In this article we will discuss about the most common crystal structures in metals:- 1. Body Centred Cubic (BCC) Crystal Structure 2. Face Centred Cubic Crystal Structure 3. Hexagonal Close-Packed (HCP) Crystal Structure.
Metals are normally crystalline, or rather in solid state, they are almost synonymous to crystalline state. But their crystalline nature is seldom apparent (from the surface outlook) in the Final metallic products like table spoon, gears, window bars, etc., whereas the surface of the minerals like galena (PbS) does show external symmetry to visualise its crystal structure.
X-ray diffraction analysis of metals provides this information. The structure and properties of the individual crystals in a metal are responsible for its ultimate engineering usefulness as they strongly effect the entire processing and shaping of the metals.
The non-directional nature of the metallic bond results in highly symmetrical close-packed structures, i.e., each metal atom (ion) in a crystal tends to surround itself with as many neighbours as possible (to minimise potential energy), though all the metals do not behave in exactly this way.
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The most common crystal structures in common metals are:
1. Body Centred Cubic (BCC) crystal structure.
2. Face Centred Cubic (FCC) crystal structure.
3. Hexagonal Close Packed (HCP) crystal structure.
1. Body Centred Cubic (BCC) Crystal Structure:
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When one atom is placed at each lattice point of unit cell of the BCC space lattice. Fig. 1.26 (a) changes to Fig. 1.26 (b). This is the unit cell of BCC crystal structure, which is often shown as in Fig. 1.26 (c) with atoms drawn small sized for better clarity. It shows an atom at each corner of the cube as well as at its body-centre.
Number of Effective Atoms Per Unit Cell:
Fig. 1.26 (d) makes it clear that 1/8 of each corner atom lies inside this unit cell as each corner atom is at the apex of eight neighbouring unit cells. Thus, 1/8 of each corner atom is the effective share of this unit cell. As there are eight corners of the cube, and the atom at the body-centre belongs entirely to this unit cell, the total effective atoms per unit cell-
where, Nc is number of effective atoms due to corner atoms, and Nb due to body-centre.
It is defined as the number of nearest neighbours to a given atom in a crystal structure. In BCC crystal structure, each body centred atom touches all the eight corner atoms (and not more) as is clear from Fig. 1.27 (a).
In a crystal structure, the co-ordination number, Z is fixed. Let us see how is this co-ordination number equal to eight for a given corner atom, say X in Fig. 1.27 (b) for the hard-ball atom model. Fig. 1.26 (b) makes it amply clear that each corner atom is in contact with the body-centered atom, and not with any other corner atom (atom x does not touch Y, P or R atoms but only atom Z.)
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But because any corner atom is at the apex of eight neighbouring unit cells (see Fig. 1.27 (b), where atom X is at the apex), i.e., X is a part of eight neighbouring unit cells, which means, this corner atom touches all the eight body-centred atoms of these eight cells. Thus, Z, the coordination number is equal to eight.
In metal crystals, it is convenient to consider atoms as being spherical (hard ball model) with a radius, r equal to one half the inter-atomic distance, or distance of closest approach between atoms. In simple cubic system, this r is half the lattice parameter.
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In BCC crystal structure, r is one-fourth of body diagonal of its unit cell, Fig. 1.28 (a). X-ray diffraction measurements give inter-atomic distances with high degree of precision. Table 1.4 gives values of some of the important metals.
If r is the radius of the atom, and ‘a’ the lattice parameter of the unit cell of BCC crystal structure, then with the help of Fig. 1.28 (a) (in which one corner atom-shown by dotted line has been removed for better clarity), the base diagonal, AB = a√2 (as angle ∠AYB = 90°, and AY = YB = a) as also shown in Fig. 1.28 (b).
The body diagonal, AC = a√3, Fig. 1.28 (b) & (c). But it is also equal to 4 r as is clear from Fig. 1.28 (a) (the body diagonal, AC includes r, and r of two corner atoms, and 2r of the body-centre atom, thus, = 4 r).
Interestingly, the interatomic distances, 2 r, are 2%, or 3% smaller when the co-ordination number is eight (in BCC), than when it is equal to twelve (in FCC and HCP). For example, interatomic distance of BCC-iron is 2.4824 A°, but is 2.540 A° in FCC-iron.
