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

The scenario matrix

Before solving anything, let's list every category of question this topic can throw. Each row is a "cell"; every worked example below is tagged with the cell it fills.

Cell	Scenario class	What makes it tricky
A	Bond-angle from geometry (dot product)	signs of components, which corners
B	Cage counting with Euler's formula (any $C_{n}$ )	generalising past $C_{60}$
C	Degenerate / limiting cages (only hexagons / only pentagons)	why one can't close and one over-closes
D	Length / spacing arithmetic (ratios)	comparing diamond vs graphite numbers
E	Density from structure (units!)	pm→cm, atoms per cell
F	Conductivity reasoning (sign of "free e⁻")	purity ≠ conducting
G	Real-world word problem	pencil / lubricant / drug cage
H	Exam twist (compare & explain)	trap: "soft = weak bonds"

Now we fill every cell — and we really do fill all eight (A–H) below.

The two ideas we lean on, summarised here (no clicking away)

Definition Hybridisation, in one picture

A carbon atom has one $2 s$ orbital and three $2 p$ orbitals. Hybridisation means the atom blends some of these into new, equal-shaped orbitals that point where bonds are needed.

Blend $s$ + three $p$ → four equal orbitals = == $s p^{3}$ ==. They spread as far apart as possible in 3-D → a tetrahedron, giving the $109.4 7^{\circ}$ diamond angle. All four hold σ-bonds, so no electron is left free.
Blend $s$ + two $p$ → three equal orbitals = == $s p^{2}$ ==, lying flat at $12 0^{\circ}$ in a plane. This uses only two of the three $p$ orbitals, so one $p$ -electron is left unhybridised, sticking out of the plane — that spare electron is what delocalises and conducts (graphite, graphene, fullerenes, CNTs).

So the only structural switch behind this whole page is: how many $p$ orbitals get blended in — three ( $s p^{3}$ , insulating tetrahedron) or two ( $s p^{2}$ , flat with one free electron).

Notation we will actually use (built from zero)

Definition Dot product — the tool that measures "same-direction-ness"

For two arrows $a = (a_{1}, a_{2}, a_{3})$ and $b = (b_{1}, b_{2}, b_{3})$ , the dot product is $a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} .$ Why this tool and not another? We want the angle between two bonds. Length alone can't give an angle; you need a quantity that grows when arrows align and shrinks/goes negative when they oppose. The dot product does exactly that, and it is tied to the angle by $cos θ = \frac{a \cdot b}{∣ a ∣ ∣ b ∣}, ∣ a ∣ = a_{1}^{2} + a_{2}^{2} + a_{3}^{2} .$ Here $∣ a ∣$ is the arrow's length (Pythagoras in 3-D), and $cos θ$ answers "how aligned?": $+ 1$ means parallel, $0$ means perpendicular, $- 1$ means opposite. The bond directions themselves come from the $s p^{3}$ hybridisation summarised just above — four equal orbitals spreading as far apart as possible.

Cell A — Bond angle in diamond (all sign cases)

Forecast: Guess before computing — is it bigger or smaller than $9 0^{\circ}$ ? (The corners are on opposite sides of the cube, so expect more than $9 0^{\circ}$ .)

Figure s01 (below): a light cube with the central red carbon at the origin, one blue arrow reaching to corner $(1, 1, 1)$ and one orange arrow reaching to corner $(1, - 1, - 1)$ ; the green label marks the $109.4 7^{\circ}$ opening between them. Notice the two arrows point to corners on opposite diagonals — that visual "leaning apart" is exactly what the negative dot product will report in step 1.

