Chapter 1: Solid State

In 1912, a German physicist named Max von Laue asked a question that sounded almost reckless: what would happen if X-rays were passed through a crystal? Nobody could see inside a solid, and nobody knew for certain how its particles were arranged. When the experiment was performed, the photographic plate behind the crystal did not show a single blurred patch. It showed a neat pattern of sharp spots, arranged with striking regularity.

That pattern was a message from inside the solid. X-rays had been scattered by particles sitting in an orderly, repeating three-dimensional arrangement — the first direct evidence that a crystal is built from a pattern that repeats itself, over and over, in every direction.

Within a couple of years, William Henry Bragg and his son William Lawrence Bragg in England turned this discovery into a tool. Working with common salt, they deduced how its particles are arranged, and reached a surprising conclusion: a crystal of sodium chloride contains no separate NaCl molecules at all. It is an endless alternating array of sodium ions and chloride ions, each surrounded by ions of the opposite charge. For this work the Braggs received the Nobel Prize in 1915.

A grain of salt on your fingertip, then, is not a lump of matter. It is architecture.

Diagram

Fig. 1.0 – A beam of X-rays passing through a small crystal and producing a regular pattern of sharp spots on a photographic plate; beside it, a magnified cut-away of the crystal showing alternating Na⁺ and Cl⁻ ions in an ordered three-dimensional array.

NoteCuriosity Corner

Q1. Why does a piece of glass break with curved, uneven edges, while a crystal of common salt splits along clean, flat faces? Q2. Diamond and graphite are both made of nothing but carbon atoms. Why is one the hardest material known, while the other is soft enough to leave a mark on paper and can even conduct electricity? Q3. If the particles of a metal are packed as tightly as spheres possibly can be, why is roughly a quarter of the crystal still empty space? Q4. Pure sodium chloride is a white solid. Yet if its crystals are heated in sodium vapour, they turn yellow. How can a colourless crystal gain colour without any coloured substance being added? Q5. Impurities usually spoil a material. Why, then, does adding a trace of phosphorus to extremely pure silicon make it far more useful than the pure silicon itself?

As you work through this chapter, you will build up exactly the ideas needed to reason out each of these questions for yourself.

1.1 Introduction

Of the three common states of matter — solid, liquid and gas — the solid state is the one in which the constituent particles are held most firmly. The solid state is characterised by strong interparticle forces of attraction, that is, the forces pulling neighbouring particles towards one another are strong enough to keep each particle anchored near a fixed position.

Because these forces are strong, most solids have a definite shape and a definite volume. Raising the temperature or applying pressure changes that shape and volume only slightly. A solid does not flow to take the shape of its container the way a liquid does, and it cannot be squeezed into a much smaller volume the way a gas can.

This does not mean the particles are motionless. They vibrate continuously about fixed mean positions, but the strong attractions do not let them wander away from those positions. The difference between the three states, shown in Fig. 1.1a, lies exactly here: in a solid the particles vibrate in place, in a liquid they are still close together but can slide past one another, and in a gas they are far apart and move almost independently.

Diagram

Fig. 1.1a – Particle picture of the three states side by side: solid (particles closely packed, small vibration arrows about fixed positions), liquid (particles close but sliding past one another), gas (particles far apart with long motion arrows); label interparticle distance in each.

NoteDefinition

A solid is a state of matter in which the constituent particles are held by strong interparticle forces of attraction at fixed mean positions, so that the substance has a definite shape and a definite volume that change only slightly with temperature and pressure.

Solid vs Liquid vs Gas

Feature Solid Liquid Gas
Shape Definite Takes the shape of the container Fills the whole container
Volume Definite Definite Not definite
Effect of temperature and pressure on volume Very small change Small change Large change
Motion of particles Vibration about fixed mean positions Particles slide past one another Free, rapid, almost independent motion
Interparticle forces Strong Moderate Very weak

Q. A solid and a liquid both have a definite volume. Why, then, can a liquid be poured while a solid cannot?

A. Definite volume only tells us that the particles are already close together, so there is little empty space to remove. Pouring needs something more — the particles must be able to change neighbours. In a liquid the attractions are strong enough to hold the particles together but weak enough to let them slip past one another, so the liquid flows. In a solid the attractions hold each particle to its own position, so the whole body keeps its shape.

What ‘Particles’ Means in This Chapter

The smallest constituent particles of solids are not all of the same kind. In some solids they are atoms, as in a piece of copper. In others they are ions, that is, atoms or groups of atoms carrying a positive or negative charge, as in common salt. In still others they are molecules, as in solid carbon dioxide.

Naming all three every time would make the discussion clumsy. Therefore, all such smallest constituent particles of solids will be referred to simply as ‘particles’ in this chapter.

The word solid comes from the Latin solidus, meaning firm, dense or whole — the same root that gives us the everyday words consolidate and solidarity.

A gas can be compressed into a cylinder until it occupies a small fraction of its original volume, but a steel rod cannot be compressed noticeably even under a heavy load. Explain this difference in terms of the interparticle distances and the interparticle forces in the two states.

1.2 Types of Solids

Two solids may look equally hard and equally rigid, and yet be built quite differently on the inside. What decides the difference is whether the particles are arranged in an orderly pattern or not. On this basis there are two types of solids, namely crystalline solids and amorphous solids.

1.2.1 Crystalline Solids

Study of many crystalline solids indicates that they possess the following characteristic properties.

  1. There is a regularity and periodicity in the arrangement of constituent particles in crystalline solids. Regularity means the particles follow a definite pattern; periodicity means that the same pattern repeats itself again and again at fixed intervals. The ordered arrangement of particles extends over a long range — that is, it continues over enormous numbers of particles in every direction, and not merely among a few neighbours.

  2. Crystalline solids have sharp melting points, that is, they melt at a definite temperature. The reason lies in the regularity itself. Every particle sits in an identical environment and is held by forces of the same strength, so when the temperature rises high enough, the whole arrangement gives way together rather than in stages.

  3. All crystalline substances except those having cubic structure are anisotropic. In other words their properties like refractive index, thermal and electrical conductivity, etc, are different in different directions. This follows directly from the ordered arrangement: as you move through the crystal along different directions, you meet different numbers of particles at different spacings, so light or heat or current does not travel equally easily along all directions.

Ice, salts such as \(\mathrm{NaCl}\), metals such as sodium, gold, copper and materials such as diamond, graphite, ceramics are examples of crystalline solids.

Fig. 1.2a shows a section through such a solid, built from two types of constituent particles.

Diagram

Fig. 1.2a – A planar array of two kinds of particles (solid circles = Particle 1, open circles = Particle 2) arranged in a regular repeating pattern; three straight arrowed lines drawn across the array in different directions, labelled AB, CD and EF.

the two-particle array of Fig. 1.2a (i) Would you call the arrangement of particles in this solid regular or irregular? (ii) Moving along the directions AB, CD and EF, is the arrangement of constituent particles the same or different? (iii) What does your answer to (ii) predict about the properties of this solid measured along these three directions?

Single Crystals and Polycrystalline Solids

A single crystal has an ordered — that is, regular and periodic — arrangement of constituent particles throughout its bulk, with no break anywhere. Such crystals are difficult to obtain; diamond is an example of a naturally formed single crystal.

The majority of crystalline solids, including metals, are polycrystalline in nature. A single granule of a polycrystalline solid is made of many single crystals, or crystallites, packed together with different orientations, as drawn in Fig. 1.2b. Each crystallite is internally ordered, but its pattern is tilted with respect to its neighbour.

Diagram

Fig. 1.2b – Left: a single crystal, one unbroken ordered array of particles; right: a polycrystalline granule showing several ordered crystallites with boundaries between them, each crystallite’s rows tilted at a different angle.

The regularity of crystals can be seen without any instrument. Collect granular sugar, common salt and blue vitriol, and observe a few granules of each under a magnifying lens or microscope. Look carefully at (i) the shape of the granules, (ii) the smoothness of the faces of the granules, and (iii) the angles between the various edges of a granule. All three of these substances are crystalline solids. Which property of crystals were you actually observing when you noticed flat faces meeting at fixed angles?

In the late eighteenth century the French mineralogist René Just Haüy examined broken fragments of calcite and noticed something remarkable: however often the mineral was cleaved, the fragments kept reappearing with the same characteristic shapes and the same angles between their faces. He argued that a crystal must be built by stacking identical tiny units in an orderly way — an idea proposed more than a century before X-rays could confirm it, and the beginning of the science of crystallography.

1.2.2 Amorphous Solids

The particles of a liquid are in constant motion. The stop action photograph of a liquid describes the amorphous state — picture that jumbled arrangement being frozen exactly where it stands. In fact, amorphous solids are supercooled liquids: liquids cooled below their freezing temperature so quickly that the particles never got the chance to settle into an ordered pattern. Amorphous solids have the following characteristics.

  1. The constituent particles in amorphous solids are randomly arranged. The particles do not have long range ordered structure, but they do have a short range order. In other words, a particle and its immediate neighbours may sit at sensible distances and angles, but that pattern is lost after a few particle diameters.

  2. Amorphous solids do not have sharp melting points. They melt gradually over a temperature interval. On heating, amorphous solids gradually and continuously soften and start to flow. Because the particles occupy a whole range of different environments, the weakest arrangements break loose first and the strongest last, so there is no single melting temperature.

  3. These solids are isotropic. In other words, their properties such as refractive index, conductivity are all independent of direction of measurement. They exhibit the same magnitude for any property in every direction. Since the disorder is the same in every direction, no direction through the solid is special.

Glass, plastic, rubber, tar, and metallic glass (metal-metalloid alloy) are a few examples of amorphous solids.

Fig. 1.2c contrasts the two kinds of arrangement directly.

Diagram

Fig. 1.2c – Two panels of the same network solid: left, the crystalline form with atoms in a regular repeating network (long range order); right, the amorphous form with the same linkages but irregular ring sizes and angles (short range order only).

Crystalline solids vs Amorphous solids

Feature Crystalline solids Amorphous solids
Arrangement of particles Regular and periodic Random
Range of order Long range order Short range order only
Melting behaviour Sharp melting point, melt at a definite temperature No sharp melting point, soften gradually over a temperature interval
Directional dependence of properties Anisotropic (except those having cubic structure) Isotropic
Examples Ice, \(\mathrm{NaCl}\), sodium, gold, copper, diamond, graphite, ceramics Glass, plastic, rubber, tar, metallic glass

The word crystal comes from the Greek krystallos, meaning ice — the ancient Greeks believed that clear quartz was water frozen so hard it could never melt again. Amorphous is built from the Greek a- (without) and morphē (form), a substance “without form” in the sense of having no ordered internal shape.

Amorphous solids are not second-rate materials; disorder is often exactly what is wanted. Ordinary glass owes its transparency and its ability to be softened and shaped to the fact that it has no sharp melting point — a glassblower works the material precisely in the interval where it flows but has not become a thin liquid. Metallic glasses, made by cooling a molten alloy so rapidly that its atoms cannot line up, have no crystal boundaries for magnetic effects to catch on, and are used in transformer cores where energy loss must be kept low.

Q1. Why does a piece of glass break with curved, uneven edges, while a crystal of common salt splits along clean, flat faces?

A. A salt crystal is built from particles arranged in a regular, periodic pattern, so it contains definite planes along which particles are neatly lined up. These planes are natural planes of weakness, and a blow makes the crystal part along them, giving flat faces meeting at fixed angles. Glass is amorphous. Its particles have only short range order, so no plane through it is any weaker than another. Being isotropic, it has no preferred direction to split along, and a crack simply wanders where the stress takes it, leaving the familiar curved, shell-like edge.

Two solids are examined in the laboratory. Solid X melts sharply at one definite temperature, and its refractive index is found to be the same when measured along three different directions. Solid Y has no definite melting point; it softens gradually over an interval of about 20 °C and then begins to flow. Classify X and Y, giving reasons.

Solid X. A sharp melting point means all the particles are held in identical environments by forces of equal strength, which happens only when the arrangement is regular and periodic. Hence X is a crystalline solid.

Its refractive index is the same in all directions, so X is isotropic. Among crystalline solids, only those having cubic structure are isotropic. Therefore X is a crystalline solid belonging to the cubic system.

Solid Y. Gradual softening over a temperature interval, with no definite melting point, is the behaviour of a solid whose particles occupy a wide range of different environments. Hence Y is an amorphous solid.

Note the trap: isotropy alone does not prove that a solid is amorphous. The melting behaviour must be checked as well, because cubic crystals are isotropic too.

Fig. 1.2d shows two arrangements, A and B, of the constituent particles of two solids. Identify each arrangement as crystalline or amorphous, and state one property of each solid that your identification allows you to predict.

Diagram

Fig. 1.2d – Two labelled panels of a two-component solid: panel A with particles in a regular repeating network; panel B with the same particles connected irregularly, with varying spacings and ring sizes.

1.2.3 Isomorphism and Polymorphism

Similarity or dissimilarity in crystal structure of different solids is described as isomorphism and polymorphism.

i. Isomorphism : Two or more substances having the same crystal structure are said to be isomorphous. In these substances the chemical composition has the same atomic ratio.

NoteDefinition

Two or more different substances that crystallise in the same crystal structure, their constituent atoms being present in the same atomic ratio, are said to be isomorphous, and the phenomenon is called isomorphism.

For example (i) \(\mathrm{NaF}\) and \(\mathrm{MgO}\) (ii) \(\mathrm{NaNO_3}\) and \(\mathrm{CaCO_3}\) are isomorphous pairs, and have the same atomic ratios, 1:1 and 1:1:3, respectively, of the constituent atoms. In the first pair one atom of a metal is matched by one atom of a non-metal in both substances; in the second pair, one metal atom, one nitrogen or carbon atom and three oxygen atoms appear in each.

ii. Polymorphism : A single substance that exists in two or more forms or crystalline structures is said to be polymorphous. Polymorphs of a substance are formed under different conditions, such as different temperatures or pressures during crystallisation — the same particles, packed in a different pattern.

NoteDefinition

A single substance that can exist in two or more different crystalline structures is said to be polymorphous, and the phenomenon is called polymorphism.

For example : Calcite and aragonite are two forms of calcium carbonate; α-quartz, β-quartz and cristobalite are three of the several forms of silica. Many crystalline forms of silica, \(\mathrm{SiO_2}\), are indeed found in nature, and these three are among them.

Polymorphism occuring in elements is called allotropy. For example: three polymorphic (allotropic) forms of carbon are diamond, graphite and fullerene.

Isomorphism vs Polymorphism vs Allotropy

Feature Isomorphism Polymorphism Allotropy
What is compared Two or more different substances One single substance One single element
Crystal structure Same for both substances Two or more different structures Two or more different structures
Composition Different substances, but same atomic ratio Same substance throughout Same element throughout
Example \(\mathrm{NaF}\) and \(\mathrm{MgO}\) Calcite and aragonite (calcium carbonate) Diamond, graphite and fullerene (carbon)

Q. Diamond and graphite are both pure carbon. Why are they called allotropes rather than polymorphs?

A. They are both. Allotropy is simply the name given to polymorphism when the substance concerned happens to be an element. Calcite and aragonite are polymorphs of a compound, calcium carbonate, so the word allotropy is not used for them; diamond and graphite are different crystalline forms of the element carbon, so they may be called either polymorphic or allotropic forms.

\(\mathrm{NaNO_3}\) and \(\mathrm{CaCO_3}\) are isomorphous. State the atomic ratio that the two share, and explain why \(\mathrm{NaNO_3}\) and \(\mathrm{CaCl_2}\) would not be expected to form an isomorphous pair.

1.3 Classification of Crystalline Solids

Knowing that a solid is crystalline tells us that its particles are ordered, but not what those particles are, nor what holds them in place. These two questions decide almost every property the solid shows. On this basis, crystalline solids are further classified into four categories : ionic solids, covalent network solids, molecular solids and metallic solids.

The four categories are separated by the simple set of questions shown in FC 1.1.

Flowchart

FC 1.1 – Classifying a crystalline solid: crystalline solid → are the particles ions? → yes → ionic solid; no → are the particles metal atoms of a single element? → yes → metallic solid; no → are the atoms joined by a continuous network of covalent bonds through the whole crystal? → yes → covalent network solid; no (separate molecules held by intermolecular forces) → molecular solid

1.3.1 Ionic Crystals

Ionic crystals have the following characteristics :

  1. The constituent particles of ionic crystals are charged ions. The cations and anions may differ in size, and this difference in size decides how the ions can pack around one another.

  2. Each ion of a given sign of charge is bonded to ions of opposite charge around it by coulomb force. In other words, the particles of ionic crystals are held by electrostatic force of attraction between oppositely charged ions. This force acts in all directions around an ion, which is why an ionic crystal has no separate molecules — every ion belongs to the crystal as a whole.

  3. Ionic crystals are hard and brittle. They have high melting points. Hardness and high melting point both follow from the strength of the electrostatic attraction, which takes a great deal of energy to overcome. Brittleness has a different cause, shown in Fig. 1.3a: if one layer of ions is made to slide even slightly over the next, ions of like charge come to face each other, the attraction turns into repulsion, and the crystal splits cleanly at that plane.

Diagram

Fig. 1.3a – Two panels of an ionic lattice of alternating cations and anions: left, the normal arrangement with each ion surrounded by ions of opposite charge; right, one layer displaced by one ion position so that like charges face each other, with repulsion arrows and a crack shown along that plane.

  1. These are nonconductors of electricity in solid state. However, they are good conductors when melted or dissolved in water. The reason is that an electric current through an ionic substance is carried by the movement of ions themselves. In the solid, every ion is locked in its lattice position and cannot travel; on melting or dissolving, the ions are set free to move towards the oppositely charged electrode.

For example : \(\mathrm{NaCl}\), \(\mathrm{K_2SO_4}\), \(\mathrm{CaF_2}\), \(\mathrm{KCl}\) are ionic crystals.

The fact that a molten ionic solid conducts is not merely a laboratory curiosity — it is the basis of an entire industry. Sodium metal and chlorine gas are manufactured by passing electricity through molten sodium chloride, because only in the molten state are the \(\mathrm{Na^{+}}\) and \(\mathrm{Cl^{-}}\) ions free to travel to the electrodes and be discharged there. A solid block of the same salt, however pure, would carry no current at all.

1.3.2 Covalent Network Crystals

Characteristics of covalent network crystals are as follows :

  1. The constituent particles in covalent network solids are atoms.

  2. The atoms in these crystals are linked by a continous system of covalent bonds. The result is a rigid three dimensional network that forms a giant molecule. The entire crystal is a single molecule. There is no point anywhere in the crystal at which one molecule ends and the next begins.

  3. As a result of rigid and strongly bonded structure, covalent network crystals are very hard. In fact they are the hardest and most incompressible of all the materials. These crystals have high melting and boiling points. To melt such a crystal, actual covalent bonds must be broken, not merely weak attractions between molecules.

