S T R Y

E L E C T R O N I C S T R U C T U R E
Atomic absorption spectroscopy, atomic emission sp ectroscopy and colorimetry for elemental analysis (pp 22, 64)
Absorption and emission forms the basis for the two analytical techniques atomic absorption spectroscopy (AAS) and atomic emission spectroscopy
(AES) (p 20). There is scope for confusion here b ecause in both techniques the sample is heated, but the reasons for the heating are different.
Atomic absorption spectroscopy
For AAS, the flame is really meant only to atomise molecules. In practice, the flame also causes emission (see Figure 3 – incidentally, if you cannot for some reason do the flame tests, the web version of this picture shows a fine lithium flame test!). However, in AAS, a beam of light from a lamp made from the element being analysed is passed through the sample (see Figure 4) and it is the absorption of this light that is measured. Some of the features that are essential for good analysis are:
• the flame must be extremely stable so that the proportion of analyte in the ground state (that which absorbs the light) is always th e same – a graphite furnace gives even better stability
• similar matrices must be used for the unknown as for the standard (so that the proportion of atoms produced is the same).
Very good instrumentation is now available and flame AAS spectrometers routinely give analyses with detection limits of 1 ppm or less. Graphite furnace instruments can be even more sensitive.
Figure 3
Flame used for AAS. If there is no analyte passing through it, the flame is blue. If the sample being analysed contains lithium, the flame is pink.
For colour, see http://chemistry.st-andrews.ac.uk/ah
C H E M I S T R Y 9

E L E C T R O N I C S T R U C T U R E
Figure 4
Schematic diagram of an atomic absorption spectrometer.
Monochromator
Selects specific Detector
Element specific
Light source li gh t sou rc e intense)
Usually most intense
Sample in
Atomiser
Sample out
Atomic emission spectroscopy
In AES, in addition to atomising the sample, the heating also ionises the atoms. Light is emitted as the electrons recombine with the atoms and the intensity of the light emitted is related to the concentration of atoms in the flame. In order to do this ionisation, the material being analysed must be heated to a very high temperature and a plasma is usually used. Again, a very stable plasma is required.
AES instruments are more expensive than AAS, but their sensitivities are often as low as 1 ppb. Often for electronic materials such as semiconductors, even 1 ppb is barely accurate enough.
Colorimetry
The more classical analytical technique of colorimetry also uses light absorption as the sensor for concentration of the analyte. Usually it also involves metal analysis and the metal or one of its ions is converted into a salt containing a coloured ion or into a coloured complex. The higher the extinction coefficient (the absorbance of 1 cm of a solution containing
1 mol l –1 ) of the ion, the more sensitive is the analysis. One of the PPAs uses this technique for the analysis of manganese in steel. It might be useful and sensible to introduce the concept of colorimetry at this stage, rather than when discussing ultra-violet/visible spectra of transition metal ions (pp 64 –
66) since the ion being analysed is manganate(VII) or MnO
4
– . This is d 0 , so does not have d–d transitions. The colour comes from charge transfer (an electron moves from an orbital predominantly on O (p orbital) to one predominantly on Mn (d orbital)). This is p –d so is an allowed transition and this accounts for the very dark colour (high extinction coefficient) of MnO
4
– . d–d transitions are formally forbidden so the colours arising from t hese t r ansi t i ons ar e much p al er .
1 0 C H E M I S T R Y

E L E C T R O N I C S T R U C T U R E
Non-visible sections of the electromagnetic spectrum (p 4, also pp 26 –28)
On page 4 there is reference to the fact that visible light is only part of the electromagnetic spectrum and examples of ultra -violet and microwave radiation are given. This would be an ideal opportunity to build on this information in a number of different ways . Two possibilities are a discussion of atmospheric chemistry (the ozone layer and the greenhouse effect) and the use of infra-red cameras. Both of these tie into other parts of this unit later on.
Atmospheric chemistry
The ozone layer (relevant to light absorption, p 4 and multiple bonding, pp
26–28)
Light from the sun is broad -band radiation as from a ‘black body’ at a temperature of 5900K (see Figure 5).
Figure 5
The spectrum of the Sun as it appears outside the Earth’s atmosphere and at the Earth’s surface compared with the emission spectrum expected for a
‘black body’ at 5900K
C H E M I S T R Y 1 1

E L E C T R O N I C S T R U C T U R E
If t hi s radi at i on al l r ea ched t he Ear t h, we wo ul d al l di e of s ki n can cer because t he ul t r a -vi ol et rays (<3 20 n m) ha ve enou gh ener gy t o brea k che mi cal bonds . If b o nds i n D NA ar e br o ke n, especi al l y i f bot h st r ands are bro ken i n a si mi l ar pl ace ( doubl e - st r and brea ks) , t hen cancer can be i ni t i at ed because t he DN A can mut at e. T he che mi st ry i n t he at mosphere, ho we ver , sa ves us. In t he hi ghest part of t he at mosphere, t he t he r mosphere (90 –10 0 km a bo ve sea l e vel ), t he concent rat i on of gases i s s mal l , but t her e i s enou gh ni t ro gen t o abs orb l i ght wi t h wa vel en gt h <200 n m. Because ni t rogen has a t ri pl e bo n d, i t wi l l onl y absor b f ar ul t ra -vi ol et l i ght , UV C ( here t he st ren gt h of t h e t r i pl e bond can be rel at ed t o t he ener gy r eq ui red t o brea k ni t ro gen i nt o at o ms; Equat i on 1). T hi s i s t he most easi l y observa bl e e vi dence t h at ni t r ogen has such a st ron g bond and so a l i n k can be made t o t he bondi n g sect i on ( p 26) . h v
N

N 2N

H 0 = 946 kJ mol – 1 
<240 n m (1)
T here i s a great deal o f ni t rogen i n t he at mos phere so al l t he ver y h ard
UV C i s absor bed; i n f act al l of i t i s absorbe d i n t he t her mospher e and t he at o ms f or med r eco mbi ne gi vi n g out heat , hence t he t er m
‘thermosphere’. Of course, ther e is still hard UVB present, which cannot be absorbe d b y ni t r ogen because i t d o es not ha ve e nou gh energy t o exci t e t he ni t r ogen mol ecul es. So me of t hi s i s a bsorbed by ox ygen i n t he st rat ospher e. It i s absor bed hi gher up, b ut t he concent rat i on of ox ygen i s so l ow t hat much of i t get s t hr ou gh t o t he st r at osphere. ( Not e t h at t he concent rat i on of gases (pressure) i ncr eases as t he al t i t ude decr eases.) Becaus e ox ygen has a do ubl e bond, i t st i l l needs hi gh -ener gy phot ons ( <24 0 n m, UV B) t o brea k t he bond (Equat i on 2). T h i s f or ms ox ygen at o ms , whi ch c an react bac k t o gi ve ox ygen mol ecul es , but because t he conc ent r at i on of ox ygen mol ecul es i s much hi gher t han t hat of ox yge n at o ms, t he pref er red r eact i on i s wi t h ox yge n mol ecul es i n t he pr e sence of a t hi r d bo d y ( M) t o gi ve o zone ( Equat i o n 3) .
O