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

Dot product. $a \cdot b = (1) (1) + (1) (- 1) + (1) (- 1) = 1 - 1 - 1 = - 1.$ Why this step? We compute the dot product first because it is the only quantity that carries directional (sign) information into the calculation. The negative result is the whole story: it tells us the arrows lean apart, so the angle exceeds $9 0^{\circ}$ . The two $- 1$ terms come from the two negative components of $b$ — that is where "all cases of sign" live.
Lengths. $∣ a ∣ = 1^{2} + 1^{2} + 1^{2} = 3$ , and $∣ b ∣ = 1^{2} + (- 1)^{2} + (- 1)^{2} = 3 .$ Why this step? The cosine formula divides by both lengths, so we must know them. Squaring inside the root kills the minus signs, so the length is the same for every corner — that is why the final $- 1/3$ never depends on which corner we pick.
Cosine. $cos θ = \frac{- 1}{3 \cdot 3} = \frac{- 1}{3} .$ Why this step? We divide the dot product by the two lengths precisely because that ratio is $cos θ$ (from the boxed formula). Dividing strips away how long the arrows are and keeps only how they are oriented — and $3 \cdot 3 = 3$ makes the denominator clean.
Undo the cosine. $θ = cos^{- 1} (- \frac{1}{3}) = 109.4 7^{\circ} .$ Why this step? We have $cos θ$ but want $θ$ itself, so we apply $cos^{- 1}$ — it asks "which angle has this cosine?" and undoes the cosine to hand us the angle.

Verify: $cos (109.4 7^{\circ}) \approx - 0.333 \approx - \frac{1}{3}$ ✓. And $109.4 7^{\circ} > 9 0^{\circ}$ matches the forecast. Units: an angle, dimensionless-degrees ✓.

$a \cdot c = (1) (- 1) + (1) (1) + (1) (- 1) = - 1 + 1 - 1 = - 1.$ Why this step? We repeat the dot product with a different sign pattern to test whether the answer really is corner-independent. Two negatives and one positive still sum to $- 1$ — the tetrahedral symmetry forces it.
$∣ c ∣ = 3$ as before. So $cos θ = - \frac{1}{3}$ , $θ = 109.4 7^{\circ}$ .

Verify: Identical to A1 — confirms every pair of tetrahedral bonds gives $109.4 7^{\circ}$ , no exceptions. ✓

Cell B — Euler counting for ANY fullerene

Forecast: More or fewer hexagons than C₆₀'s 20? (Bigger cage → guess more.)

Edges. $E = \frac{3 V}{2} = \frac{3 \times 70}{2} = 105.$ Why this step? We need $E$ before we can use Euler, and we get it from the $s p^{2}$ fact: each atom sends out 3 bonds, but each bond is shared by 2 atoms, so we multiply $V$ by 3 and divide by 2 to avoid double-counting.
Total faces. $F = 2 - V + E = 2 - 70 + 105 = 37.$ Why this step? Euler's formula $V - E + F = 2$ is rearranged to solve for the one unknown left, $F$ . Now we know the total number of rings before splitting them into shapes.
Split into rings. Let $p$ = pentagons, $h$ = hexagons: $p + h = 37$ . Why this step? The 37 faces are only ever pentagons or hexagons in a fullerene, so their counts must add to the total. Using the "12 pentagons always" rule (proved in B2), $p = 12$ , so $h = 37 - 12 = 25.$

Verify: Count edges another way — a pentagon has 5 sides, a hexagon 6, each edge shared by 2 faces: $E = \frac{5 p + 6 h}{2} = \frac{5 ( 12 ) + 6 ( 25 )}{2} = \frac{60 + 150}{2} = 105$ ✓. And $25 > 20$ matches the forecast. ✓

Vertices from faces. $V = \frac{5 p + 6 h}{3} .$ Why this step? We want everything written in terms of $p$ and $h$ so Euler becomes one equation in those unknowns. Counting corners face-by-face gives $5 p + 6 h$ corner-slots, but each vertex is shared by 3 faces, so we divide by 3.
Edges from faces. $E = \frac{5 p + 6 h}{2} .$ Why this step? Same face-counting trick for edges: $5 p + 6 h$ side-slots, but each edge is shared by 2 faces, so we divide by 2.
Plug into Euler. $V - E + F = \frac{5 p + 6 h}{3} - \frac{5 p + 6 h}{2} + (p + h) = 2.$ Why this step? We substitute the two expressions above and $F = p + h$ into $V - E + F = 2$ , turning geometry into a single algebraic equation we can solve.
Combine the first two terms. $(\frac{1}{3} - \frac{1}{2}) (5 p + 6 h) = - \frac{1}{6} (5 p + 6 h) .$ So $- \frac{5 p + 6 h}{6} + p + h = 2 \Rightarrow \frac{- 5 p - 6 h + 6 p + 6 h}{6} = 2 \Rightarrow \frac{p}{6} = 2 \Rightarrow p = 12.$ Why this step? $\frac{1}{3} - \frac{1}{2} = - \frac{1}{6}$ over a common denominator, so the two fractions merge. Putting everything over $6$ lets the $h$ terms ( $- 6 h + 6 h$ ) completely cancel. That is the beautiful fact: no matter how many hexagons, you always need exactly 12 pentagons to close the ball.

