Diamond and Graphite: Lattice, Motif and Structure

1Why This Topic Matters for GATE

Under Section XE2.1 (Classification and Structure of Materials), the GATE 2027 XE2 syllabus lists the crystalline forms of carbon and asks for "Diamond and Graphite in terms of lattice and motif", followed by graphene, fullerenes and carbon nanotubes. That wording tells you what a complete answer looks like. Saying that diamond has "the diamond structure" or that graphite is "layered" is not enough. You are expected to name the underlying lattice, state the motif and where its atoms sit, and then use that description to count atoms, find neighbours, compute a packing factor or a density, and explain properties.

Three things make this lesson worth mastering carefully. First, diamond is the standard case in which the shortcut "FCC means four atoms per cell, coordination number 12 and packing factor 0.74" fails, and questions can exploit that shortcut. Second, the lattice-plus-motif habit you build here transfers to the rest of the structure syllabus: zinc blende is diamond with two kinds of atom, wurtzite and HCP metals need the same hexagonal-lattice reasoning that graphite needs, and NaCl, CsCl, perovskite and spinel are all lattice-plus-motif problems. Third, carbon gives the cleanest illustration of the structure–property theme that runs through the syllabus: the same element is a transparent electrical insulator of extreme hardness in one structure and an opaque, in-plane electrical conductor that shears easily in the other.

The topic also touches parts of the syllabus outside XE2.1. Structure factor and the indexing of cubic diffraction patterns (XE2.4) decide which X-ray reflections diamond shows. Unary pressure–temperature diagrams (phase diagrams of a single component, XE2.2) explain why graphite is the stable form of carbon at ambient pressure. Atomic bonding, band theory and thermal conduction by phonons (XE2.3) explain the contrast in properties. This lesson therefore includes the short pieces of diffraction and thermodynamics you need, and no more.

Expect counting questions (atoms per cell, coordination number), geometry questions (bond length, radius, packing factor), density and lattice-parameter numericals, X-ray line-position and selection-rule questions, and conceptual comparisons between diamond and graphite. No claim is made here about how many marks any of these carried in past papers. Section 10 gives practice questions in all three GATE formats (MCQ, MSQ and NAT). Every one is labelled a GATE-style practice question, because no previous-year question was reproduced whose year and wording could be verified.

2Prerequisite Concepts, Explained Briefly

This section gives the minimum background needed to follow the rest of the lesson. If you are comfortable with lattices, unit cells and hybridisation, skim it; if not, read it slowly, because every later section leans on it.

Crystal, lattice and motif. A crystal is a solid in which the arrangement of atoms repeats regularly in three dimensions. To describe that repetition we separate two ideas. The lattice is a mathematical array of points, every one with identical surroundings, generated by three translation vectors a, b, c: each lattice point sits at ua + vb + wc, where u, v, w are integers. The motif (also called the basis) is the group of atoms attached in exactly the same way to every lattice point. Together they give the central statement of crystallography: crystal = lattice + motif. Think of wallpaper. The grid on which the pattern repeats is the lattice, and the design printed at each grid point is the motif. A lattice point is only a location. It may carry one atom, several atoms, or (as you will see in a honeycomb) no atom at all.

Unit cells and Bravais lattices. A unit cell is a small box that fills space when repeated by the lattice translations. A primitive cell contains exactly one lattice point. A conventional cell is chosen to display the symmetry of the lattice and may contain more than one lattice point; the conventional face-centred cubic (FCC) cell, with lattice points at the corners and the centres of the six faces, contains four. The edge lengths of the cell are the lattice parameters: a single length a for a cube, and two lengths, a (in the layer) and c (along the six-fold axis), for the hexagonal cell, with 120° between the two a axes. There are 14 Bravais lattices; this lesson uses two, the face-centred cubic lattice and the simple hexagonal lattice.

Positions and counting. We describe where an atom sits by its fractional coordinates (x, y, z), the position written as fractions of the cell edges. The point (½, ½, 0) is the centre of the bottom face of a cube, and (¼, ¼, ¼) lies a quarter of the way along the body diagonal. To count the atoms that belong to one cell, weight each atom by the fraction of it lying inside the cell: an atom inside counts 1, on a face ½ (shared by two cells), on an edge ¼ (shared by four) and at a corner ⅛ (shared by eight).