O h v
2O

H 0 = 500 kJ mol – 1
O + O
2
M
O
3

<240 n m (2)
(3)
O
3 h v
O
2
+ O

H 0 = 946 kJ mol – 1

<320 n m (4)
T he ozone i n t he o zo n e l ayer i s f or med b y t h e che mi st r y of Equat i o n
3. It s l e vel i s kept c o nst ant b y react i ons 2 a nd 3 bot h occurr i n g. It i s r eal l y i mport ant t hat t hi s happens beca use o zone , w hi ch has b onds of or der 1.5 (see p 28) , c an absor b l i ght wi t h w avel en gt hs up t o 320 n m
( UV A) .
1 2 C H E M I S T R Y

E L E C T R O N I C S T R U C T U R E
When this happens, it breaks down to give oxygen atoms and molecules
(Equation 4). In this way, all of the hard UVC (nitrogen, triple bond), UVB
(oxygen, double bond) and UVA (ozone, 1.5 bonds) are removed from the
Sun’s radiation and that which reaches Earth is relatively harmless to the molecules of life. It is worth noting that the fact that the atmosphere is largely made up of compounds that have multiple bonds allows the removal of the damaging radiation, but allows through the radiation that gives us
(visible) light and heat.
The greenhouse effect
Infra-red radiation (heat, short wavelength) is also important to life on earth.
Most of the radiation that reaches the Earth’s surface is visible light. This is absorbed by plants, rocks, etc. Various processes occur and some of the energy is released as heat. If this simply radiated out from the Earth, calculations suggest that the temperature of the Earth would be only –18 o C.
All water would be frozen and life as we know it could not have developed.
However, infra-red radiation is the type that leads to the excitation of vibrations, provided that the vibrations lead to a change in the dipole moment of the molecule. This will not happen for nitrogen or oxygen (no change in dipole moment), but will for carbon dioxide and water (change in dipole moment). Carbon dioxide and water are present in the part of the atmo sphere closest to the Earth’s surface (the troposphere), so they absorb some of the emitted infra-red radiation. They then re-emit this radiation in all directions and some of it returns to the Earth’s surface, heating the Earth’s surface to c.
25 o C and making it habitable. This is the greenhouse effect (see Figure 6), so it is worth noting that without it we humans would not exist. If the levels of carbon dioxide and related gases in the troposphere increase, global warming above that which we have alr eady will occur and this will lead to climate change and possibly devastating consequences.
C H E M I S T R Y 1 3

E L E C T R O N I C S T R U C T U R E
Figure 6
Schematic representation of the greenhouse effect, showing how carbon dioxide helps to heat the troposphere by absorbing escaping infra -red radiation and re-emitting it back towards the Earth’s surface.
Infr a-r ed r adiation re -emitt ed fro m
CO
2 in all dir ect ions
Ultra -viol et r adiation, visib le infr a-r ed radi ation fro m sun Some goes b ack to Earth
Infr a-r ed r adiation re -emitt ed fro m Earth
Earth’s surface
Police Camera Action! (relevant to light absorption (p 4) and semiconductors (pp 38–43)
Many of us have seen Police Camera Action!
on television, and marvelled at the ability of the police to ‘see in the dark’. They do this by using infra -red radiation (heat). Small differences in temperature can be detected at long distances because heat is really like light, but in the infra -red region of the spectrum. This links nicely into the semi conductor part of the unit because the detectors are just like solar cells, except that the band gap of the semiconductor is in the infra-red region of the spectrum, rather than in the visible. Because infra-red radiation has low energy, the band gap must be small. The material used for these infra -red sensors is cadmium mercury telluride.
We know that elements of Group 14 (Si and Ge) are semiconductors. They are known as elemental semiconductors. Semiconducting behaviour is often also found in compounds where the average of the Groups that the elements are in is 14. These are called compound semiconductors and examples include GaAs and InP, both made from a Group 13 and a Group 15 element, and ZnS, made from elements of Groups 12 and 16. The band g ap decreases down a group, i.e. behaviour becomes more metal -like. Cadmium mercury telluride (Cd x
Hg
1– x
Te) is a compound semiconductor of heavy elements of
Groups 12 and 16 so it is a semiconductor with a narrow band gap (in the infra-red region of the spectrum).
The fact that the band gap is so small means that the conduction band contains many electrons at room temperature because heat (infra -red
1 4 C H E M I S T R Y

E L E C T R O N I C S T R U C T U R E radiation) can promote them from the valence band (Figure 7b). This makes it quite conducting without any infra-red radiation from the target and makes the detectors very difficult to operate because they are very ‘noisy’. That is why the detectors are cooled with liquid nitrogen (many more of the electrons are in the valence band and few are in the conducti on band, Figure 7a) to give a good signal-to-noise ratio. A temperature difference of 0.1
o C can be detected at a distance of at least a mile.
Figure 7
Band structure of a narrow band gap semiconductor. (a) Ideal for use as a detector so that any electrons in the valence band (causing conductivity) are there because they have been promoted from the conduction band by photons from the target being imaged. (b) Actual band structure at room temperature because heat from the surroundings has promoted many ele ctrons to the conduction band. A few extra photons from the target to be imaged make little difference to the conductivity so the signal is difficult to detect. The idealised situation, (a), is approached by cooling the detector to liquid nitrogen temperature.
( a) (b)
Conduct i on band
V al ence band
C H E M I S T R Y 1 5

C H E M I S T R Y 1 6

C H E M I C A L B O N D I N G
Multiple bonding (pp 26–28)
We have already discussed the importance to life on earth of the O=O double bond in oxygen and the N

N triple bond in nitrogen because they remove far ultra-violet rays from the Sun’s radiation before it arrives on Earth. We have also discussed ozone and the fact that it has 1.5 bonds means that it requires less energetic light to dissociate an O atom than is required for O
2
, so it removes UVA. Without it, much of the UVA would get through to ground level and there would be an increase in DNA damage and skin cancer. That is why there is so much concern about the hole in the ozone layer. A discussion of the role of chlorofluorocarbons in the destruction of ozone and their banning under the Montreal protocol would be of interest, but may be beyond the scope of this unit.
Dative covalent bonds (p 27)
Experiment 6: Reaction of ammonia with water
You may wish to demonstrate the formation of a dative covalent bond in forming the ammonium ion. You can do this by dissolving some ammonia in water and then, using Universal indicator paper, showing the rise in pH because NH
4
+ (aq) and OH – (aq) ions are formed. You could also take a solution of ammonia and test it with universal indicator paper, either by dipping the paper into the ammonia, or by holding a damp strip of Universal indicator paper just above an open bottle of ammonia solution. Concentrated ammonia is preferable, but even 5 mol l –1 solution makes the paper turn deep blue almost immediately.
The fountain experiment is also an excellent way to demonstrate this reaction.
Experiment 7: Formation and dissociation of ammonium chloride
An alternative way of showing the formation of a dative covalent bond is to allow the fumes from ammonia solution to mingle with the fumes from concentrated hydrochloric acid (fume cupboard). Clouds of white smoke
(ammonium chloride) form (Equation 5).
NH
3
+ HCl NH
4
+ Cl – (5)
C H E M I S T R Y 1 7

C H E M I C A L B O N D I N G
Another way to show this, as well as the reversibilty of the reaction, is to heat some ammonium chloride crystals (50 mg) in a test -tube over a bunsen burner. If you only heat the bottom of the tube, a white solid will form at the cold neck of the tube. This is again ammoni um chloride.
This could lead well into a discussion of covalent and ionic bonding.
Ammonium chloride is ionic so should not be volatile. It dissociates to ammonia and hydrogen chloride on heating. These are covalent and volatile so move towards the entrance of the tube. In the cooler region they recombine to form the crystalline ammonium chloride. Here again a link could be made to the unit involving thermodynamics. Clearly, the position of the equilibrium shown in Equation 5 is being altered by the temperature.
The value of