  4. The electrons are localised in covalent bonds and hence are not mobile. As a result, covalent solids are poor conductors of heat and electricity.

For example : diamond, quartz (\(\mathrm{SiO_2}\)), boron nitride, carborandum are covalent network solids. Diamond is the hardest known material.

Graphite is a covalent solid yet soft and good conductor of electricity. Explain.

Diamond and Graphite: Same Atoms, Different Networks

Q. What are the structures of diamond and graphite, and what kind of covalent bonds link the carbon atoms in each?

A. In diamond, every carbon atom is joined to four other carbon atoms by single covalent bonds directed towards the corners of a tetrahedron, and this pattern repeats through the whole crystal in three dimensions. In graphite, each carbon atom is joined to only three others in the same plane, giving flat sheets of linked hexagons. Each carbon in graphite therefore has one valence electron left over, and these electrons are not tied to any single bond — they are delocalised over the whole sheet. The sheets themselves are held to one another only by weak forces.

The two structures are drawn side by side in Fig. 1.3b. Because diamond’s bonds run in every direction, there is no easy way to displace one part of the crystal past another, and no electron is free to move — hence extreme hardness and poor electrical conductivity. In graphite, the sheets can slide over one another easily, which makes the solid soft and slippery, while the delocalised electrons within each sheet can drift under an applied potential, which makes it a good conductor of electricity.

Diagram

Fig. 1.3b – Left: diamond, each carbon bonded tetrahedrally to four others in a continuous three-dimensional network; right: graphite, flat hexagonal sheets of carbon atoms with weak forces between the layers marked, and the delocalised electron cloud shaded above and below each sheet.

Q2. Diamond and graphite are both made of nothing but carbon atoms. Why is one the hardest material known, while the other is soft enough to leave a mark on paper and can even conduct electricity?

A. Being allotropes, they contain the same atoms but joined in different networks, and it is the network that decides the properties. Diamond is a rigid three-dimensional covalent network in which every carbon uses all four valence electrons in bonds pointing in four different directions; deforming it means breaking covalent bonds, so it is exceptionally hard, and since no electron is free, it does not conduct electricity. Graphite is built of flat sheets held to each other by weak forces, so the sheets slip apart under the slight pressure of a pencil on paper, making it soft; and because each carbon uses only three of its valence electrons in bonds, the remaining electrons are delocalised over the sheet and are free to carry a current.

Carborundum — silicon carbide — was made by the American inventor Edward Acheson in the early 1890s while he was heating clay and carbon in an attempt to produce artificial diamonds. What he obtained instead were hard, glittering crystals that could scratch glass. Recognising he had made one of the hardest substances then known, he named it carborundum and turned it into the first successful synthetic abrasive.

Distinguishing Ionic from Covalent Crystals

Both ionic and covalent crystals are hard and have high melting and boiling points, so hardness alone will not tell them apart. Electrical properties will.

Both are insulators at low temperature. Ionic solids become good conductors only at high temperature, above their melting points, where the ions are finally free to move — the rise in conductivity on melting is sudden and dramatic. The conductivity of covalent solids is in general low and increases with temperature; however, there is no abrupt rise in conductivity when the substance is melted, because melting a covalent network does not release any charged particle that was not already there.

1.3.3 Molecular Crystals

Substances such as \(\mathrm{Cl_2}\), \(\mathrm{CH_4}\), \(\mathrm{H_2}\), \(\mathrm{CO_2}\), \(\mathrm{O_2}\) on solidfication give molecular crystals. Crystalline organic compounds are also molecular solids.

  1. The constituent particles of molecular solids are molecules (or unbonded single atoms) of the same substance.

  2. The bonds within the molecules are covalent. The molecules are held together by various intermolecular forces of attraction. It is worth being clear about this double structure: strong covalent bonds hold the atoms inside each molecule, but only weak forces hold one molecule to the next, and it is these weak forces that break when the solid melts. For example :

  1. Weak dipole-dipole interactions in polar molecules such as solid \(\mathrm{HCl}\), \(\mathrm{H_2O}\), \(\mathrm{SO_2}\), which possess permanent dipole moment. In such molecules one end is slightly negative and the other slightly positive, so neighbouring molecules line up positive end to negative end.

  2. Very weak dispersion or London forces in nonpolar molecules such as solid \(\mathrm{CH_4}\), \(\mathrm{H_2}\). These forces are also involved in monoatomic solids like argon, neon. (These substances are usually gases at room temperature.) They arise from momentary, accidental uneven distributions of electrons that induce matching distortions in a neighbour.

  3. Intermolecular hydrogen bonds in solids such as \(\mathrm{H_2O}\) (ice), \(\mathrm{NH_3}\), \(\mathrm{HF}\) and so forth.

Q. What is a hydrogen bond?

A. It is the attraction between a hydrogen atom that is covalently bonded to a small, strongly electronegative atom — fluorine, oxygen or nitrogen — and a lone pair of electrons on such an atom in a neighbouring molecule. It is much weaker than a covalent bond, but distinctly stronger than ordinary dipole-dipole attraction, which is why ice, ammonia and hydrogen fluoride melt at higher temperatures than their molecular sizes alone would suggest.

  1. Because of weak intermolecular attractive forces, molecular solids are usually soft substances with low melting points.

  2. These solids are poor electrical conductors and are good insulators, since the molecules are electrically neutral as a whole and there are no free electrons or ions to carry a current.

1.3.4 Metallic Crystals

These are crystalline solids formed by atoms of the same metallic element, held together by a metallic bond.

Metallic bond : In a solid metal, the valence electrons are delocalised over the entire crystal leaving behind positively charged metal ions. Therefore, metallic crystals are often described as an array of positive ions immersed in a sea of mobile electrons, as pictured in Fig. 1.3c. The attractive interactions between cations and mobile electrons constitute the metallic bonds. (For more details refer to section 1.9.2)

Diagram

Fig. 1.3c – A regular array of positive metal ions drawn as circles marked with a plus sign, immersed in a shaded cloud of freely moving electrons shown as small dots with motion arrows.

NoteDefinition

A metallic bond is the attractive interaction between the positively charged metal ions fixed at the lattice sites and the valence electrons that are delocalised over the entire crystal.

Metallic crystals have the following properties:

  1. Metals are malleable, that is, they can be hammered into thin sheets.

  2. Metals are ductile, that is, they can be drawn into wires.

  3. Metals have good electrical and thermal conductivity.

Malleability and ductility are direct consequences of the electron sea. When one layer of metal ions is pushed over another, the mobile electrons simply flow along and continue to hold the shifted ions together, so the metal changes shape instead of shattering. Conductivity follows just as directly: the delocalised electrons are already free to travel through the whole crystal, so they move at once when a potential is applied.

Examples : metals such as Na, K, Ca, Li, Fe, Au, Ag, Co, etc.

Malleability and ductility are the reason metals are shaped rather than moulded. Copper is drawn out into the thin wire that carries electricity through every building, and gold is beaten into leaf so thin that light passes through it, used for gilding domes and manuscripts. Try the same treatment on a crystal of common salt and it will only turn to powder — the difference is entirely in how the two kinds of crystal respond when one layer is pushed past another.

The properties of different types of crystalline solids are summarized in Table 1.1.

Table 1.1 : Properties of four types of crystalline solids

Property Ionic solids Covalent network solids Molecular solids Metallic solids
1. Particles of unit cell Cations and anions Covalently bonded atoms Monoatomic or polyatomic molecules Metallic ions in a sea of electrons
2. Interparticle forces Electrostatic Covalent bonds London, dipole-dipole forces and/or hydrogen bonds Metallic bonds (attraction between cations and mobile valence electrons)
3. Hardness Hard and brittle Very hard Soft Variable from soft to very hard
4. Melting points High, 600 °C to 3000 °C High, 1200 °C to 4000 °C Low, (−272 °C to 400 °C) Wide range, (−39 °C to 3400 °C)
5. Thermal and electrical conductivity Poor electrical conductors in solid state. Good conductors when melted or dissolved in water Poor conductors. Exceptions : i. Graphite : good conductor of electricity. ii. Diamond : good conductor of heat Poor conductor of heat and electricity Good conductor of heat and electricity
6. Examples \(\mathrm{NaCl}\), \(\mathrm{CaF_2}\) diamond, silica ice, benzoic acid Na, Mg, Cu, Au

Classify each of the following crystalline solids into one of the four categories, stating in each case what the constituent particles are and what force holds them together:

(i) solid carbon dioxide (dry ice) (ii) carborandum (iii) potassium (iv) \(\mathrm{CaF_2}\)

  1. Solid \(\mathrm{CO_2}\). The particles are \(\mathrm{CO_2}\) molecules. Within each molecule the bonding is covalent, but the molecules themselves are nonpolar, so they are held to one another only by weak dispersion (London) forces. Hence dry ice is a molecular solid, and this is why it is soft and passes into vapour so readily.

  2. Carborandum. It is silicon carbide, in which the atoms are joined by a continuous system of covalent bonds extending through the whole crystal. The particles are atoms and the force is the covalent bond. Hence it is a covalent network solid, which is consistent with its extreme hardness.

  3. Potassium. It is a single metallic element whose atoms release their valence electrons into a common sea, leaving \(\mathrm{K^{+}}\) ions at the lattice points. The particles are metal ions in a sea of electrons and the force is the metallic bond. Hence it is a metallic solid.

  4. \(\mathrm{CaF_2}\). The particles are \(\mathrm{Ca^{2+}}\) and \(\mathrm{F^{-}}\) ions, held by electrostatic force of attraction between opposite charges. Hence it is an ionic solid, hard and brittle, non-conducting as a solid but conducting when molten.

Arrange the following three solids in increasing order of melting point and justify your order in terms of the particles present and the force that must be overcome on melting: solid \(\mathrm{CO_2}\), \(\mathrm{NaCl}\), quartz (\(\mathrm{SiO_2}\)).

A crystal of \(\mathrm{NaCl}\) shatters when struck sharply with a hammer, whereas a block of copper of the same size only flattens. Both solids are described as hard. Explain the difference in behaviour in terms of what happens when one layer of particles is displaced over the next.

1.4 Crystal Structure

We now know that the particles of a crystalline solid are arranged in an orderly, repeating way. To describe that arrangement precisely, two separate ideas are needed: where the repeating positions are, and what sits at each of those positions. The ordered three dimensional arrangement of particles in a crystal is therefore described using two terms, namely, lattice and basis.

1.4.1 Crystal, Lattice and Basis

Lattice is a geometrical arrangement of points in a three dimensional periodic array. Note carefully that these points are not particles. They are imaginary positions in space, marked out so that every point has exactly the same surroundings as every other point.

NoteDefinition

A lattice is a geometrical arrangement of points in a three dimensional periodic array, every point having identical surroundings.

A crystal structure is obtained by attaching a constituent particle to each of the lattice points. Such constituent particles that are attached to the lattice points form the basis of the crystal lattice. The basis may be a single atom, a single ion, or a whole group of atoms — whatever unit is being repeated.

NoteDefinition

The constituent particle, or group of particles, attached to each lattice point is called the basis of the crystal lattice.

Crystal lattice is also called space lattice of crystal. Thus, crystal is the structure that results by attaching a basis to each of the lattice points. It is represented by the following equation.

\[\text{Lattice} + \text{Basis} = \text{Crystal}\]

The idea is shown in Fig. 1.4a: the bare array of points is the lattice, the particle placed at every point is the basis, and the two together give the crystal.

Diagram

Fig. 1.4a – Three panels joined by a plus sign and an equals sign: (i) a regular array of bare dots labelled Lattice, (ii) a single two-particle group labelled Basis, (iii) the same array with that group attached at every point, labelled Crystal.

Q. Why separate the lattice from the basis at all — why not simply say “the particles are arranged in a pattern”?

A. Because two quite different substances can share the same lattice while having entirely different particles at each point, and one substance can have a complicated group of particles repeating at each point. Splitting the description into “where the repetition happens” (lattice) and “what is repeated” (basis) lets us handle both cases with one small set of geometrical patterns instead of a separate description for every substance.

1.4.2 Unit Cell

The space lattice of a crystal is built up of a three dimensional basic pattern. This basic pattern is repeated in three dimensions to generate the entire crystal. The smallest repeating structural unit of a crystalline solid is called unit cell.

NoteDefinition

The unit cell is the smallest repeating structural unit of a crystalline solid, which on repetition in three dimensions generates the entire crystal.

When the unit cells are stacked together to generate the crystal, each unit cell shares its faces, edges and corners with neighbouring unit cells. Because of this sharing, a particle sitting at a corner or on a face does not belong to one unit cell alone — a point we shall use in the next section to count particles.

It is important to understand that the geometric shape of a unit cell is same as that of the macroscopic crystal. For example, if the crystal has cubic shape the unit cell will also have its constituent particles arranged to form a tiny cube. The shape you see with your eyes is the shape of the unit cell, repeated an enormous number of times.

Describing a Unit Cell by Its Parameters

Since a unit cell is a small box, it can be described completely by giving the lengths of its three edges and the angles between them. The dimensions of unit cell along the three axes are denoted by the symbols a, b and c. The angles between these axes are represented by the symbols α, β and γ as shown in Fig 1.1.

Diagram

Fig. 1.1 – Unit cell parameters: a general parallelepiped unit cell with the three edge lengths labelled a, b and c meeting at one corner, and the three angles between the axes marked α (between b and c), β (between a and c) and γ (between a and b).

These six quantities — three edge lengths and three angles — are called the unit cell parameters. Give them, and the shape of the box is fixed.

the unit cell parameters shown in Fig. 1.1 (i) Which pair of edges does each of the angles α, β and γ lie between? (ii) What conditions on a, b, c and on α, β, γ would make this box a cube? (iii) If instead a = b ≠ c while all three angles remain 90°, what shape would the unit cell take?

1.4.3 Types of Unit Cell

There are four types of unit cells.

i. Primitive or simple unit cell : In primitive unit cell, the constituent particles are present at its corners only.

ii. Body-centred unit cell : In this type of unit cell, one constituent particle is present at the centre of its body in addition to the corner particles.

iii. Face-centred unit cell : This unit cell consists of particles at the centre of each of the faces in addition to the corner particles.

iv. Base-centred unit cell : This unit cell consists of particles at the centre of any two of its opposite faces in addition to the corner particles.

All four are drawn together in Fig. 1.4b. In every case the corner particles are present; the types differ only in what extra positions are occupied.

Diagram

Fig. 1.4b – Four skeleton boxes side by side, each with particles at all eight corners: (i) primitive, corners only; (ii) body-centred, with one particle at the centre of the box; (iii) face-centred, with a particle at the centre of each of the six faces; (iv) base-centred, with a particle at the centre of only two opposite faces.

Primitive vs Body-centred vs Face-centred vs Base-centred unit cell

Feature Primitive (simple) Body-centred Face-centred Base-centred
Particles at corners Yes Yes Yes Yes
Particle at centre of the body No Yes No No
Particles at centres of faces No No At all six faces At two opposite faces only

A unit cell has particles at all eight corners and one particle at the centre of each of two opposite faces, with no particle inside the body. Name this type of unit cell, and state how it differs from a face-centred unit cell.

1.4.4 Crystal Systems

By mathematical analysis, it has been proved that only fourteen different kinds of space lattices are possible. In other words, there are only 14 ways in which similar points can be arranged in a three dimensional order. These 14 lattices, which describe the crystal structure, are called Bravais lattices.

NoteDefinition

The fourteen distinct space lattices that describe all possible crystal structures are called Bravais lattices.

Fourteen Bravais lattices are divided into seven crystal systems. The possible combinations of lattice point spacings (a, b and c) along three axes and the angles (α, β and γ) between these axes give rise to seven crystal systems. In other words, seven crystal systems are associated with 14 Bravais lattices also called 14 unit cells.

The seven crystal systems are named as cubic, tetragonal, orthorhombic, rhombohedral, monoclinic, triclinic and hexagonal system. Cubic system will be discussed in the following section.

It may seem surprising that so few arrangements exist, when crystals in nature look so varied. The restriction is geometrical: a pattern must be able to fill space completely, leaving no gaps, and repeat identically in every direction. Only these fourteen arrangements can do so.

The names of the fourteen Bravais lattices (unit cells) for each of the seven crystal systems are shown in Table 1.2.

Table 1.2 : The fourteen Bravais lattices in the seven crystal systems

Crystal system Bravais lattices (names) Number
1. Cubic i. simple or primitive, ii. body-centred, iii. face-centred 3
2. Orthorhombic i. simple or primitive, ii. body-centred, iii. face-centred, iv. base-centred 4
3. Tetragonal i. simple or primitive, ii. body-centred 2
4. Monoclinic i. simple or primitive, ii. base centred 2
5. Rhombohedral i. simple or primitive 1
6. Triclinic i. simple or primitive 1
7. Hexagonal i. simple or primitive 1

Diagram

Fig. 1.4c – Skeleton unit cell drawings for Table 1.2, grouped by crystal system: cubic (simple, body-centred, face-centred), orthorhombic (simple, body-centred, face-centred, base-centred), tetragonal (simple, body-centred) with edges a, a, c marked, monoclinic (simple, base-centred) with angle β marked, rhombohedral (simple), triclinic (simple) with α, β, γ marked, hexagonal (simple) with 120° and 60° angles at the base marked.

These lattices carry the name of Auguste Bravais, a French physicist who worked on the geometry of crystals around the middle of the nineteenth century. Rather than examining crystals in the laboratory, he treated the problem as pure geometry — in how many distinct ways can identical points be arranged so that each point has the same surroundings as every other? His answer, fourteen, has stood ever since, and every crystal structure discovered since then has fallen into one of his fourteen patterns.

The link between unit cell shape and visible shape explains a familiar sight. Ice crystallises in the hexagonal system, and a growing snowflake extends the same six-fold pattern outwards as more water molecules attach to it — which is why every snowflake, however elaborate, has six arms and never five or seven. In the same way, a grain of common salt shows flat square faces because sodium chloride crystallises in the cubic system.

For a certain crystalline solid, X-ray measurement gives the unit cell parameters as a = b = c = 412 pm and α = β = γ = 90°. The unit cell is found to have a particle at each corner and one particle at the centre of each of its six faces.

(i) To which crystal system does this solid belong? (ii) Name the type of unit cell. (iii) Name the Bravais lattice. (iv) How many Bravais lattices does this crystal system have in all?

  1. All three edge lengths are equal and all three angles are 90°. A box whose three edges are equal and mutually perpendicular is a cube. Hence the solid belongs to the cubic system.

  2. Particles are present at the corners and, in addition, at the centre of each of the six faces. By the classification in section 1.4.3 this is a face-centred unit cell.