Coordination and packing. The coordination number (CN) is the number of nearest neighbours of an atom, and the nearest-neighbour distance d is the distance to them. In the hard-sphere model, atoms are rigid spheres of radius r that touch their nearest neighbours, so d = 2r. The atomic packing factor (APF) is the fraction of the cell volume occupied by these spheres: the number of atoms per cell times the volume of one sphere, divided by the volume of the cell. For simple cubic, BCC and FCC structures the APF values are 0.52, 0.68 and 0.74. The density follows directly from the structure: ρ = nM/(V_cN_A), where n is the number of atoms per cell, M the molar mass, V_c the cell volume and N_A Avogadro's number. Keep the units straight: 1 Å = 10⁻⁸ cm, so 1 ų = 10⁻²⁴ cm³.

Carbon bonding. A carbon atom has four valence electrons (2s² 2p²). In a covalent bond two atoms share a pair of electrons, and the number and direction of bonds are set by hybridisation, the mixing of atomic orbitals. When the 2s orbital mixes with all three 2p orbitals we get four equivalent sp³ orbitals pointing to the corners of a regular tetrahedron, 109.47° apart. When the 2s orbital mixes with only two of the 2p orbitals we get three sp² orbitals in a plane, 120° apart, and one unhybridised 2p orbital perpendicular to that plane. Head-on overlap of orbitals gives a strong σ bond; sideways overlap of parallel p orbitals gives a π bond, whose electrons can spread over many atoms. Finally, van der Waals (dispersion) forces are the weak attractions between neutral atoms or molecules caused by fluctuating electron clouds. Different structural forms of the same element are called allotropes; diamond and graphite are allotropes of carbon.

3Learning Outcomes

After working through this lesson you should be able to do the following.

1. State crystal = lattice + motif and apply it to diamond and graphite, saying clearly which points are lattice points and which atoms belong to the motif.
2. Write the fractional coordinates of all eight atoms in the diamond conventional cell and explain how they arise from an FCC lattice with the motif (0,0,0) + (¼,¼,¼).
3. Determine the atoms per conventional cell (8), atoms per primitive cell (2), coordination number (4), nearest-neighbour distance (√3 a/4), atomic radius, packing factor (π√3/16 ≈ 0.34) and density for any diamond-cubic crystal (C, Si, Ge, grey tin).
4. Describe graphite as a hexagonal lattice with a four-atom motif and AB stacking, and use a ≈ 2.46 Å and c ≈ 6.71 Å to obtain the in-plane bond length, the layer spacing, the density and the areal atom density.
5. Compare sp³ diamond with sp² graphite (bonding, bond length, dimensionality, van der Waals interlayer forces) and connect the difference to hardness, electrical conduction, thermal conduction and optical behaviour.
6. Predict the allowed and forbidden X-ray reflections of a diamond-cubic crystal, identify the structure from sin²θ ratios, and locate the graphite (002) reflection.
7. Explain why graphite is the stable form of carbon at ambient pressure, why diamond nevertheless persists, and estimate the pressure at which diamond becomes stable at room temperature.
8. Solve MCQ, MSQ and NAT problems on all of the above, avoiding the common traps listed in Section 9.

4Complete Concept Explanation

4.1 Carbon: one element, two bonding patterns

Carbon can satisfy its four valence electrons in two ways that matter for this lesson. If every carbon atom forms four σ bonds to four neighbours using sp³ orbitals, the only way to keep all four bonds equivalent is a three-dimensional network in which each atom sits at the centre of a regular tetrahedron of neighbours. That network is diamond. If instead every carbon atom forms three σ bonds in one plane using sp² orbitals, one electron per atom is left over in a p orbital pointing out of the plane. The result is a flat sheet of hexagons held together by σ bonds, with the leftover electrons shared across the whole sheet. Stack such sheets and you have graphite. Everything else in this lesson (lattice, motif, coordination number, density, properties) follows from these two bonding choices.

4.2 Lattice and motif made visible: a two-dimensional warm-up

Before building diamond in three dimensions, look at a single graphite layer (called graphene when it is isolated). It shows in a flat picture why a lattice and a motif are different things. In the honeycomb net of Figure 1c every atom has three neighbours at 120°. Is the honeycomb itself a lattice? No. Pick an atom of type A (blue) and note the directions in which its three neighbours lie: one straight up and two pointing down and outwards, like the letter Y. Now pick an atom of type B (orange): its three neighbours point in exactly the opposite directions, like an upside-down Y. No translation can carry an A atom onto a B atom and preserve its surroundings, because the two are related by a 180° rotation, not by a translation. So the atoms of a honeycomb are not all lattice-equivalent.

