Unit 9-1

Unit 9 The p-Block Elements

Section 9.1 The Halogens （1）Characteristic properties of the halogens

The Group VII elements, the halogens, are reactive non-metals. They have the electronic structure ns2 np5 for their valence shells. Halogens exist as diatomic molecules, X2. The first element in the group, fluorine, is anomalous, just as lithium and beryllium are in Group I and II respectively. Astatine at the bottom, is radioactive, as are francium and radium in the s-block. Not much is known about this element except that it seems to be typical of its group as far as its chemical properties are concerned. Element atomic

number electronic

configuration colour and

state atomic

radius nmionic radius

X- nm electronegativity electron

affinity kJmol-1

Fluorine 9 [He]2s22p5 pale yellow g. 0.072 0.136 4.0 -348 Chlorine 17 [Ne]3s23p5 yellow green g. 0.099 0.181 3.0 -364 Bromine 35 [Ar]3d104s24p5 red brown l. 0.114 0.195 2.8 -342 Iodine 53 [Kr] violet black s. 0.133 0.216 2.5 -314

Astatine 85 [Xe] 0.140 ? 2.2 ? 1. High electronegativity and electron affinity All halogens have high electronegativities and highly negative values of electron affinity, indicating that they have a high tendency to attract an additional electron to complete the octet, which has extra stability. 2. Ionic and covalent bonding in oxidation state -1 When halogens react with other elements, they can complete their octet by either gaining an additional electron to form the halide ion, or by sharing their p electron to form a covalent bond, depending on the electronegativity of the atom with which the halogen is bonded. Ionic bonding All the Group VII elements form ionic halides with electropositive metals, i.e. the alkali and alkaline earth metals. The halogen atom, X, achieves a noble gas configuration by gaining one electron to form a halide ion, X-, with oxidation state -1. Covalent bonding All the halogen atoms may achieve a noble gas electronic configuration by forming a single covalent bond with non-metals, e.g. in the hydrogen halides, HX :

Except for fluorine, the halogens can also show covalencies of 3,5 and 7, by promoting electrons from p orbitals into empty d orbitals in the valence shell. Iodine, for example, shows covalencies 1,3,5 and 7:

5s 5p 5p 5p 5d 5d 5d 5d 5d covalency example 1 ICl 3 ICl3 5 IF5 7 IF7

Why can fluorine atoms form only a single covalent bond whereas all the other halogens have maximum covalencies greater then one ? Fluorine cannot expand its octet as there are no low lying empty d-orbitals available and the energy required to promote electrons into the third quantum shell is prohibitively high. Since it is the most electronegative of all elements and only one of its unpaired p-orbital is available for completion of the octet, its only possible oxidation states are limited to -1 with other atom and 0 with itself in F2 .

Unit 9-2 （2）Variation in properties of the halogens and their compounds 1. Variations in melting and boiling points, electronegativities and electron affinities

Element atomic radius nm

melting point K

boiling point K

electronegativity electron affinity kJmol-1

Fluorine 0.072 53 85 4.0 -348 Chlorine 0.099 172 239 3.0 -364 Bromine 0.114 266 331 2.8 -342 Iodine 0.133 387 456 2.5 -314

(a) Describe and explain the trends in melting points and boiling points of the halogen group.

Melting points and boiling points increase down the group due to the increase in van der Waals forces. Van der Waals forces increase as the number of electrons of a molecule increases. The greater the number of electrons, the greater the possibility that temporary dipoles will be large, and the stronger the attractive forces. Moreover, as the molecular size increases so the possible area of contact between molecules increases, with a consequent increase in the attractive power of van der Waals forces.

(b) Describe and explain the trends in electronegativity of the halogen group.

Electronegativity decreases as the group is descended. This is because the outer electrons become progressively better shielded from the nucleus as the atomic size increases. Thus, electrons in a covalent bond are attracted less and less to the halogen as its atomic number increases.

(c) Describe and explain the trends in electron affinity from chlorine to iodine.

Electron affinity is defined as : X(g) + e- → X-(g) E.A. = ∆H0

Electron affinity decreases numerically with increasing atomic number. This is because the outer electrons become more shielded from the nucleus as the atomic size increases, so the tendency to attract another electron decreases as the group is descended.

(d) Suggest a reason why the electron affinity of fluorine is anomalous.