  3. Combining the two answers, the Bravais lattice is the face-centred cubic lattice.

  4. From Table 1.2, the cubic system has three Bravais lattices: simple (primitive), body-centred and face-centred.

Notice that the numerical value of the edge, 412 pm, was not needed for any part of the answer. What mattered was only the equality of the three edges and of the three angles — the parameters, not their size.

1.5 Cubic System

Of the seven crystal systems, the cubic system is the simplest to picture and the one most metals adopt, so it is the system we shall work with in detail. There are three kinds of unit cells in cubic system : primitive or simple cubic (sc), body-centred cubic (bcc) and face-centred cubic (fcc) (Fig. 1.2).

  1. Simple cubic unit cell (sc) has a particle at each of the eight corners of a cube.

  2. Body-centred cubic unit cell (bcc) has particles at its eight corners and an additional particle in the center of the cube.

  3. Face-centred cubic unit cell (fcc) has particle at the centre of each of six faces in addition to the particles at eight corners of the cube.

Diagram

Fig. 1.2 – Cubic unit cell: three columns labelled sc, bcc and fcc, each shown in three ways — as touching space-filling spheres, as a skeleton cube with small spheres at the occupied positions, and as a cut-away cube showing the portions of the spheres that actually lie inside the cell.

1.5.1 Number of Particles in Cubic Unit Cells

A unit cell never sits alone. It is one box in an endless stack, and a particle lying on a corner or a face is shared with the neighbouring boxes that meet there. So when we ask how many particles a unit cell contains, we must count only the share that belongs to that cell.

i. Primitive or simple cubic unit cell (sc) : A simple cubic cell has particles at its eight corners. When these unit cells are stacked together, particle at each corner of a given unit cell is shared with seven other neighbouring cubes that come together at that corner. As a result the corner particle contributes its 1/8th part to the given unit cell. Thus, a simple cubic cell has

\[\frac{1}{8} \times 8 = 1 \text{ particle per unit cell}\]

ii. Body-centred cubic unit cell (bcc) : A bcc unit cell has eight corner particles and an additional particle at the centre of the cube. One eighth of each particle from eight corners belongs to the given unit cell as mentioned in simple cubic unit cell.

The particle at the centre of a given cube is not shared by any other cube. Hence, it belongs entirely to the given unit cell. Thus bcc unit cell has one particle from eight corners plus one particle in the centre of the cube, making total of 2 particles per bcc unit cell.

iii. Face-centred cubic unit cell (fcc) : A fcc unit cell has particles at the eight corners plus particles at the centre of its six faces. As described in simple cubic unit cell, one particle from eight corners belongs to the given unit cell.

Each particle at the centre of the six faces is shared with one neighbouring cube. Thus, 1/2 of each face particle belongs to the given unit cell. From six faces, 1/2 × 6 = 3 particles belong to the given unit cell.

Therefore, fcc unit cell has one corner particle plus 3 face particles, total 4 particles per unit cell.

The three sharing rules used above are worth memorising as a set, and are illustrated in Fig. 1.5a: each corner particle of a cube is shared by 8 cubes, each face particle is shared by 2 cubes and each edge particle is shared by 4 cubes.

Diagram

Fig. 1.5a – Sharing of particles between neighbouring unit cells: (a) eight cubes meeting at a common corner with one sphere at that corner, marked 1/8; (b) two cubes meeting across a common face with a sphere at the face centre, marked 1/2; (c) four cubes meeting along a common edge with a sphere on that edge, marked 1/4.

Q. Why is a corner particle shared by exactly eight cubes and a face particle by exactly two?

A. Count how many boxes meet at that position. At a corner point, cubes meet from left and right, from front and back, and from above and below — two choices in each of three directions, giving 2 × 2 × 2 = 8 cubes, so each takes one-eighth. A face lies between just two cubes, one on either side, so each takes one-half. An edge is shared by four cubes arranged around it, so each takes one-quarter.

Simple cubic vs Body-centred cubic vs Face-centred cubic

Feature sc bcc fcc
Particles at eight corners Yes Yes Yes
Particle at body centre No Yes (1, unshared) No
Particles at face centres No No Yes (6, each shared by 2)
Contribution from corners 1/8 × 8 = 1 1/8 × 8 = 1 1/8 × 8 = 1
Contribution from other positions — 1 1/2 × 6 = 3
Total particles per unit cell 1 2 4

A cubic unit cell of a hypothetical solid has particles at all eight corners, a particle at the middle of each of its twelve edges, and one particle at the body centre. Calculate the number of particles that belong to this unit cell.

1.5.2 Relationship Between Molar Mass, Density of the Substance and Unit Cell Edge Length

We can now connect something we can measure in the laboratory — the density of a solid — with something measured by X-rays, the edge length of its unit cell. The relationship is deduced in the following steps.

i. If edge length of cubic unit cell is \(a\), the volume of unit cell is \(a^3\).

ii. Suppose that mass of one particle is \(m\) and that there are \(n\) particles per unit cell.

\[\text{Mass of unit cell} = m \times n \qquad \ldots (1.1)\]

iii. The density of unit cell (\(\rho\)), which is same as density of the substance, is given by

\[\rho = \frac{\text{mass of unit cell}}{\text{volume of unit cell}} = \frac{m \times n}{a^3} = \text{density of substance} \qquad \ldots (1.2)\]

The density of one unit cell is the same as the density of the whole crystal because the crystal is nothing but the same cell repeated over and over, with no gaps in between.

iv. Molar mass (\(M\)) of the substance is given by

\[M = \text{mass of one particle} \times \text{number of particles per mole} = m \times N_A\]

where \(N_A\) is Avogadro number. Therefore,

\[m = \frac{M}{N_A} \qquad \ldots (1.3)\]

This step is what lets us swap the mass of a single particle, which we cannot weigh, for the molar mass, which we can.

v. Combining Eq. (1.1) and (1.3), gives

\[\rho = \frac{n M}{a^3 N_A} \qquad \ldots (1.4)\]

By knowing any four parameters of Eq. (1.4), the fifth can be calculated. This single equation therefore works in several directions: it gives the density from the structure, the molar mass from the density, or the number of particles per unit cell — and hence the type of cubic lattice — when everything else is known.

density, \(\rho\) (SI unit kg m⁻³; in crystal calculations it is almost always used as g cm⁻³, where 1 g cm⁻³ = 10³ kg m⁻³. Density is always positive. Since edge lengths are quoted in picometres, convert before substituting: 1 pm = 10⁻¹² m = 10⁻¹⁰ cm, so an edge of 408 pm becomes 4.08 × 10⁻⁸ cm, and only then is \(a^3\) in cm³ consistent with a density in g cm⁻³.)

When gold crystallizes, it forms face-centred cubic cells. The unit cell edge length is 408 pm. Calculate the density of gold. Molar mass of gold is 197 g/mol.

\[\rho = \frac{M n}{a^3 N_A}\]

\(M\) = 197 g mol⁻¹, \(n\) = 4 atoms for fcc, \(N_A\) = 6.022 × 10²³ atoms mol⁻¹,

\(a\) = 408 pm = 408 × 10⁻¹² m = 4.08 × 10⁻⁸ cm

Substitution of these quantities in the equation gives

\[\rho = \frac{197\ \text{g mol}^{-1} \times 4\ \text{atom}}{(4.08\times10^{-8})^3\ \text{cm}^3 \times 6.022\times10^{23}\ \text{atom mol}^{-1}}\]

\[= 19.27\ \text{g/cm}^3,\ \ 19.27 \times 10^3\ \text{kg/m}^3\]

Note why \(n\) = 4 was used: gold is face-centred cubic, and an fcc unit cell contains 4 particles, as counted in section 1.5.1. Using the wrong value of \(n\) is the commonest error in this type of problem.

The density obtained in the worked example above, about 19.3 g cm⁻³, is one of the highest of any common metal, and it is the quickest way to expose a counterfeit gold bar. A bar plated with gold over a cheaper metal has the right colour and very nearly the right mass, but not the right volume — measure the volume by displacement, divide the mass by it, and a false bar gives a density well below the value that gold’s face-centred cubic packing demands.

Copper crystallizes in a face-centred cubic lattice with unit cell edge length 361 pm. Calculate the density of copper. (Molar mass of copper = 63.5 g mol⁻¹, \(N_A\) = 6.022 × 10²³ mol⁻¹)

For an fcc unit cell, the number of particles per unit cell is \(n\) = 4.

Convert the edge length to centimetres so that the density comes out in g cm⁻³:

\[a = 361\ \text{pm} = 361 \times 10^{-10}\ \text{cm} = 3.61 \times 10^{-8}\ \text{cm}\]

\[a^3 = (3.61 \times 10^{-8})^3\ \text{cm}^3 = 4.705 \times 10^{-23}\ \text{cm}^3\]

Substituting in Eq. (1.4),

\[\rho = \frac{n M}{a^3 N_A} = \frac{4 \times 63.5\ \text{g mol}^{-1}}{4.705 \times 10^{-23}\ \text{cm}^3 \times 6.022 \times 10^{23}\ \text{mol}^{-1}}\]

\[\rho = \frac{254}{28.34}\ \text{g cm}^{-3} = 8.96\ \text{g cm}^{-3}\]

The density of copper is 8.96 g cm⁻³, that is, 8.96 × 10³ kg m⁻³.

An element crystallizes in a body-centred cubic lattice with unit cell edge length 287 pm. The density of the element is 7.87 g cm⁻³. Calculate its molar mass. (\(N_A\) = 6.022 × 10²³ mol⁻¹)

1.6 Packing of Particles in Crystal Lattice

Constituent particles of a crystalline solid are close packed. While describing the packing of particles in a crystal, the individual particles are treated as hard spheres — rigid balls of fixed size that can touch one another but cannot overlap or squash. This simple picture turns out to describe metals remarkably well.

Why should the particles crowd together at all? Because the closeness of particles maximize the interparticle attractions. Every extra neighbour that a particle touches is one more attraction holding the crystal together, so the arrangement that packs the spheres most tightly is also the arrangement of lowest energy.

The number of neighbouring spheres that touch any given sphere is its coordination number. Magnitude of the coordination number is a measure of compactness of spheres in close-packed structures. The larger the coordination number, the closer are the spheres to each other.

NoteDefinition

The coordination number of a particle in a crystal is the number of neighbouring particles that touch it.

The question of how spheres pack most tightly is far older than crystallography. In the early 1600s Johannes Kepler, better known for his laws of planetary motion, considered the stacking of cannonballs and asserted that no arrangement of equal spheres could be denser than the familiar pyramid-like stack — the arrangement we shall meet in this section as cubic close packing. Simple as the claim sounds, a complete proof was not achieved until the very end of the twentieth century.

1.6.1 Close Packed Structures

The three dimensional close packed structure can be understood conveniently by looking at the close packing in one and two dimensions. We shall build it up step by step: first a row, then a layer, then a stack of layers. The whole scheme is summarised in FC 1.2.

Flowchart

FC 1.2 – Building close packed structures: row of touching spheres (1D) → stack rows aligned → square close packed layer (2D, coordination number 4) → stack layers AAAA → simple cubic structure (coordination number 6); OR stack rows staggered → hexagonal close packed layer (2D, coordination number 6) → second layer in depressions (B layer) → branch: third layer above first (ABAB…) → hcp structure (coordination number 12); third layer over octahedral voids (ABCABC…) → ccp/fcc structure (coordination number 12)

a. Close Packing in One Dimension

A close packed one dimensional structure results by arranging the spheres to touch each other in a row (Fig. 1.3 (a)). In such a row every sphere touches two neighbours, one on each side.

b. Close Packing in Two Dimensions

A close packed two dimensional (planar) structure results by stacking the rows together such that they are in contact with each other. There are two ways to obtain close packing in two dimensions, shown in Fig. 1.3 (b) and Fig. 1.3 (c).

Diagram

Fig. 1.3 – (a) Close packing in one dimension: a single row of touching spheres. (b) Square close packing: rows placed directly above one another so spheres align vertically and horizontally; mark one sphere and the square formed by joining the centres of its four closest neighbours. (c) Hexagonal close packing in two dimension: rows staggered so each sphere sits in the depression of the row below; mark one sphere and the regular hexagon formed by joining the centres of its six closest neighbours, and mark two triangular voids with apices pointing in opposite directions.

i. Square close packing : One dimensional rows of close packed spheres are stacked over each other such that the spheres align vertically and horizontally (Fig. 1.3 (b)). If the first row is labelled as ‘A’ type, being exactly same as the first row, the second row is also labelled as ‘A’ type. Hence this arrangement is called A, A, A, A….. type two dimensional arrangement.

In this arrangement, every sphere touches four neighbouring spheres — one on the left, one on the right, one above and one below. Hence, two dimensional coordination number, here, is 4. A square is obtained by joining the centres of these four closest neighbours (Fig. 1.3(b)). Therefore, this two dimensional close packing is called square close packing in two dimension.

ii. Hexagonal close packing : Close packed one dimensional row (Fig. 1.3 (a)) shows that there are depressions between the neighbouring spheres. If the second row is arranged in such a way that its spheres fit in the depressions of the first row, a staggered arrangement results.

If the first row is called ‘A’ type row, the second row, being different, is called ‘B’ type. Placing the third row in staggered manner in contact with the second row gives rise to an arrangement in which the spheres in the third row are aligned with the spheres in the first row. Hence the third row is ‘A’ type. Similarly spheres in the fourth row will be alligned with the spheres in the second row and hence the fourth row would be ‘B’ type. The resulting two dimensional arrangement is ‘ABAB…’ type (Fig. 1.3 (c)).

In this arrangement each sphere touches six closest neighbours — two in its own row, two in the row above and two in the row below. Thus, the two dimensional coordination number in this packing is 6. A regular hexagon is obtained by joining the centres of these six closest spheres (Fig. 1.3 (c)). Hence, this type of two dimensional close packing is called hexagonal close packing in two dimensions.

Compared to the square close packing in two dimensions, the coordination number in hexagonal close packing in two dimensions is higher. Moreover the free space in this arrangement is less than in square packing, making it more efficient packing than square packing. Slipping each row into the hollows of the row below simply wastes less floor space than lining the rows up.

From Fig. 1.3(c) it is evident that the free spaces (voids) are triangular in shape. These triangular voids are of two types. Apex of the triangular voids in alternate rows points upwards and downwards. This small difference in orientation will matter greatly when a second layer is placed on top.

Square close packing vs Hexagonal close packing in two dimensions

Feature Square close packing Hexagonal close packing
Arrangement of rows Aligned, spheres directly above one another Staggered, spheres in the depressions of the row below
Row pattern A, A, A, A… ABAB…
Two dimensional coordination number 4 6
Shape obtained by joining centres of closest neighbours Square Regular hexagon
Free space More Less
Efficiency of packing Less efficient More efficient

the triangular voids in the hexagonal close packed layer of Fig. 1.3(c) (i) Locate two triangular voids whose apices point in opposite directions. (ii) Counting around one sphere, how many triangular voids touch it? (iii) Why can a sphere of the next layer not sit in every one of these triangular voids at the same time?

c. Close Packing in Three Dimensions

Stacking of two dimensional layers gives rise to three dimensional crystal structures. Two dimensional square close packed layers are found to stack only in one way to give simple cubic lattice. Two dimensional hexagonal close packed layers are found to stack in two distinct ways. Accordingly two crystal structures, namely, hexagonal close packed (hcp) structure and face centred cubic (fcc) structure are formed.

i. Stacking of square close packed layers: Stacking of square close packed layers generates a three dimensional simple cubic structure. Here, the second layer is placed over the first layer so as to have its spheres exactly above those of the first layer (Fig. 1.4). Subsequent square close packed layers are placed one above the other in the same manner.

In this arrangement, spheres of all the layers are perfectly aligned horizontally as well as vertically. Hence, all the layers are alike, and are labelled as ‘A’ layers. This arrangement of layers is described as ‘AAAA…’ type. The structure that results on stacking square close packed layers is simple cubic. Its unit cell is the primitive cubic unit cell (Fig. 1.4).

Diagram

Fig. 1.4 – Stacking of square closed packed layers: four square close packed layers drawn one above the other, all spheres vertically aligned, each layer labelled ‘A Layer’, with one primitive cubic unit cell outlined within the stack.

It can be seen that in the simple cubic structure, each sphere touches six neighbouring spheres, four in its own layer, one in the layer above and one in the layer below. Hence, coordination number of each sphere is 6. Polonium is the only metal that crystallizes in simple cubic closed packed structure.

ii. Stacking of two hexagonal close packed layers : To generate a close packed three dimensional structure, hexagonal close packed layers are arranged in a particular manner. In doing so, spheres of the second layer are placed in the depression of the first layer (Fig. 1.5). If the first layer is labelled as ‘A’ layer, the second layer is labelled as ‘B’ layer because the two layers are aligned differently.

Diagram

Fig. 1.5 – Two layers of closed packed spheres: a first layer of spheres in one shade with a second layer in a contrasting shade resting in its depressions, so that only some of the first layer’s triangular depressions are covered.

It is evident from the Fig. 1.5 that all triangular voids of the first layers are not covered by the spheres of the second layer. Only half of them can be, because the depressions of the two layers point in opposite directions and cannot all be occupied at once. Two different kinds of hole are therefore left behind.

The triangular voids that are covered by spheres of the second layer generate tetrahedral void (Fig. 1.6). A tetrahedral void is surrounded by four spheres. On joining the centres of these four spheres a tetrahedron is formed which encloses the tetrahedral voids (Fig. 1.6).

Diagram

Fig. 1.6 – Tetrahedral void: three touching spheres of the lower layer with a fourth sphere resting in their depression, the small space between them labelled ‘Tetrahedral hole’; beside it, the tetrahedron obtained by joining the centres of the four spheres.

NoteDefinition

A tetrahedral void is the vacant space enclosed by four spheres in contact, whose centres, when joined, form a tetrahedron.

The remaining triangular voids of the first layer have above them the triangular voids of the second layer. The overlapping triangular voids from the two layers together form an octahedral void which is surrounded by six spheres (Fig. 1.7).

Diagram

Fig. 1.7 – Octahedral void: three spheres of the lower layer and three of the upper layer arranged so that their triangular depressions overlap with apices in opposite directions, the enclosed space labelled ‘Octahedral hole’; beside it, the octahedron obtained by joining the centres of the six spheres, with vertices labelled M, N, O, P, Q, R.

NoteDefinition

An octahedral void is the vacant space enclosed by six spheres in contact, whose centres, when joined, form an octahedron.

It is important to note that the triangular shapes of depressions in A and B layer do not overlap. The apices of two triangular depressions in A and B layer point in opposite directions. The depressions in which spheres of second layer rest are tetrahedral voids, while the depressions in which no sphere rests are octahedral voids.