What repeats by translation is the pattern. If we place lattice points on the A atoms, those points form a two-dimensional hexagonal lattice (Figure 1a): two translation vectors of equal length a at 120° to each other. To rebuild the crystal we attach the same two-atom motif to every lattice point (Figure 1b): atom A at (0, 0) and atom B at (1/3, 2/3), in fractional coordinates of the two-dimensional cell. In one line, crystal = lattice + motif. Notice also that you could have placed the lattice points at the centres of the hexagons, where there is no atom at all. That is perfectly legitimate, because lattice points are labels we assign, not atoms. This is the distinction between atoms and lattice points that the diamond structure exposes in three dimensions.

4.3 Building diamond cubic: an FCC lattice with a two-atom motif

Now to three dimensions. Start with the face-centred cubic lattice: lattice points at (0,0,0), (½,½,0), (½,0,½) and (0,½,½) in the conventional cube, repeated by all lattice translations. Attach to every lattice point the same motif of two carbon atoms: one at the lattice point itself, and one displaced from it by (¼,¼,¼), a quarter of the way along the body diagonal of the cube. Applied to the four lattice points of one cell, this gives eight atoms.

The corner atom at (0,0,0) is repeated by translation at the other seven corners, and the face-centre atom at (½,½,0) reappears on the opposite face at (½,½,1). That is why four positions in the first row account for every corner and face-centre atom in Figure 2.

There are three equivalent ways to picture this one structure, and you should be comfortable switching between them. The first is the description above, an FCC lattice with a two-atom motif. The second is two identical interpenetrating FCC lattices, the second displaced from the first by (¼,¼,¼). The third is an FCC arrangement of atoms in which half of the tetrahedral voids (the small gaps at the centre of four neighbouring atoms whose centres form a tetrahedron) are also occupied. An FCC cell has eight tetrahedral voids, at (¼,¼,¼), (¾,¾,¼), (¾,¼,¾), (¼,¾,¾) and at (¾,¾,¾), (¼,¼,¾), (¼,¾,¼), (¾,¼,¼); diamond fills the first four and leaves the other four empty. Filling all eight with a second kind of atom gives the fluorite arrangement (the structure of CaF₂), and filling the same four with a different element gives zinc blende (Section 13).

Why must the motif contain two atoms? Look at the neighbours. The four nearest neighbours of the atom at the origin lie at displacements (¼,¼,¼), (¼,−¼,−¼), (−¼,¼,−¼) and (−¼,−¼,¼): the vectors of the form (±¼,±¼,±¼) with an even number of minus signs. The four nearest neighbours of the atom at (¼,¼,¼) lie at (−¼,−¼,−¼), (−¼,¼,¼), (¼,−¼,¼) and (¼,¼,−¼): the vectors with an odd number of minus signs. The two tetrahedra point in opposite directions, so no translation turns one atom's surroundings into the other's. The two atoms are not lattice-equivalent.

4.4 Counting the atoms in the conventional cell

Apply the counting rule from Section 2 to the eight atoms of Figure 2.

The same answer follows from the lattice-plus-motif description: the conventional FCC cell has 4 lattice points, each carrying a two-atom motif, so 4 × 2 = 8. A primitive FCC cell contains one lattice point, so a primitive cell of diamond contains 2 atoms and has volume a³/4. When a question asks for the atoms per conventional unit cell of diamond, silicon or germanium, the answer is 8; when it says primitive cell, the answer is 2.

4.5 Coordination number and bond geometry

Every atom in diamond, whichever atom of the motif it is, has four nearest neighbours at the corners of a regular tetrahedron, so CN = 4 (Figure 4). Any two of the four bonds make the tetrahedral angle, arccos(−⅓) = 109.47°. A coordination number of 4 is the signature of directional covalent bonding; compare CN = 12 for FCC and HCP metals, where bonding is non-directional and atoms pack as closely as possible.

The nearest neighbours lie at a distance d = √3a/4, which is the C–C bond length: 1.545 Å for carbon with a = 3.567 Å. Beyond the first shell the neighbours are 12 atoms at a/√2, 12 atoms at √11a/4 and 6 atoms at a. The second shell is worth remembering: its 12 atoms are the FCC lattice neighbours of the atom, which is why the number 12 reappears.

4.6 Nearest-neighbour distance, atomic radius and packing factor

In the hard-sphere model the atoms touch along the bonds, so the atomic radius is half the bond length: r = √3a/8, which is 0.772 Å for carbon. The volume of the eight atoms divided by the volume of the cube gives the atomic packing factor, APF = π√3/16 = 0.340 (the derivation is in Section 5). Only 34% of the cell volume is filled by spheres; 66% is empty. This is well below simple cubic (0.52), even though diamond is built on an FCC lattice. The price of four-fold tetrahedral bonding is an open structure. It is the single most common source of error on this topic: students see "FCC" and write 0.74.