The low value of fluorine is probably due to the repulsion between the incoming electron and the electrons in the outer shell. This incoming electron must get closer to the shell electrons in the case of F because of its very small size, compared to the other halogens.

(e) Consider the bond dissociation enthalpies of the halogens shown in table :

Element F-F Cl-Cl Br-Br I-I Bond length nm 0.148 0.198 0.228 0.266 Bond dissociation enthalpy kJ mol-1 158 242 193 151

(1) Explain the trend from chlorine to iodine. Bond dissociation enthalpy decreases in the order Cl2, Br2, I2, owing to the increasing length of

the halogen-halogen bond. The increasing bond length is due to the increasing atomic size of the halogen with increasing atomic number.

(2) Suggest a reason for the low bond dissociation enthalpy of fluorine.

This is probably due to the repulsion of non-bonding pairs of electrons on each fluorine atom at each end of the F-F bond. The fluorine atoms are so small that the repulsion between these non-bonding electrons are much greater than in the case of other halogens, because of their closer proximity.

Unit 9-3 2. Relative oxidizing power of halogens When the halogens combine with metals or non-metals, they normally act as oxidizing agents :

1 X2(standard) + e → X-(aq)

The half-reaction can be separated into three stages as shown in figure : 1 X2(standard) + e X-

(aq) X(g) + e X-

(g)

F Cl Br I ∆H0

at / kJ mol-1 79 121 97 76 Electron affinity /kJ mol-1 -348 -364 -342 -314 ∆H0

hyd / kJ mol-1 -457 -381 -351 -307 ∆H0 / kJ mol-1 -726 -624 -596 -545 E0 / V +2.87 +1.36 +1.07 +0.54

Fluorine is the most reactive halogen and the most powerful oxidizing.

Order of oxidizing power : F2 > Cl2 > Br2 > I2

The high reactivity of fluorine is explained partly in terms of its low bond dissociation enthalpy, which means that little energy is required to break the F-F bond in the initial stages of a reaction. Another factor is the tendency to form strong bonds with other elements. Thus, fluorine tends to bring out the highest oxidation states of other elements. Reactions with iron (II) ion The reaction between fluorine and aqueous solution of Fe2+ is unsuitable to carry out because water is oxidized to oxygen and HF gas : 2 F2(g) + 2 H2O(l) → 4 HF(aq) + O2(g) (Traces of H2O2 and ozone are also formed.)

Chlorine and bromine will both oxidize Fe2+

(aq) to Fe3+(aq) :

Cl2(aq) + 2 Fe2+(aq) → 2 Cl-

(aq) + 2 Fe3+(aq)

Br2(aq) + 2 Fe2+(aq) → 2 Br-

(aq) + 2 Fe3+(aq)

Iodine, however, is such a weak oxidizing agent that it cannot remove electrons from iron (II) ions to

form iron (III) ions. These observations are exactly as we might have predicted since E0 for Fe3+

(aq), Fe2+(aq) / Pt is + 0.77 V.

Reaction with sodium and phosphorus Reactant Fluorine Chlorine Bromine Iodine Sodium Reacts explosively

to form NaF Continues to burn in chlorine to form NaCl

Continues to burn in vapour to form NaBr

Continues to burn in vapour to form NaI

Red phosphorus Ignites spontaneously to form PF5

On heating forms PCl3 and PCl5

Reacts spontaneously to form PBr3

Solid spontaneously forms PI3

Examples: 2 Na(s) + F2(g) → P4(s) + 6 Cl2(g) → P4(s) + 10 Cl2(g) →

Unit 9-4 3. Disproportionation of the halogens in alkalis

Disproportionation is a redox reaction in which one particular reactant is simultaneously oxidized and reduced.