When the co-ordination number is less, i.e., when number of nearest neighbouring atoms is less, closer approach of atoms become possible as there are less electronic repulsions due to less number of nearest ions.
Body centred cubic structure represents non-close packing of atoms (less densely packed). Fig. 1.29 (a) represents in two dimensions a square array of atoms not in contact. A second layer of the same pattern fits
into the valleys of atoms of the first layer, Fig. 1.29 (b).
Repetition of the first layer on top of the second layer results in BCC arrangement. Fig. 1.29 (c). The atoms are in contact only in the directions of the cube diagonals, i.e. only the body centred atom touches the rest neighbouring eight atoms, the latter do not with each other.
Atomic Packing Factor (APF):
It is the fraction of the volume occupied by the atoms in a crystal structure. Let us take a unit cell of BCC crystal as the base for calculation, i.e., let us find out in a BCC crystal unit cell, the volume of this unit cell, and the volume occupied by the atoms in this unit cell (based on hard-ball model)-
As there are two effective atoms per a BCC unit cell, and if r is the radius of the atom in BCC unit cell with ‘a’ as its lattice parameter, then,
Thus, 68% of the volume of BCC crystal is occupied by atoms, and the rest is almost void. It is confirmed that BCC crystal structure is an open structure as compared to FCC and HCP crystal structures, each having a better clock-packing of 74%.
Theoretical Density of an Element having BCC Crystal Structure:
Let us calculate the density of BCC-iron (at 20°C) if the radius of a BCC-iron atom is 1.241 A (mass of an atom of iron is 55.85 atomic mass unit (amu), and 1 amu = 1.66 x 10-27 kg)
As density, ρ = mass/volume
take a unit cell of BCC-iron, and then try to find out the mass of that unit cell, or the mass of the effective atoms present in that unit cell, and also find out the volume of that unit cell as radius of iron atom is given as 1.241 A°, thus, from equation 1.3,
Problem:
Calculate the APF for simple cubic structure.
Solution:
A simple cubic structure has one atom per unit cell, and also, a = 2 r, thus,
Geometry of Close-Packed Crystal Structures in Metals:
As the metallic bond is non-directional, many metals tend to form highly symmetrical close-packed structures, which result in their relatively high densities. The most common and important are face-centred cubic (FCC) and hexagonal close-packed (HCP) structures. To get a clear picture of arrangements of atoms in these two crystal structures, it is necessary to examine the geometry of possible close-packing of atoms.
It is convenient to take atoms as hard, incompressible spheres-as hard balls (This assumption is truly a gross approximation, because the electron-probability distribution around a nucleus of an atom never becomes exactly zero even at large distances from the centre of nucleus, which are being assumed to be bound within the radius of the sphere). Such an assumption enables us to easily imagine the arrangement of atoms in the crystal.
Let us take a large number of such spheres and place them on a plane surface to get a close packed plane of them. The maximum possible close-packing is obtained when this plane is packed as illustrated in Fig. 1.30, where (a) illustrates the front view, (b) the top view of the plane. The Fig. 1.30 (c) and (d) do not represent closest possible packing as there are much larger gaps in between spheres.
Let us call the closest-packing of Fig. 1.30 (b) as the first layer, or ‘A’ layer. In building a three-dimensional close-packed crystal structure, another similar close-packed plane of spheres (call it second layer, or ‘B’ layer) is to be put on the top of the plane ‘A’ of spheres. Fig. 1.31 (a) illustrates that a sphere of plane ‘B’ shall have maximum close-packing when it sits in the valley space produced by putting three spheres of plane ‘A’.
Fig. 1.31 (b) illustrates a part of Fig. 1.30 (b) taken for convenience. This Fig. shows clearly two sets of valleys, one of which is called as B, and the other as C, but at a time only one type of them can be covered by the spheres of the second close-packed plane. If the spheres of second plane cover the B type of valleys of Fig. 1.31 (b), the two stacked layers of close-packed planes would look like as in Fig. 1.31 (c).
Here, spheres, or atoms of the first layer A are shown full circles, while of the second plane with dotted circles. Had they covered the ‘C’ positions, the picture would look like as in Fig. 1.31 (d), which in three dimensions would look like Fig. 1.31 (e). The crystal structure of the two layers would look just the same in either case, i.e., 1.31 (c) or 1.31 (d).