An open structure does not mean a low atomic density. The number of atoms per unit volume in diamond is 8/a³ = 1.76 × 10²³ atoms cm⁻³, among the highest of any solid, because carbon atoms are small and the C–C bond is short. APF measures how much of the volume the spheres fill; atomic density measures how many atoms there are per unit volume. They are different quantities.

4.7 Graphite, step 1: the graphene layer

In graphite each carbon atom forms three σ bonds with neighbours 120° apart (sp²). The layer is the honeycomb net of Figure 1, with a C–C distance of 1.42 Å, noticeably shorter than the 1.54 Å of diamond. The shorter bond has a simple explanation: the leftover p electrons form π bonding spread over the layer, so each C–C link is more than a single bond. Each atom supplies one π electron, which is half a π bond per atom, and there are 1.5 C–C links per atom, so each link carries one σ bond plus one-third of a π bond: an average bond order (the number of electron-pair bonds joining two atoms) of 4/3.

The two-dimensional lattice parameter is a = √3d = 2.46 Å (derivation in Section 5). The cell area is (√3/2)a² = 5.24 Ų and holds two atoms, so a single layer contains about 38 atoms per square nanometre.

4.8 Graphite, step 2: stacking the layers in AB order

In graphite the layers are 3.35 Å apart, more than twice the 1.42 Å bond length, and they are held together only by van der Waals forces. They stack in the sequence ABAB…, called AB or Bernal stacking. Layer B is layer A shifted sideways by one C–C bond length along a bond direction (Figure 5). The consequence is that half of the atoms in a layer lie directly above atoms of the layer beneath, while the other half lie above the centres of hexagons. The third layer repeats the first, so the true repeat distance along the stacking direction is two layers: c = 2 × 3.35 Å ≈ 6.71 Å. The letters A and B here name layers; they are unrelated to the atoms A and B of Figure 1.

The lattice of graphite is therefore the simple hexagonal lattice (Pearson symbol hP, meaning hexagonal primitive) with a ≈ 2.46 Å, c ≈ 6.71 Å and γ = 120°, and the motif consists of four atoms. In the standard setting of space group P6₃/mmc (the space group is the complete set of symmetry operations of the crystal, and 194 is its number in the International Tables for Crystallography) the four atoms sit at the positions below; crystallographers label these two sets of equivalent positions Wyckoff sites 2b and 2c. Atoms of the first kind have a neighbour directly above and below in the adjacent layers; atoms of the second kind sit above and below the centres of hexagons.

The conventional cell is the rhombic prism spanned by a₁, a₂ and c. Count its atoms with the rule from Section 2. The two atoms at (0,0,1/4) and (0,0,3/4) lie on the vertical edges of the prism, each edge shared by four cells, so 4 edges × 2 atoms × ¼ = 2. The atoms at (1/3,2/3,1/4) and (2/3,1/3,3/4) lie wholly inside and count 2. The total is 4 atoms.

Why is c/2 not a lattice translation? Going from the layer at height 1/4 to the next layer at height 3/4 needs a vertical shift of c/2 and a sideways shift of one bond length. A vertical shift alone would put a B layer where an A layer belongs, and the crystal would not map onto itself. Only a shift of a full c restores the pattern. That is why the motif has four atoms rather than two: two layers, two atoms per layer.

Graphite also occurs as a rhombohedral polytype (a form built from the same layers in a different stacking order) with ABCABC stacking (space group R-3m), usually as a minor component of natural graphite. In this course, graphite means the hexagonal ABAB form unless stated otherwise.

4.9 Bonding compared: an sp³ network versus sp² layers

The energy scales explain everything that follows in Section 7. The van der Waals binding between layers of graphite is about 0.05 eV per atom (a thermal-desorption measurement gives 52 ± 5 meV per atom, and other experiments scatter between roughly 0.035 and 0.06 eV), whereas the covalent cohesion within a layer is several electron-volts per atom. That difference of about two orders of magnitude is the physical origin of graphite's strong anisotropy (the dependence of a property on direction): a graphite crystal is very strong along its layers and very weak between them.

4.10 Stability: why does diamond exist at all?

At 298 K and 1 atm, graphite is the thermodynamically stable form of carbon, meaning the form with the lowest Gibbs energy, the quantity whose minimum decides stability at constant temperature and pressure. Standard thermochemical tables give the standard Gibbs energy of formation of diamond, taking graphite as the reference, as about +2.9 kJ mol⁻¹. The enthalpy contribution is about +1.9 kJ mol⁻¹, and the entropy makes things worse rather than better: the standard molar entropy of graphite (5.74 J mol⁻¹ K⁻¹) exceeds that of diamond (2.38 J mol⁻¹ K⁻¹), so for graphite → diamond ΔG = ΔH − TΔS ≈ 1.9 + 1.0 = 2.9 kJ mol⁻¹. It is a common surprise that the layered, "looser" structure has the higher entropy. The stiffer diamond lattice has higher vibrational frequencies and therefore lower vibrational entropy.

