Covalent bonding in silicon crystals: Worked examples

Before we start, three symbols must be earned. If any of these feels unfamiliar, read the box; we never use a symbol you haven't met.

The scenario matrix

Every question this topic can throw sits in one of these cells. The examples below are tagged with the cell they cover — the example numbers match the cell order exactly.

Cell	Case class	What makes it tricky	Example
A	Counting electrons per atom	confusing bonds vs electrons	Ex 1
B	Coordination / geometry	tetrahedral 3D vs flat cartoon	Ex 2 (figure)
C	Temperature raised	ratio of exponentials	Ex 3 (figure)
D	Temperature lowered toward 0 K	limiting/degenerate value	Ex 4
E	Different material, different $E_{g}$	insulator vs semiconductor decision	Ex 5
F	Absolute count from proportionality	needs a reference point (anchor)	Ex 6
G	Real-world word problem	translate English → equation	Ex 7
H	Exam twist / trap	spot the wrong assumption	Ex 8

We also flag the degenerate inputs so no scenario is left uncovered:

$T \to 0$ K → exponent $\to - \infty$ → $n_{i} \to 0$ (perfect insulator). Worked in Ex 4.
$T \to \infty$ → exponent $\to 0$ → $n_{i}$ saturates toward its prefactor. Worked in Ex 4, Step 4.
$E_{g} \to 0$ → no gap → conducts like a metal at any $T$ . Worked in Ex 5, Step 4.

Cell A — counting electrons

Worked example Example 1 (Cell A) — how many electrons does one Si atom "feel"?

Q: Inside the crystal, how many outer-shell electrons does a single silicon atom effectively surround itself with, and how many bonds is that?

Forecast: Guess now — is the answer 4, 8, or 16? And is the bond count the same number?

Step 1. Own valence electrons $= 4$ . Why this step? Silicon is group IV → its $n = 3$ shell holds $3 s^{2} 3 p^{2} = 4$ electrons. This is the atom's own contribution.

Step 2. Add one shared electron from each neighbour: $+ 4$ . Why this step? Coordination number is $4$ — four nearest neighbours, each lending one electron into a shared pair.

Step 3. Total electrons felt $= 4 + 4 = 8$ . Why this step? Eight = a full octet → stable, low-energy, no free carriers at 0 K.

Step 4. Count bonds: $4$ , not $8$ . Why this step? Each bond is a pair (2 electrons). Four bonds × two electrons = eight electrons, but only four bonds. This is the classic trap.

Verify: Electrons $=$ (own 4) $+$ (borrowed 4) $= 8$ ✓. Bonds $=$ 4 bonds each holding 1 of Si's own electrons + 1 neighbour's = 4 bonds, 8 electrons. Octet satisfied, sanity holds.

The figure below is the whole point of this example: on the left is the flat textbook cartoon (bonds drawn at 90°), on the right is the true 3D tetrahedron (bonds at 109.5°). As you read Example 2, look at the right panel — the orange bonds fan out into space, and the red arrow marks the angle we are about to compute. Notice the left panel's four bonds sit on a plane, which is why its 90° is only a drawing convenience, not physics.

Figure — Covalent bonding in silicon crystals

Worked example Example 2 (Cell B) — what angle sits between two Si–Si bonds?

Q: Four bonds point from one silicon atom to the corners of a regular tetrahedron. What is the angle between any two of them, and why is the flat 90° grid a lie?

Forecast: Guess: is the angle 90°, 109.5°, or 120°?

Step 1. Place the four bond directions at the corners of a cube's alternating vertices — the standard tetrahedron construction (the right panel of the figure). Two such directions are $(1, 1, 1)$ and $(1, - 1, - 1)$ . Why this step? The diamond-cubic lattice literally builds each atom's four bonds toward alternate cube corners; using coordinates lets us compute the angle instead of guessing.

Step 2. The dot product of two direction vectors gives the cosine of the angle between them: $cos θ = \frac{a \cdot b}{∣ a ∣ ∣ b ∣} .$ Why this tool? The dot product is the one operation that turns two directions into a single number encoding "how aligned" they are — exactly the angle question we're asking. No other simple operation reads off an angle directly.

Step 3. Compute: $a \cdot b = (1) (1) + (1) (- 1) + (1) (- 1) = - 1$ , and each length is $3$ , so $cos θ = \frac{- 1}{3 \cdot 3} = - \frac{1}{3} .$ Why this step? Plugging the actual bond vectors in turns the abstract formula into a real number.

Step 4. $θ = arccos (- \frac{1}{3}) \approx 109.47°$ . Why this step? $arccos$ answers "which angle has this cosine?" — it undoes the cosine to hand us the angle in degrees.

Verify: $cos (109.47°) \approx - 0.333$ ✓. The flat 90° drawing is a cartoon: it correctly shows 4 bonds, 8 electrons but distorts the true $109.5°$ tetrahedral geometry (left vs right panel). See Diamond cubic crystal structure.