H does not change much with temperature, but the value of
T

S must change with T . In this case,

S is positive from left to right since ammonia and hydrogen chloride is the more disordered system. At low temperature,

H , which is positive, is greater than T

S so that the equilibrium lies to the left (positive

G ), whereas at high T , T

S >


G is negative and the equilibrium lies to the right. The PPA on the
H so that thermodynamic prediction (decomposition of sodium hydrog en carbonate) also demonstrates a similar effect.
Experiment 8: Comparison between ammonia, water and methane
One other demonstration that could be done here if you have access to liquid ammonia (you really have to buy it) is to look at the properties of liquid ammonia, compared to those of water. Some of these are listed in Table 4.
Table 4
Property
Melting point/ o C
Boiling point/ o C
Density/gm cm –3
Self-ionisation constant
Water
0
100
1
10 –14
Ammonia
–78
–33
0.7
10 –27
IMPORTANT SAFETY NOTE: Experiments involving liquid ammonia must be performed in an efficient fume cupboard.
Liquid ammonia boils away quietly in uninsulated vessels, and a frosty coating of ice crystals forms on the outside of the vessel as moisture is condensed out of the air. This demonstrates the difference in boiling point between ammonia and water, and shows that the boiling poi nt of a mmon i a i s bel ow t he fr ee zi n g poi nt of wat er.
1 8 C H E M I S T R Y

C H E M I C A L B O N D I N G
The density difference can be demonstrated by pouring a little toluene into both water and liquid ammonia. Use a ratio of no more than 1 volume of toluene to 4 of ammonia or water so it is obvious which layer is which. The toluene (density 0.87 g cm –3 ) floats on the water but sinks in the ammonia.
With ammonia, this is best done in an unsilvered Dew ar vessel, but if one is not available it is still possible to see the lower layer of toluene in a boiling tube or measuring cylinder before the layer of frost becomes too thick.
Figure 8 cotton wool test tube boiling tube cotton wool or calcium chloride granules
If an unsilvered Dewar is not available, a simple air -insulated test-tube, as shown in Figure 8, can be used instead. Condensation of ice on the outside of the test-tube containing liquid ammonia is very much reduced by restricting access to the atmosphere in this way. Support the test -tube on a wad of cotton wool at the bottom of a boiling tube, and secure the test -tube by packing a ring of cotton wool between it and the top of the boiling tube.
Very little frosting now occurs when liquid ammonia is poured into the test tube. Resting the test-tube on a bed of anhydrous calcium chloride granules to remove moisture from the air between the tubes is even more effective.
N.B. Do not use this arrangement for liquid nitrogen since liquid oxygen will be condensed from the air between the tubes.
* * *
These properties can be explained in terms of the polarity of the
O(N)–H bond and the number of lone pairs. Oxygen is more electronegative than nitrogen so the O–H bonds are more polar. This means that the O –H bond breaks more easily than the N–H bond, hence Equation 7 lies further to the right than Equation 6 and the self -ionisation constant of water ( K eq
of
Equation 7) is hi gher t han t hat for a mmoni a ( K e q
f or Equat i on 6) .
C H E M I S T R Y 1 9

C H E M I C A L B O N D I N G
K = 1 0
– 2 7
2NH
3
NH
4
+ + NH
2
– (6)
K = 1 0
– 1 4
2H
2
O H
3
O + + OH – (7)
The higher bond polarity also means that hydrogen bonding is stronger in water than in ammonia (more hydrogen bonds occur in water than in ammonia because each water molecule can have both H atoms and both lone pairs involved in H bonding, whilst for ammonia only one lone pair is ava ilable).
This leads to a higher density for water as well as a higher melting point and boiling point. Note that the trend continues for methane, where hydrogen bonding is not possible because of the lack of lone pairs. Self -ionisation is negligible because there is no lone pair onto which to transfer H + , and methane is a gas at most accessible temperatures (bp –164 o C, mp –182 o C).
Ionic lattices (p 34)
We shall demonstrate some of the properties of ionic compounds in Section 3
(Section 3, pp 44 and following).
Superconductors (pp 35–38)
It is quite easy to make YBa
2
Cu
3
O
7– x
(note that it is a non-stoichiometric oxide) by heating the three binary oxides together.
Experiment 9: Synthesis of superconducting YBa
2
Cu
3
O
7 – x
This procedure is adapted from a met hod originally described by Goodfellow
Technology Limited who reserve the copyright. All information given by
Goodfellow Cambridge Limited (formerly Goodfellow Technology Limited) is given as a guide only. Although every effort has been made to ensure th at the information is correct, no warranty is given as to its completeness or accuracy.
Grind Y
2
O
3
(1.13 g), CuO (2.39 g) and BaCO
3
(3.95 g) together with a mortar and pestle until a homogeneous brown powder is obtained containing no white lumps. (If you have a three or four decimal place balance, it is better to reduce the amounts by a factor of 5 so that grinding is easier.) Place enough mixture in a crucible to fill it about 3/4 full and tap gently to ensure that the powder settles and there is very little air trapped between the particles (you may wish to press it gently with the end of a spatula). Place the crucible in an oven set at 940 o C and leave it heating overnight. If you have an automatic oven, set it for 12 h. Otherwise, switch the oven off the next morning and monitor the temperature drop. This should not be faster than
100 o C h –1 . If it is faster than that, use the heater controls to slow it down.
The important temperature range for this cooling rate is 800 –400 o C. The
2 0 C H E M I S T R Y

C H E M I C A L B O N D I N G cooled sample will have shrunk to about half its size and should appear as a black mass. Transfer it from the crucible into the mortar and grind it to a fine powder.
You now need to make the material into a pellet. A pellet press is best, but a much cheaper alternative is to use the apparatus shown in Figure 9, which your technical department might be persuaded to make for you. It consists of a cubic metal (preferably stainless steel) block about 3 cm in each dimension. Into this is drilled a hole about 0.5 cm i n diameter. There is a plunger which fits snugly into the hole and a plate. Put the plunger into the hole, leaving a space of about 0.5 cm depth at the top. Fill this with the black powder and smooth off. Press the pellet using a vice and then remove t he pellet from the press gently as shown. The pellet will be very fragile, so be very careful. Broken pellets can be reground and recast.
Figure 9
Making a superconductor pellet. (a) The powder is put into the block containing the plunger. (b) The steel plate is placed over the top and the apparatus is squeezed in a vice. (c) The pellet is removed by gently squeezing at an angle with the vice.
(a) (b) (c)
P o w d e r
P e l l e t
V i c e
S t e e l p l a t e
B l o c k
P l u n g e r
You now need to heat the pellet again in the crucible (actually three nested crucibles, with the inner one made from alumina and the outer two from silica is better as it cuts down the thermal gradients) exactly as before (940 o C, 12 h, slow cooling). The pellet is then the superconductor YBa
2
Cu
3
O
7– x
.
It is difficult to show that it is a s uperconductor because there are grain boundaries in the solid which have quite a high resistance. You could try to measure the resistance using a simple ohm meter. At room temperature, the resistance will be very high. However, if you can obtain some li quid nitrogen, you could cool the pellet and the resistance should go down because the critical temperature ( T c
) is 90 K and the temperature of liquid nitrogen is
77 K. (Actually this is very difficult to achieve.)
C H E M I S T R Y 2 1