Diamond nevertheless persists indefinitely at ambient conditions, because turning it into graphite requires breaking strong σ bonds and rearranging the whole framework. That is a large activation barrier (an energy hump that must be climbed before a change can occur), so diamond is metastable: thermodynamically unfavoured, kinetically frozen. Pressure changes the balance. Because (∂G/∂P)_T = V and diamond has the smaller molar volume (3.42 cm³ mol⁻¹ against 5.29 cm³ mol⁻¹ for graphite), raising the pressure raises the Gibbs energy of graphite faster than that of diamond. Above a boundary pressure of roughly 1.6 GPa at 298 K, diamond becomes the stable phase, and the boundary rises steeply with temperature (Figure 6). Worked Example 8.8 turns this argument into a number.

5Equations, Symbols, Assumptions, Units and Derivations

5.1 Symbols and units

5.2 Assumptions used in this lesson

5.3 Counting atoms and the geometry of diamond

The atoms belonging to one cell are counted by weighting each atom by the fraction of it inside the cell.

For diamond, 4 interior atoms + 6 face atoms × ½ + 8 corner atoms × ⅛ = 4 + 3 + 1 = 8.

Nearest-neighbour distance. Take the atom at (0,0,0) and its neighbour at (¼,¼,¼). The separation vector is (a/4, a/4, a/4), so its length is

This is also half the body diagonal of the sub-cube of edge a/2 in Figure 4, which is the quickest way to recall it: the body diagonal of a cube of edge e is √3e, so half of it for e = a/2 is √3a/4.

Atomic radius. Touching spheres along the bond give 2r = d, so

Atomic packing factor. The cell contains 8 atoms, each a sphere of volume (4/3)πr³, and the cell volume is a³. Substituting (5.3) step by step:

Tetrahedral angle. From the atom at the origin, two of its bonds point along (1,1,1) and (1,−1,−1) (the factor ¼a is common and cancels). Each vector has length √3, so the angle θ between them follows from the dot product:

5.4 Density from the structure

If a cell contains n atoms of molar mass M, its mass is nM/N_A, and its density is that mass divided by the cell volume:

For diamond cubic, n = 8 and V_c = a³:

Unit check. g mol⁻¹ divided by (mol⁻¹ × cm³) leaves g cm⁻³. The most frequent numerical slip is forgetting that a is given in Å: convert with 1 ų = 10⁻²⁴ cm³ before dividing. For diamond, a³ = 45.38 ų = 4.538 × 10⁻²³ cm³.

5.5 Geometry of graphite

Bond length and lattice parameter. In the honeycomb, two neighbouring lattice points (say two A atoms) are joined through a B atom by two bonds of length d meeting at the C–C–C angle of 120°. By the cosine rule their separation, which is the lattice parameter a, satisfies a² = d² + d² − 2d² cos 120° = 3d², so

For a = 2.46 Å this gives d = 1.42 Å.

Cell volume. The two in-plane vectors have equal length a and enclose 120°, so the area of the cell is a² sin 120° = (√3/2)a². Multiplying by the height c:

Density and layer spacing. With four atoms per cell, equation (5.6) becomes

Two layers fit in one repeat, so the spacing between adjacent layers is half of c:

Areal density of a layer. One layer contains two atoms in a cell of area (√3/2)a², so

A packing measure for graphite (model-dependent). If the atoms of a layer are treated as touching spheres, r = d/2 = a/(2√3), and four such spheres in the cell of volume (5.9) give

This number is smaller than the value for diamond, but treat it with care: the layers are 3.35 Å apart while touching spheres would be only 1.42 Å apart along the stacking direction, so the model describes the openness of the structure rather than any real contact between the layers.

5.6 Diffraction relations and selection rules

X-rays of wavelength λ are diffracted by a set of planes of spacing d_hkl when Bragg's law is satisfied:

For a cubic crystal the spacing depends on the cell edge and the sum of squared indices, and eliminating d_hkl gives the relation used to index powder patterns:

Because the prefactor is the same for every line, the sin²θ values of a cubic pattern are in the ratio of the integers N = h² + k² + l². For a hexagonal crystal such as graphite the spacing is

so the basal reflections (h = k = 0) have d = c/l.