Verify: For C₆₀: $p = 12$ (independent of $h$ ) ✓, matching the parent note's answer.

Cell C — Both degenerate cages: hexagons-only AND pentagons-only

Figure s02 (below): two panels. Left (blue): three hexagons meeting at one red vertex; their interior angles $12 0^{\circ}$ each add to $36 0^{\circ}$ , so the surface lies perfectly flat — this is graphene. Right (orange): three pentagons meeting at a vertex; their angles $10 8^{\circ}$ each add to only $32 4^{\circ}$ , leaving a $3 6^{\circ}$ wedge missing, so the surface must pucker up to close the gap. Compare the two panels: the missing wedge on the right is what forces curvature; the flat left panel never curls.

From B2, $p = 12$ always for a closed cage. If we force $p = 0$ , the equation $\frac{p}{6} = 2$ becomes $0 = 2$ — a contradiction. Why this step? A contradiction means "no such closed object exists." Pure hexagons can only tile a flat infinite sheet — that is exactly graphene, a single unbroken $s p^{2}$ honeycomb. It never curves shut.
Geometry reason (angular defect). At every vertex three faces meet. With three hexagons the interior angles are $3 \times 12 0^{\circ} = 36 0^{\circ}$ — a full turn, so the sheet lies flat (zero curvature). Replace one hexagon-corner by a pentagon-corner and the angles become smaller ( $10 8^{\circ}$ instead of $12 0^{\circ}$ ), so the sum at a pentagon-containing vertex falls short of $36 0^{\circ}$ : the surface has to bend inward to close the missing wedge. That "shortfall from $36 0^{\circ}$ " is called the angular defect at a vertex.

Verify (Descartes' theorem, done correctly): Descartes' theorem says the angular defects of all vertices add up to $72 0^{\circ}$ for any convex cage. A pure-hexagon vertex has defect $36 0^{\circ} - 36 0^{\circ} = 0$ , contributing nothing — so a hexagons-only surface has total defect $0 \neq = 72 0^{\circ}$ , again proving it cannot close ✓. In C₆₀ each of the $60$ vertices touches exactly one pentagon and has defect $36 0^{\circ} - (2 \times 12 0^{\circ} + 10 8^{\circ}) = 36 0^{\circ} - 34 8^{\circ} = 1 2^{\circ}$ ; total $= 60 \times 1 2^{\circ} = 72 0^{\circ}$ ✓. (Note: the $1 2^{\circ}$ here is the defect per vertex, not "per pentagon" — the two must not be conflated.) This is the limiting/degenerate answer: zero pentagons ⇒ zero total defect ⇒ never closes.

Forecast: We already proved $p = 12$ always. So with $h = 0$ we expect exactly 12 pentagons and nothing else — guess whether that is a real solid.

Faces. With $h = 0$ and the always-true $p = 12$ : total faces $F = p + h = 12.$ Why this step? The general law from B2 fixed $p = 12$ regardless of $h$ ; setting $h = 0$ just drops the hexagons, leaving 12 pentagonal faces — the smallest possible fullerene-type cage.
Edges and vertices. $E = \frac{5 p + 6 h}{2} = \frac{5 ( 12 )}{2} = 30,$ and $V = \frac{5 p + 6 h}{3} = \frac{5 ( 12 )}{3} = 20.$ Why this step? Same face-counting as B2, now with $h = 0$ : 12 pentagons contribute $5 \times 12 = 60$ side-slots (÷2 for shared edges → 30 edges) and $60$ corner-slots (÷3 for shared vertices → 20 vertices).
Euler check. $V - E + F = 20 - 30 + 12 = 2$ ✓ — it closes. This is the pentagonal dodecahedron (a 12-faced Platonic solid), the theoretical cage " $C_{20}$ ". Why this step? A valid $V - E + F = 2$ with every vertex degree-3 means a genuine convex polyhedron exists. So the pentagon-only extreme is the complementary boundary to C1: hexagons-only over-flattens and never closes; pentagons-only over-curves into the tiniest possible closed cage.