Q. Both voids begin as triangular depressions. What exactly makes one tetrahedral and the other octahedral?

A. What happens above the depression. If a sphere of the upper layer settles into the depression, the space is enclosed by the three spheres below plus that one sphere above — four spheres in all, giving a tetrahedral void. If no sphere settles there, the depression of the lower layer meets an oppositely pointing depression of the upper layer, and the space is enclosed by three spheres below and three above — six in all, giving the larger octahedral void.

iii. Placing third hexagonal close packed layer : There are two ways of placing the third hexagonal close packed layer on the second. This single choice is what separates the two most common metal structures.

One way of doing this is to align the spheres of the third layer with the spheres of the first layer. The resulting pattern of the layers will be ‘ABAB….’. This arrangement results in hexagonal close packed (hcp) structure (Fig. 1.8(a)). Metals such as Mg, Zn, have hcp crystal structure.

The second way of placing the third hexagonal close packed layer on the second is to cover the octahedral voids by spheres of the third layer. In such placing, the spheres of the third layer do not align with the spheres of the second or the spheres of the first layer. The third layer is, therefore, called ‘C’ layer. The spheres of the fourth layer get aligned with the spheres of the first layer. Hence, the fourth layer is called ‘A’ layer.

This pattern of stacking hexagonal close packed layers is called ‘ABCABC….’. This arrangement results in cubic close packed (ccp) structure (Fig. 1.8(b)). This is same as fcc structure. Metals such as copper and Ag have ccp (or fcc) crystal structure.

Diagram

Fig. 1.8 – Formation of hexagonal closed packed structures: (a) layers stacked ABAB… with an expanded view showing the third layer directly above the first; (b) layers stacked ABCABC… with an expanded view showing the third layer displaced from both the first and the second, and the fcc unit cell outlined within it.

hcp vs ccp (fcc)

Feature hcp ccp / fcc
Layers used Hexagonal close packed layers Hexagonal close packed layers
Position of the third layer Directly above the first layer Over the octahedral voids, aligned with neither the first nor the second layer
Stacking pattern ABAB… ABCABC…
Coordination number 12 12
Unit cell Hexagonal Face-centred cubic
Examples Mg, Zn Copper, Ag

1.6.2 Coordination Number in Close Packed Structure

a. In the simple cubic (sc) crystal structure, that results from stacking of square close packed layers, each sphere is surrounded by 6 neighbouring spheres, 4 in its own layer, 1 above and 1 below. Hence coordination number of any sphere in sc is 6.

b. In both hcp and ccp/fcc structures that result from stacking of hexagonal close packed layers in two different ways, each sphere is surrounded by 12 neighbouring spheres, 6 in its own layer, 3 above and 3 below. Hence, the coordination number of any sphere in hcp or ccp/fcc structure is 12.

Notice that hcp and ccp differ in how the layers repeat, yet each sphere has exactly the same 12 neighbours in the same contact positions. This is why the two structures share so many properties, as we shall see again when packing efficiency is calculated.

1.6.3 Number of Voids per Atom in hcp and ccp

The tetrahedral and octahedral voids occur in hcp and ccp/fcc structures. There are two tetrahedral voids associated with each atom. The number of octahedral voids is half that of tetrahedral voids. Thus, there is one octahedral void per atom.

Stated as a rule that can be applied to any quantity of substance: if N denotes number of particles, then number of tetrahedral voids is 2N and that of octahedral voids is N. The total number of voids is therefore 3N.

These voids are not wasted space in a chemical sense. In many ionic compounds, one kind of ion forms the close packed arrangement while the other kind occupies some fraction of the voids, and the fraction occupied fixes the formula of the compound.

The voids of a close packed metal are large enough to lodge very small atoms. Palladium takes up hydrogen in exactly this way, the hydrogen atoms slipping into the interstitial voids between the close packed metal atoms, so that a piece of the metal can hold a surprising volume of the gas and release it again on warming. Metal hydrides of this kind are studied as a way of storing hydrogen fuel as a solid rather than as a compressed gas.

A metal crystallizes in the cubic close packed structure. A sample of this metal contains 4.0 × 10²² atoms. Calculate (a) the number of tetrahedral voids, (b) the number of octahedral voids, and (c) the total number of voids in the sample.

For a close packed structure, if the number of atoms is N, then the number of tetrahedral voids is 2N and the number of octahedral voids is N.

Here N = 4.0 × 10²² atoms.

  1. Number of tetrahedral voids = 2N = 2 × 4.0 × 10²² = 8.0 × 10²²

  2. Number of octahedral voids = N = 4.0 × 10²²

  3. Total number of voids = 2N + N = 3N = 3 × 4.0 × 10²² = 1.2 × 10²³

The relation holds for hcp as well as ccp, since both are close packed structures with coordination number 12.

In a compound, atoms of element Q form a cubic close packed lattice and atoms of element P occupy one half of the octahedral voids. Derive the formula of the compound, showing your reasoning in terms of the number of voids per atom.

Zinc crystallizes in the hcp structure and silver in the ccp structure. State the coordination number in each, and explain why both structures leave each sphere with the same number of touching neighbours even though their layer sequences are different.

1.7 Packing Efficiency

Like coordination number, the magnitude of packing efficiency gives a measure of how tightly particles are packed together. Coordination number counts neighbours; packing efficiency answers the more direct question — of the space inside the unit cell, how much is actually filled by matter?

Packing efficiency is the fraction or a percentage of the total space occupied by the spheres (particles).

\[\text{Packing efficiency} = \frac{\text{volume occupied by particles in unit cell}}{\text{total volume of unit cell}} \times 100 \qquad \ldots (1.5)\]

NoteDefinition

Packing efficiency is the percentage of the total volume of a unit cell that is actually occupied by the constituent particles, treated as hard spheres.

The method is the same for all three cubic lattices, and it is worth fixing it in mind before starting: first relate the radius \(r\) of the sphere to the edge length \(a\) by finding the direction along which the spheres actually touch; then find the volume of one sphere in terms of \(a\); then multiply by the number of particles in the unit cell; and finally divide by \(a^3\).

1.7.1 Packing Efficiency of Metal Crystal in Simple Cubic Lattice

The packing efficiency of a metal crystal in a simple cubic lattice is obtained by the following steps.

Step 1 : Radius of sphere : In simple cubic unit cell, particles (spheres) are at the corners and touch each other along the edge. A face of simple cubic unit cell is shown in Fig. 1.9. It is evident that

\[a = 2r \qquad \text{or} \qquad r = a/2 \qquad \ldots (1.6)\]

where \(r\) is the radius of atom and ‘a’ is the length of unit cell edge. In other words, the edge is made up of one full radius from the sphere at each end of it.

Diagram

Fig. 1.9 – Face of simple cubic unit cell: four touching spheres at the corners of a square, the square of edge \(a\) drawn through their centres, with the radius \(r\) marked on two adjacent spheres along one edge.

Step 2 : Volume of sphere : Volume of a sphere = \((4/3\pi)(r^3)\). Substitution for \(r\) from Eq. (1.6) gives

Volume of one particle

\[= (4/3\pi)(a/2)^3 = \frac{\pi a^3}{6} \qquad \ldots (1.7)\]

Step 3 : Total volume of particles : Because simple cubic unit cell contains only one particle,

\[\text{volume occupied by particle in unit cell} = \frac{\pi a^3}{6}\]

Step 4 : Packing efficiency

\[\text{Packing efficiency} = \frac{\text{volume occupied by particle in unit cell}}{\text{total volume of unit cell}} \times 100\]

\[= \frac{\pi a^3/6}{a^3} \times 100 = \frac{100\pi}{6} = \frac{100 \times 3.142}{6} = 52.36\%\]

Thus, in simple cubic lattice, 52.36 % of total space is occupied by particles and 47.64% is empty space, that is, void volume.

Q. The edge length \(a\) cancelled out at the last step. Does packing efficiency then depend at all on how large the atoms are?

A. No, and that is the point. Because the spheres touch, the radius is always a fixed fraction of the edge length, so \(a\) appears in the numerator and the denominator to the same power and cancels. Packing efficiency depends only on the geometry of the arrangement, not on the size of the atom. Every simple cubic metal, whatever its atomic radius, wastes exactly 47.64% of its volume.

1.7.2 Packing Efficiency of Metal Crystal in Body-Centred Cubic Lattice

Step 1 : Radius of sphere (particle) : In bcc unit cell, particles occupy the corners and in addition one particle is at the centre of the cube. Figure 1.10 shows that the particle at the centre of the cube touches two corner particles along the diagonal of the cube. Note carefully that the corner spheres do not touch each other along the edge here; contact happens only along the body diagonal, so it is that diagonal we must express in terms of \(a\). To obtain radius of the particle (sphere) Pythagorus theorem is applied.

Diagram

Fig. 1.10 – bcc unit cell: a cube with corners labelled A, D, E, F, spheres at the corners and one sphere at the body centre C, with the face diagonal FD and the body diagonal AF drawn as dashed lines and the right angles at ∠FED and ∠ADF marked.

  • For triangle FED, ∠ FED = 90°.

\[\therefore \ \mathrm{FD}^2 = \mathrm{FE}^2 + \mathrm{ED}^2 = a^2 + a^2 = 2a^2 \quad (\text{because } \mathrm{FE} = \mathrm{ED} = a) \qquad \ldots (1.8)\]

  • For triangle AFD, ∠ ADF = 90°

\[\therefore \ \mathrm{AF}^2 = \mathrm{AD}^2 + \mathrm{FD}^2 \qquad \ldots (1.9)\]

Substitution of Eq. (1.8) into Eq. (1.9) yields

\[\mathrm{AF}^2 = a^2 + 2a^2 = 3a^2 \quad (\text{because } \mathrm{AD} = a)\]

\[\text{or} \quad \mathrm{AF} = \sqrt{3}\,a \qquad \ldots (1.10)\]

The Fig. 1.10 shows that AF = 4\(r\). Along the body diagonal we meet one radius of the first corner sphere, a full diameter of the central sphere, and one radius of the far corner sphere — that is, \(r + 2r + r = 4r\).

Substitution for AF from equation (1.10) gives

\[\sqrt{3}\,a = 4r \quad \text{and hence,} \quad r = \frac{\sqrt{3}}{4}\,a \qquad \ldots (1.11)\]

Step 2 : Volume of sphere : Volume of sphere particle = \(4/3\ \pi r^3\). Substitution for \(r\) from Eq. (1.11), gives

\[\text{volume of one particle} = \frac{4}{3}\pi \left(\frac{\sqrt{3}}{4}a\right)^3 = \frac{4}{3}\pi \times \frac{(\sqrt{3})^3}{64}\,a^3 = \frac{\sqrt{3}\,\pi a^3}{16}\]

Step 3 : Total volume of particles : Unit cell bcc contains 2 particles. Hence, volume occupied by particles in bcc unit cell

\[= 2 \times \frac{\sqrt{3}\,\pi a^3}{16} = \frac{\sqrt{3}\,\pi a^3}{8} \qquad \ldots (1.12)\]

Step 4 : Packing efficiency

\[\text{Packing efficiency} = \frac{\text{volume occupied by particles in unit cell}}{\text{total volume of unit cell}} \times 100\]

\[= \frac{\sqrt{3}\,\pi a^3}{8a^3} \times 100 = 68\ \%\]

Thus, 68% of the total volume in bcc unit lattice is occupied by atoms and 32 % is empty space or void volume.

1.7.3 Packing Efficiency of Metal Crystal in Face-Centred Cubic Lattice (or ccp or hcp Lattice)

Step 1 : Radius of particle/sphere : The corner particles are assumed to touch the particle at the centre of face ABCD as shown in Fig. 1.11. Here the line of contact is the face diagonal, not the edge and not the body diagonal.

Diagram

Fig. 1.11 – fcc unit cell: a cube with one face ABCD shaded, spheres at the corners A, B, C, D of that face and one sphere at the centre of the face, with the face diagonal AC drawn as a dashed line and the right angle at ∠ABC marked.

The triangle ABC is right angled with ∠ABC = 90°. According to Pythagorus theorem,

\[\mathrm{AC}^2 = \mathrm{AB}^2 + \mathrm{BC}^2 = a^2 + a^2 = 2a^2\]

(because AB = BC = \(a\))

\[\text{Hence,} \quad \mathrm{AC} = \sqrt{2}\,a \qquad \ldots (1.13)\]

Figure 1.11 shows that AC = 4 \(r\) — again one radius, a full diameter of the face-centre sphere, and one radius. Substitution for AC from Eq. (1.13) gives

\[\sqrt{2}\,a = 4r \quad \text{or} \quad r = \frac{\sqrt{2}}{4}a = \frac{a}{2\sqrt{2}} \qquad \ldots (1.14)\]

Step 2 : Volume of sphere : Volume of one particle = \(\dfrac{4}{3}\pi r^3\). Substitution for \(r\) from Eq. (1.14) gives

\[\text{Volume of one particle} = \frac{4}{3}\pi\left(\frac{a}{2\sqrt{2}}\right)^3 = \frac{4}{3}\pi a^3 \times \left(\frac{1}{2\sqrt{2}}\right)^3 = \frac{\pi a^3}{12\sqrt{2}}\]

Step 3 : Total volume of particles : The unit cell of fcc lattice contains 4 particles. Hence, volume occupied by particles in fcc unit cell

\[= 4 \times \frac{\pi a^3}{12\sqrt{2}} = \frac{\pi a^3}{3\sqrt{2}}\]

Step 4 : Packing efficiency

\[\text{Packing efficiency} = \frac{\text{volume occupied by particles in unit cell}}{\text{total volume of unit cell}} \times 100\]

\[= \frac{\pi a^3}{3\sqrt{2}\,a^3} \times 100 = \frac{\pi}{3\sqrt{2}} \times 100 = 74\%\]

Thus in fcc/ccp/hcp crystal lattice, 74% of the total volume is occupied by particles and 26% is void volume or empty space.

Because hcp and ccp have the same coordination number and the same close packed layers, they share this same efficiency of 74% — the highest possible for identical spheres.

Which of the three lattices, sc, bcc and fcc has the most efficient packing of particles? Which one has the least efficient packing?

Q3. If the particles of a metal are packed as tightly as spheres possibly can be, why is roughly a quarter of the crystal still empty space?

A. Because spheres cannot fill space completely. However closely you arrange rigid balls of equal size, gaps remain wherever three or four curved surfaces meet, and these gaps are exactly the tetrahedral and octahedral voids met in section 1.6. The calculation above puts a number on it: in the closest possible arrangement, fcc/ccp/hcp, the spheres occupy 74% of the volume and 26% is unavoidably void. The only way to do better would be to deform the particles so they are no longer spheres — which is why the void percentage is a property of the geometry, not of the metal.

Table 1.3 shows the expressions for various parameters of particles in terms of unit cell dimension for cubic systems.

Table 1.3 : Edge length and particle parameters in cubic system

Unit cell Relation between \(a\) and \(r\) Volume of one particle Total volume occupied by particles in unit cell
1. sc \(r = a/2\) = 0.5000\(a\) \(\pi a^3/6\) = 0.5237 \(a^3\) \(\pi a^3/6\) = 0.5237 \(a^3\)
2. bcc \(r = \sqrt{3}a/4\) = 0.4330\(a\) \(\sqrt{3}\,\pi a^3/16\) = 0.34\(a^3\) \(\sqrt{3}\,\pi a^3/8\) = 0.68\(a^3\)
3. fcc/ccp \(r = \sqrt{2}a/4\) = 0.3535\(a\) \(\pi a^3/12\sqrt{2}\) = 0.185\(a^3\) \(\pi a^3/3\sqrt{2}\) = 0.74\(a^3\)

Table 1.4 shows the summary of coordination number of particles and packing efficiency in various cubic systems.

Table 1.4 : Coordination number and packing efficiency in systems

Lattice Coordination number of atoms Packing efficiency
1. sc 6 : four in the same layer, one directly above and one directly below 52.4 %
2. bcc 8 : four in the layer below and four in the layer above 68 %
3. fcc/ccp/hcp 12 : six in its own layer, three above and three below 74 %

Read the two tables together and the pattern is clear: as the coordination number rises from 6 to 8 to 12, the packing efficiency rises from 52.4% to 68% to 74%. More touching neighbours means less wasted space, exactly as expected.

Not every solid tries to pack tightly. Water molecules in ice are held by hydrogen bonds that point in fixed directions, and to satisfy those directions the molecules must stand apart in an open framework with far more empty space than a close packed arrangement would leave. That open structure is less dense than liquid water, which is why ice floats — and why lakes freeze from the top down, leaving liquid water and living organisms below the surface.

A metal crystallizes in a body-centred cubic lattice with unit cell edge length 316 pm. Calculate the radius of the metal atom.

In a bcc lattice the spheres touch along the body diagonal, and from Eq. (1.11),

\[r = \frac{\sqrt{3}}{4}a = 0.4330\,a\]

Substituting \(a\) = 316 pm,

\[r = 0.4330 \times 316\ \text{pm} = 136.8\ \text{pm}\]

The radius of the metal atom is 136.8 pm.

Note that had the same edge length belonged to an fcc metal, the relation would have been \(r = 0.3535\,a\), giving a smaller radius. Identifying the lattice first, and hence the correct direction of contact, is therefore essential.

1.7.4 Number of Particles and Unit Cells in x g of Metallic Crystal

The number of particles and the number of unit cells in given mass of a metal can be calculated from the known parameters of unit cell, namely, number of particles ‘n’ per unit cell and volume ‘\(a^3\)’ of unit cell. Density (\(\rho\)) and molar mass (\(M\)) of a metal are related to each other through unit cell parameters as shown below :

\[\rho = \frac{\text{mass}}{\text{volume}} = \frac{\text{number of particles in unit cell}}{\text{volume of unit cell}} \times \frac{M}{N_A}\]

\[\therefore \rho = \frac{n}{a^3} \times \frac{M}{N_A}\]

\[\therefore M = \rho\,\frac{a^3 N_A}{n}\]

where ‘n’ is the number of particles in unit cell and ‘\(a^3\)’ is the volume of unit cell.

Number of particles in ‘x’ g metal :

Molar mass, \(M\), contains \(N_A\) particles.

\[\therefore \ x \text{ g of metal contains } \frac{xN_A}{M} \text{ particles.}\]

Substitution of \(M\) gives

\[\text{Number of particles in 'x' g} = \frac{xN_A}{\rho a^3 N_A/n} = \frac{xn}{\rho a^3}\]

Notice how \(N_A\) cancels here — the enormous Avogadro number disappears from the final expression, leaving only quantities that can be measured directly.