Which reflections appear? A reflection is observed only if the structure factor, the sum of the waves scattered by all atoms in the cell, is not zero:

For diamond, use the lattice-plus-motif description. The motif atom at (¼,¼,¼) contributes a phase 2π(h + k + l)/4 = π(h + k + l)/2 relative to the atom at the lattice point, and the four FCC lattice points contribute the phases 0, π(h + k), π(h + l) and π(k + l). The result is a product of a motif factor and an FCC factor:

The second bracket is 4 when h, k, l are all odd or all even ("unmixed") and 0 otherwise; that is the FCC rule. The first bracket then removes further reflections.

So diamond shows a reflection only if the indices are all odd, or all even with a sum divisible by 4. Compare the first allowed lines of the cubic lattices in terms of N = h² + k² + l²:

Simple cubic and BCC give the same first three ratios; they are told apart by the seventh line, because N = 7 cannot be written as a sum of three squares, so simple cubic has no line there, while BCC continues 1, 2, 3, 4, 5, 6, 7. For graphite, the four atoms at (0,0,1/4), (0,0,3/4), (1/3,2/3,1/4) and (2/3,1/3,3/4) give, for the basal reflections (00l),

which is zero for odd l. The first basal reflection of graphite is therefore (002), at d = c/2, not (001).

5.7 Thermodynamic estimate of the graphite–diamond boundary

At constant temperature the Gibbs energy of a phase changes with pressure according to

For the change graphite → diamond, with a molar volume change ΔV = V_diamond − V_graphite that we take as constant, integrating from the reference pressure P° (1 bar) gives

Diamond becomes the stable phase when ΔG(P) falls to zero. Since ΔV is negative and P° is negligible in comparison, the equilibrium pressure is

Use SI units throughout: ΔG° in J mol⁻¹ and ΔV in m³ mol⁻¹ give P in Pa. Remember 1 cm³ = 10⁻⁶ m³ and 1 GPa = 10⁹ Pa.

6Essential Diagrams and Infographic Instructions

Figures 1 to 6 appear in Section 4 where you first need them. This section says what each figure must contain, so that you can redraw it in your own notes or build an infographic from it, and gives a fast hand-sketch recipe for exam scratch work.

Suggested Figure: lattice and motif in two dimensions (Figure 1). Three panels side by side. Panel (a): a hexagonal array of points with the rhombic primitive cell shaded, equal vectors a₁ and a₂ drawn as arrows, and the 120° angle marked. Panel (b): one cell with atom A at (0,0) and atom B at (1/3,2/3) and a cross marking the lattice point. Panel (c): the honeycomb net with A atoms in one colour, B atoms in another, rings on the lattice points and a note that B is not a lattice point. Quick sketch: draw a regular hexagon and colour alternate corners with two colours; the two colours are the two atoms of the motif, and the hexagon centres form the lattice.

Suggested Figure: diamond cubic cell (Figure 2). A cube in oblique view with the eight corner and six face-centre atoms in one colour family and the four interior atoms at (¼,¼,¼), (¾,¾,¼), (¾,¼,¾), (¼,¾,¾) in a contrasting colour, hidden edges dashed, and sixteen bonds drawn from the interior atoms to their four blue neighbours. The legend should read corner 8 × ⅛ = 1, face 6 × ½ = 3, interior 4 × 1 = 4. Quick sketch: draw a cube and mark its corners and face centres; divide the cube mentally into eight small cubes (octants) and put one dot at the centre of every second octant, choosing four that do not share a face; join each interior dot to the four corners of its own small cube that coincide with a corner or a face centre of the big cube.

Suggested Figure: plan view with heights (Figure 3). A square seen down [001]. Label each atom with its height as a fraction of a: 0 at the four corners and at the centre, ½ at the four edge midpoints, ¼ at (¼,¼) and (¾,¾), and ¾ at (¾,¼) and (¼,¾). Quick sketch: the ¼ and ¾ atoms lie on the two diagonals, one diagonal each.

Suggested Figure: tetrahedral coordination (Figure 4). One octant cube of edge a/2 with the central atom joined to four alternate corners, labelled with the bond length √3a/4 and the angle 109.47°. Quick sketch: draw a cube, put a dot at its centre, and join the dot to one corner and to the three corners that lie diagonally opposite that corner across each of the three faces meeting at it; the four corners you have used are the vertices of a regular tetrahedron.