C H E M I C A L B O N D I N G
You should be able to demonstrate the Me issner effect. Tie the pellet carefully to a piece of thread and suspend it from a clamp. Place liquid nitrogen into an expanded polystyrene cup and bring it up so that the suspended pellet is in the nitrogen. Wait until the violent boiling stops.
Remove the liquid nitrogen and bring a magnet close to the pellet. The pellet should be repelled, but will soon fall back as it warms above room temperature.
Another nice demonstration of the Meissner effect is as follows. Take a block of expanded polystyrene and cut a depression in the top. Place several
(5–7) pellets of superconductor in the depression and then fill the depression with liquid nitrogen. If you now place a small button magnet above the cooled pellets of superconductor, the magnet should levitate and spin.
If you do not have easy access to liquid nitrogen, you may be able to get some from the local health centre or chiropodist. It is used for treating warts and veruccas.
* * *
It is interesting to note that YBa
2
Cu
3
O
7– x
is a green powder. This is not the most expected form for a superconductor. Indeed, it is reported that a
Japanese group discovered it before Bendorz and Muller, but they did not believe that a green powder could have interesting electrical properties!
Semiconductors (pp 38–43)
There is a fabulous web-site on semiconductors. It is part of a Britney Spears web-site at http://www.britneyspears.ac and is called ‘Britney Spears’ guide to semiconductor physics’. The title page, which contains the photograph i n
Figure 10, claims that Britney is an expert on semiconductor physics and that she likes to explain the technology behind the sound storage and reproduction on CDs. The rest of this part of the web -site is serious semiconductor physics. Everything, such as p–n junctions, is explained qualitatively, with excellent pictures, and the mathematics behind everything is described. It goes into much more detail than you will ever need, but is liberally sprinkled with pictures of the superstar and makes learning about semiconductors qui t e f un.
2 2 C H E M I S T R Y

C H E M I C A L B O N D I N G
Fi gure 10
Photograph from ‘Britney Spears guide to semiconductor physics’, ht t p:/ / www.bri t neyspe ars.ac
Se mi condu ct or s are al l ar ound us , most co mmonl y i n c o mp ut er s, al t hough i t i s di f fi cul t t o see t he m t her e. M or e ob vi ous are t he sol ar cel l s on sol ar -p owered cal cul at or s and you may ha ve a model bui l di ng ki t at ho me t hat has a sol ar panel for r unni n g t he hel i copt er bl ades , wi nd mi l l sai l s or what ever. If not , pi ct ures of sol ar -po wered ve hi cl es can be found on t he w eb , e. g. at ht t p: / / ww w. kyocera.de
C H E M I S T R Y 2 3

2 4 C H E M I S T R Y

S O M E C H E M I S T R Y O F T H E P E R I O D I C T A B L E
The second and third short periods
This is an ideal opportunity to do some simple demonstrations.
Oxides (pp 44–47)
Experiment 10: Properties of oxides
If you can get hold of the oxides, you can examine their behaviour by dissolving them in water and measuring their pH. The easiest ones are the solids, but some will not dissolve. For those that do, you could measure the pH of the solution (using universal indicator pa per) and show which ones give acidic solutions and which ones give basic solutions. In some cases the compounds are very toxic, so you will not be able to use them, while in others extreme care must be taken. Ones that can be easily demonstrated are the oxides of Li, Na, Mg, B, Al, C, Si, and S. The common forms of the oxides are shown in the table below, together with how to generate them if they are not readily available and any hazards associated with their preparation or use.
For each of these experiments, 50 mg of solids and 1–2 cm 3 of liquids should be enough, unless otherwise stated.
Oxide Generation
Li
2
O Can be purchased
Na
2
O Cut some sodium under dry air (desiccator containing KOH pellets). Leave it to stand; the white solid that slowly forms on the freshly cut surfaces is sodium oxide (in wet air, NaOH forms). Try to scrape some solid off and dissolve it in water.
MgO
Burn magnesium ribbon (50 mg) in air (nice bright flame). The powder that forms (you may wish to grind it) is largely MgO. It is not very soluble in water, but does give a mildly alkaline solution. It should dissolve in dilute HCl(aq).
Hazards
Sodium reacts quite violently with water.
Na
2
O gives a strongly alkaline aqueous solution. It will burn holes in clothes, and attack the skin.
2 5 C H E M I S T R Y

S O M E C H E M I S T R Y O F T H E P E R I O D I C T A B L E
Oxide Generation
B
2
O
3
Can be purchased. Again only sparingly soluble in cold water. More soluble in hot. Dissolves in dilute NaOH(aq). B
2
O
3
is converted to B(OH)
3
(boric acid) on dissolving in water. Solutions of boric acid in water do not affect Universal indicator paper – the paper itself is of comparable acidity. However, a solution of
1 cm 3 of universal indicator in 100 cm 3 of distilled water may be boiled for a few minutes
CO
2 until it turns green, indicating that all the dissolved CO
2
has been removed. Addition of 1 g of analytical grade boric acid to the hot water turns it yellow-orange – about as strong an acid as aqueous CO
2
.
Al
2
O
3
Purchased forms are often very difficult to dissolve. You can make it by adding dilute
NaOH(aq) (1 drop) to a solution of aluminium sulphate (50 mg) in water (1.5 cm 3 ). Once you have the gelatinous precipitate, you can show that it redissolves on dropwise addition of dil ute
HCl(aq) to excess (2 drops, 5 mol l –1 , should be enough), or on addition of NaOH(aq) (again, 2 drops, 5 mol l –1 , should be enough).
Add dilute HCl(aq) to sodium bicarbonate (solid or solution) in a conical flask fitted with a gas delivery tube. To show the acidity of CO
2
it is necessary to pass it into water that has previously been well flushed with nitrogen or air from a cylinder, or boiled for some time. Do this with water containing 1 cm 3 of Universal indicator. It will turn green once all the CO
2
has been removed, and turn orange on passing in the
CO
2
. Even blowing through a straw is sufficient.
SiO
2
Water in contact with normal air has a pH of c.
5.2 because of the dissolved CO
2
.
You can buy this as silica, but it is often very difficult to dissolve. Precipitated silica or silica gel for chromatography (0.2–0.5 mm grain size) are best and do dissolve in hot NaOH(aq) (2 cm 3 ,
5 mol l –1 ).
Hazards
2 6 C H E M I S T R Y

