Exercises — Group 14 (Carbon family) — allotropes of C (diamond, graphite, fullerenes, graphene, CNTs)

Allotropes. Same element, same state, different atomic arrangement. What we did: matched the definition. Why it matters: the whole chapter is "one element, many structures" — see Giant covalent vs molecular solids.

• Diamond → $s{p}^3$ (4 σ-bonds, tetrahedral)
• Graphite → $s{p}^2$
• Graphene → $s{p}^2$
• Fullerene C₆₀ → $s{p}^2$
• CNT → $s{p}^2$

Why the four are all $s{p}^2$ (self-contained reasoning): each carbon in graphite, graphene, C₆₀ and a CNT makes exactly 3 σ-bonds to 3 neighbouring carbons arranged in a flat (or gently curved) hexagon pattern. Making 3 σ-bonds requires mixing one $s$ + two $p$ orbitals = three $s{p}^2$ orbitals pointing $120^{\circ }$ apart. That uses 3 of carbon's 4 valence electrons; the 4th electron stays in an unhybridised p-orbital perpendicular to the sheet, forming the delocalised π-system. Diamond is the odd one out because each carbon there bonds to 4 neighbours, which needs four $s{p}^3$ orbitals (one $s$ + three $p$) and leaves no leftover p-electron. The rule: count the neighbours — 3 bonds ⇒ $s{p}^2$, 4 bonds ⇒ $s{p}^3$. See Hybridisation sp sp2 sp3.

Fullerene C₆₀. It is a closed cage of 60 atoms — a finite molecule, so it dissolves in non-polar solvents like benzene. Diamond, graphite, graphene and CNTs are extended networks and do not dissolve.

What we compute: the fractional shortening. $\frac{154-141.5}{154}\times 100\%=\frac{12.5}{154}\times 100\%\approx 8.12\%$ Answer: graphite's in-plane bond is about $8.1\%$ shorter. Why shorter: in graphite the leftover p-electrons form a delocalised π-cloud, giving each C–C bond partial double-bond character. Double bonds pull atoms closer, so the length drops below the pure single-bond value of diamond.

• Ice: you overcome weak hydrogen bonds / van der Waals between whole water molecules — the O–H covalent bonds stay intact.
• Diamond: there are no separate molecules; the entire crystal is one giant covalent network, so melting means breaking actual C–C covalent bonds everywhere — enormously harder, hence ~$3550{}^{\circ }\text{C}$.

The larger distance, $335$ pm, is between layers and corresponds to weak van der Waals forces — see Van der Waals forces. The short $141.5$ pm distance is a strong covalent bond within a layer. Rule used: weaker force ⇒ larger equilibrium separation.

Diamond ($s{p}^3$): all 4 valence electrons are locked into 4 σ-bonds. Every electron is localised between two nuclei → no mobile charge → insulator. Graphite ($s{p}^2$): only 3 electrons go into σ-bonds; the 4th sits in a p-orbital and delocalises across the sheet, forming a mobile π-cloud → free charge → conductor in-plane. Key idea: conductivity needs mobile charge, and purity is irrelevant. Figure s01 (below) shows this side by side: on the left, a diamond carbon and its 4 neighbours with each bonding electron (pink dot) pinned between two atoms; on the right, a row of graphite carbons with the π-electrons drawn as a smeared blue cloud floating above and below the sheet, free to slide.

Softness comes from layers sliding, not from weak covalent bonds. Within a layer the bonds are short ($141.5$ pm) and strong (partial double-bond character). But between layers there are only weak van der Waals forces. When you press a pencil, whole layers shear off and slide — that flaking is the graphite mark. So "soft" describes the inter-layer weakness, while the in-plane bonds remain very strong.

Heat and charge travel by different carriers.

• Electricity needs mobile electrons; diamond has none free → insulator.
• Heat in diamond travels as lattice vibrations (atoms jiggling and passing the jiggle on through stiff, light C–C bonds). Stiff strong bonds + light atoms transmit vibrations extremely efficiently → excellent thermal conductor.

So "conductor" must always be qualified: of what? Diamond conducts vibrations, not charge.

Step 1 — count edges. Each vertex has degree 3, so summing degrees gives $3V$. Every edge is shared by 2 vertices, so it is counted twice: $E=\frac{3V}{2}=\frac{3\times 60}{2}=90.$ Why this step: we cannot use Euler until we know $E$; the degree-3 fact converts vertices into edges. Step 2 — Euler for faces. $V-E+F=2\Rightarrow 60-90+F=2\Rightarrow F=32.$ Step 3 — split into pentagons ($p$) and hexagons ($h$). Two equations: $p+h=32(\text{total faces})$ Count edges by faces: pentagons contribute 5 edges, hexagons 6, each edge shared by 2 faces: $\frac{5p+6h}{2}=E=90\Rightarrow 5p+6h=180.$ Substitute $h=32-p$: $5p+6(32-p)=180\Rightarrow 5p+192-6p=180\Rightarrow -p=-12\Rightarrow p=12,h=20.$ Result: exactly 12 pentagons and 20 hexagons. Figure s02 (below) shows why: on the left, a tiling of hexagons alone stays perfectly flat (that is graphene); on the right, inserting pentagons introduces curvature and bends the sheet closed into the C₆₀ ball.