Fluorine

Fluorine , being the strongest oxidizing agent, will oxidize the hydroxide ions : OF2 is unstable and at 70℃, it decomposes to liberate oxygen gas : 2 F2(g) + 4 OH-

(aq) → 4 F-(aq) + O2(g) + 2 H2O(l)

70℃

Chlorine, Bromine and iodine undergo very similar reactions with alkalis. Chlorine

With cold, dilute alkali at 15℃, chlorine disproportionates to form a mixture of chloride and chlorate(I) ions :

At 70℃, chlorate (I) further disproportionates to form chloride and chlorate (V) ions : Therefore, sodium chlorate (V) is obtained by passing chlorine gas into hot concentrated solution of

sodium hydroxide : Bromine

As the decomposition of bromate (I) is rapid at 15℃, it may be made only at temperature about 0℃ :

Br2(l) + 2 OH-(aq) → Br-

(aq) + BrO-(aq) + H2O(l)

0℃ At 15℃, bromate (V) and bromide is formed by the disproportionation of bromine with alkalis :

Iodine With iodine, decomposition of Iodate (I) occurs rapidly even at 0℃, so iodate (V) and iodide being

formed immediately upon dissolving iodine in alkali :

Stability of halate(I) ion : ClO- > BrO- > IO-

(a) The disproportionation reactions of bromine and iodine with alkalis are reversible. Bromine or

iodine can be regenerated by the reaction of the corresponding halate (V) and halide ions in acidic medium. Write the ionic equations involved.

(b) Sodium chlorate (III) and sodium chlorate (V) are formed when chlorine dioxide is dissolved in

sodium hydroxide solution. Deduce the ionic equation for the disproportionation reaction.

(c) Potassium iodate (VII) is sufficiently powerful to oxidize Mn2+(aq) to MnO4

-(aq) in acidic medium.

The resulting solution turns dark blue with the addition of starch solution. Deduce the ionic equation.

4. Comparative study of the reactions of halide ions

Unit 9-5 Reactions with halogens Chlorine water Bromine water Iodine solution Initial colour almost none orange brown

Colour after shaking with KI solution brown brown Colour after shaking Upper layer yellow yellow with CCl3CH3(l) lower layer violet violet

1. I-

(aq)

Conclusion I- oxidized I- oxidized Colour after shaking with KBr solution orange unchanged Colour after shaking Upper layer almost none yellow with CCl3CH3(l) lower layer orange brown violet

2. Br-

(aq)

Conclusion Br- oxidized no reaction Colour after shaking with KCl solution unchanged unchanged Colour after shaking Upper layer almost none yellow with CCl3CH3(l) lower layer orange brown violet

3. Cl-

(aq)

Conclusion no reaction no reaction

Write ionic equations for the reactions taking place. Cl2(aq) + 2 Br-

(aq) → 2 Cl-(aq) + Br2(aq) EO

cell = Cl2(aq) + 2 I-

(aq) → 2 Cl-(aq) + I2(aq) EO

cell = Br2(aq) + 2 I-

(aq) → 2 Br-(aq) + I2(aq) EO

cell = Reactions with silver ions Reagent F-

(aq) Cl-(aq) Br-

(aq) I-(aq)

AgNO3(aq) acidified with HNO3(aq)

no reaction white precipitate of AgCl

pale yellow precipitate of AgBr

yellow precipitate of AgI

Solubility of AgX(s) in excess NH3(aq)

soluble soluble slightly soluble insoluble

Effect of sunlight on AgX(s)

no effect white AgCl turns purple-grey

pale yellow AgBr turns green-yellow

no effect

Except fluoride, an aqueous solution of halide precipitates the corresponding silver ion from silver nitrate acidified with dilute nitric acid. Excess nitric acid must be present to prevent the precipitation of other insoluble silver compounds such as sulphite and carbonate. Ag+

(aq) + X-(aq) → AgX(s) X = Cl, Br and I

If excess ammonia solution is added, silver chloride dissolves readily due to the formation of a soluble and colourless complex, diamminesilver (I) ion : Silver bromide is slightly soluble in excess ammonia solution :

The darkening effect of light on the silver chloride and silver bromide precipitates is because the light converts some of the silver halide to small particles of metallic silver, which darkens the precipitates. For example, eye-glasses that darken in the sunlight contain small amounts of white, finely dispersed silver chloride, AgCl. Exposure to sunlight converts some of the Ag+ ions to metallic silver, which is dark. The reaction is reversible: in a dark room, AgCl is reformed and the glass becomes clear again. Reactions with phosphoric (V) acid and sulphuric (VI) acid

Unit 9-6 Reagent added Fluoride Chloride Bromide Iodide 1. Action of conc. H3PO4(aq)

Steamy fumes on warming.

Steamy fumes on warming.

Steamy fumes on warming.

Steamy fumes on warming.