S O M E C H E M I S T R Y O F T H E P E R I O D I C T A B L E
Oxide Generation
P
2
O
5
Buy it as phosphorus pentoxide. It dissolves in water to give an acidic solution. A suitable scale is 50 mg in 5 cm 3 water.
SO
2
Use dilute HCl(aq) (5 mol l –1 ) rather than concentrated to avoid contaminating the gas with hydrogen chloride and add it dropwise onto solid
Na
2
S
2
O
5
(Na
2
S
2
O
5
generates more SO
2
than does
Na
2
SO
3
in reacting with acid) in a conical flask fitted with a magnetic stirrer, dropping funnel and gas delivery system. Care must be taken to avoid suck-back when passing SO
2
into water – conduct the gas into an inverted funnel just touching the surface of the water, or bubble it through the water using a 20 cm 3 bulb pipette, which can be withdrawn immediately if any suck-back occurs. The solution produced is pH 2 or 3 to Universal indicator paper and a sample, after neutralisation with ammonia (no harm in adding an excess), gives a white precipitate with barium chloride solution (0.1 mol l –1 ) redissolves on adding excess dilute nitric acid (5 mol l –1 ).
Oxygen from the air gradually oxidises the solution to sulphuric acid, but adding ‘20 vols’
H
2
O
2
(1.8 mol l –1 ) is faster (use 1 cm 3 peroxide
Hazards
The reaction of P
2
O with water is very
5 violent so it is best to use very small amounts at a time and to stir the water in a beaker with a magnetic stirrer.
Wear gloves and use a fume cupboard.
SO
2
is highly toxic and should always be handled in a fume cupboard for every 2 cm 3 of sulphurous acid). The temperature rockets and the acidity increases to pH 1. On treating the solution now with barium chloride, the white precipitate produced is insoluble in nitric acid. Alternatively, add
H
2
O
2
(aq) to the solution of barium sulphite in dilute nitric acid, when a white precipitate of barium sulphate is produced.
It is probably not worthwhile to try to demonstrate the melting behaviour of the solids nor their conductivities. This is probably best done for the chlorides (see later).
C H E M I S T R Y 2 7

S O M E C H E M I S T R Y O F T H E P E R I O D I C T A B L E
Chlorides (pp 48–50)
Experiment 11: Melting and conductivity of metal halides
You should be able to melt LiCl (mp 605 o C) and NaI (mp 661 o C), but not
NaCl (mp 801 o C), MgCl
2
(mp 714 o C) or MgBr
2
(mp 700 o C) in a crucible over a bunsen burner. A mixture of equal masses of MgCl
2
and MgBr
2
is also fusible.
A multi-meter with steel wire electrodes (unfolded paper clips) will demonstrate that the melts conduct electricity. Al ternatively, you can set up a circuit with a light bulb and d.c. power supply and make the circuit through the molten salt using graphite electrodes. The bulb should light if the voltage is turned up.
In principle, you could also electrolyse the chlorid es to show that chlorine is released at the positive electrode and the metal deposited at the negative electrode, but this is not recommended as chlorine is pretty unpleasant and the metals are very reactive with air or water . This electrolysis is carried out commercially in the production of sodium and chlorine. The sodium floats to the top of the melt, where it forms a crust of sodium hydroxide through reaction with the air. It is said that operators can walk around on this crust, but this is not recommended! Electrolysis of aqueous sodium chloride solution produces chlorine and hydrogen. There is a build -up of hydroxide ions in solution and this is the basis of the important chlor -alkali industry.
Experiment 12: Halides and water
You could dissolve the halide salts of Group 1 and 2 metals (50 mg) in water (10 cm 3 ) and measure their pH (should be neutral) and conductivity. There is little that can be done with the covalent halides of the second short period as they are all too reactive to handle in a laboratory except CCl
4
, which is a highly potent carcinogen so should not be handled . It is also not very reactive.
AlCl
3
, SiCl
4
and PCl
3
are readily hydrolysed, but the reactions can be quite violent. For SiCl
4
and PCl
3
, which are volatile liquids, the best way to demonstrate that they react with water is to pour some of the liquid into a beaker. It will give off white fumes. Note that the fumes that you can see are the oxide, SiO
2
, and the acid, phosphorous acid (Equations 8 and 9).
SiCl
4
+ 2H
2
O SiO
2
+ 4HCl (8)
PCl
3
+ 3H
2
O
HO
HO
P
O
H
+ 3HCl (9)
2 8 C H E M I S T R Y

S O M E C H E M I S T R Y O F T H E P E R I O D I C T A B L E
To show that HCl is also given off, you could put some damp Universal indicator paper into the fumes and it will turn pink. This is largely due to the presence of HCl. You can also add a small amount of the liquid dropwise to water using a Pasteur pipette. The reaction is quite vigorous and for SiCl
4
a white precipitate of SiO
2
forms. For PCl
3
, a clear solution should be obtained. In both cases, the solution will be acid. Yo u could titrate with sodium hydroxide (methyl orange indicator). In this case, you will have to measure the amount of chloride added, quickly because of the fuming. It is suggested that you use a dropping pipette and add 1 g dropwise to about 25 cm 3 of water. You can then titrate this with standard NaOH(aq) (1.0 mol l –1 ).
You should be able to show that 4 mol of acid are produced per mole of SiCl
4
(all HCl), while 5 mol of acid are produced from PCl
3
, 3 of HCl and 2 from the H
3
PO
3
, according to Equations 8 and 9. Note the structure of H
3
PO
3
, which is not P(OH)
3 as might be expected for hydrolysis of PCl
3
, but rather
HP(O)(OH)
2
. This means that H
3
PO
3 is a dibasic acid (pK
1
= 2.0, pK
2
= 6.6) because the H atom directly attached to P is not acidic.
The titration can be carried out as follows. Using a strong base, neutralisation of the first proton can be followed using methyl orange (pK a about 3) and neutralisation of the second proton can be followed using phenolphthalein (pK a
about 9). The three free hydroxonium ions are neutralised first (hydrochloric acid is a strong acid), then the weak phosphorous acid is neutralised in two stages.
H
[H
3
PO
2
PO
3H
3
O + + 3OH
3
+ OH
3
] –
–
+ OH
– 6 H

[H
2
PO
– 
[HPO
3
] –
3
] 2–
2
O
+ H
+ H
2
O
2
O
Thus, at the first end-point four protons have been neutralised and at the second end-point all five have been neutralised, so the second titre should be
1.25 times the first.
PCl
3
reacts quietly with cold water. About 1 g of PCl
3
(i.e. a ‘pipetteful’ – one squeeze of the bulb) is squirted into a tared conical flask containing about 25 cm 3 of cold distilled water. The flask is re -weighed to determine the mass of PCl
3
added. After swirling to dissolve the PCl
3
, the solution is titrated against standardised NaOH(aq) using first methyl orange then (in the same flask) phenolphthalein as indicators.
C H E M I S T R Y 2 9

S O M E C H E M I S T R Y O F T H E P E R I O D I C T A B L E
NaOH solution
First titre (methyl orange)
Second titre (phenolphthalein)
Ratio of titres (1.25 expected)
PCl
3
First titre
1.015 mol l –1
31.53 cm 3
39.40 cm 3
1.2496
137.33 g mol –1
 –3 mol PCl
3
3
Second titre
 –3 mol PCl
3
Mass of PCl
3
actually added 1.10 g
3
Experiment 13: Properties of AlCl
3
Now we will look at AlCl
3
. This does not really fume in air although it does bubble up. This is because AlCl
3
in solid form has an ionic lattice structure, so AlCl
3 is not very volatile. On heating it sublimes at 180 o C. You can demonstrate this by heating a small amount (50 mg) in a test -tube over a bunsen burner. Crystals of sublimed AlCl
3
should grow near the neck of the test-tube. Alternatively, you can put a small amount of AlCl
3
(0.1 g) into a petrie dish and put the cover on. You can then heat the petrie dish on a hotplate, gradually increasing the temperature until crystals start to grow on the cooler cover.
A small amount of AlCl
3
can be dissolved in water ( CARE: This reaction is very exothermic and can be quite violent ) and the pH tested using universal indicator paper (pH c.
4). In this case, the chemistry is slightly different, as shown in Equations 10 and 11.
AlCl
3
+ 6H
2
O [Al(H
2
O)
6
] 3 + + 3Cl –
(10)
[Al(H
2
O)
6
] 3 + [Al(H
2
O)
5
(OH)] 2 + + H +
(11)
The acidity arises from the loss of protons from the [Al(H
2
O)
6
] 3+ ion because, being small, Al does not like to have such a high positive charge. One way of reducing this charge is to lose a proton to form [Al(H
2
O)
5
(OH)] 2+ .
Unfortunately, this is much more difficult to demonstrate quantitatively because titration with NaOH(aq) removes more protons and forms Al(OH)
3
. A nice way to demonstrate the acidity
3 0 C H E M I S T R Y