Cell C — heating up (with a figure)

The figure below plots $n_{i}$ against temperature on a log vertical axis (each gridline up is ×10). Follow the blue curve as you read Example 3: it climbs steeply, and the three coloured dots mark the temperatures we compute (green 300 K, red 350 K, orange 400 K). The gray arrow labels the physical meaning — more heat breaks more bonds. Because the vertical axis is logarithmic, a straight-looking rise is actually an exponential explosion in carrier count.

Worked example Example 3 (Cell C) — heat 300 K → 400 K

Q: By roughly what factor does $n_{i}$ grow when silicon is heated from 300 K to 400 K? ( $E_{g} = 1.12$ eV.)

Forecast: Guess the factor: about ×2, ×20, or ×200?

Step 1. Exponent at 300 K: $\frac{E _{g}}{2 k _{B} T} = \frac{1.12}{2 ( 8.617 \times 1 0 ^{- 5} ) ( 300 )} \approx 21.66$ . Why this step? The whole $T$ -dependence of the exponential part lives here; evaluate it at each temperature.

Step 2. Exponent at 400 K: $\frac{1.12}{2 ( 8.617 \times 1 0 ^{- 5} ) ( 400 )} \approx 16.25$ . Why this step? Same formula, hotter $T$ → smaller exponent → less suppression.

Step 3. Take the ratio. Because $n_{i} \propto e^{- exponent}$ (dropping the gentle $T^{3/2}$ prefactor), $\frac{n _{i} ( 400 )}{n _{i} ( 300 )} \approx e^{- 16.25} / e^{- 21.66} = e^{21.66 - 16.25} = e^{5.41} .$ Why this step? Dividing exponentials subtracts exponents — the unknown constant cancels, so we get a clean ratio without knowing it.

Step 4. $e^{5.41} \approx 224$ . Why this step? Convert the exponent to a plain multiplier.

Verify: $ln (224) \approx 5.41$ ✓. So $n_{i}$ jumps roughly ×220 — about two orders of magnitude — for just a 100 K rise. (Including the $T^{3/2}$ correction $(400/300)^{1.5} \approx 1.54$ would nudge this to ~×340, still the same order.) More heat → more broken bonds → more conduction. Ties to Electron-hole pair generation and recombination.

Cell D — the degenerate limits, T → 0 K and T → ∞

Worked example Example 4 (Cell D) — what happens at absolute zero (and at infinite heat)?

Q: As $T \to 0$ K, what does $n_{i}$ approach, and what does that mean physically? Also, what happens as $T \to \infty$ ?

Forecast: Guess $n_{i} (0 K)$ : infinite, some fixed number, or zero?

Step 1. Look at the exponent $- \frac{E _{g}}{2 k _{B} T}$ as $T \to 0^{+}$ . Why this step? The behaviour of $n_{i}$ is dictated by what the exponent does at the extreme (the $T^{3/2}$ prefactor only makes $n_{i}$ vanish faster, so it doesn't change the conclusion).

Step 2. As $T \to 0^{+}$ , the denominator $\to 0^{+}$ , so $\frac{E _{g}}{2 k _{B} T} \to + \infty$ , hence the exponent $\to - \infty$ . Why this step? Dividing a fixed positive number by a shrinking positive number blows up; the minus sign sends it to negative infinity.

Step 3. $e^{- \infty} = 0$ , so $n_{i} \to 0$ . Why this step? $e$ raised to a very negative power collapses to zero — no bonds have the energy to break.

Step 4. Opposite limit: as $T \to \infty$ , the exponent $\to 0$ , so $e^{0} = 1$ and the exponential part saturates. (The $T^{3/2}$ prefactor keeps growing, but the exponential — the dominant physics — no longer suppresses carriers.) Why this step? Covers the other degenerate end so no scenario is missing — the exponential can't grow past 1; the crystal is limited by how many bonds exist to break.

Verify: $n_{i} (0 K) = 0$ ✓ — matches the parent claim "silicon is a perfect insulator at 0 K." Every electron is locked in a bond; there is literally nothing to move.

Cell E — a different material (and the E_g → 0 limit)

Worked example Example 5 (Cell E) — diamond vs silicon, and what if the gap vanished?

Q: Carbon (diamond) shares the same bonding and lattice but has $E_{g} \approx 5.5$ eV. Compare its $n_{i}$ to silicon's at 300 K. Insulator or semiconductor? Then: what would $n_{i}$ do if $E_{g} \to 0$ ?

Forecast: Guess how many orders of magnitude fewer carriers diamond has: 10, 50, or 90?

Step 1. Exponent for diamond at 300 K: $\frac{5.5}{2 ( 8.617 \times 1 0 ^{- 5} ) ( 300 )} \approx 106.4$ . Why this step? Same master equation, just swap in the larger gap.

Step 2. Compare to silicon's exponent $21.66$ . Ratio of carrier counts: $\frac{n _{i}^{Si}}{n _{i}^{C}} \approx e^{106.4 - 21.66} = e^{84.7} .$ Why this step? Subtracting exponents again cancels the prefactor, isolating the pure effect of $E_{g}$ .

Step 3. Convert to base 10: $e^{84.7} = 1 0^{84.7/2.303} \approx 1 0^{36.8}$ . Why this step? Orders of magnitude are easiest to grasp in base 10; divide the natural exponent by $ln 10 \approx 2.303$ .