Suspected product HF(g) HCl(g) HBr(g) HI(g) Confirmatory test

White smoke with ammonia.

White smoke with ammonia.

White smoke with ammonia.

White smoke with ammonia.

2. Action of conc. H2SO4(l)

Steamy fumes on heating.

Steamy fumes in cold. No green gas even on heating.

Steamy fumes in cold. Brown gas on warming.

Steamy violet fumes in cold. More on warming. Bad egg smell.

Suspected product HF(g) HCl(g) HBr(g) and Br2(g) HI(g) , I2(g) and H2S(g)

Confirmatory test

White smoke with ammonia.

White smoke with ammonia. Blue litmus turns red but not bleached.

White smoke with ammonia. Orange-brown colour in CCl3CH3(l)

Violet colour in CCl3CH3(l)

(a) List the hydrogen halides in order of decreasing ease of oxidation by concentrated sulphuric acid.

Explain the sequence using bond energy terms. HI, HBr, HCl, HF (HI most readily oxidized). The oxidation of a hydrogen halide by sulphuric acid can be represented by the equation : 2 H-X(g) + H2SO4(l) → X-X + SO2(g) + 2H2O(l) (except HF and HCl) In energy terms, the difference between one halide and another lies in the differences in E(H-X)

and E(X-X). Since two H-X bonds have to be broken for each X-X bond formed, differences in E(H-X) are more important. The group trend in E(H-X) is very pronounced; the H-I bond is the weakest and HI is therefore most readily oxidized.

(b) Outline the laboratory preparation of each of the four hydrogen halides, HF, HCl, HBr and HI. Preparation of HF and HCl : Both of these gases may be prepared by heating a halide salt with

concentrated sulphuric (VI) acid. CaF2(s) + H2SO4(l) → CaSO4(s) + 2 HF(g) NaCl(s) + H2SO4(l) → NaHSO4(s) + HCl(g) HCl may be collected by downward delivery. HF is usually made in small quantities only and not

collected since it attacks glassware. Preparation of HBr and HI : Both of these gases are prepared by heating the appropriate halide salt

with a non-oxidizing, non-volatile acid such as phosphoric (V) acid. NaBr(s) + H3PO4(l) → NaI(s) + H3PO4(l) → Both gases may be collected by downward delivery. (c) Explain the observations and write the equations for the reaction between conc.H2SO4(l). and (i) bromide Brown fumes are formed. (ii) iodide Violet fumes are seen and smell of bad eggs is noticed. 5. Acidic properties of hydrogen halide and the anomalous behaviour of hydrogen fluoride

Unit 9-7 Hydrogen fluoride Hydrogen chloride Hydrogen

bromide Hydrogen iodide

Boiling point / ℃ 20 -85 -69 -35 Bond dissociation enthalpy / kJ mol-1

+567 +431 +366 +298

Acid dissociation constant /mol dm-3

7 x 10-4 107 >107 >107

All the hydrogen halides are extremely soluble in water. The resulting solutions are strongly acidic

(except HF(aq) ) because of the reaction with water to form hydroxonium ion, HX(g) + H2O(l) → X-

(aq) + H3O+(aq)

(a) With the aid of a labelled diagram, explain why HF is a liquid at s.t.p. whereas all the other

hydrogen halides are gases. Measurement of molar mass show that HF is associated in the liquid state, i.e. it exists as (HF)n,

where n is at least 2. This association is due to the presence of hydrogen bonds which operate between the HF molecules.

Therefore, more energy is required to separate individual HF molecules than other hydrogen halides

molecules, and hence boiling point of HF is the greatest amongst hydrogen halides. (b) Why is hydrofluoric acid ,HF(aq) ,a weak acid in dilute solution whereas the others are strong ? HF(aq) + H2O(l) = F-

(aq) + H3O+(aq) Ka = 7 x 10-4 mol dm-3

Hydrofluoric acid is a weak acid (only about 10% of the HF molecules ionize in 0.1 M solution) for

the following reasons : (i) The H-F bond is very strong. (ii) Hydrogen bonds between HF molecules make ionization more difficult. The other hydrogen halides have weaker hydrogen-halide bonds and do not form hydrogen bonds. (c) Amongst the three strong acids, hydrochloric acid, hydrobromic acid and hydroiodic acid, there are

small differences in acid strength. State and explain the trend which exist. Acid strength decreases slightly in the order HI(aq) > HBr(aq) > HCl(aq) . This is mainly due to the

increase in bond enthalpy in passing from HI to HCl. (d) Comment on the following statement. Hydrofluoric acid forms two salts with potassium hydroxide : potassium fluoride (KF) and

potassium hydrogen difluoride (KHF2). HF is a weak acid, i.e. it is only partially ionize into F- and H3O+. Some of the F- ions react with