If the spheres of the second layer have covered ‘B’ valleys, then look at Fig. 1.31 (c), the placing of third layer introduces a slight complication because there are two different ways in which it can be stacked above the second layer.
One way is to put the centres of the spheres of the third layer exactly vertically above ‘C’ positions, Fig. 1.31 (c) and have the close-packing. The placing of the atoms of this third plane is neither exactly above ‘A’ position, nor ‘B’ positions of the Fig. 1.31 (c). To describe the stacking of this structure, the first layer be classed ‘A’, the second layer as ‘B’ and the third as ‘C’.
Thus, the stacking sequence of close-packed layers is then ABC. As all the possible alternative positions for close- packed layers have been used up in this stack of three layers of spheres, the fourth layer can be added in the ‘A’ position again. Thus proceeding, this sequence becomes ABCABC… to build up to any desired thickness as illustrated in Fig. 1.32 (a).
It is a stacking sequence of atoms in FCC crystal structure. Fig. 1.32 (b) also illustrates FCC crystal structure just like the one in Fig. 1.23 (a) for copper (FCC) except that one corner has been ‘cut’ off to show a close packed plane (shown as a triangle). These close packed planes stacked one above another are at an angle to the cube faces. Thus, when close packed planes are stacked by ABCABC… stacking at an angle, face centred cubic crystal structure is created.
There is an alternative way in which the close packed layers of Fig. 1.31 (c) can be stacked to make a close-packed structure. The third layer of spheres is now added so that it is directly over the first ‘A’ positions. This structure too is close packed, but the stacking sequence is ABABA…, as illustrated in Fig. 1.33 (a).
Hexagonal close-packed crystal structure has stacking sequence of ABABA… It we put the third plane of atoms in Fig. 1.31 (d), the result shall be similar because the stacking sequence from above two alternatives shall be ACBACB… and ACAC… respectively.
Thus, there can be other methods of stacking close-packed planes of spheres, the only restriction being that no two adjacent close-packed planes can have same symbol (i.e., two atoms of two planes cannot be just vertically above each other). A stacking ABBA… is not permissible. The packing efficiency of all close packed stackings is 0.74 and has a coordination number of 12.
The Fig. 1.33 (a) illustrates an exploded view, when the three close-packed planes are coming closer for ABAB… stacking. The middle plane is also a continuous plane but only the three full drawn spheres shall form part of the unit cell of HCP crystal structure, the assembled view of which is illustrated in Fig. 1.33 (b).
It is clear that the spheres of the top third plane are exactly vertically over the spheres of the first layer (bottom layer), and is thus designated again ‘A’. The three middle plane spheres fit into the valleys between the top and the bottom planes.
2. Face Centred Cubic Crystal Structure:
The unit cell of the FCC space lattice is illustrated in Fig. 1.34 (a), whereas the Fig. 1.34 (b) illustrates unit cell of its crystal structure (basis is one atom at each lattice point). It shows that there are eight atoms at the eight corners of the cube, and six atoms, one each at the centres of the six faces of the cube. Fig. 1.34 (c) illustrates a unit cell of FCC as commonly drawn (atoms, or ions are drawn small sized) for better clarity.
Effective Number of Atoms per Unit Cell:
The Fig. 1.35. illustrates a schematic representation of a unit cell after the eight neighbouring unit cells at each corner have shared the corner atoms and also after each face centred atom has been shared by two neighbouring unit cells. Each corner atom contributes only 1/8 of itself to a unit cell, whereas each face-centred atom (there are six such atoms) contributes only 1/2 of itself to a unit cell.
Thus, the total effective atoms per unit cell in FCC crystal structure
Co-Ordination Number, Z:
In close-packed structure like FCC, close packed planes are stacked over one another. Fig. 1.31 (b) clearly indicates that any one atom in a close packed plane is in contact with six other atoms of its own plane.