Why the dot product? The angle $\theta$ between two vectors is exactly the quantity the dot product isolates: $\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \theta \Rightarrow \cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}.$ This is the right tool because the tetrahedral bond directions are naturally written as cube-corner vectors, and their angle is what we want. Step 1 — top (dot product): $\vec{a}\cdot \vec{b}=(1)(1)+(1)(-1)+(1)(-1)=1-1-1=-1.$ Step 2 — bottom (lengths): $|\vec{a}|=\sqrt{1^2+1^2+1^2}=\sqrt{3},|\vec{b}|=\sqrt{3},|\vec{a}||\vec{b}|=3.$ Step 3 — the cosine and undo it: $\cos \theta =\frac{-1}{3}\Rightarrow \theta ={\cos }^{-1}\left(-\frac{1}{3}\right)\approx 109.47^{\circ }.$ What the negative sign means: $\cos \theta <0$ tells us the angle is obtuse (more than $90^{\circ }$) — the bonds splay apart to keep the four electron pairs as far from one another as possible.

With $V=70$, degree 3 gives $E=\frac{3\times 70}{2}=105$, and Euler gives $F=2-V+E=2-70+105=37$. Let $p$ = pentagons, $h$ = hexagons. Then $p+h=37$ and $\frac{5p+6h}{2}=105\Rightarrow 5p+6h=210$. Substitute $h=37-p$: $5p+222-6p=210\Rightarrow -p=-12\Rightarrow p=12,h=25.$ Conclusion: $p=12\ne 0$, so an all-hexagon cage is impossible — you always need exactly 12 pentagons to close any such fullerene cage. (This "always 12" is a deep consequence of Euler's formula.)

1. Start with graphene. A single $s{p}^2$ sheet: 3 σ-bonds per atom in a honeycomb, plus one delocalised π-electron per atom forming a continuous conducting cloud — see Conductivity and delocalised electrons. 2. Roll it into a tube. A CNT is that sheet wrapped into a cylinder. Because the sheet is now closed around its circumference, an electron travelling around the tube must come back in phase with itself — a standing-wave (periodic) condition. 3. The rolling direction (chirality) sets which electron waves fit. Depending on the angle/direction of the roll, the allowed circular waves may or may not include the special momentum states that graphene needs to conduct. 4. Result: if those conducting states survive the wrapping → the tube is metallic; if the wrapping forbids them and opens a small energy gap → the tube is semiconducting. One-line takeaway: same atoms, same bonds, but geometry of the roll selects the electrons' allowed states — structure decides property, again.

Answer: graphene.

• (a) Conductor: $s{p}^2$ carbons each donate one delocalised π-electron → mobile π-cloud across the sheet → excellent in-plane conduction.
• (b) Transparent: it is a single layer of atoms — almost nothing there to absorb light, so it lets ~98% of visible light through.
• (c) One atom thick: graphene is by definition a single one-atom-thick honeycomb sheet — the thinnest possible 2-D material.

Diamond fails (a); graphite fails (b) and (c) (many stacked layers); C₆₀ fails (c) — so graphene is the unique fit.

A stable graphite/graphene needs the 4th electron of each atom to sit in a p-orbital and overlap sideways with its neighbours' p-orbitals to build the delocalised π-system. Whether that works depends on atom size:

• Silicon is a larger atom than carbon, so its valence p-orbitals are bigger and more spread out (diffuse), and two bonded Si atoms sit further apart than two carbons.
• Sideways (π) overlap needs the p-orbitals to reach across the gap and merge; with large, diffuse p-orbitals at a longer distance the sideways overlap is poor. So no strong, stable π-system can form.
• Because π-bonding is weak, silicon strongly prefers to use all four electrons in σ-bonds → $s{p}^3$ tetrahedral (diamond-type) structure. That is why silicon's normal crystal is "silicon-diamond" but there is no silicon-graphite.
• Additionally, catenation is weaker for silicon: Si–Si bonds are longer and weaker than C–C, so long self-linked frameworks are less stable to begin with.

One-line takeaway: bigger atom → poor π-overlap and weaker catenation → σ-only $s{p}^3$ structures, no graphite/graphene analogue. See Silicon and its differences from carbon and Catenation in Group 14.

Name the allotrope ::: by hybridisation and dimensionality — can you do all 5 from memory? Why diamond insulates but graphite conducts ::: localised σ-electrons vs delocalised π-electrons. Edge count for a degree-3 cage ::: $E=3V/2$ (remember to halve!). Faces of C₆₀ by Euler ::: $F=2-60+90=32$ = 12 pentagons + 20 hexagons. Tetrahedral angle ::: ${\cos }^{-1}(-1/3)\approx 109.47^{\circ }$. Why no silicon graphite ::: large Si → poor π-overlap and weaker catenation.