S O M E C H E M I S T R Y O F T H E P E R I O D I C T A B L E of [Al(H
2
O)
6
] 3+ is to dissolve aluminium sulphate (1 –2 g) in water
(50 cm 3 ) and add sodium bicarbonate (3–4 g). Vigorous effervescence occur s because of the reaction of the acidic protons with the bicarbonate to give carbonic acid, which is unstable with respect to the formation of CO
2
. A milky white precipitate of Al(OH)
3
also forms. The chemistry here is described by Equations 11–12.
H + + HCO
3
– H
2
CO
3
H
2
O + CO
2
(12)
Hydrides (pp 51–53)
Apart from methane, ammonia, water, HCl and perhaps NaH, MgH
2
and
CaH
2
, the hydrides are all too reactive or toxic to handle in a school or college laboratory. Nevertheless, they do show very well t he difference in the electronegativity of elements. Where the electronegativity of the element attached to H is similar to that of H (CH
4
), the C–H bond is unreactive towards water. If the element is more electronegative than H, the polarity of the bond is such that the H is positive and the element is negative. Water, ammonia, HCl and HF are examples. The reaction of hydrides with water depends on the electronegativity of the element with respect to O. N is less electronegative than O so ammonia is ba sic, water is neutral and, because Cl and F are more electronegative than water, HCl and HF are acidic. Because we tend to think of H as being acidic (H + ), the more surprising compounds are the metal hydrides. In these compounds the element attached to H is less electronegative than H so that they contain H – . For school demonstrations, the best one to use is CaH
2
(although it is in the fourth short period, it is fortunately mentioned on page 51). It is relatively stable so can be stored in a closed bottle for a long time. Nevertheless, it reacts with water to give hydrogen and an alkaline solution (contrast HCl).
Experiment 14: Reaction of calcium hydride with water
If you can obtain CaH
2
, you might demonstrate the hydridic nature of the H atoms by putting a small amount of CaH
2
(a small spatula end, 50 mg) into a test-tube, adding about 1 cm 3 of water from a wash bottle and swirling.
Vigorous effervescence will be observed and a lighted splint will give a pop.
( CARE: The reaction is very exothermic .) When the reaction has subsided, testing the solution with universal indicator paper will show that it is alkaline
(Equation 13).
CaH
2
+ H
2
O

Ca 2 + + 2OH – + H
2
(13)
C H E M I S T R Y 3 1

S O M E C H E M I S T R Y O F T H E P E R I O D I C T A B L E
The hydrogen economy
Because fossil fuels are running out and because the burning of fossil fuels leads to the production of CO
2
, which has been associated with the greenhouse effect (see above), global warming and climate change , there is considerable interest in alternative energy strategies. One strategy that is being seriously considered is to use hydrogen as a fuel. This is a Utopian dream because it could provide for all our energy needs in a totally renewable manner. When hydrogen burns it gives heat (energy) and water. If the hydrogen could be produced from water, then we would have a totally recyclable system. Of course, the thermodynamics are against us because to produce hydrogen from water requires energy. Being t hermodynamically uphill, we might think it cannot be done. However, we know that water can be split into hydrogen and oxygen by electrolysis. Solar light, even at the red end of the spectrum (800 nm), has sufficient energy to split water into hydrogen and oxygen. The fact that when you are having a barbeque on the beach and the sun comes out you do not have an enormous explosion, means that the water splitting reaction does not happen under normal conditions.
However, researchers are developing catalyst s to put into the water so that this photochemical decomposition will occur. This chemistry is summarised in Equation 14.
(15) h 
H
2
O
H
H
+ O
- 14
Alternatively, solar cells could be used to electrolyse water. Even if the efficiency is only 10%, all the world’s energy needs could be supplied by harvesting the light that falls on an area the size of Libya. Perhaps Colonel
Gadhafi would not like this, but better distributed scenarios can be envisaged.
If we are to implement this wonderful renewable and permanent so lution to our energy problems, we have to find ways of storing and releasing hydrogen, since we would probably not like to drive around with compressed hydrogen cylinders in our cars. The Hindenberg and RR2 airships have given hydrogen a bad press. Actually, hydrogen is not as bad as it appears. Before North Sea gas, we used town gas, which was predominantly hydrogen. Feasibility studies have shown that hydrogen can be stored in underground caverns and transported in existing pipelines. Furthermore, be cause hydrogen is so light, it burns upwards, away from where people are.
3 2 C H E M I S T R Y

S O M E C H E M I S T R Y O F T H E P E R I O D I C T A B L E
Experiment 15: Exploding hydrogen balloons
A nice demonstration of hydrogen’s upward burning is to fill two balloons with hydrogen and tie them to the same piece of string with 6 inches between the top of one balloon and the bottom of the other. You can then apply a lighted taper (on the end of a long stick) to the top balloon. It will burst with a rather satisfying ‘bang’, but the other balloon will remain intact. You can then show the effect again by turning the lights off and applying the taper to the other balloon and demonstrating that the flame burns upwards. Any other fuel tends to be of similar density to, or heavier than, air so the flame burns downwards and incinerates everyone on the ground around the point of the explosion.
It is still probably not desirable to have compressed hydrogen in your car or aeroplane, so an alternative is to absorb the hydrogen into a metal or alloy to make a hydride. One of the earliest metals to be investigated was magnesium because it absorbs hydrogen at atmospheric pressure and room temperature to give magnesium hydride (Equation 15).
(15)
1 atm 25°C
Mg + H
2
MgH
2
80°C
Because of the low atomic mass of magnesium, it can s tore a relatively large amount of hydrogen (7.7 wt %). Furthermore, the hydrogen can be released on heating (entropy again, the reverse of Equation 15). In fact magnesium will not be used because the temperature required to release the hydrogen is too high (80 o C), but the unusual properties of MgH
2
(reversible formation from Mg and H
2
under reasonable conditions) spurred the research into alternatives, which are currently being vigorously pursued.
Transition metal chemistry (pp 59–68)
Because of the colour of transition metal complexes, there are many attractive experiments that can be carried out to demonstrate transition metal chemistry.
Many of these can be carried out in a small beaker (the flatter the bottom, the better) on an overhead projector. T he whole class will then easily see the colour changes.
Experiment 16: Colour in transition metal salts
It is perhaps instructive to start by showing the class a series of transition metal salts, either as solids or solutions ( c.
0.25 g in c.
10 cm 3 of water), and compare them with salts of, for example, K + , Ca 2 + , Zn 2+ and Al 3+ . The difference is very obvious. It is useful to include
C H E M I S T R Y 3 3