This atom sits (or is in contact with) on the top of three atoms of the lower plane, Fig. 1.31 (a), and is also in contact with three atoms of upper close packed plane of atoms. Thus, the number of atoms in contact with one particular atom, or the co-ordination number of FCC (as well as the other close packed structure, HCP) crystal structure,
= Z = 6 + 3 + 3= 12
If we look at the FCC unit cell, Fig. 1.36, atom ‘A’, the face centred one of the left side face is clearly seen to be in contact with total twelve atoms (four corner atoms of its face, four face-centred atoms on right side of it and four face-centred atoms on left of it). Thus, co-ordination number of FCC crystal structure is 12 and is fixed for this structure.
Based on the definition that the radius of an atom is equal to one half the closest distances between atoms, Fig. 1.37 illustrates this in a partly drawn unit cell of FCC crystal structure. If ‘a’ is the lattice parameter and r is the radius of an atom, then Fig. 1.37 (b) illustrates that,
Table 1.4 illustrates the distance of closest approach between atoms of some FCC metals.
Atomic Packing Factor of FCC Crystal:
Based on hard-ball model, we shall take a unit cell of FCC and then find out the effective volume occupied by the atoms in a volume of that unit cell.
As, APF = Volume of effective atoms per unit cell / Volume of unit cell of FCC
As in FCC crystal structure, there are four effective atoms per unit cell and with ‘r’ as radius of an atom, the total effective volume of atoms per unit cell is = 4 x 4/3 π r3 and also in FCC crystal structure-
Thus, atoms occupy 74% of the volume of the crystal in FCC, and the rest is almost void. This packing factor is higher than in BCC crystals (68%). FCC (as well as HCP) has the best close packing of atoms. It is impossible to achieve 100% filling of space with same sized spheres. Thus, 74% value for FCC (also for HCP) crystal structure represents the theoretical maximum density of packing for same sized atoms.
Problem:
Calculate the density (theoretical) of copper, a FCC metal if lattice parameter, a = 3.61 A, and the atomic weight of copper is 63.5.
Solution:
As, density = mass / volume
Let us take a unit cell of FCC copper crystal, and try to find out the mass of the unit cell, i.e., mass of the effective atoms present in the unit cell, and also the volume of the unit cell.
There are 4 atoms / cell in FCC. As atomic weight of copper is 63.5 gram weight, and there are 6.02 x 1023 atoms of copper (Avagadro’s number) in this 63.5 gram weight. Thus,
3. Hexagonal Close-Packed (HCP) Crystal Structure:
One of the Bravais lattice is called simple hexagonal space lattice (it is not hexagonal close-packed). The primitive cell as well as the unit cell of simple hexagonal space lattice is a parallelopiped as illustrated in Fig. 1.38 (a), also shown by thick lines in Fig. 1.38 (b). It has one point per unit cell.
Fig. 1.38 (b) illustrates how a unit cell with hexagonal symmetry can be generated within a set of four unit cells of simple hexagonal space lattice. This hexagonal prism is another way of representing simple hexagonal Bravais space lattice.
Let us try to create a simple hexagonal crystal structure by placing one atom at each lattice point of the hexagonal prism space lattice. Then looking at Fig. 1.38 (b) with atoms placed at lattice sites, the atom X sits vertically on top of atom Y of the bottom plane, i.e., we are placing atoms of upper (second) close- packed plane vertically on top of atoms of bottom close packed plane.
It is an unstable state. Fig. 1.38 (d) illustrates this point. Here atoms of the front row of the planes have been shown. The atoms of the second (upper) plane are not stable at these points. If these atoms sit in the valleys of the bottom plane atoms as illustrated in Fig. 1.38 (e), these are more close-packed and stable. The forces of attraction of atoms of bottom plane pull them to valleys.
In this process, this close-packed plane of atoms becomes the intermediate plane of atoms B of hexagonal close- packed crystal structure. The third plane of atoms can be then put in valleys of B plane of atoms, but fall vertically above atoms of plane one (or A). This is illustrated by hexagonal close-packed crystal structure in Fig. 1.38 (c), where three atoms of the intermediate plane are shown placed at mid-point of C-axis. Fig. 1.33 (a) also illustrates it.
Thus, stability makes the simple hexagonal crystal structure to become close packed hexagonal crystal structure. This is what happens in actual metals like magnesium, where magnesium atoms not only occupy the lattice points of the simple hexagonal space lattice, but also the intermediate points.