HF molecules to form HF2- ions :

The HF2

- ion is hydrogen bonded : Thus, hydrofluoric acid contains two anions, F- and HF2

-, and when it reacts with KOH, two salts are formed, i.e. KF and KHF2 .

Unit 9-8 6. Laboratory preparation of chlorine and hydrogen chloride gases Laboratory preparation of chlorine gas :

Laboratory preparation of hydrogen chloride gas :

Section 9.2 Group IV Elements

Unit 9-9 （1）Characteristic properties of the Group IV elements The similarity between elements in the same family which was so obvious in Group I, II and VII is much less apparent in Group IV. Here there is a considerable change in the character of the elements as atomic number rises. 1. Variation in melting point and boiling point The following table shows a list of physical properties for the elements in Group IV :

Element atomic number

Electronic configuration

Atomic radius / nm

Melting point / oC

Boiling point / oC

Conductivity

Carbon

6 [He]2s22p2 0.077 3730(d.) 4830(d.) fairly good(g.)non-cond.(d.)

Silicon

14 [Ne]3s23p2 0.117 1410 2680 semi-cond.

Germanium

32 [Ar]3d104s24p2 0.122 937 2830 semi-cond.

Tin

50 [Kr] 0.141 232 2270 good

Lead

82 [Xe] 0.154 327 1730 good

d. = diamond g. = graphite These changes in property are related to the increasing metallic character and the decreasing non-metallic character as atomic number rises.

Element Electronegativity First ionization enthalpy＊ / kJ mol-1

Type of structure

C 2.5 1086 Giant covalent Si 1.8 787 Giant covalent (similar to diamond) Ge 1.8 760 Giant covalent (similar to diamond) Sn 1.8 707 Giant metallic Pb 1.8 715 Giant metallic

The structural changes from giant covalent structures in carbon and silicon to giant metallic

structures in tin and lead help to explain the changes in physical properties. Giant covalent structures :

Unit 9-10 In diamond, every carbon atom can be imagined to be at the center of a regular tetrahedron surrounded by four carbon atoms whose centers are at the corners of the tetrahedron. Within the structure, every carbon atom forms four covalent bonds by sharing electrons with each of its four nearest neighbours. Silicon and germanium crystallize in the same giant covalent structure as diamond. In graphite, the carbon atoms are arranged in flat, parallel layers. Each layer contains millions of hexagonally arranged carbon atoms. Each carbon atom is covalently bonded to three other atoms in its layer. These are formed by three of the four valence electrons of carbon, whilst the fourth electron is delocalised over the whole layer. This delocalisation of electrons similar to that in metals results in graphite conducting electricity and in its shiny appearance. In the giant covalent structures of diamond, graphite, silicon and germanium, atoms are linked by localized electrons in strong covalent bonds throughout the whole three-dimensional lattice. It is therefore very difficult to distort a covalently bonded crystal, because this would involve breaking many covalent bonds. Consequently, diamond, graphite, silicon and germanium have very high melting points and very high boiling points. Giant metallic structures : Tin and lead have distorted close-packed metal structures. As the atoms get larger and the atomic radius increases, the metallic bonding becomes weaker. As a result the attraction of mobile valence electrons for the cationic lattice gets less. Hence there is a decrease in melting point and boiling point from tin to lead. ＊ The first ionization enthalpy decreases considerably from carbon to silicon. After that, it falls relatively little. The reason for this is that, after silicon, there is a large increase in nuclear charge. This is associated with the filling of 'd' and 'f' sub-shells and it counterbalances the increase in atomic radius. 2. Dissimilarity in properties of elements as illustrated by the composition of oxides and chlorides