Number of unit cells in ‘x’ g metal :

Since ‘n’ particles correspond to 1 unit cell,

\[\therefore \ \frac{xn}{\rho a^3} \text{ particles correspond to } \frac{xn}{\rho a^3} \times \frac{1}{n} \text{ unit cells.}\]

\[\therefore \text{Number of unit cells in 'x' g metal} = \frac{x}{\rho a^3}\]

Number of unit cells in volume ‘V’ of metal

\[= \frac{V}{a^3}\]

This last one needs no density at all: if you know the volume of the piece of metal and the volume of one cell, dividing one by the other counts the cells.

A compound made of elements C and D crystallizes in fcc structure. Atoms of C are present at the corners of the cube. Atoms of D are at the centres of faces of the cube. What is the formula of the compound?

  1. C atoms are present at the 8 corners. The contribution of each corner atom to the unit cell is 1/8 atom. Hence, the number of C atom that belongs to the unit cell = 8 × (1/8) = 1

  2. D atoms are present at the centres of six faces of unit cell. Each face-centre atom is shared between two cubes. Hence, centribution of each face centre atom to the unit cell is 1/2 atom.

The number of D atoms that belong to unit cell = 1/2 × 6 = 3

There are one C atom and three D atoms in the unit cell.

\[\therefore \text{Formula of compound} = \mathrm{CD_3}\]

The unit cell of metallic silver is fcc. If radius of Ag atom is 144.4 pm, calculate (a) edge length of unit cell (b) volume of Ag atom, (c) the percent of the volume of a unit cell, that is occupied by Ag atoms, (d) the percent of empty space.

  1. For fcc unit cell, \(r\) = 0.3535 \(a\)

\(r\) = 144.4 pm = 144.4 × 10⁻¹² m = 144.4 × 10⁻¹⁰ cm

\[a = \frac{r}{0.3535} = \frac{144.4 \times 10^{-10}\ \text{cm}}{0.3535} = 4.085 \times 10^{-8}\ \text{cm}\]

  1. Volume of Ag atom = \(\dfrac{4}{3}\pi r^3\)

\[= \frac{4}{3} \times 3.142 \times (144.4 \times 10^{-10}\ \text{cm})^3 = 1.261 \times 10^{-23}\ \text{cm}^3\]

  1. In fcc unit cell, there are 4 Ag atoms

Volume occupied by 4 Ag atoms

\[= 4 \times 1.26 \times 10^{-23}\ \text{cm}^3 = 5.044 \times 10^{-23}\ \text{cm}^3\]

Total volume of unit cell = \(a^3\)

\[= (4.085 \times 10^{-8}\ \text{cm})^3 = 6.817 \times 10^{-23}\ \text{cm}^3\]

Percent of volume occupied by Ag atoms

\[= \frac{\text{volume occupied by atoms in unit cell}}{\text{total volume of unit cell}} \times 100 = \frac{5.044 \times 10^{-23}\ \text{cm}^3}{6.817 \times 10^{-23}\ \text{cm}^3} = 74\%\]

  1. Percent empty space = 100 − 74 = 26%

Parts (c) and (d) are a useful check on section 1.7.3: worked out from real measured data for silver, the answer comes back to the same 74% and 26% obtained from pure geometry.

A compound is formed by two elements A and B. The atoms of element B forms ccp structure. The atoms of A occupy 1/3rd of tetrahedral voids. What is the formula of the compound ?

The atoms of element B form ccp structure. The number of tetrahedral voids generated is twice the number of B atoms.

Thus, number of tetrahedral voids = 2B

The atoms A occupy (1/3) of these tetrahedral voids.

Hence, number of A atoms = 2B × 1/3

Ratio of A and B atoms = 2/3 B : 1B

\[= 2/3 : 1 = 2 : 3\]

\[\text{Formula of compound} = \mathrm{A_2B_3}\]

Niobium forms bcc structure. The density of niobium is 8.55 g/cm³ and length of unit cell edge is 330.6 pm. How many atoms and unit cells are present in 0.5 g of niobium?

  1. Number of atoms in x g niobium = \(\dfrac{xn}{\rho a^3}\)

x = 0.5 g, \(n\) = 2 (for bcc structure), \(\rho\) = 8.55 g/cm³,

\(a\) = 330.6 pm = 3.306 × 10⁻⁸ cm.

Number of atoms in 0.5 g of niobium

\[= \frac{0.5\ \text{g} \times 2}{8.55\ \text{g cm}^{-3} \times (3.306 \times 10^{-8}\ \text{cm})^3} = 3.25 \times 10^{21}\]

  1. Number of unit cells in x g = \(\dfrac{x}{\rho a^3}\)

Number of unit cells in 0.5 g of niobium

\[= \frac{0.5\ \text{g}}{8.55\ \text{g cm}^{-3} \times (3.306 \times 10^{-8}\ \text{cm})^3} = 1.62 \times 10^{21}\]

The second answer is exactly half the first, as it must be: a bcc cell holds 2 atoms, so the atoms are counted twice over when cells are counted.

A compound forms hcp structure. What is the number of (a) octahedral voids (b) tetrahedral voids (c) total voids formed in 0.4 mol of it.

Number of atoms in 0.4 mol = 0.4 × \(N_A\)

\[= 0.4 \times 6.022 \times 10^{23} = 2.4098 \times 10^{23}\]

  1. Number of octahedral voids = number of atoms = 2.4098 × 10²³

  2. Number of tetrahedral voids

\[= 2 \times \text{number of atoms} = 2 \times 2.4098 \times 10^{23} = 4.818 \times 10^{23}\]

  1. Total number of voids

\[= 2.409 \times 10^{23} + 4.818 \times 10^{23} = 7.227 \times 10^{23}\]

A metal crystallizes in a simple cubic lattice with unit cell edge length 334 pm. Calculate (a) the radius of the metal atom and (b) the percentage of empty space in the crystal.

The hcp and ccp structures have different unit cells, yet both have a packing efficiency of 74%. Explain why the two efficiencies must be equal, referring to the way the close packed layers are built in each case.

1.8 Crystal Defects or Imperfections

Everything said so far describes a perfect crystal, in which every particle sits exactly where the pattern demands. Real crystals are not like that. The real, naturally occurring crystalline substances do not have perfect crystal structures. They have some disorders or irregularities in the stacking of atoms. Such irregularities in the arrangement of constituent particles of a solid crystal are called defects or imperfections.

NoteDefinition

An irregularity in the regular arrangement of the constituent particles of a crystalline solid is called a defect or imperfection.

Defects are created during the process of crystallization. The imperfections are more if the crystallization occurs at a faster rate. It means that the defects can be minimized by carrying out crystallization at a slower rate — when particles are given time to find their proper positions, fewer of them are trapped in the wrong place.

In fact ideal crystals with no imperfections are possible only at the absolute zero of temperature. Above this temperature no crystalline materials are 100 % pure. They contain defects. The reason is that above absolute zero the particles have thermal energy, and some of them will always have enough of it to leave their sites.

Whatever be the nature of a crystal defect, electrical neutrality of the solid is maintained. This is a rule worth remembering, because it explains why so many defects come in matched pairs, as we shall see.

It is important to note that sometimes defects are to be intentionally created for manipulating the desired properties in crystalline solids. A defect is not always a flaw; it is often the reason a material is useful at all.

There are three types of defects: point defects, line defects and plain defects. Only point defects will be discussed in this chapter.

1.8.1 Point Defects

These defects are irregularities produced in the arrangement of basis at lattice points in crystalline solids. In other words, the disorder is confined to a point in the lattice — one site, or a small group of sites, rather than a whole line or plane.

There are three major classes of point defects: stoichiometric point defects, impurity defects and nonstoichiometric point defects. The complete classification is set out in FC 1.3.

Flowchart

FC 1.3 – Classification of point defects: point defect → branch 1: stoichiometric point defect (stoichiometry unchanged) → vacancy defect, self interstitial defect, Schottky defect, Frenkel defect; branch 2: impurity defect (foreign atoms present) → substitutional impurity defect (including vacancy through aliovalent impurity), interstitial impurity defect; branch 3: nonstoichiometric point defect (stoichiometry changed) → metal deficiency defect, metal excess defect → metal excess splits further into neutral atom or extra positive ion at an interstitial site, and anion vacancy occupied by an electron (F-centre)

a. Stoichiometric Point Defects

Chemical formula of a compound shows fixed ratio of number of atoms or number of cations and anions. This fixed ratio is the stoichiometry of the compound.

In stoichiometric defect, the stoichiometry remains unchanged. In other words, the ratio of number of atoms or number of cations and anions of compound remains the same as represented by its chemical formula. The crystal is imperfect, but its formula is still the formula written on the bottle.

There are four types of stoichiometric point defects: vacancy defect, self interstitial defect, Schottky defect and Frenkel defect.

i. Vacancy defect : During crystallization of a solid, a particle is missing from its regular site in the crystal lattice. The missing particle creates a vacancy in the lattice structure. Thus, some of the lattice sites are vacant because of missing particles as shown in Fig. 1.12. The crystal is, then, said to have a vacancy defect. The vacancy defect can also be developed when the substance is heated.

Diagram

Fig. 1.12 – Vacancy defect: a regular two-dimensional array of identical particles with two lattice sites left empty, arrows pointing to the empty sites labelled ‘Vacancy’.

Due to the absence of particles, the mass of the substance decreases. However, the volume remains unchanged. As a result the density of the substance decreases. The crystal keeps its size because the surrounding particles stay where they are; only matter has gone missing from inside it.

ii. Self interstitial defect in elemental solid : Interstitial sites in a crystal are the spaces or voids in between the particles at lattice points. When some particles of a crystalline elemental solid occupy interstitial sites in the crystal structure, it is called self interstitial defect. The word ‘self’ matters here: the particle in the void is of the same kind as those at the lattice points, not a foreign atom.

This defect occurs in the following two ways :

Firstly, an extra particle occupies an empty interstitial space in the crystal structure as shown in Fig. 1.13. This extra particle is same as those already present at the lattice points.

Diagram

Fig. 1.13 – Self interstitial defect: a regular array of identical particles with one additional particle of the same kind squeezed into the space between the lattice sites.

The extra particles increase the total mass of substance without increasing volume. Hence its density increases.

Secondly, in an elemental solid a particle gets shifted from its original lattice point and occupies an interstitial space in the crystal as shown in the Fig. 1.14.

Diagram

Fig. 1.14 – Self interstitial defect of the second kind: a regular array with one lattice site empty, labelled ‘Vacancy’, and the displaced particle shown occupying a nearby interstitial space, labelled ‘Interstitial atom’, with an arrow from the empty site to the new position.

It is interesting to note that in this second case, because of the displacement of a particle a vacancy defect is created at its original regular lattice site. At the same time interstitial defect results at its new position. We can, therefore, say that in this defect there is a combination of vacancy defect and self interstitial defect.

This defect preserves the density of the substance because there is neither loss nor gain in mass of a substance. Nothing has entered the crystal and nothing has left it; a particle has merely moved house.

iii. Schottky defect : In an ionic solid, equal number of cations and anions are missing from their regular positions in the crystal lattice creating vacancies as shown in Fig. 1.15. It means that a vacancy created by a loss of cation is always accompanied by a vacancy formed by a loss of anion.

Diagram

Fig. 1.15 – Schottky defect: an ionic lattice of alternating cations and anions with one cation site and one anion site left empty, arrows labelling them ‘Cation vacancy’ and ‘Anion vacancy’.

Thus, there exist two holes per ion pair lost, one created by missing cation and the other by a missing anion. Such a paired cation-anion vacancy defect is a Schottky defect. The pairing is not accidental — losing a cation alone would leave the crystal with a net negative charge, and the rule of electrical neutrality does not permit that.

NoteDefinition

A Schottky defect is the point defect in which an equal number of cations and anions are missing from their regular lattice sites in an ionic crystal, so that the crystal remains electrically neutral.

Conditions for the formation of Schottky defect

  1. Schottky defect is found in ionic compounds with the following characteristics :
  • High degree of ionic character.
  • High coordination number of anion
  • Small difference between size of cation and anion. The ratio \(r_{cation}/r_{anion}\) is not far below unity.

The last condition is the key one: if the two ions are nearly the same size, neither of them can fit into an interstitial space, so the only way a displaced ion can go is out of the lattice altogether, leaving a vacancy behind.

Consequences of Schottky defect

  • As the number of ions decreases, mass decreases. However, volume remains unchanged. Hence, the density of a substance decreases.
  • The number of missing cations and anions is equal, the electrical neutrality of the compound is preserved.

This defect is found in ionic crystals such as \(\mathrm{NaCl}\), \(\mathrm{AgBr}\) and \(\mathrm{KCl}\).

iv. Frenkel defect : Frenkel defect arises when an ion of an ionic compound is missing from its regular lattice site and occupies interstitial position between lattice points as shown in Fig. 1.16.

Diagram

Fig. 1.16 – Frenkel defect: an ionic lattice of alternating cations and anions showing two cation sites vacant, labelled ‘Vacancy of cation’, and the corresponding cations shown squeezed into interstitial positions between lattice points, labelled ‘Cation at interstitial site’.

The cations are usually smaller than anions. It is, therfore, more common to find the cations occupying interstitial sites. It is easier for the smaller cations to accomodate the interstitial spaces.

Frenkel defect is not found in pure alkali metal halides because cations of allkali metals due to large size cannot occupy interstitial space.

It is important to note that the smaller cation is displaced from its normal site to an interstitial space. It, therefore, creates a vacancy defect at its original position and interstitial defect at its new location in the same crystal. Frenkel defect can be regarded as the combination of vacancy defect and interstitial defect.

NoteDefinition

A Frenkel defect is the point defect in which an ion leaves its regular lattice site and occupies an interstitial position in the same crystal, creating a vacancy at the site it has left.

Conditions for the formation of Frenkel defect

  • Frenkel defect occurs in ionic compounds with large difference between sizes of cation and anion.
  • The ions of ionic compounds must be having low coordination number.

Consequences of Frenkel defect

  • As no ions are missing from the crystal lattice as a whole, the density of solid and its chemical properties remain unchanged.
  • The crystal as a whole remains electrically neutral because the equal numbers of cations and anions are present.

This defect is found in ionic crystals like \(\mathrm{ZnS}\), \(\mathrm{AgCl}\), \(\mathrm{AgBr}\), \(\mathrm{AgI}\), \(\mathrm{CaF_2}\).

Schottky defect vs Frenkel defect

Feature Schottky defect Frenkel defect
What happens Equal numbers of cations and anions leave the crystal An ion leaves its lattice site and moves into an interstitial space in the same crystal
Size relation of ions Small difference between sizes of cation and anion Large difference between sizes of cation and anion
Coordination number High coordination number of anion Low coordination number
Effect on density Density decreases Density remains unchanged
Effect on electrical neutrality Preserved Preserved
Regarded as A paired cation–anion vacancy A combination of vacancy defect and interstitial defect
Examples \(\mathrm{NaCl}\), \(\mathrm{AgBr}\), \(\mathrm{KCl}\) \(\mathrm{ZnS}\), \(\mathrm{AgCl}\), \(\mathrm{AgBr}\), \(\mathrm{AgI}\), \(\mathrm{CaF_2}\)

Both defects carry the names of the physicists who first described them in the earlier part of the twentieth century — Walter Schottky in Germany, whose name is also attached to a well-known type of semiconductor contact, and Yakov Frenkel in the Soviet Union, who worked on the theory of disorder in solids. Naming defects after people is a reminder of how recent this understanding is: crystals were being used and studied for centuries before anyone explained what their flaws actually were.

Silver bromide, \(\mathrm{AgBr}\), is listed as showing both Schottky and Frenkel defects, while \(\mathrm{NaCl}\) shows only the Schottky defect. Account for this difference.

The type of defect an ionic solid can support is decided mainly by the relative sizes of its ions.

For a Schottky defect, the cation and the anion should be of comparable size, so that neither can fit into an interstitial space and ions can only be lost as pairs from the lattice. Both \(\mathrm{NaCl}\) and \(\mathrm{AgBr}\) satisfy this reasonably well, so both show the Schottky defect.

For a Frenkel defect, there must be a large difference between the sizes of cation and anion, so that the smaller cation can be accommodated in an interstitial space. In \(\mathrm{AgBr}\) the \(\mathrm{Ag^{+}}\) ion is considerably smaller than the large \(\mathrm{Br^{-}}\) ion, and can slip into an interstitial position; hence \(\mathrm{AgBr}\) also shows the Frenkel defect.

In \(\mathrm{NaCl}\), which is an alkali metal halide, the \(\mathrm{Na^{+}}\) ion is too large to occupy an interstitial space, and therefore the Frenkel defect is not found in it.

Conclusion: \(\mathrm{AgBr}\) shows both defects because its ions satisfy the size condition for each; \(\mathrm{NaCl}\) shows only the Schottky defect.

b. Impurity Defect

Impurity defect arises when foreign atoms, that is, atoms different from the host atoms, are present in the crystal lattice. There are two kinds of impurity defects : Substitutional and interstitial impurity defects.

i. Substitutional impurity defect : In this defect, the foreign atoms are found at the lattice sites in place of host atoms. The regular atoms are displaced from their lattice sites by impurity atoms.

For example :

  • Solid solutions of metals (alloys) : Brass is an alloy of Cu and Zn. In brass, host Cu atoms are replaced by impurity of Zn atoms. The Zn atoms occupy regular sites of Cu atoms as shown in Fig. 1.17.

Diagram

Fig. 1.17 – Brass: a close packed array of copper atoms in one shade with a few zinc atoms of a contrasting shade occupying regular lattice sites among them; label ‘Copper’ and ‘Zinc’.

  • Vacancy through aliovalent impurity : Vacancies are created by the addition of impurities of aliovalent ions (that is, ions with oxidation state (o.s.) different from that of host ions) to an ionic solid.

Suppose that a small amount of \(\mathrm{SrCl_2}\) impurity is added to \(\mathrm{NaCl}\) during its crystallization. The added \(\mathrm{Sr^{2+}}\) ions (O.S. + 2) occupy some of the regular sites of \(\mathrm{Na^{+}}\) host ions (O.S. +1).

In order to maintain electrical neutrality, every \(\mathrm{Sr^{2+}}\) ion removes two \(\mathrm{Na^{+}}\) ions. One of the vacant lattice sites created by removal of two \(\mathrm{Na^{+}}\) ions is occupied by one \(\mathrm{Sr^{2+}}\) ion. The other site of \(\mathrm{Na^{+}}\) ion remains vacant as shown in Fig. 1.18.

Diagram

Fig. 1.18 – Vacancy through aliovalent ion: a NaCl lattice of alternating Na⁺ and Cl⁻ ions in which one Sr²⁺ ion occupies a sodium site and a nearby sodium site is empty, with an arrow labelled ‘Vacancy of Na⁺ ion’.

Q. Why must a vacancy appear at all? Could the \(\mathrm{Sr^{2+}}\) ion not simply take one \(\mathrm{Na^{+}}\) site and leave the rest of the lattice untouched?