Suggested Figure: AB stacking (Figure 5). A top view of two honeycomb layers, one solid and one dashed, offset by one bond length so that half of the atoms coincide in projection; beside it an edge-on view with three layers labelled A, B, A, a dimension of 3.35 Å between layers and c = 6.71 Å over two layers. Quick sketch: draw three horizontal lines, mark atoms on each, shift the middle line's atoms sideways relative to the outer two, and write c over the A-to-A distance.

Suggested Figure: carbon phase boundary (Figure 6). Axes of pressure (GPa) and temperature (K), a rising line separating a lower graphite field from an upper diamond field, the ambient point marked deep in the graphite field, and the boundary pressure at 298 K labelled. Quick sketch: the line has positive slope because diamond is denser; the region of higher pressure belongs to the denser phase.

Suggested infographic: one-page comparison poster. Two columns, diamond in blue and graphite in slate grey, using the same colours as the figures. Each column shows the unit cell or layer, the lattice and motif in one line ("FCC + 2 atoms" and "hexagonal + 4 atoms"), coordination number, bond length, density and one property with its structural reason. A central strip lists the three quantities students most often confuse: atoms per cell (8 and 4), coordination number (4 and 3), and packing factor (0.34 for diamond).

7Structure–Property Connections and Material Examples

The two structures of Section 4 differ in three respects: the dimensionality of the strong bonding (a three-dimensional network in diamond, two-dimensional sheets in graphite), the state of the valence electrons (all localised in diamond, one per atom delocalised in graphite) and the symmetry (cubic against hexagonal). Each property below follows from one or more of these.

7.1 Mechanical behaviour

In diamond every atom is held by four strong, short, directional σ bonds in a continuous three-dimensional network, so there is no weak direction. The high density of strong bonds gives an exceptionally high stiffness (a Young's modulus of the order of 1 TPa, roughly 1100 to 1200 GPa depending on direction and source) and the greatest hardness of any naturally occurring material, 10 on the Mohs scale. Because the bonds are directional and the structure is open, diamond at room temperature fractures rather than flows: it is hard but brittle. It cleaves (splits along a preferred crystal plane) on the {111} family of planes, meaning all planes equivalent to (111), and the simple bond-breaking model of surface energy shows why. Counting the bonds that must be cut to create a unit area of new surface gives 4/a² on {100}, 2.83/a² on {110} and 2.31/a² on {111}, so {111} planes are the cheapest to separate.

In graphite the σ bonds within a layer are short and strong, so a layer resists stretching very strongly, but between layers only van der Waals forces act. Layers slide over one another under very small shear stresses, which is why graphite is soft, marks paper and is used as a solid lubricant. (Practical lubrication by graphite also depends on adsorbed vapours, which is beyond this syllabus.)

7.2 Electrical behaviour

In diamond all four valence electrons of each atom are used in localised σ bonds, which fill the valence band (the highest band of occupied electron states) completely. The energy gap between the valence band and the empty conduction band, called the band gap, is about 5.5 eV, so pure diamond is an excellent electrical insulator (resistivity of the order of 10¹³ to 10¹⁶ Ω cm) and is transparent to visible light. In graphite three electrons per atom are used in σ bonds, but the fourth, in the p orbital perpendicular to the layer, is delocalised over the whole sheet. The resulting π bands overlap slightly, so graphite is a semimetal (a material whose valence and conduction bands overlap only slightly, leaving few charge carriers, though mobile ones) with high conductivity along the layers. Across the layers electrons must hop between sheets, and the conductivity is lower by a factor of the order of 10³ or more; published anisotropy ratios depend strongly on crystal quality, so quote the order of magnitude, not a single figure.

7.3 Thermal conduction

In a non-metal, heat is carried mainly by lattice vibrations, whose quanta are called phonons. Diamond combines light atoms, very strong bonds and a rigid, defect-poor three-dimensional network, so phonons travel fast and far. The room-temperature thermal conductivity of high-quality diamond is about 2000 W m⁻¹ K⁻¹, roughly five times that of copper, even though diamond is an electrical insulator. It is the standard example that heat conduction and electrical conduction need not go together. The same reasoning applies within a graphite layer, where the in-plane thermal conductivity of high-quality graphite is of the same order, about 2000 W m⁻¹ K⁻¹. Across the layers the weak bonding gives only about 6 W m⁻¹ K⁻¹, a ratio of several hundred. Graphite is therefore an excellent heat spreader in the plane of its layers and a poor conductor perpendicular to them.

7.4 Optical behaviour

The band gap of diamond, about 5.5 eV (an ultraviolet wavelength of roughly 225 nm), is larger than the energy of any visible photon (about 1.8 to 3.1 eV). Visible light is therefore not absorbed and diamond is transparent; its high refractive index, about 2.4, gives its brilliance. The delocalised π electrons of graphite absorb light across the visible range, so graphite is opaque and black with a metallic lustre.