Verify: Angular defect check — each dodecahedron vertex meets three pentagons: defect $= 36 0^{\circ} - 3 \times 10 8^{\circ} = 36 0^{\circ} - 32 4^{\circ} = 3 6^{\circ}$ ; total $= 20 \times 3 6^{\circ} = 72 0^{\circ}$ ✓, exactly Descartes' requirement. Real chemistry footnote: $C_{20}$ is highly strained (the sharp $3 6^{\circ}$ curvature crowds the bonds), so it is extremely unstable — which is why stable fullerenes add hexagons to spread the curvature out. ✓

Cell D — Bond-length arithmetic (diamond vs graphite)

Difference. $154 - 141.5 = 12.5$ pm. Why this step? "How much shorter" is literally a subtraction — the absolute gap between the two lengths, in the same unit (pm), so nothing needs converting.
Fractional shortening. $\frac{12.5}{154} = 0.08117 \dots$ Why this step? A raw $12.5$ pm gap means little without a reference; dividing by the diamond length rescales it to "fraction of the diamond bond," the fair baseline because the question asks "shorter than diamond."
Percent. $0.08117 \times 100 = 8.12%.$ Why this step? Multiplying a fraction by 100 just re-expresses it as parts-per-hundred, the form the question wants. The shorter bond signals partial double-bond character from graphite's delocalised π-electrons.

Verify: $141.5 = 154 \times (1 - 0.0812) = 154 \times 0.9188 = 141.5$ pm ✓. Units: pm cancels in the ratio, answer is a pure $%$ ✓.

Cell E — Density from structure (units matter most)

Forecast: Diamond is famously dense-ish (~3.5 g/cm³). Let's see if the structure predicts it.

Convert length to cm. $357 pm = 357 \times 1 0^{- 12} m = 357 \times 1 0^{- 10} cm = 3.57 \times 1 0^{- 8} cm .$ Why this step? Density in g/cm³ needs the volume in cm³; if we left the side in pm the answer would be off by enormous powers of ten. ( $1 pm = 1 0^{- 12} m$ and $1 m = 100 cm$ , so $1 pm = 1 0^{- 10} cm$ .)
Cell volume. $V = a^{3} = (3.57 \times 1 0^{- 8})^{3} = 4.55 \times 1 0^{- 23} cm^{3} .$ Why this step? A cube's volume is side cubed; we cube the already-converted cm length so the volume comes out directly in cm³, ready for density.
Mass of 8 atoms. $m = \frac{8 \times 12.0}{6.022 \times 1 0 ^{23}} = 1.594 \times 1 0^{- 22} g .$ Why this step? One mole is $N_{A}$ atoms weighing $12.0$ g, so one atom weighs $12.0/ N_{A}$ g; the cell holds 8 atoms, hence the factor 8. This gives the mass sitting inside our one cube.
Density. $ρ = \frac{m}{V} = \frac{1.594 \times 1 0 ^{- 22}}{4.55 \times 1 0 ^{- 23}} = 3.50 g/cm^{3} .$ Why this step? Density is by definition mass per unit volume, so we divide the cell's mass by the cell's volume — and because the cell tiles the whole crystal identically, this ratio is the bulk density. Final answer: $ρ \approx 3.50 g/cm^{3}$ .

Verify: Measured diamond density $\approx 3.51 g/cm^{3}$ ✓ — the structure predicts reality. Units: $g / cm^{3}$ ✓, matches the forecast.