Unit 9-11 The most striking feature of the compounds of Group IV elements is the existence of two oxidation states, +2 and +4. All the Group IV elements have four electrons in their outermost shell (ns2 np2), so it is not surprising that they show a well-defined oxidation state of +4. However, none of the elements forms an M4+ cation in its solid compounds. This is due to the high ionization enthalpies involved in removing four successive electrons from an atom. Consequently, the bonding in the tetravalent compounds is predominantly covalent. Compounds of tin and lead in which the Group IV element has an oxidation state +2 are normally regarded as ionic. In these compounds, the Sn2+ and Pb2+ ions are believed to form by loss of the two 'p' electrons in the outer shell. The two 's' electrons remain relatively stable and unreactive in the filled 's' sub-shell. This is sometimes referred to as the inert pair effect. Composition of oxides of Group IV elements :

Oxides in the +4 oxidation state :

Oxide

CO2 SiO2 GeO2 SnO2 PbO2

Structure

simple molecular

giant covalent intermediate between giant covalent and ionic

Boiling point / oC

- 78 2590 1200 1900 decomposes on heating

Thermal stability

stable even high temperatures decomposes to PbO

Oxides in the +2 oxidation state :

Oxide

CO SiO GeO SnO PbO

Structure

simple molecular

predominantly ionic

Boiling point / oC

- 191 / / / 1470

Thermal stability

readily oxidized to dioxide stable

It is interesting to consider the relative stabilities of the +2 and +4 oxidation states for the oxides of

different elements. In carbon and silicon oxides, the +4 state is very stable relative to +2. The +2 state is rare and easily oxidized to +4. Thus, CO reacts very exothermically to form CO2. SiO is too unstable to exsist under normal conditions, although it has been obtained at 2000 oC. Germanium forms oxides in both +4 and +2 states. However, GeO2 is rather stable than GeO. GeO is readily converted to GeO2. In tin compounds, the +4 state is only slightly more stable than the +2 state. In lead compounds, the +2 state is more stable. PbO2 is a strong oxidizing agent, whilst PbO is relatively stable. Thus, PbO2 can oxidize hydrochloric acid to chlorine : Composition of chlorides of Group IV elements :

Unit 9-12 All the elements of Group IV form tetrachlorides of formula ECl4. All tetrachlorides are volatile covalent compounds. Tetrachloride

CCl4 SiCl4 GeCl4 SnCl4 PbCl4

Structure

Simple molecular Molecular shape : tetrahedral

Volatility Low melting point and boiling point.

All are volatile liquids at room temperature.

Thermal stability

As the E-Cl bonds become longer and weaker down the group, the tetrachlorides become less stable. Thus, CCl4, SiCl4 and GeCl4 are stable even at high temperatures. SnCl4 decomposes on heating to form SnCl2 and Cl2. PbCl4 decomposes readily to form PbCl2 and Cl2.

Dichloride

CCl2 SiCl2 GeCl2 SnCl2 PbCl2

Structure

Too unstable to exist Unstable Predominantly ionic;

in solution, Sn2+ is formed.

Predominantly ionic;

in solution, Pb2+ is formed.

（2）Silicon and silicates

Unit 9-13 1. Uses of silicon Occurrence :

As a tetravalent metalloid, silicon is less reactive than its chemical analog carbon. It is the second most abundant element in the Earth's crust, making up 25.7% of it by weight. It occurs in feldspar, mica, quartz and sand, mainly in the form of silicon dioxide (also known as silica) and silicates (compounds containing silicon, oxygen and metals). Major uses of silicon :

Semiconductor - Semiconductor is a special class of materials which have an electrical resistivity between those of electrical conductors and electrical insulators. The semiconductors which are pure elements and compounds are described as intrinsic semiconductors. Silicon is the intrinsic semiconductor which is used for the production of the integrated circuits known as silicon chips. Ultrapure silicon can be doped with other elements to adjust its electrical response by controlling the number and charge (positive or negative) of current carriers. Such semiconductor is called extrinsic semiconductor. There are two types of extrinsic semiconductors : n-type and p-type. In n-type semiconductors, the dopants are Group V elements, such as phosphorus and antimony, which have five electrons in the outer shell. When a dopant atom replaces an atom of silicon in the structure, the fifth electron is supplied to the material, creating a negative charge. In p-type semiconductors, the dopants are Group III elements, such as boron and alumjinium, which have three electrons in the outer shell. When a dopant atom replaces a silicon atom, it forms three electron-pair bonds with silicon atoms, but the fourth bond is incomplete : it has only one electron. The vacancy is called an electron hole and is positively charge. Transistors, solar cells and other semiconductor devices which are used in electronics and other high-tech applications are often composed of the semiconductors of silicon.