7.5 Summary of structure–property links

7.6 Material examples

Materials with the diamond structure. Silicon and germanium share the diamond structure, so every result of Section 5 applies to them with the appropriate lattice parameter and molar mass. Grey tin (α-Sn), stable below about 13 °C, also adopts it. In zinc blende compounds such as ZnS, GaAs and cubic SiC, the two atoms of the motif are different elements; the lattice is still FCC and the coordination is still tetrahedral, but the crystal loses the inversion symmetry of diamond and its space group becomes F-43m instead of Fd-3m.

The densities were calculated with equation (5.7) and the molar masses of Section 5.2; measured values agree with them to better than 0.2%.

Diamond in use. Cutting and grinding tools exploit its hardness, heat spreaders for high-power electronics exploit its thermal conductivity, and optical windows exploit its transparency over a wide range of wavelengths.

Graphite in use. Electrodes, electrical brushes and the anodes of lithium-ion cells exploit in-plane conduction; crucibles and furnace parts exploit stability at high temperature; pencils and dry lubricants exploit easy shear; and heat spreaders exploit in-plane thermal conductivity.

8Fully Worked Conceptual and Numerical Examples

Unless stated otherwise, N_A = 6.022 × 10²³ mol⁻¹ and M = 12.01 g mol⁻¹ for carbon. Cover the solution and try each problem yourself before reading it.

9Common Student Mistakes and GATE Traps

Most lost marks on this topic come from a small set of predictable slips. Read the table once now and again just before you attempt Section 10.

10MCQ, MSQ and NAT Practice Questions

The twenty questions below are original GATE-style practice questions written for this lesson. None is a reproduction of a previous-year GATE question. In an MCQ exactly one option is correct. In an MSQ one or more options are correct and all of them must be chosen. A NAT answer is a number to the stated precision; the solution gives the range that would be accepted. Unless a question says otherwise, use N_A = 6.022 × 10²³ mol⁻¹, M = 12.01 g mol⁻¹ for carbon and λ = 1.5406 Å for Cu Kα1. Attempt every question before opening its solution in Section 11.

10.1 Multiple-choice questions (MCQ)

10.2 Multiple-select questions (MSQ)

10.3 Numerical answer type questions (NAT)

11Detailed Solutions

Open a solution only after you have committed to an answer. Each one explains why the correct option is correct and, for MCQ and MSQ, why the others are not.

11.1 Solutions to the MCQ

11.2 Solutions to the MSQ

11.3 Solutions to the NAT

12One-Minute Revision Summary, Formula Sheet and Comparison Table

12.1 Key concepts to remember

12.2 Formula sheet

12.3 Comparison table: diamond and graphite

12.4 Numbers worth memorising

13Connections to Prerequisite and Next Lessons

What this lesson assumed. You should be secure in the idea of a lattice with a motif and a unit cell; the FCC and hexagonal Bravais lattices; atoms per cell, coordination number and packing factor for simple cubic, BCC, FCC and HCP; close-packed structures with their tetrahedral and octahedral voids; Miller and Miller–Bravais indices; and the basics of covalent and van der Waals bonding. If Section 2 felt unfamiliar, revisit those topics before continuing.

What comes next. The rest of the carbon line in the syllabus is graphene, fullerenes and carbon nanotubes. There you will roll and curve the honeycomb layer of Section 4.7, and you will need the lattice vectors a₁ and a₂ and the 120° geometry of Figure 1. After that come compound structures built on the same ideas: zinc blende (diamond with two different elements), wurtzite (its hexagonal counterpart), NaCl, CsCl, perovskite and spinel. Diffraction lessons will extend the structure-factor reasoning of Section 5.6, for which diamond is the model example, and the phase-diagram lessons will treat the carbon boundary of Figure 6 as a unary pressure–temperature diagram.

A preview of the next step. Suppose the two atoms of the diamond motif are different, zinc at (0,0,0) and sulfur at (¼,¼,¼). The lattice is still FCC, the cell contains 4 Zn and 4 S atoms, and the coordination is 4:4, which is zinc blende. The structure factor becomes 4[f_Zn + f_S e^iπ(h+k+l)/2] for unmixed indices. For (200) this equals 4(f_Zn − f_S), which is not zero, so (200) is allowed but weak. Diamond's forbidden reflections are forbidden only because the two atoms of its motif are identical.

References and Verified Sources

Advanced Materials Lab. GATE 2027 XE2 Materials Science course, Lesson 4.2: Diamond and Graphite, Lattice, Motif and Structure.