S O M E C H E M I S T R Y O F T H E P E R I O D I C T A B L E manganese sulphate since this is almost colourless (very pale pink in the solid). This is discussed in more detail below.
* * *
There are some nice experiments that can be used to demonstrate how transition metals exhibit different oxidation states and how these different oxidation states have different colours.
Experiment 17: Different oxidation state s of manganese
(CARE: This experiment involves concentrated (then molten) NaOH, which is highly corrosive. The flask will be destroyed.)
Prepare a saturated solution by dissolving NaOH (10 g) in water (10 cm 3 ) in a totally clean 100 cm 3 conical flask (it is best to use a new one). Cool to room temperature. Add a small crystal (the smaller, the better) of potassium permanganate. This will dissolve to give a purple solution (MnO
4
– , manganese is in an oxidation state of +7). Place the flask on a gauze o ver a bunsen burner and heat gently. The colour will rapidly change to green
(MnO
4
2– , manganese is in an oxidation state of +6). Carry on heating and the colour will change to blue (MnO
4
3– , manganese is in an oxidation state of +5) and then if you are lucky to straw colour (Mn 3+ ) and colourless (Mn 2+ ). On cooling, you will find that the NaOH solidifies and is pale blue. This is because the stable oxidation state of manganese in fused alkali is MnO
4
2– .
Note that the reducing agent in these reactions is OH – and it is oxidised to oxygen. An example of the chemistry going on is shown in Equations 16 –18:
MnO
4
– + e – MnO
4
2 –
(16)
OH – 1/2O
2
+ H + + e –
(17)
Overall MnO
4
– + OH – MnO
4
2 – + 1/2O
2
+ H +
(18)
Experiment 18: Different oxidation states of vanadium
In a reagent bottle (50 or 100 cm 3 , glass or plastic stopper) or conical flask (50 or 100 cm 3 ), dissolve ammonium metavanadate (0.1 g) in dilute
HCl(aq) (20 cm 3 of 5 mol l –1 ) by shaking for a minute or two. The pale yellow solid is first converted by the acid to brick-red V
2
O
5
, which then dissolves giving a yellow solution containing the [VO
2
(H
2
O)
4
] + ion
(vanadium is in an oxidation state of +5). The flask is placed on a white tile, granulated zinc (2 g in fairly large pieces) is added then the stoppe r is placed loosely in the neck of the flask so that H
2
can still escape. Rapid effervescence of H
2
occurs and the yellow solution quickly turns
3 4 C H E M I S T R Y

S O M E C H E M I S T R Y O F T H E P E R I O D I C T A B L E sky blue in colour, due to [VO(H
2
O)
5
] 2+ (vanadium is in an oxidation state of
+4). This slowly changes to emerald green, due to [V(H
2
O)
6
] 3+ (vanadium is in an oxidation state of +3), then, after 15 –20 minutes, violet, due to
[V(H
2
O)
6
] 2+ (vanadium is in an oxidation state of +2).
Using 4 g of zinc completes the reduction to vanadium in an oxidation state of +2 in half the time, but the intermediate oxidation state changes are less distinct. Aqueous vanadium in an oxidation state of +2 is slowly oxidised by air back to green vanadium in an oxidation state of +3 – the solution turns blue (mixture of violet vanadium in an oxidation state of +2 and green vanadium in an oxidation state of +3) on standing in air for an hour with occasional swirling.
Experiment 19: Different oxidation states of chromium
The same method as above is used with 0.1 g of potassium dichroma te, dilute
HCl(aq) (20 cm 3 , 5 mol l –1 ) and zinc (4 g). The orange solution contains
K
2
Cr
2
O
7
(Cr VI ). Stopper the bottle and shake vigorously, occasionally pausing and removing the stopper to prevent a build -up of pressure from the hydrogen being released by the reaction of HCl with zinc. The initially orange solution gradually changes to emerald green Cr 3+ (in fact the green colour arises from a complex containing co-ordinated water and chloride, [Cr(H
2
O)
4
Cl
2
] + ), then slowly to sky blue. The blue solutio n (chromium in an oxidation state of +2) is rapidly oxidised to green (chromium in an oxidation state of +3) on exposure to air.
[Cr(H
2
O)
6
] 3+ is violet. To show the colour of [Cr(H
2
O)
6
] 3+ , use 0.2 g
KCr(SO
4
)
2
.12H
2
O in 10 cm 3 of water. The solution appears violet under incandescent lighting conditions, but blue -grey under fluorescent lighting and blue-green by daylight. [Cr(H
2
O)
6
] 3+ is also obtained by dissolving chromium(III) chloride in water to give a green solution of [Cr(H
2
O)
4
Cl
2
] + and leaving it to stand overnight. Slow hydrolysis gives [Cr(H
2
O)
6
] 3+ .
* * *
Ligands (pp 59–60)
EDTA forms very stable complexes with transition elements. This can be demonstrated by carrying out the following experiment.
Experiment 20: Complexation by EDTA
Make up a solution of copper(II) sulphate (0.1 g) in water (2 cm 3 ) in a test-tube. Pour half the solution into another test -tube. To one of the solutions add concentrated ammonia solution dropwise. A pale blue precipitate of Cu(OH)
2
forms because ammonia is a base and Cu(OH)
2
,
C H E M I S T R Y 3 5

S O M E C H E M I S T R Y O F T H E P E R I O D I C T A B L E being neutral, is insoluble in water. Continue to add ammonia dropwise and the precipitate will redissolve to give a deep blue solution containing
[Cu(NH
3
)
4
(H
2
O)
2
] 2+ (see pp 65–66).
Take the other copper(II) sulphate solution and add EDTA (0.2 g). Heat over a bunsen burner to boiling point. Allow to cool. Note the intensification of the blue colour as a copper EDTA complex forms. Now add concentrated ammonia solution dropwise to excess. No precipitate forms and no dark blue colour is observed because the EDTA has formed a stable complex and is not easily displaced by ammonia.
* * *
This ability to co-ordinate so strongly to metal ions, and the fact that the complexes are charged (2 – if the metal is 2 + , since EDTA is a 4 – ligand;
Figure 36, p 60 does not show the charges but there is one negative charge on each O which is not double bonded to C), means that EDTA is used for treating heavy metal poisoning, e.g. by lead. The EDTA co -ordinates to the lead making a soluble complex, which can readily be excreted. Actually,
EDTA also co-ordinates very strongly to Ca 2+ , so, if used alone, it would do some damage to bones. In practice it is administered as the calcium complex
Na
2
[Ca(EDTA)]. This exchanges with the lead sufficiently to solubilise the lead. EDTA is also found in solutions of ‘sequestered iron’ used to feed house plants. [Fe(EDTA)] + (iron in an oxidation state of +3) is present and this provides a slow release of Fe 3+ over time.
Experiment 21: Changes in co-ordination number of copper(II)
To demonstrate the effect of halides on co -ordination number, a copper(II) sulphate solution (10 mg in 5 cm 3 of water) can be treated with concentrated
HCl(aq) dropwise to excess. The pale blue colour changes to light green
(mixture of blue, octahedral [Cu(H
2
O)
6
] 2+ and yellow, tetrahedral [CuCl
4
] 2– ) after addition of 0.3–0.4 cm 3 and to the yellow solution of [CuCl
4
] 2– after addition of 1 cm 3 . Addition of water restores the blue colour of [Cu(H
2
O)
6
] 2+ .
Experiment 22: Changes in co-ordination number of cobalt(II) (1)
A better demonstration is carried out in exactly the same way but using cobalt(II) sulphate (10 mg) in place of copper(II) sulphate. The original solution is pale pink (octahedral [Cu(H
2
O)
6
] 2+ ) but turns dark blue (tetrahedral
[CoCl
4
] 2– ) on addition of Cl – . Again, this reaction can be reversed by diluting with water. These reactions are what occur in ‘self -indicating’ silica gel.
The gel contains blue [CoCl
4
] 2– when it is dry. On absorption of water,
[Co(H
2
O)
6
] 2+ forms and the gel turns pink. On heating, the water is driven off and [CoCl
4
] 2– is regenerated.
3 6 C H E M I S T R Y