Hexagonal close packed crystal structure, thus, has a basis of two identical atoms labelled A and B at each point of the simple hexagonal space lattice, whether the parallelopiped one, or the hexagonal prism like. A pair of atoms A and B, connected by dashed line in Fig. 1.36 (c) can be associated to a lattice point of the space lattice.
Problem:
Can the Hexagonal Close Packed Unit Cell be called a Space Lattice?
Solution:
No, a space lattice requires every point to be identical, i.e., when one traverses a distance from A atom to B atom in Fig. 1.39, it should take us to another similar atom by traversing the same distance ahead of B, but it takes us to X, where there is no atom.
Thus, it fails the test as a space lattice. But with a basis of a pair of atoms, it is a unit cell of hexagonal close-packed crystal structure. Fig. 1.40 (a) shows unit cell of simple hexagonal space lattice which changes to (b) unit cell of HCP crystal structure. Fig. 1.41 (a) illustrates parallelopiped unit cell of space lattice of simple hexagonal, which changes to unit cell of closed packed hexagonal by putting the basis of two atoms (parallelopiped type).
Number of Effective Atoms per Unit HCP Cell:
There are six atoms, one each at the six corners of each hexagonal face (there are two such in a unit cell) and one atom each at the two face centres (top and bottom) as illustrated in Fig. 1.40 (b). It is clear here that only three atoms of the middle layer (which is also a continuous close-packed plane) are part of this unit cell.
Now, each corner atoms of the hexagonal prism is shared by six prisms, and thus 1/6 of each corner atom is the effective part of this unit cell, Fig. 1.40 (c), and as there are 12 corner atoms, the effective contribution of the corner atoms to this unit cell is 12 x 1/6 = 2 atoms.
Each face-centred atom is shared by two hexagonal prisms, and there are two such atoms. Thus, the effective atoms due to the face-centred atoms to this unit cell is 2 x 1/2 = 1 atom. Looking carefully at the Fig. 1.40 (b) and (c), it is correct to assume that three atoms of the middle layer effectively form part of this unit cell.
Thus, total effective atoms per unit HCP cell = 2 + 1 + 3 = 6
Co-Ordination Number, Z:
While discussing the close packing of atoms, it had become clear that an atom in its close packed plane is in contact with six atoms of its plane. It is in direct contact with three atoms of the plane above it and three atoms of the plane below it as this atom sits in the valley of three atoms of these planes. Thus, the coordination number of the hexagonal close packed crystal structure-
= 6 + 3 + 3
Z = 12
which is equal to that of FCC crystal structure.
Atomic Radius:
It ‘a’ is the lattice parameter, which is also equal to the side of the hexagon, and is also equal to the distance between two nearest neighbours, then, Fig. 1.40 (b).
r = a / 2
Ideal c/a Ratio:
The nearest neighbour distance in HCP cell is ‘a’. The height of the unit cell ‘c’ is not necessarily equal to ‘a’. And, thus, two lattice parameters ‘a’ and ‘c’ are needed to specify an HCP unit cell. The c/a ratio is determined experimentally and is found to differ for different elements.
Because of the close-packing of planes in the HCP crystal structure, the lattice parameter ‘c’ is not independent of ‘a’ in the ideal case. The value of c/a, calculated on the basis of close-packed planes arranged in ABABA… sequence is called its ideal c/a ratio. The c/a ratio is also called axial ratio.
The ABA… packing sequence of HCP is illustrated in Fig. 1.42 (a), which has three mid-plane atom and one each from upper and lower planes, which fit in valleys of the three mid-plane atoms. The centres of three mid-plane atoms are joined together and are joined to centres of atoms of upper and lower planes.
The top and the bottom atoms are centred at two lattice points, one above the other on the two hexagonal basal planes of the unit cell. Thus, the distance between them, DD’ is the unit distance along the c-axis, i.e.,
DD’ = c. As ‘a’ the lattice parameter is the distance between centres of two neighbouring atoms, thus, each side of the tetrahedron is equal to ‘a’ here.
DD’ = c = 2 DF, where DF is perpendicular from top apex D to the base PQH. Thus,
The c/a ratio of some of the common HCP metals is given in Table 1.5. As calculated above, c/a ratio for truly spherical atoms comes out as 1.633. This ideal ratio does not occur in any HCP metal. In Mg, the c/a ratio is close to ideal (1.624), while in Zn and Cd, it is higher than ideal ratio, whereas, Be, Ti, Zr have smaller than ideal. The smaller c/a ratio means ‘c’, the distance between basal planes is smaller, and atomic diameter is smaller than ‘a’ (if c/a is greater, then d = a), i.e., atoms are compressed slightly in the direction along c-axis.