Steel - Silicon is an important constituent of some steels.

Bronze - Most bronze produced is an alloy of copper and silicon.

Photonics - Silicon can be used in lasers to produce coherent light with a wavelength of 456 nm.

LCDs and solar cells - Hydrogenated amorphous silicon has shown promise in the production of low-cost, large-area electronics in applications such as LCDs. It has also shown promise for large-area, low-cost solar cells.

Glass - Silica from sand is a principal component of glass. Glass can be made into a great variety of shapes and with a many different physical properties. Silica is used as a base material to make window glass, containers, and insulators, and many other useful objects.

Construction - Silica is a major ingredient in bricks because of its low chemical activity.

Abrasives - Silicon carbide is one of the most important abrasives. 2. Bonding and structures of silicates

Unit 9-14 Silicon and oxygen make up most of the Earth’s crust. They form the basis of a class of minerals called silicates. All silicates and analogues are derived from the silicate ions, SiO4

2-. The silicon atoms can be replaced by other metals to form analogous compounds, notably the aluminosilicates in which aluminium atoms partially replace the silicon atoms. (a) The builiding block of the silicate minerals All silicate minerals are buil up from the basic unit of silicate(IV) ion, SiO4

2-, which hs the following structural representations :

The Si atoms is covalently bonded to 4 oxygen atoms. Each oxygen atom possesses a formal negative charge. Hence each tetrahedral unit has a formal charge of –4. when linked together, the extended units are also negative charged. Presence of other metallic ions such ad Ca2+ or Mg2+ are necessary for electrical neutrality. The covalent Si-O bond, having a bond enthapy of 466 kJ mol-1, is particularly strong compared with the C-C bond which has a bond enthalpy of 347 kJ mol-1. The linkage –Si-O-Si-O- is very stable and instead of existing as discrete units of SiO4

2- ions, the silicates tend to form infinite chains, sheets or networks. (b) The silicate structures

Structure

Figure Corners shared at each tetrahedron

Repeat unit Example

Tetrahedra

A 0 SiO44- Olivines,

(Mg/Fe)2SiO4

Pairs of tetrahedra

B 1 Si2O76- Thortveitite,

Sc2(Si2O7) Closed rings

C 2 SiO32- Beryl,

Be3Al2(Si6O18) Infinite single chains

D 2 SiO32- / Si2O6

4- Pyroxenes, CaMg(Si2O6)

Infinite double Chains

E 2.5 Si4O116- Amphibole (asbestos),

Ca2Mg5(Si4O11)2(OH)2

Infinite sheets

F 3 Si2O52- Talc,

Mg3(Si4O10)(OH)2

Infinite three-dimensional network

G 4 SiO2 Quartz,

(SiO2)n

(i) Tetradedra / pairs of tetrahedra SiO4

4- Si2O76-

Unit 9-15

The cations are arranged around the tetrahedral units on a regular crystalline lattice. (ii) Closed rings

(iii) Infinite single chains

(iv) Infinite double chains

Asbestos is a generic term for a group of naturally occurring hydrated silicates that can be processed

mechanically into long fibres. Asbestos refers to two kinds of minerals : amphiboles and chrysotile. Amphibole, Ca2Mg5(Si4O11)2(OH)2, possesses the infinite double chain structure of Fig. E. Chrysotile, Mg3Si2O5(OH)4, accounts for almost 90% of worldwide asbestos production. It has a

sheet structure of Fig. F. Both chain and sheet structures are held together by the electrostatic attraction between the cations

and the negative silicates structures. Crystal cleavage is expected to occur along the chain direction because the non-directional ionic links are comparatively weaker than the strong Si-O covalent bonds. Hence asbestos minerals are fibrous. (v) Infinite sheets

Unit 9-16

Talc, Mg3(Si4O10)(OH)2 has structure consisting of layers and parallel infinite sheets. All of the

strong bonding interactions among the atoms occur within the layers. Two parallel infinite sheets have the unshared oxygen atoms of the tetrahedral pointing towards each other. In the middle of this sandwich are the magnesium and hydroxide ions, which serve to bind the two silicate sheets together. Only weak van der Waals forces hold the sandwiches (layers) together. Thus layers can slip easily across one another and accounts for the ease for it to be pulverized to make talcum powder, a soft and fine powder t make one's skin feel smooth and dry. (vi) Infinite three-dimensional network