A. No, because the charges would not balance. A single \(\mathrm{Sr^{2+}}\) ion brings two positive charges into a site that previously held one, so the crystal would carry a net positive charge. To keep the solid electrically neutral, two \(\mathrm{Na^{+}}\) ions must go for every \(\mathrm{Sr^{2+}}\) that comes in — one of the two sites is filled by the \(\mathrm{Sr^{2+}}\) ion and the other is left empty. The vacancy is the price of neutrality.

ii. Interstitial impurity defect : In this defect, the impurity atoms occupy interstitial spaces of lattice structure. For example in steel, Fe atoms occupy normal lattice sites. The carbon atoms are present at interstitial spaces, as shown in Fig. 1.19.

Diagram

Fig. 1.19 – Stainless steel: a close packed array of large iron atoms with much smaller carbon atoms lodged in the interstitial spaces between them; label ‘Iron’ and ‘Carbon’.

A trace impurity in the right lattice site is what turns a plain mineral into a gemstone. Pure aluminium oxide is colourless; when a small fraction of its aluminium ions is replaced by chromium ions at regular lattice sites, the crystal absorbs part of the visible spectrum and becomes the deep red of ruby. Change the impurity and the colour changes with it — the same host lattice with different substitutional impurities gives the blues of sapphire. In each case the crystal structure is untouched; only a few sites in a million have a different occupant.

c. Nonstoichiometric Defects

Nonstoichiometric defect arises when the ratio of number of atoms of one kind to that of other kind or the ratio of number of cations to anions becomes different from that indicated by its chemical formula. In short, stoichiometry of the compound is changed.

It is important to note that the change in stoichiometry does not cause any change in the crystal structure. The lattice pattern stays exactly what it was; only the census of who occupies it has altered.

There are two types of nonstoichiometric defects.

i. Metal deficiency defect : This defect is possible only in compounds of metals that show variable oxidation states. The reason is that the missing positive charge has to be made up somewhere, and only a metal that can exist in a higher oxidation state can do so.

In some crystals, positive metal ions are missing from their original lattice sites. The extra negative charge is balanced by the presence of cation of the same metal with higher oxidation state than that of missing cation.

For example, in the compound \(\mathrm{NiO}\) one \(\mathrm{Ni^{2+}}\) ion is missing creating a vacnacy at its lattice site. The deficiency of two positive charges is made up by the presence of two \(\mathrm{Ni^{3+}}\) ions at the other lattice sites of \(\mathrm{Ni^{2+}}\) ions as shown in Fig. 1.20. The composition of \(\mathrm{NiO}\) then becomes \(\mathrm{Ni_{0.97}O_{1.0}}\)

Diagram

Fig. 1.20 – Nonstoichiometric Ni₀.₉₇O₁.₀: a lattice of alternating Ni²⁺ and O²⁻ ions with one nickel site empty, labelled ‘Vacancy’, and two of the remaining nickel sites occupied by Ni³⁺ ions shown in a contrasting colour.

A sample of nickel oxide has the nonstoichiometric composition \(\mathrm{Ni_{0.97}O_{1.0}}\). Calculate the percentage of the nickel present as \(\mathrm{Ni^{3+}}\) ions.

In the formula \(\mathrm{Ni_{0.97}O_{1.0}}\), the total nickel present is 0.97 mol per 1 mol of oxide ion, and the oxide ion carries a charge of −2. Since the crystal as a whole must be electrically neutral, the total positive charge on nickel must be +2.

Let the amount of \(\mathrm{Ni^{3+}}\) be \(x\) mol. Then the amount of \(\mathrm{Ni^{2+}}\) is (0.97 − \(x\)) mol.

Equating the total positive charge to +2,

\[3x + 2(0.97 - x) = 2\]

\[3x + 1.94 - 2x = 2\]

\[x = 0.06\]

Therefore the amount of \(\mathrm{Ni^{3+}}\) is 0.06 mol out of a total of 0.97 mol of nickel.

\[\text{Percentage of } \mathrm{Ni^{3+}} = \frac{0.06}{0.97} \times 100 = 6.19\ \%\]

About 6.19% of the nickel is present as \(\mathrm{Ni^{3+}}\) and the remaining 93.81% as \(\mathrm{Ni^{2+}}\). Note that the whole calculation rests on one idea only — the crystal must stay electrically neutral.

ii. Metal excess defect : There are two types of metal excess defects.

  • A neutral atom or an extra positive ion occupies interstitial position : \(\mathrm{ZnO}\) presents two ways of metal excess defect. In the first case in \(\mathrm{ZnO}\) lattice one neutral Zn atom is present in the interstitial space as shown in Fig. 1.21(a)

Diagram

Fig. 1.21 (a) – Neutral Zn atom at interstitial site: a lattice of alternating Zn²⁺ and O²⁻ ions with one extra neutral Zn atom drawn in the space between lattice points, labelled ‘Excess neutral Zn atom’.

In the second case, when \(\mathrm{ZnO}\) is heated it decomposes as :

\[\mathrm{ZnO} \longrightarrow \mathrm{Zn^{2+}} + \tfrac{1}{2}\,\mathrm{O_2} + 2e^{-}\]

The excess \(\mathrm{Zn^{2+}}\) ions are trapped in interstitial site in the lattice. The electrons also diffuse in the crystal to occupy interstitial sites as shown in Fig. 1.21(b).

Diagram

Fig. 1.21 (b) – Zn²⁺ ions and electrons at interstitial sites: the same ZnO lattice with an extra Zn²⁺ ion in an interstitial position and two electrons, marked as dots, occupying nearby interstitial spaces; label ‘Electron’.

In both the cases, nonstoichiometric formula of \(\mathrm{ZnO}\) is \(\mathrm{Zn_{1+x}O_{1.0}}\). In either case the metal is in excess of the amount the formula \(\mathrm{ZnO}\) demands, and the extra electrons that come with it keep the crystal neutral.

When \(\mathrm{ZnO}\) is heated it turns yellow and returns back to original white colour on cooling. What could be the reason ?

  • By anion vacancies (Colour or F-centres) : This type of defect imparts colour to the colourless crystal. For example, when \(\mathrm{NaCl}\) crystal is heated in the atmosphere of sodium vapour, sodium atoms are deposited on the crystal surface.

\(\mathrm{Cl^{-}}\) ions diffuse to the crystal surface creating vacancies at their regular sites. These \(\mathrm{Cl^{-}}\) ions combine with Na atoms on the surface to form \(\mathrm{NaCl}\), by releasing electron from sodium atom.

\[\mathrm{Na} + \mathrm{Cl^{-}} \longrightarrow \mathrm{NaCl} + e^{-}\]

The electrons released diffuse into the crystal and occupy vacant sites of anions as shown in Fig. 1.22. The anion vacant sites occupied by electrons are F-centres or colour-centres.

Diagram

Fig. 1.22 – An F-centre in a crystal: an ionic lattice of alternating positive and negative ions in which one anion site is occupied instead by an electron, marked e⁻ and shown in a contrasting colour.

NoteDefinition

An F-centre is an anion vacancy in a crystal that is occupied by an electron, which gives colour to an otherwise colourless crystal.

\(\mathrm{NaCl}\) shows yellow colour due to the formation of F-centre. The crystal of \(\mathrm{NaCl}\) has excess Na. The nonstoichiometric formula of \(\mathrm{NaCl}\) is the \(\mathrm{Na_{1+x}Cl_{1.0}}\)

The letter F in “F-centre” is not an initial of a scientist’s name. It comes from the German word Farbe, meaning colour — the effect was first studied and named by German workers on alkali halide crystals, who called such a site a Farbzentrum, a colour centre.

Q4. Pure sodium chloride is a white solid. Yet if its crystals are heated in sodium vapour, they turn yellow. How can a colourless crystal gain colour without any coloured substance being added?

A. Nothing coloured is added — a defect is created. Sodium atoms deposit on the surface, chloride ions migrate out to meet them, and the electrons released by the sodium atoms move into the empty anion sites left behind. An electron trapped in such a vacancy is an F-centre. Being loosely held, it can absorb energy from visible light and be promoted to a higher energy level; the light of the absorbed colour is removed and the crystal appears in the remaining colour, yellow in the case of sodium chloride. The colour therefore comes from trapped electrons in anion vacancies, and the crystal is now slightly metal-rich, \(\mathrm{Na_{1+x}Cl_{1.0}}\).

\(\mathrm{NaCl}\) is doped with \(10^{-3}\) mol % of \(\mathrm{SrCl_2}\). Calculate the number of cation vacancies produced per mole of \(\mathrm{NaCl}\). (\(N_A\) = 6.022 × 10²³ mol⁻¹)

State, with a reason in each case, how the density of the crystal is affected by (i) a vacancy defect, (ii) a self interstitial defect in which an extra particle enters an interstitial space, (iii) a Schottky defect, and (iv) a Frenkel defect.

1.9 Electrical Properties of Solids

Q. What is electrical conductivity, and what do the terms ‘insulator’ and ‘semiconductor’ mean?

A. Electrical conductivity measures how easily a material lets an electric current pass through it; the higher the conductivity, the smaller the opposition offered to the flow of charge. An insulator is a material of extremely low conductivity, which effectively stops a current. A semiconductor lies in between — it conducts far less than a metal but far more than an insulator, and, as we shall see, its conductivity can be altered enormously by temperature or by adding traces of another element.

Solids show very wide range of electrical conductivity — wider than for almost any other physical property, spanning more than twenty powers of ten. Accordingly solids are classified into the following three categories : conductors, insulators and semiconductors.

i. Conductors : Solids having electrical conductivities in the range 10⁴ to 10⁷ Ohm⁻¹m⁻¹ are called conductors. Metals and electrolytes (ionic solids) are examples of electrical conductors. Metals conduct electricity by movement of electrons while electrolytes conduct electricity by movement of ions. In this section we will study some aspects of electronic conduction of electricity.

ii. Insulators : Solids having low electrical conductivities in the range 10⁻²⁰ to 10⁻¹⁰ Ohm⁻¹m⁻¹ are called insulators. Most nonmetals and molecular solids belong to this category.

iii. Semiconductors : Solids having electrical conductivities in the range 10⁻⁶ to 10⁴ Ohm⁻¹ m⁻¹ are semiconductors. This range is intermediate between conductors and insulators. Metalloids like silicon, germanium belong to this category.

electrical conductivity (SI unit Ω⁻¹ m⁻¹, also written S m⁻¹, where S is the siemens; conductivity is always positive and is the reciprocal of resistivity. Band gap energies of solids are quoted in electron volt, eV, where 1 eV = 1.602 × 10⁻¹⁹ J — a convenient unit because the gaps involved are far too small to be written comfortably in joules.)

1.9.1 Band Theory

Electrical conductivities of solid metals, nonmetals and metalloids are explained in terms of band theory. A band is made of closely spaced electronic energy levels. Band formation can be correlated to formation of molecular orbitals (MOs) by interaction of atomic orbitals.

Q. How many molecular orbitals are formed by the interaction of two atomic orbitals?

A. Two. The number of molecular orbitals formed is always equal to the number of atomic orbitals that combine — one bonding molecular orbital, lower in energy than the original atomic orbitals, and one antibonding molecular orbital, higher in energy. This rule is the whole basis of what follows: if two atomic orbitals give two molecular orbitals, then the huge number of atoms in a crystal must give an equally huge number of molecular orbitals.

According to MO theory interaction of atomic orbitals of combining atoms results in formation of equal number of MOs which spread over the entire molecule. Similar to this, interaction of energy levels of electrons in the closely spaced constituent atoms in solids result in formation of bands.

Since a crystal contains an enormous number of atoms, the molecular orbitals formed are enormously many and lie so close together in energy that they can no longer be told apart. They merge into what is effectively a continuous range of allowed energies — a band.

Band theory considers formation of two types of bands, namely, conduction band and valence band. Another important concept of band theory is the band gap.

i. Conduction band : The highest energy band containing electrons is the conduction band. It is formed by interaction of the outermost energy levels of closely spaced atoms in solids. Conduction band may be partially occupied or vacant. Electrons in conduction band are mobile and delocalized over the entire solid. They conduct electricity when electrical potential is applied.

ii. Valence band : The band having lower energy than conduction band is the valence band.

The electrons in valence band are not free to move because they are tightly bound to the respective nuclei.

iii. Band gap : The energy difference between valence band and conduction band is called band gap. Size of the band gap decides whether electrons from valence band can be promoted to vacant conduction band or not. When band gap is too large to promote electrons from valence band to vacant conduction band by thermal energy, it is called forbidden zone.

When band gap is small, electrons from higher energy levels in valence band can be promoted to conduction band by absorption of energy (such as thermal, electromagnetic).

NoteDefinition

The band gap is the energy difference between the valence band and the conduction band of a solid; when it is too large for thermal energy to promote an electron across it, it is called the forbidden zone.

The band gap energy values of a few solids are as shown in Table 1.5.

Table 1.5 : Band gap energies of some solids

Solid \(E_{\text{gap}}\) eV
Diamond 5.47
Sodium 0
Silicon 1.12
Germanium 0.67

Diagram

Fig. 1.9a – A simple bar chart of the band gap energies in Table 1.5: bars for diamond (5.47 eV), sodium (0), silicon (1.12 eV) and germanium (0.67 eV), with the vertical axis labelled band gap energy / eV.

the band gap energies of Table 1.5 and Fig. 1.9a (i) Using only these values, classify each of the four solids as a conductor, a semiconductor or an insulator. (ii) Which of silicon and germanium should conduct better at room temperature, and why? (iii) Diamond is a form of carbon, which lies in the same group as silicon and germanium — suggest why its band gap is so much larger.

The electrical properties of metallic conductors, insulators and semiconductors are explained in terms of band theory as follows.

1.9.2 Metals

Metals are good conductors of electricity. The outermost electrons of all the atoms in the metallic crystal occupy conduction band. The number of electrons in conduction band of metals is large. Hence metals are good conductors of electricity.

The conduction bands in metals can be further labelled as ‘s’ band (Fig. 1.23 (a)), overlapping s and p bands (Fig. 1.23(b)) and so on. This depends on the atomic orbitals involved in band formation.

Diagram

Fig. 1.23 – Metallic conductor: (a) a single ‘s band’ drawn as a vertical block, its lower part shaded to show it is filled with electrons and its upper part unshaded, with no gap; (b) an s band and a p band drawn as two blocks that overlap in energy, the filled portion shaded.

Band formation in metallic conductors, thus, results in delocalization of the outermost electrons of all the metal atoms leaving behind metal ions. This is described as ‘cations of metal are immersed in the sea of electrons’.

The cations of metal atoms occupying lattice sites vibrate about their mean positions. At higher temperatures, metal cations undergo increased vibrational motion about their lattice sites. The flow of electrons is interrupted by increased vibrational motion. As a result conductivity of metals decreases with increase in temperature. The electrons are already free; heating simply puts more obstacles in their path.

Two Metals, Two Kinds of Band Picture

Metallic sodium is an example of conductor where the conduction band is partially filled and there is no band gap. Electronic configuration of Na is [Ne]3s¹. Interaction of the partially filled 3s AOs of all the Na atoms gives rise to same number of MOs. All these closely spaced MOs together form a continuous band of energies which is called 3s band.

Lower half of 3s band corresponds to BMOs and is filled while the upper half of 3s band corresponds to AMOs and is empty. There is no gap between these two halves. The 3s band in sodium is the conduction band which contains same number of electrons as the sodium atoms. This is responsible for the high conductivity of metallic sodium.

Metallic magnesium is an example of conductor with overlapping bands. Electronic configuration of Mg is [Ne]3s²3p⁰. Interaction of completely filled 3s AOs of all the Mg atoms gives rise to the same number of MOs all of which are filled. These together form the 3s band which is a completely filled band.

Interaction of vacant 3p AOs of all the Mg atoms gives rise to the same number of vacant MOs together called 3p band. This is the vacant band. The filled 3s band and vacant 3p band overlap each other. As a result, higher energy electrons move from 3s band to 3p band.

Magnesium is worth pausing over: a completely filled band by itself would allow no conduction, since a full band has no vacant level for an electron to move into. It is the overlap with the empty 3p band that rescues the situation and makes magnesium a conductor.

1.9.3 Insulators

In insulators the valence band is completely filled with electrons and the conducation band is empty.

The valence band and conduction band in insulators are separated by a large energy gap called forbidden zone as shown in Fig. 1.24. Here, thermal energy is insufficient to promote electrons from valence band to conduction band.

Diagram

Fig. 1.24 – Insulators: a shaded block at the bottom labelled ‘Filled valence band’ and an unshaded block well above it labelled ‘Empty conduction band’, with a double-headed arrow between them labelled ‘Forbidden zone (Large energy gap)’.

As a result the conduction band remains vacant. The material is, therefore, an insulator.

1.9.4 Semiconductors

Electrical conductivity of a semiconductor material is intermediate between that of metals and insulators. The metalloids Si and Ge are semiconductors.

Like insulators, the valence band in semiconductor is completely filled with electrons and conduction band is empty. However, the energy gap between the two bands is smaller than that in an insulator. (Fig. 1.25)

Diagram

Fig. 1.25 – Semiconductors: a shaded block labelled ‘Filled valence band’ and an unshaded block labelled ‘Empty conduction band’ placed close above it, with a short double-headed arrow between them labelled ‘Small energy gap’.

At a temperature above absolute zero a few electrons in the valence band have enough thermal energy to jump through the small band gap and occupy higher energy conduction band. The conduction band, thus, becomes partially filled and the valence band becomes partially empty.

The electrons in conduction band are free to move. When electric potential is applied to a semiconductor, it conducts a small amount of electicity.

Such a pure semiconductor material which has a very low but finite electrical conductivity is called intrinsic semiconductor.

NoteDefinition

A pure semiconductor material, which has a very low but finite electrical conductivity arising only from electrons thermally promoted across its small band gap, is called an intrinsic semiconductor.

The electrical conductivity of a semiconductor increases with increasing temperature. This is because, the number of electrons with sufficient energy so as to get promoted to the conduction band increases as temperature rises. Thus, at higher temperatures, there are more mobile electrons in the conduction band and more vacancies in the valence band than at lower temperature. In fact semiconductors are insulators at low temperatures and conductors at high temperatures.

Electrical conductivity of metals decreases and that of semiconductor increases with increasing temperature. The two behave oppositely because heating does different jobs in the two cases: in a metal the charge carriers already exist and heating only obstructs them, while in a semiconductor heating actually creates the charge carriers.