Step 4. Degenerate case $E_{g} \to 0$ : the exponent $- \frac{E _{g}}{2 k _{B} T} \to 0$ , so $e^{0} = 1$ — the exponential no longer suppresses anything and carriers are abundant at any temperature, i.e. metal-like conduction. Why this step? Closes the $E_{g} \to 0$ degenerate input promised in the matrix — a zero gap means no barrier to breaking bonds, exactly what a metal looks like.

Verify: Diamond has roughly $1 0^{37}$ times fewer free carriers than silicon → effectively zero → insulator ✓. Same bonding, but the size of $E_{g}$ decides the fate; and $E_{g} \to 0$ gives metal-like behaviour ✓. This is why bond type alone does not settle conductivity — the gap does.

Cell F — an absolute number from proportionality

Cell G — a real-world word problem

Worked example Example 7 (Cell G) — a chip in the desert

Q: A sensor chip runs at 300 K indoors but reaches 350 K in a hot desert enclosure. Its leakage current scales with $n_{i}$ . By what factor does intrinsic leakage roughly increase? ( $E_{g} = 1.12$ eV.)

Forecast: Guess: barely changes (×1.5), or noticeably (×15)?

Step 1. Translate "leakage scales with $n_{i}$ " into "find $n_{i} (350) / n_{i} (300)$ ." Why this step? The physics word "leakage" hides the same ratio calculation — spotting that is the whole skill of a word problem.

Step 2. Exponent at 350 K: $\frac{1.12}{2 ( 8.617 \times 1 0 ^{- 5} ) ( 350 )} \approx 18.57$ . Why this step? Evaluate the exponent at the new temperature.

Step 3. Ratio $\approx e^{21.66 - 18.57} = e^{3.09} \approx 22$ . Why this step? Subtract exponents; the prefactor cancels (a $(350/300)^{1.5} \approx 1.26$ correction leaves the order unchanged).

Verify: $ln (22) \approx 3.09$ ✓. A modest 50 K rise multiplies leakage by ~20× — a genuine engineering headache, and why chips need cooling. Units cancel (dimensionless ratio) ✓.

Cell H — the exam trap

Worked example Example 8 (Cell H) — spot the wrong assumption

Q: A student writes: "Each Si atom shares 8 electrons, one with each of its 8 neighbours." Two errors are hidden here. Find and fix both.

Forecast: Can you spot both before reading on?

Step 1. Error 1 — neighbour count. Silicon's coordination number is 4, not 8. Why this step? The tetrahedral geometry (Ex 2) allows exactly four nearest neighbours.

Step 2. Error 2 — electron vs bond. Silicon feels 8 electrons but forms only 4 bonds, each a shared pair. Why this step? The number 8 refers to the octet of electrons, never to a count of neighbours or bonds.

Step 3. Corrected sentence: "Each Si atom forms 4 bonds with 4 neighbours, sharing a pair (2 electrons) per bond, so it feels 8 electrons total." Why this step? Restating cleanly cements the distinction the exam is testing.

Verify: $4 bonds \times 2 electrons = 8$ electrons, from $4$ neighbours ✓. Consistent with Ex 1.

Wrap-up — the whole matrix, closed

Recall Quick self-test

As $T \to 0$ K, what does $n_{i}$ approach and why? ::: $n_{i} \to 0$ ; the exponent $- E_{g} / (2 k_{B} T) \to - \infty$ so $e^{- \infty} = 0$ — no bond has energy to break. Heating Si from 300 K to 400 K multiplies $n_{i}$ by roughly what? ::: About ×220 ( $e^{5.41}$ ), exponential-only estimate. Why does diamond ( $E_{g} = 5.5$ eV) stay an insulator with the same bonding as Si? ::: Its far larger gap makes the exponent ~106 vs ~22; carriers are ~ $1 0^{37}$ × fewer → effectively zero. Bonds vs electrons for one Si atom? ::: 4 bonds, 8 electrons felt (own 4 + borrowed 4). What angle sits between two Si–Si bonds, and how do we get it? ::: $109.47°$ , computed from $arccos (- 1/3)$ using the dot product of two tetrahedral bond vectors. What two temperature corrections did we drop from the master equation? ::: The $T^{3/2}$ density-of-states prefactor and the Varshni band-gap narrowing $E_{g} (T)$ — both mild, both raise $n_{i}$ .

Connections

Parent topic (Hinglish)
Band theory of solids
Intrinsic vs extrinsic semiconductors
Doping with donors and acceptors
Electron-hole pair generation and recombination
Diamond cubic crystal structure
Ionic vs covalent vs metallic bonding

Worked examples — Covalent bonding in silicon crystals

The scenario matrix

Cell A — counting electrons

Cell B — geometry

Cell C — heating up (with a figure)

Cell D — the degenerate limits, T → 0 K and T → ∞

Cell E — a different material (and the E_g → 0 limit)

Cell F — an absolute number from proportionality

Cell G — a real-world word problem

Cell H — the exam trap

Wrap-up — the whole matrix, closed

Connections