Atoms should be considered spheroidal in shape and not spherical. The smaller separation between basal planes, probably is the cause of increased critical resolved shear stress in these metals, i.e. deformation becomes difficult in them.
A parallelopiped unit cell of HCP has eight corner atoms and one atom inside the cell. Thus, there are two effective atoms per unit cell. In the ideal case,
c/a = 1.633
Some Other Important Crystal Structures:
The three simple crystal structures i.e., BCC, FCC and HCP are very commonly found in the ‘true metals’ (with one, two, or occasionally three valence electrons). Some metals have much more complicated structures. For example, one form of manganese has crystal structure like BCC, but the basis is 29 atoms, i.e., there are 29 atoms at each cube corner and the body centre.
The less metallic elements (and non-metals) generally form more complicated crystal structures. This is because the metallic (free electron) bond is partly replaced by the covalent (electron pair) bond, i.e. atoms are bound to some neighbours by covalent bonds and others by metallic bonds.
The atoms are then bonded covalently into chains, or sheets, and these are then held by metallic and van der Waals bonds. The coordination number can be found by the 8-N rule, where, N is the number of the group to which the element belongs in the periodic table.
Arsenic, antimony and bismuth form layer structures, where in each layer an atom has three close neighbours, in agreement with the chemical valency of the elements, or applying the 8-N rule to these elements of group V B, have a coordination number of 3.
Some elements of high valency, however, do behave as true metals. The metal Pb has FCC crystal structure, but has four valence electrons per atom. This is explained on the basis of incomplete ionization, i.e., some of the electrons are not released to form the free electron cloud, or used for bonding but are held back within the parent atom.
Carbon occurs in two crystalline forms- Diamond and graphite. Each atom of carbon in diamond forms four covalent bonds, which results in a three-dimensional network. Though diamond exists in two crystal forms, Cubic and hexagonal, and both have regular tetrahedrons with an interbond angle of 109.5°.
The better known crystal called Diamond cubic (DC) structure is illustrated in Fig. 1.4. The space lattice is FCC with two atoms per lattice point as the basis (one at the corner and other at a point quarter away along the body diagonal). If ‘a’ is the lattice parameter, then distance of separation between the two atoms of the basis is a √3 /4 = 1.54A°, and is also equal to diameter of the carbon atom in the DC structure.
The coordination number is four, which is low, because of covalent bonding which causes inefficient packing of carbon atoms in the crystal. Diamond being the hardest naturally occurring element, is used in wire drawing dies, and also as an abrasive in polishing and grinding.
Fig. 1.43. Crystal structure of graphite.
Graphite forms three covalent bonds between carbon atoms to give a sheet like structure with interbond angle of 120°, as illustrated in Fig. 1.43. These sheets are held together by van der Waals bonds. The fourth electron resonates between the bonds, and its mobility is the cause of more than 100 times increase in electrical and thermal conductivity in a direction parallel to the sheets as compared to in the perpendicular direction.
This weak intersheet bond makes the graphite to be soft, and acts as a solid lubricant due to the movement of sheets over each other. ‘Lead’ pencils are graded from 4B to 4 H (very soft to very hard) depending on varying proportions of graphite to clay.
Graphite is used for making electrical arc electrodes due to its high electrical conductivity and high sublimation temperature, which make it suitable as an electrical resistance heating element. ‘Carbon fibre’ materials have very high strength.
In the crystalline form of graphite, as illustrated in Fig. 1.43, the carbon atoms in each sheet of graphite have very orderly arrangement and the sheets are perfectly oriented with respect to each other. If the sheets are rotated randomly with respect to each other, it becomes non-crystalline, and is then called ‘amorphous’ carbon.
In between these two extreme structures, there are structures of carbon black, lamp black, etc. Thus, all graphite except the crystalline, are non-crystalline, but each has a different degree or type of order. The same carbon atoms when arranged in different crystalline order give diamond, the hardest, and graphite so soft, and still further the lamp-black.