Conductors vs Semiconductors vs Insulators

Feature Conductors Semiconductors Insulators
Conductivity range (Ohm⁻¹ m⁻¹) 10⁴ to 10⁷ 10⁻⁶ to 10⁴ 10⁻²⁰ to 10⁻¹⁰
Band picture Conduction band partially filled, or filled band overlapping a vacant band; no band gap Filled valence band and empty conduction band separated by a small band gap Filled valence band and empty conduction band separated by a large forbidden zone
Effect of rise in temperature Conductivity decreases Conductivity increases Remains an insulator
Examples Metals such as sodium, magnesium, copper Si, Ge Most nonmetals and molecular solids, diamond

The band gap of silicon is 1.12 eV. Calculate the longest wavelength of electromagnetic radiation that can promote an electron from the valence band of silicon to its conduction band.

(\(h\) = 6.626 × 10⁻³⁴ J s, \(c\) = 3.0 × 10⁸ m s⁻¹, 1 eV = 1.602 × 10⁻¹⁹ J)

The longest wavelength corresponds to the smallest energy that will just bridge the gap, so the photon energy must equal the band gap energy.

First convert the band gap into joules:

\[E = 1.12 \times 1.602 \times 10^{-19}\ \text{J} = 1.794 \times 10^{-19}\ \text{J}\]

The energy of a photon is related to its wavelength by

\[E = h\nu = \frac{hc}{\lambda} \qquad \therefore \ \lambda = \frac{hc}{E}\]

\[\lambda = \frac{6.626 \times 10^{-34}\ \text{J s} \times 3.0 \times 10^{8}\ \text{m s}^{-1}}{1.794 \times 10^{-19}\ \text{J}}\]

\[\lambda = \frac{1.988 \times 10^{-25}}{1.794 \times 10^{-19}}\ \text{m} = 1.108 \times 10^{-6}\ \text{m} = 1108\ \text{nm}\]

Radiation of wavelength 1108 nm or shorter can excite an electron across the gap. This lies just beyond the red end of the visible region, which is why silicon absorbs visible light strongly enough to be useful in a solar cell, while radiation of longer wavelength passes through without exciting any electron.

1.9.5 Extrinsic Semiconductors and Doping

The conductivity of a semiconductor can be increased by doping. The process of addition of minute quantity of impurities to a semiconductor to increase its conductivity is called doping. The added impurity is called dopant.

NoteDefinition

Doping is the deliberate addition of a minute quantity of an impurity, called the dopant, to a semiconductor in order to increase its electrical conductivity.

A doped semiconductor, having higher conductivity than pure intrinsic semiconductor, is an extrinsic semiconductor.

There are two types of extrinsic semiconductors, namely, n-type and p-type semiconductors.

i. n-type Semiconductor

n-type semiconductor contains increased number of electrons in the conduction bond.

An n-type semiconductor is obtained by adding group 15 element to intrinsic semiconductor which belongs to group 14.

Let a small quantity of phosphorus be doped into pure silicon. Will the resulting material contain the same total number of electrons as the original pure silicon ? Will the material be electrically neutral or charged ?

Consider, for example, doping of Si with phosphorus. Si has a crystal structure in which each Si atom is linked tetrahedrally to four other Si atoms. When small quantity of phosphorous is added to pure Si, the P atoms occupy some vacant sites in the lattice in place of Si atoms, as shown in Fig. 1.26. The overall crystal structure of Si remains unchanged.

Diagram

Fig. 1.26 – P atom occupying regular site of Si atom: a two-dimensional network of silicon atoms each joined to four neighbours by electron pairs, with one site occupied by a phosphorus atom that has a fifth, unpaired electron drawn beside it and labelled ‘Mobile electron’.

Four of the five valence electrons of P are utilized in bonding the closest to four Si atoms. Thus, P has one extra electron than needed for bonding. Therefore, Si doped with P has more number of electrons in the conduction band than those in the conduction band in pure Si as shown in Fig. 1.27. It is thus transperent that the conductivity of Si doped with P is higher than that of pure Si. The electrons in conduction band move under the influence of an applied potential and conduct electricity.

Because the charge carriers are the increased number of electrons, Si or Ge doped with group 15 elements such as P, As, Sb or Bi is an n-type semiconductor. The letter n stands for the negative charge of the carriers, not for any charge on the material as a whole.

Diagram

Fig. 1.27 – n-type and p-type semiconductor: (a) n-type — a filled valence band below and a conduction band above containing a few extra electrons marked with arrows, labelled ‘extra electrons’; (b) p-type — a conduction band that is empty and a valence band containing a few vacancies marked and labelled ‘Holes’.

ii. p-type Semiconductor

A p-type semiconductor is produced by doping a pure semiconductor material (Si or Ge) with an impurity of group 13 elements. These elements contain less number of valence electrons than that of the pure semiconductor.

Consider, for example, pure Si doped with boron. The B atoms occupy normal positions of some of the Si atoms in the lattice as shown in Fig. 1.28. Boron atom has only three valence electrons. It does not have enough electrons to form bonds with its four Si neighbours.

Diagram

Fig. 1.28 – B atom occupying regular site of Si atom: the same silicon network with one site occupied by a boron atom joined to only three neighbours, the fourth position drawn as an empty space and labelled ‘hole (no electron)’.

B atom forms bonds with three Si atoms only. The missing fourth electron creates an electron vacancy. It is called a hole.

Fig. 1.28 shows the holes in the valence band of p-type semiconductor.

A hole has a tendency to accept electron from its close vicinity. Thus, a hole behaves as if it has a positive charge. The electrons in partially filled valence band move under the influence of an applied potential. The holes move in the opposite direction.

An easy way to picture this: when an electron from a neighbouring bond moves into the hole, the hole itself has effectively shifted to where that electron came from. Nothing positive is really travelling, yet the vacancy migrates as though it were a positive particle — which is why it is counted as the charge carrier.

Because the charge carriers are holes which behave like positive charge, the Si or Ge doped with group 13 elements like B, Ga or In, is a p-type semiconductor.

Whether intrinsic or extrinsic, the semiconductor material is electrically neutral. An n-type semiconductor such as Si doped with P has more electrons than those needed for bonding and thus has electrons in the partially filled conduction band. A p-type semiconductor such as Si doped with B has the less electrons than needed for bonding and thus has vacancies (holes) in the valence band. In neither case is any charge added or removed — each dopant atom brings its own nucleus along with its electrons.

n-type vs p-type semiconductor

Feature n-type p-type
Group of the dopant Group 15 (P, As, Sb, Bi) Group 13 (B, Ga, In)
Valence electrons of dopant compared with Si or Ge One more than needed for bonding One less than needed for bonding
Defect created Extra electron in the conduction band Electron vacancy, called a hole, in the valence band
Majority charge carrier Electrons (negative) Holes (behaving as positive)
Net charge on the material Neutral Neutral

Q5. Impurities usually spoil a material. Why, then, does adding a trace of phosphorus to extremely pure silicon make it far more useful than the pure silicon itself?

A. Because in a semiconductor the useful property is not purity but the availability of charge carriers, and pure silicon has very few of them — only the handful of electrons that thermal energy can push across its 1.12 eV band gap. Phosphorus belongs to group 15 and brings five valence electrons to a site that needs only four for bonding. The fifth electron is left over and is easily promoted into the conduction band, so each phosphorus atom donates a mobile carrier without disturbing the crystal structure at all. A few such atoms per million raise the conductivity enormously, giving an n-type extrinsic semiconductor. The material remains electrically neutral throughout; what the impurity supplies is not charge, but mobility.

The control of conduction in doped semiconductors is the foundation of modern electronics. Working at Bell Laboratories in the late 1940s, John Bardeen, Walter Brattain and William Shockley demonstrated the transistor, a device in which a small signal applied to doped semiconductor material controls a much larger current. The three shared the Nobel Prize in Physics for the work, and the vacuum tubes that had filled entire rooms were replaced within a generation by pieces of doped silicon smaller than a grain of rice.

Joining a piece of n-type material to a piece of p-type material makes a p–n junction, and almost every semiconductor device is built from it. In a solar cell, light absorbed near such a junction excites electrons across the band gap and the junction drives them in one direction, producing a current — which is why the band gap of silicon, 1.12 eV, matters so much: it is well matched to the energy of sunlight. In a light-emitting diode the process runs in reverse, with electrons falling back across the gap and releasing the energy as light of a colour fixed by the size of the gap.

Classify each of the following as an intrinsic semiconductor, an n-type semiconductor or a p-type semiconductor, and name the majority charge carrier in each case:

(i) pure germanium (ii) silicon doped with indium (iii) germanium doped with arsenic

  1. Pure germanium. No dopant has been added, so the only carriers are electrons promoted across the small band gap by thermal energy, together with the vacancies they leave behind. It is an intrinsic semiconductor.

  2. Silicon doped with indium. Indium belongs to group 13 and has three valence electrons, one fewer than the four needed to bond with four silicon neighbours. Each indium atom therefore creates a hole in the valence band. It is a p-type semiconductor, and the majority carriers are holes.

  3. Germanium doped with arsenic. Arsenic belongs to group 15 and has five valence electrons, one more than needed for bonding with four germanium neighbours. The extra electron is easily promoted to the conduction band. It is an n-type semiconductor, and the majority carriers are electrons.

In all three cases the material as a whole is electrically neutral.

Germanium is doped with a trace of gallium. State (i) the type of extrinsic semiconductor formed, (ii) the majority charge carrier, and (iii) whether the doped material carries a net electrical charge. Justify each answer.

The conductivity of copper falls when it is heated, while that of pure silicon rises. Explain both observations using band theory.

The band gap of germanium is 0.67 eV. Calculate the longest wavelength of radiation that can promote an electron from its valence band to its conduction band.

(\(h\) = 6.626 × 10⁻³⁴ J s, \(c\) = 3.0 × 10⁸ m s⁻¹, 1 eV = 1.602 × 10⁻¹⁹ J)

1.10 Magnetic Properties of Solids

Magnetic properties of solids can be understood easily in terms of classical picture of electron. The electrons spin about their own axis. The spinning electrons act like tiny magnets because their spinning action generates induced magnetic field. Every electron in every substance is therefore a minute magnet; whether the substance as a whole behaves as one depends entirely on whether these tiny magnets cancel out.

If an orbital contains one electron, the unbalanced spin exhibits magnetism. However, when electrons are paired their spin is balanced and no magnetic property is observed. Two electrons in the same orbital spin in opposite senses, so their induced magnetic fields point in opposite directions and cancel each other exactly.

On the basis of magnetic properties solids are classified into three major classes : diamagnetic, paramagnetic and ferromagnetic. The behaviour of each in a magnetic field is drawn in Fig. 1.10a.

Diagram

Fig. 1.10a – Behaviour of the three classes in a magnetic field: (a) diamagnetic — all electron spins shown in cancelling pairs, sample shown being weakly pushed out of the field; (b) paramagnetic — individual magnetic moments drawn as small arrows pointing randomly with the field off, and loosely aligned with the field on, sample weakly drawn in; (c) ferromagnetic — arrows within domains, randomly oriented domains with the field off and all aligned with the field on, sample strongly drawn in and remaining aligned after the field is removed.

i. Diamagnetic Solids

The substances with all electrons paired, are weakly repelled by magnetic fields. These substances are said to be diamagnetic.

Pairing of electrons balances the spins and hence, cancels their magnetic moments. With no net magnetic moment of its own, such a substance does not merely ignore a magnetic field — it is pushed out of it weakly.

NoteDefinition

Substances in which all the electrons are paired, so that their magnetic moments cancel completely, are weakly repelled by a magnetic field and are called diamagnetic.

\(\mathrm{N_2}\), \(\mathrm{F_2}\), \(\mathrm{NaCl}\), \(\mathrm{H_2O}\) and benzene are some examples of diamagnetic substances.

ii. Paramagnetic Solids

The substances with unpaired electrons are weakly attracted by magnetic field. These substances are called paramagnetic substances.

The spinning of unpaired electron gives rise to a magnetic moment. The substance is attracted by magnetic field because of magnetic moment.

It is important to understand that these substances exhibit magnetism in presence of external magnetic field only. They lose magnetism when the external magnetic field is removed. The reason is that the individual magnetic moments are independent of one another: an applied field lines them up, but once it is withdrawn, thermal motion scatters their directions again at random.

NoteDefinition

Substances that contain unpaired electrons and are therefore weakly attracted by a magnetic field, but lose their magnetism as soon as the external field is removed, are called paramagnetic.

Oxygen, \(\mathrm{Cu^{2+}}\), \(\mathrm{Fe^{3+}}\), \(\mathrm{Cr^{3+}}\) are some examples of paramagnetic substances.

iii. Ferromagnetism

The substances containing large number of unpaired electrons are attracted strongly by magnetic field. These substances are said to be ferromagnetic.

These substances can be permanently magnetised. They retain magnetism even after the removal of external magnetic field. Here the magnetic moments of neighbouring atoms do not act independently; they influence one another and stay locked in the same direction once aligned, which is why the alignment survives the removal of the field.

NoteDefinition

Substances containing a large number of unpaired electrons whose magnetic moments align together, so that they are strongly attracted by a magnetic field and retain magnetism even after the field is removed, are called ferromagnetic.

Some example of ferromagnetic substances are Fe, Co, Ni, Gd, \(\mathrm{CrO_2}\).

Q. Both paramagnetic and ferromagnetic substances contain unpaired electrons. Why does only the ferromagnetic one stay magnetised after the field is switched off?

A. The difference lies in whether the tiny magnets act alone or together. In a paramagnetic substance each magnetic moment responds to the applied field by itself, and the moment its neighbour does nothing to hold it in place, so random thermal motion undoes the alignment as soon as the field is gone. In a ferromagnetic substance the moments of neighbouring atoms are strongly coupled and align as whole groups; once such a group has been turned into line by an external field, the coupling keeps it there, and the substance remains a permanent magnet.

Diamagnetic vs Paramagnetic vs Ferromagnetic

Feature Diamagnetic Paramagnetic Ferromagnetic
Electrons All paired Contains unpaired electrons Contains a large number of unpaired electrons
Net magnetic moment Zero, moments cancel Present Present, and moments align together
Behaviour in a magnetic field Weakly repelled Weakly attracted Strongly attracted
On removing the field No magnetism at any stage Magnetism is lost Magnetism is retained; can be permanently magnetised
Examples \(\mathrm{N_2}\), \(\mathrm{F_2}\), \(\mathrm{NaCl}\), \(\mathrm{H_2O}\), benzene Oxygen, \(\mathrm{Cu^{2+}}\), \(\mathrm{Fe^{3+}}\), \(\mathrm{Cr^{3+}}\) Fe, Co, Ni, Gd, \(\mathrm{CrO_2}\)

The very word magnet is thought to come from Magnesia, a district of ancient Greece where naturally magnetic lodestone was found, and ferromagnetism is built on the Latin ferrum, iron, the metal in which the effect is most familiar. The terms diamagnetic and paramagnetic are much more recent: they date from the 1840s, when Michael Faraday showed that substances usually thought of as non-magnetic are in fact pushed out of a strong magnetic field, and needed names for the two opposite kinds of weak response.

Magnetic classification is the basis of data storage. Chromium dioxide, \(\mathrm{CrO_2}\), is ferromagnetic and was widely used as the coating of high-quality recording tapes, information being stored as the direction in which small regions of the coating are magnetised; the same principle, with other ferromagnetic materials, records data on the magnetic disk of a computer. Paramagnetism has its uses too: because oxygen is one of the few common gases attracted by a magnetic field, medical and industrial oxygen analysers can measure the oxygen content of a gas mixture simply by measuring the force the mixture experiences in a magnetic field.

Predict whether each of the following is diamagnetic or paramagnetic, and state the number of unpaired electrons in each:

(i) \(\mathrm{Zn^{2+}}\) (ii) \(\mathrm{Cr^{3+}}\) (iii) \(\mathrm{Cu^{2+}}\)

The test in each case is simply whether any electron is left unpaired.

  1. \(\mathrm{Zn^{2+}}\) has the configuration [Ar]3d¹⁰. All five d orbitals are completely filled, so every electron is paired and the magnetic moments cancel. Number of unpaired electrons = 0. Hence \(\mathrm{Zn^{2+}}\) is diamagnetic.

  2. \(\mathrm{Cr^{3+}}\) has the configuration [Ar]3d³. The three d electrons occupy three separate orbitals with parallel spins. Number of unpaired electrons = 3. Hence \(\mathrm{Cr^{3+}}\) is paramagnetic.

  3. \(\mathrm{Cu^{2+}}\) has the configuration [Ar]3d⁹. Four of the d orbitals are filled and one holds a single electron. Number of unpaired electrons = 1. Hence \(\mathrm{Cu^{2+}}\) is paramagnetic.

Since \(\mathrm{Cr^{3+}}\) has more unpaired electrons than \(\mathrm{Cu^{2+}}\), it is the more strongly paramagnetic of the two — though both are only weakly attracted compared with a ferromagnetic substance such as iron.

Classify each of the following as diamagnetic or paramagnetic and give the number of unpaired electrons in each: \(\mathrm{Sc^{3+}}\) ([Ar]3d⁰), \(\mathrm{Ti^{3+}}\) ([Ar]3d¹), \(\mathrm{Ni^{2+}}\) ([Ar]3d⁸), \(\mathrm{Zn}\) ([Ar]3d¹⁰4s²).

Both oxygen and iron contain unpaired electrons, yet a bar of iron can be made into a permanent magnet while oxygen cannot. Explain the difference, and state how each behaves when an external magnetic field is applied and then removed.

1.11 Summary

  1. The solid state is characterised by strong interparticle forces of attraction. The particles vibrate about fixed mean positions, so a solid has a definite shape and a definite volume that change only slightly with temperature and pressure. The smallest constituent particles — atoms, ions or molecules — are referred to throughout as ‘particles’.

  2. Solids are of two types. In crystalline solids the particles show regularity and periodicity with long range order; these solids melt sharply at a definite temperature and are anisotropic, except those having cubic structure. In amorphous solids the particles are randomly arranged with short range order only; these are supercooled liquids, they soften gradually over a temperature interval, and they are isotropic.

  3. Examples of crystalline solids are ice, \(\mathrm{NaCl}\), sodium, gold, copper, diamond, graphite and ceramics; examples of amorphous solids are glass, plastic, rubber, tar and metallic glass. A single crystal is ordered throughout its bulk, while most crystalline solids, including metals, are polycrystalline.

  4. Two or more substances having the same crystal structure and the same atomic ratio are isomorphous, for example \(\mathrm{NaF}\) and \(\mathrm{MgO}\) (1:1), and \(\mathrm{NaNO_3}\) and \(\mathrm{CaCO_3}\) (1:1:3). A single substance existing in two or more crystalline forms is polymorphous, for example calcite and aragonite, or α-quartz, β-quartz and cristobalite. Polymorphism in an element is called allotropy — diamond, graphite and fullerene are allotropes of carbon.