Cell F — Conductivity reasoning (the sign of "mobile charge")

Forecast: Which has free electrons? (Insulator has none; conductors have delocalised π.)

Diamond: all 4 valence electrons locked in $s p^{3}$ σ-bonds → zero mobile electrons → insulator. Why this step? Conductivity needs mobile charge, not purity. Diamond is pure carbon yet insulates — purity is irrelevant.
Graphite: each $s p^{2}$ carbon keeps one delocalised p-electron; these roam within each layer → conducts in-plane. Why this step? The $s p^{2}$ blend leaves exactly one un-hybridised $p$ -electron per atom (see the summary), and those form the mobile π-cloud that carries current sideways along the sheet.
Graphene: a single such layer → the same free π-cloud, unobstructed → excellent conductor.
Ranking: graphene ≳ graphite (in-plane) $≫$ diamond ( $= 0$ ).

Verify: Consistency check — the only difference between the three is hybridisation ( $s p^{3}$ vs $s p^{2}$ ), and $s p^{2}$ is exactly what leaves a spare mobile electron. The ranking follows purely from "does a free electron exist? sign yes/no." ✓

Cell G — Real-world word problem

Pencil "lead" is graphite. Its layers are held by weak van der Waals forces; drag it across paper and whole layers shear off and stick to the fibres — that dark streak is transferred graphite sheets. Why this step? Writing = leaving material behind, which needs a solid whose layers detach easily → weak interlayer forces, not weak bonds.
Diamond is a 3-D giant covalent network with no slip planes; nothing shears off. Instead it cuts whatever is softer. Why this step? No weak plane exists to peel, so force is transmitted through unbroken covalent bonds in every direction — the hallmark of a giant covalent solid.
Trap avoided: the in-plane graphite bonds are stronger/shorter than diamond's (D1). Graphite is soft only because of interlayer sliding.

Verify: Dimensionality check — graphite is "2-D layers + vdW", diamond "3-D giant covalent"; the behaviour follows directly from that one structural difference ✓.

Cell H — Exam twist (steel-man the wrong idea)

Forecast: Is dissolving about the bonds inside a molecule, or the forces between molecules?

Steel-man: dissolving feels like "bonds giving way," so weaker bonds ⇒ dissolves. Sounds reasonable.
The truth: dissolving does not break the strong C–C bonds inside C₆₀ at all. C₆₀ is a discrete molecule (a molecular solid); the solvent only overcomes the weak van der Waals attractions between separate C₆₀ balls. Diamond has no separate molecules — the whole crystal is one covalent network, so there is nothing to peel apart short of breaking bonds → insoluble. Why this step? Solubility is about inter-molecular forces, not intra-molecular bond strength — exactly the molecular-vs-giant-covalent distinction from the summary. C₆₀'s internal bonds are as strong as graphite's.

Verify: Matches parent's Forecast-Verify drill: "C₆₀ dissolves in benzene — yes, discrete non-polar molecule." ✓

Recall One-line recap of every cell

A angle ::: $cos^{- 1} (- 1/3) = 109.4 7^{\circ}$ for any tetrahedral bond pair. B Euler ::: $C_{70}$ : $E = 105, F = 37$ , 12 pentagons + 25 hexagons. C degenerate ::: hexagons-only gives $0 = 2$ (never closes = flat graphene); pentagons-only gives the 12-face dodecahedron $C_{20}$ (over-curved, strained). D length ::: graphite in-plane bond is $8.12%$ shorter than diamond's. E density ::: structure predicts $ρ \approx 3.50 g/cm^{3}$ for diamond. F conductivity ::: graphene ≳ graphite (in-plane) ≫ diamond (mobile π-electrons vs none). G word ::: pencil writes because weak-vdW layers shear off; diamond scratches (3-D covalent network). H twist ::: C₆₀ solubility is about weak vdW between molecules, not weak internal bonds.

Connections

Parent topic
Hybridisation sp sp2 sp3
Euler's formula for polyhedra
Catenation in Group 14
Van der Waals forces
Giant covalent vs molecular solids
Conductivity and delocalised electrons
Silicon and its differences from carbon