The mineral Quartz is one form of silica. In quartz, all the four vertices of each tetrahedron are linked to other tetrahedral. The quartz network carries no charge and there are no cations in its structure. The three dimensional network silicates such as quartz are much harder than their linear or layer counterparts. As the silica structure consists of a giant network of strong covalent bonds. Melting points of quartz and silica (sand) are very high. Example Predict the structure of each of the following silicate minerals : (a) Andradite, Ca3Fe2(SiO4)3 (b) Vlasovite, Na2ZnSi4O10

(c) Hardystonite, Ca2ZnSi2O7 (d) Chrysotile, Mg3Si2O5(OH)4 (c) The aluminosilicates

Unit 9-17 Very often the silicon atoms are replaced by aluminium atoms to form the silicate analogue, the aluminosilicates. When an aluminium atom replaces a silicon atom, it contributes only three electrons to the bonding framework in place of the four electrons of silicon atoms. The additional required electron is supplied by the ionization of a metal atom such as sodium or potassium. (i) Infinite sheets Mica belongs to a family in which one of the four silicon atoms in the structural unit of talc is replaced by an aluminium atom and inserting a potassium atom to supply the fourth electron needed for electrical neutrality. Mica has a composition of KMg3(AlSi3O10)(OH)2. Micas are harder than talc and their layers slide less readily over one another. Like talc, crystals of mica cleave easily into sheets. The cations occupy sites between the infinite sheets. The van der Waals attraction between sheets is increased by the presence of extra ionic charge and accounts for the overall hardness of mica over talc. (ii) Infinite three dimensional network The feldspars, in which albite NaAlSi3O8 is an example, are the most abundant aluminosilicate minerals in the Earth's surface. The silicon atoms and aluminium atoms occupy the centers of interlinked tetrahedral of SiO4

4- and AlO45-. These tetrahedral connect at each corner to other tetrahedral

forming an intricate, three dimensional, negatively charged framework. The sodium cations sit within the voids in this structure. Example Predict the structure of each of the following aluminosilicate minerals. In each case, the aluminium atoms grouped with the silicon and oxygen in the formula substitute for Si in the tetrahedral sites. (a) Biktatite, Li(AlSi2O6) (b) Muscovite, KAl2(AlSi3O10)(OH)2

(c) Cordierite, Al3Mg2(AlSi5O18) 3. Daily life importance of the silicate minerals (a) Silicates

Unit 9-18 (i) Asbestos (Infinite double chains or infinite sheets)

Asbestos is an excellent thermal insulator that is non-combustible, acid-resistant, and strong. In the past, it was used extensively in construction work to make cement floor tiles, roof covers and ducts.

It can also be woven into fabric to make fire-resistant blankets. Its use has been decreased greatly in recent years because inhalation of small asbestos

fibres during mining and manufacturing or during the removal of grayed and crumbled building materials can cause the lung disease called asbestosis.

(ii) Talc (Infinite sheets)

The common use of talc crystals is to make talcum powder which is a soft and fine powder to make one's skin feel smooth and dry.

Its resistance to heat and electricity makes it ideal for surfacing benches and switchboards. It is also an important filler for paint and rubber.

(iii) Quartz (Infinite three dimensional network)

The hardness of quartz is widely made use of in construction work as sandstones. Quartz / silica tubing are used for high temperature heating. Quartz possesses piezoelectric property which enables it to make crystal oscillators used in

watches and electronic circuits and also as pressure sensor in electronic balances. Quartz is also widely used in jewelry and ornamental decorations.

(b) Aluminosilicates (i) Feldspar (Infinite three dimensional network)

In glassmaking, feldspar provides alumina for improving hardness, durability, and resistance to chemical corrosion.

In ceramic, feldspar is used as flux, lowering the vitrifying temperature of a ceramic body during firing and forming a glassy phase.

(ii) Mica (Infinite sheets)

Mica possesses excellent electrical insulation and is widely used in electronic products such as capacitors, washers for transistors and radar high tension coils.

It also has excellent heat insulation and is used in soldering irons and jet engines.