  5. Crystalline solids are of four types. Ionic crystals contain ions held by electrostatic force; they are hard, brittle, high melting, non-conducting as solids but conducting when molten or dissolved. Covalent network crystals contain atoms linked by a continuous system of covalent bonds forming one giant molecule; they are very hard, incompressible, high melting and poor conductors. Molecular crystals contain molecules held by dipole-dipole forces, London forces or hydrogen bonds; they are soft, low melting and good insulators. Metallic crystals contain metal ions in a sea of delocalised valence electrons; they are malleable, ductile and good conductors of heat and electricity.

  6. A lattice is a periodic three dimensional array of points; the particle attached at each point is the basis. Thus Lattice + Basis = Crystal.

  7. The unit cell is the smallest repeating structural unit of a crystalline solid, described by the edge lengths a, b, c and the angles α, β, γ. Unit cells are of four types — primitive, body-centred, face-centred and base-centred.

  8. Only fourteen space lattices are geometrically possible. These 14 Bravais lattices are grouped into seven crystal systems: cubic, tetragonal, orthorhombic, rhombohedral, monoclinic, triclinic and hexagonal.

  9. The cubic system has three unit cells. A corner particle is shared by 8 cubes, a face particle by 2 cubes and an edge particle by 4 cubes. Hence a simple cubic cell contains 1 particle, a body-centred cubic cell 2 particles and a face-centred cubic cell 4 particles.

  10. Density, molar mass and edge length are connected by

\[\rho = \frac{n M}{a^3 N_A}\]

where n is the number of particles per unit cell, M the molar mass, a the edge length and \(N_A\) the Avogadro number. Knowing any four quantities, the fifth can be calculated. In x g of a metal there are \(\dfrac{xn}{\rho a^3}\) particles and \(\dfrac{x}{\rho a^3}\) unit cells, and a volume V of metal contains \(\dfrac{V}{a^3}\) unit cells.

  1. Particles are treated as hard spheres that pack closely, because closeness maximises interparticle attraction. The number of neighbouring spheres touching a given sphere is its coordination number.

  2. Close packing in one dimension gives a row; in two dimensions it gives either square close packing (AAAA type, coordination number 4) or hexagonal close packing (ABAB type, coordination number 6, more efficient). Stacking square layers gives the simple cubic structure, in which polonium alone among metals crystallises.

  3. Stacking hexagonal close packed layers gives two structures. Placing the third layer above the first gives ABAB… stacking, the hcp structure, found in Mg and Zn. Placing the third layer over the octahedral voids gives ABCABC… stacking, the ccp structure, which is the same as fcc, found in copper and Ag.

  4. Where a sphere of the upper layer rests in a depression, the space enclosed by four spheres is a tetrahedral void; where two oppositely pointing depressions overlap, the space enclosed by six spheres is an octahedral void. For N particles there are 2N tetrahedral voids and N octahedral voids.

  5. Packing efficiency is the percentage of the unit cell volume occupied by particles.

\[\text{Packing efficiency} = \frac{\text{volume occupied by particles in unit cell}}{\text{total volume of unit cell}} \times 100\]

For simple cubic, \(r = a/2\) and the efficiency is 52.36%; for bcc, \(r = \sqrt{3}a/4\) and the efficiency is 68%; for fcc/ccp/hcp, \(r = \sqrt{2}a/4\) and the efficiency is 74%. Coordination numbers are 6, 8 and 12 respectively — the higher the coordination number, the higher the packing efficiency.

  1. Real crystals contain defects. Perfect crystals are possible only at absolute zero, defects are fewer when crystallisation is slow, electrical neutrality is always maintained, and defects are sometimes created deliberately to obtain desired properties. Point defects fall into three classes.

  2. In stoichiometric point defects the formula is unchanged. A vacancy defect lowers the density; a self interstitial defect raises it when an extra particle enters a void, and leaves it unchanged when a lattice particle merely shifts into a void. A Schottky defect is the loss of equal numbers of cations and anions, which lowers the density and occurs when the ions are of similar size; it is found in \(\mathrm{NaCl}\), \(\mathrm{AgBr}\) and \(\mathrm{KCl}\). A Frenkel defect is the shift of an ion, usually the smaller cation, into an interstitial site, which leaves the density unchanged and occurs when cation and anion differ greatly in size; it is found in \(\mathrm{ZnS}\), \(\mathrm{AgCl}\), \(\mathrm{AgBr}\), \(\mathrm{AgI}\) and \(\mathrm{CaF_2}\).

  3. Impurity defects arise from foreign atoms. In a substitutional defect the foreign atom takes a lattice site, as zinc does in brass; adding an aliovalent impurity such as \(\mathrm{SrCl_2}\) to \(\mathrm{NaCl}\) creates cation vacancies, since every \(\mathrm{Sr^{2+}}\) ion replaces two \(\mathrm{Na^{+}}\) ions and occupies only one of the two sites. In an interstitial impurity defect the foreign atom occupies a void, as carbon does in steel.

  4. Nonstoichiometric defects change the ratio of the ions without changing the crystal structure. Metal deficiency occurs in metals of variable oxidation state, as in \(\mathrm{Ni_{0.97}O_{1.0}}\), where missing \(\mathrm{Ni^{2+}}\) ions are compensated by \(\mathrm{Ni^{3+}}\) ions. Metal excess occurs when a neutral atom or an extra cation with its electrons occupies interstitial sites, giving \(\mathrm{Zn_{1+x}O_{1.0}}\), or when electrons occupy anion vacancies as F-centres, giving colour to the crystal and the formula \(\mathrm{Na_{1+x}Cl_{1.0}}\) for yellow \(\mathrm{NaCl}\).

  5. By electrical conductivity, solids are conductors (10⁴ to 10⁷ Ohm⁻¹m⁻¹), semiconductors (10⁻⁶ to 10⁴ Ohm⁻¹m⁻¹) or insulators (10⁻²⁰ to 10⁻¹⁰ Ohm⁻¹m⁻¹).

  6. Band theory explains this range. Closely spaced energy levels merge into bands; the conduction band is the highest energy band containing electrons and its electrons are mobile, the valence band lies lower and its electrons are tightly bound, and the energy difference between them is the band gap. In metals the conduction band is partially filled or overlaps a vacant band, so conduction is easy, and conductivity falls on heating because vibrating cations obstruct the electrons. In insulators a large forbidden zone separates a filled valence band from an empty conduction band. In semiconductors the gap is small, so a few electrons cross it thermally and conductivity rises on heating.

  7. A pure semiconductor is an intrinsic semiconductor. Doping — adding a minute quantity of a dopant — gives an extrinsic semiconductor. Silicon or germanium doped with a group 15 element such as P, As, Sb or Bi has extra electrons in the conduction band and is n-type; doped with a group 13 element such as B, Ga or In it has holes in the valence band and is p-type. Both types are electrically neutral as a whole.

  8. By magnetic behaviour, solids are diamagnetic when all electrons are paired and are weakly repelled (\(\mathrm{N_2}\), \(\mathrm{F_2}\), \(\mathrm{NaCl}\), \(\mathrm{H_2O}\), benzene); paramagnetic when unpaired electrons are present, being weakly attracted but losing magnetism once the field is removed (oxygen, \(\mathrm{Cu^{2+}}\), \(\mathrm{Fe^{3+}}\), \(\mathrm{Cr^{3+}}\)); and ferromagnetic when a large number of unpaired electrons align together, giving strong attraction and permanent magnetisation (Fe, Co, Ni, Gd, \(\mathrm{CrO_2}\)).

Flowchart

FC 1.4 – The chapter at a glance: solid → crystalline or amorphous → crystalline classified as ionic, covalent network, molecular or metallic → structure described as lattice + basis → unit cell (primitive, body-centred, face-centred, base-centred) → 14 Bravais lattices in 7 crystal systems → cubic system: sc (1 particle, CN 6, 52.36%), bcc (2 particles, CN 8, 68%), fcc/ccp/hcp (4 particles, CN 12, 74%) → real crystals carry point defects: stoichiometric (vacancy, self interstitial, Schottky, Frenkel), impurity (substitutional, interstitial), nonstoichiometric (metal deficiency, metal excess/F-centre) → properties that follow: electrical (conductor, semiconductor, insulator; intrinsic and extrinsic n-type/p-type) and magnetic (diamagnetic, paramagnetic, ferromagnetic)

1.12 Objective Questions

  1. Molecular solids are
    1. crystalline solids
    2. amorphous solids
    3. ionic solids
    4. metallic solids
  2. Which of the follwong is n-type semiconductor?
    1. Pure Si
    2. Si doped with As
    3. Si doped with Ga
    4. Ge doped with In
  3. In Frenkel defect
    1. electrical neutrality of the substance is changed.
    2. density of the substance is changed.
    3. both cation and anion are missing
    4. overall electical neutrality is preserved.
  4. In crystal lattice formed by bcc unit cell the void volume is
    1. 68 %
    2. 74 %
    3. 32 %
    4. 26 %
  5. The coordination number of atoms in bcc crystal lattice is
    1. 2
    2. 4
    3. 6
    4. 8
  6. Which of the following is not correct?
    1. Four spheres are involved in the formation of tetrahedral void.
    2. The centres of spheres in octahedral voids are at the apices of a regular tetrahedron.
    3. If the number of atoms is N the number of octahedral voids is 2N.
    4. If the number of atoms is N/2, the number of tetrahedral voids is N.
  7. A compound forms hcp structure. Number of octahedral and tetrhedral voids in 0.5 mole of substance is respectively
    1. 3.011×10²³, 6.022×10²³
    2. 6.022×10²³, 3.011×10²³
    3. 4.011×10²³, 2.011×10²³
    4. 6.011×10²³, 12.022×10²³
  8. Pb has fcc structure with edge length of unit cell 495 pm. Radius of Pb atom is
    1. 205 pm
    2. 185 pm
    3. 260 pm
    4. 175 pm

1.13 Check Your Concepts

  1. A solid is found to be isotropic and to melt sharply at one definite temperature. A student concludes that it must be amorphous. Is the conclusion correct? Justify your answer.

  2. Assertion : Amorphous solids do not have a sharp melting point. Reason : The constituent particles of an amorphous solid occupy a wide range of different environments. State whether the assertion and reason are true, and whether the reason correctly explains the assertion.

  3. Diamond and graphite are both allotropes of carbon. Explain, in terms of bonding and structure, why graphite is used as a solid lubricant while diamond is used as an abrasive.

  4. Match the following: | Column A | Column B | |—|—| | a. \(\mathrm{CaF_2}\) | i. simple cubic structure | | b. brass | ii. substitutional impurity defect | | c. ice | iii. Frenkel defect | | d. polonium | iv. metal deficiency defect | | e. \(\mathrm{Ni_{0.97}O_{1.0}}\) | v. molecular solid held by hydrogen bonds |

  5. Both ionic and covalent network crystals are hard and have high melting points. Describe how their electrical behaviour on heating and on melting can be used to tell them apart.

  6. Explain why crystallisation carried out at a slow rate produces a crystal with fewer defects than crystallisation carried out rapidly.

  7. State, with reasons, the effect of each of the following on the density of the crystal: (i) a vacancy defect, (ii) a Schottky defect, (iii) a Frenkel defect, (iv) a self interstitial defect in which an extra particle enters a void.

  8. Why is the Frenkel defect not found in pure alkali metal halides, although it is found in silver halides?

  9. When a small amount of \(\mathrm{SrCl_2}\) is added to \(\mathrm{NaCl}\) during crystallisation, cation vacancies appear in the crystal. Explain why the vacancies are necessary, and state how many \(\mathrm{Na^{+}}\) ions are removed for every \(\mathrm{Sr^{2+}}\) ion introduced.

  10. Arrange the simple cubic, body-centred cubic and face-centred cubic lattices in increasing order of (i) coordination number and (ii) packing efficiency. State the relationship you observe between the two orders and explain it.

  11. When zinc oxide is heated it turns yellow, and on cooling it becomes white again. Explain this observation in terms of the defect produced, and write the nonstoichiometric formula of the heated solid.

  12. Explain what an F-centre is and why sodium chloride heated in sodium vapour appears yellow, although no coloured substance has been added to it.

  13. Assertion : A p-type semiconductor carries a net positive charge. Reason : Its majority charge carriers are holes, which behave as though they were positive. State whether the assertion and reason are true, and whether the reason correctly explains the assertion.

  14. The conductivity of a metal decreases on heating while that of an intrinsic semiconductor increases. Explain both observations using band theory.

  15. Magnesium has a completely filled 3s band, yet it is a good conductor of electricity. Explain how this is possible.

  16. Distinguish between isomorphism and polymorphism, giving one example of each. State which of the two is called allotropy and when.

  17. Oxygen and iron both contain unpaired electrons, but only iron can be made into a permanent magnet. Explain the difference in their magnetic behaviour, and classify each substance.

  18. A single granule of a metal is said to be polycrystalline. Explain what this means, and state how it differs from a single crystal.

  19. What are the types of particles in each of the four main classes of crystalline solids ?

  20. The following pictures show population of bands for materials having different electrical properties. Classify them as insulator, semiconductor or a metal.

Diagram

Fig. 1.E1 – Three band pictures side by side for question 20: (i) a filled valence block with a partly filled block immediately above it and no gap; (ii) a filled valence block with an empty block far above it, separated by a large gap; (iii) a filled valence block with an empty block a short distance above it, separated by a small gap.

  1. What is the unit cell?

  2. The picture represents bands of MOs for Si. Label valence band, conduction band and band gap.

Diagram

Fig. 1.E2 – Band picture for question 22: a shaded lower block and an unshaded upper block separated by a short gap marked with a double-headed arrow, all three features left unlabelled for the student to label.

  1. A solid is hard, brittle and electrically nonconductor. Its melt conducts electricity. What type of solid is it?

  2. Mention two properties that are common to both hcp and ccp lattices.

  3. Sketch a tetrahedral void.

  4. What are ferromagnetic substances?

  5. What are valence band and conduction band?

  6. Distinguish between ionic solids and molecular solids.

  7. Calculate the number of atoms in fcc unit cell.

  8. How are the spheres arranged in first layer of simple cubic close-packed structures? How are the successive layers of spheres placed above this layer?

  9. Calculate the packing efficiency of metal crystal that has simple cubic structure.

  10. What are paramagnetic substances? Give examples.

  11. What are the consequences of Schottky defect?

  12. Cesium chloride crystallizes in cubic unit cell with \(\mathrm{Cl^{-}}\) ions at the corners and a \(\mathrm{Cs^{+}}\) ion in the centre of the cube. How many CsCl molecules are there in the unit cell?

  13. Obtain the relationship between density of a substance and the edge length of unit cell.

  14. How are tetrahedral and octahedral voids formed?

  15. Third layer of spheres is added to second layer so as to form hcp or ccp structure. What is the difference between the addition of third layer to form these hexagonal close-packed structures?

  16. Distinguish with the help of diagrams metal conductors, insulators and semiconductors from each other.

  17. What are n-type semiconductors? Why is the conductivity of doped n-type semiconductor higher than that of pure semiconductor? Explain with diagram.

  18. Explain with diagram, Frenkel defect. What are the conditions for its formation? What is its effect on density and electrical neutrality of the crystal?

  19. What is an impurity defect? What are its types? Explain the formation of vacancies through aliovalent impurity with example.

1.14 Practice With Numericals

  1. Silver crystallizes in a face-centred cubic lattice with unit cell edge length 409 pm. Calculate the density of silver. (Molar mass of Ag = 108 g mol⁻¹, \(N_A\) = 6.022 × 10²³ mol⁻¹)

  2. An element crystallizes in a body-centred cubic lattice with unit cell edge length 288 pm and density 7.2 g cm⁻³. Calculate the molar mass of the element.

  3. Nickel crystallizes in a face-centred cubic lattice with unit cell edge length 352 pm. Calculate (a) the radius of a nickel atom and (b) the density of nickel. (Molar mass of Ni = 58.7 g mol⁻¹)

  4. A metal crystallizes in a face-centred cubic lattice and the radius of its atom is 144 pm. Calculate (a) the edge length of the unit cell and (b) the volume of the unit cell in cm³.

  5. A compound crystallizes in a cubic unit cell in which atoms of A are present at all the corners and one atom of B is present at the body centre. Deduce the formula of the compound.

  6. In a compound, atoms of element X form a face-centred cubic lattice and atoms of element Y occupy all the octahedral voids. Deduce the formula of the compound.

  7. A substance forms a cubic close packed structure. Calculate the number of (a) octahedral voids, (b) tetrahedral voids and (c) total voids present in 0.2 mol of the substance. (\(N_A\) = 6.022 × 10²³ mol⁻¹)

  8. \(\mathrm{NaCl}\) is doped with \(10^{-2}\) mol % of \(\mathrm{SrCl_2}\). Calculate the number of cation vacancies produced per mole of \(\mathrm{NaCl}\).

  9. A metal crystallizes in a body-centred cubic lattice with unit cell edge length 400 pm. Calculate (a) the radius of the metal atom and (b) the total volume occupied by the particles in one unit cell.

  10. Copper crystallizes in a face-centred cubic lattice with unit cell edge length 361 pm and density 8.9 g cm⁻³. Calculate the number of atoms and the number of unit cells present in 5.0 g of copper.

  11. A sample of iron oxide has the nonstoichiometric composition \(\mathrm{Fe_{0.93}O_{1.0}}\). Calculate the percentage of the iron present as \(\mathrm{Fe^{3+}}\) ions.

  12. The band gap of diamond is 5.47 eV. Calculate (a) the band gap in joules and (b) the longest wavelength of radiation that could promote an electron from its valence band to its conduction band. (\(h\) = 6.626 × 10⁻³⁴ J s, \(c\) = 3.0 × 10⁸ m s⁻¹, 1 eV = 1.602 × 10⁻¹⁹ J)

  13. A metal crystallizes in a simple cubic lattice with unit cell edge length 300 pm. Calculate (a) the radius of the metal atom, (b) the volume occupied by the particle in one unit cell and (c) the percentage of void space in the crystal.

  14. Cu crystallizes in fcc unit cell with edge length of 495 pm. What is the radius of Cu atom?

  15. The density of iridium is 22.4 g/cm³. The unit cell of iridium is fcc. Calculate the radius of iridium atom. Molar mass of iridium is 192.2 g/mol.

  16. Aluminium crystallizes in cubic close packed structure with unit cell edge length of 353.6 pm. What is the radius of Al atom? How many unit cells are there in 1.00 cm³ of Al?

  17. In an ionic crystalline solid atoms of element Y form hcp lattice. The atoms of element X occupy one third of tetrahedral voids. What is the formula of the compound ?

  18. An element with molar mass 27 g/mol forms cubic unit cell with edge length of 405 pm. If density of the element is 2.7 g/cm³, what is the nature of cubic unit cell?

  19. An element has a bcc structure with unit cell edge length of 288 pm. How many unit cells and number of atoms are present in 200 g of the element?