S O M E C H E M I S T R Y O F T H E P E R I O D I C T A B L E
Experiment 23: Changes in co-ordination number of cobalt(II) (2)
A rather similar example is provided if cobalt(II) chloride (10 mg) is dissolved in ethanol (5 cm 3 ). Ethanol is bigger than water and only four ethanol molecules fit around the cobalt, so blue, tetrahedral [Co(EtOH)
4
] 2+ forms. Addition of water to this solution produces octahedral [Co(H
2
O)
6
] 2+ so it turns pink.
* * *
When the very first breathalysers were being proposed, it was suggested that tubes filled with cobalt sulphate could be used. The more alcohol one had in one’s breath, the more salt would turn blue and the length of blue material in the tube would be an indication of the breath alcohol level. Unfortunately, the equilibrium constant for displacement of water is rather low and anyone who had enough alcohol in his or her breath to make a significant change to the colour would be so hopelessly dru nk that they would probably not be able to blow into the tube at all!
Experiment 24: Changes in co-ordination number of nickel(II)
Another good example of a change in colour with co -ordination number is carried out as follows. Dissolve nickel(II) sulphat e (50 mg) in water (5 cm 3 ) in a beaker. Add a solution of dimethylglyoxime (50 mg) in ethanol (3 cm 3 ).
This gives a brilliant red precipitate of the complex shown in Figure 12, which is square planar. It is neutral, so insoluble in water, hence the precipitate formation.
Figure 12
Structure of dimethylglyoxime complex of nickel.
H
O O
H
3
C N N
CH
3
Ni
H
3
C N N
CH
3
O O
H
Colour from light absorption (p 62)
For a good way to demonstrate the principles of light emission and absorption shown on p 62, see the earlier discussion on light emiss ion and absorption
(Experiment 1) .
C H E M I S T R Y 3 7

S O M E C H E M I S T R Y O F T H E P E R I O D I C T A B L E
The spectrochemical series (p 63)
It is quite difficult to get a good example to illustrate the effects of the spectrochemical series, i.e. the effect of the ligand on the value of

, because other things often happen when exchanging ligands, like the metal changes its co-ordination number or oxidation state, the symmetry of the molecules changes (partial substitution) or charge -transfer bands start to dominate the spectrum. Perhaps the best example is the reaction of [Ni(H
2
O)
6
] 2+ with ammonia to give [Ni(NH
3
)
6
] 2+ .
Experiment 25: The spectrochemical series using Ni complexes (oxidation state +2)
Dissolve nickel(II) sulphate (50 mg) in water (1 cm 3 ) in a test-tube. Add concentrated ammonia solution ( fume cupboard ) dropwise to excess. The solution changes from green [Ni(H
2
O)
6
] 2+ to purple [Ni(NH
3
)
6
] 2+ (note the precipitation of green Ni(OH)
2
and the formation of blue [Ni(NH
3
)
4
(H
2
O)
2
] 2+ on the way).
The light absorption is at 8500, 13800 and 25300 cm –1 for [Ni(H
2
O)
6
] 2+ and at
10750, 17500 and 28200 cm –1 for [Ni(NH
3
)
6
] 2+ . The shift to shorter wavelength (higher energy) is because ammonia causes a larger d orbital splitting (higher

) than water. The reason why there is more tha n one d–d transition is beyond the scope of Advanced Higher, but is explained in most textbooks on inorganic chemistry.
Colour in manganese complexes (p 64)
We have already noted that [MnO
4
] – , d 0 is very dark purple because of charge transfer, as described on p 64. We have also noted that manganese(II) salts are essentially colourless in solution. This is because [Mn(H
2
O)
6
] 2+ contains
Mn 2+ , which is 3d 5 . The crystal field energy level diagram is as shown in
Figure 13.
Figure 13
Crystal field energy level diagram for Mn 2+ , 3d 5 . Note that each orbital contains one electron and all the electrons have the same spin.
All the orbitals contain one electron and all the electrons have the same spin. An electron cannot move from one of the lower levels to one of the higher levels and keep the same spin because the Pauli exclusion
3 8 C H E M I S T R Y

S O M E C H E M I S T R Y O F T H E P E R I O D I C T A B L E principle (p 14) would be broken. It can move if it changes spin, but absorptions of this kind have a very low probability so are extremely weak.
Catalysis (pp 66–68)
The example involving Rochelle’s salt given in the book is a good one to demonstrate.
Experiment 26: Cobalt complexes in catalysis
Take Rochelle’s salt (potassium sodium tartrate, 25 g) and dissolve it in water (250 cm 3 ) in a tall-form 1 litre beaker. Add H
2
O
2
solution (30 cm 3 of
100 volume or 100 cm 3 of 30 volume). Nothing much happens. Heat to about
60 o C (just too hot to touch) over a bunsen burner. Again not much happens (a few bubbles start to form). Meanwhile, dissolve cobalt(II) chloride (1 g) in water (25 cm 3 ). Add it to the hot solution in the beaker. The pale pink solution goes darker pink then green and there is vigorous effervescence.
Once the effervescence subsides (all the H
2
O
2
is used up), the solution turns pink. The active catalyst is Co 3+ (green). In this experiment, the reaction occurring is oxidation of tartrate to CO
2
by H
2
O
2
and the effervescence is the evolution of CO
2
.
Using these quantities, the reaction can overflow and be quite messy so you may wish to scale down the quantities.
Experiment 27: Autocatalysis of a permanganate oxidation
Another example that students may have come across is in the titration of oxalic acid (ethanedioic acid) with potassium permanganate. This reaction is slow at room temperature, so the titration is usually carried out at 60 o C.
However, the reaction is catalysed by Mn 2+ ions. This can be demonstrated as follows.
Make up a solution of oxalic acid by dissolving sodium oxalate (2 g) in sulphuric acid (50 cm 3 , 1 mol l –1 ). Add KMnO
4
solution (1 cm 3 ,
0.2 mol l –1 ). A purple solution should form. Divide this solution into two portions. Heat one on a bunsen burner and it will decolourise. To the other, add a solution of manganese(II) sulphate (10 mg) in water
(1 cm 3 ). The solution should rapidly tur n pale brown and then sl owl y decol ouri se.
C H E M I S T R Y 3 9

S O M E C H E M I S T R Y O F T H E P E R I O D I C T A B L E
This experiment shows a number of different points:
(i) Mn 2+ catalyses the reduction of MnO
4
– by ethanedioic acid at room temperature
(ii) the reaction does not go in one step but goes through a detectable intermediate, the brown [Mn(C
2
O
4
)
2
] – (manganese is in an oxidation state of +3)
(iii) Mn 2+ only catalyses the reduction to manganese in an oxidation state of
+3 effectively; the reduction of manganese from an oxidation state of
+3 to +2 is still slow at room temperature
(iv) because Mn 2+ is the product of the overall reaction, once the reaction without added Mn 2+ has started it is not necessary to heat the solution during the rest of the titration because the reaction is catalysed thereafter. In fact this is normally not done.
(v) when the product of the reaction is also a catalyst for the reaction, this is called autocatalysis .
4 0 C H E M I S T R Y