Covalent bonding in silicon crystals: Question bank

Before you start, pin down the words and symbols you will keep meeting:

Valence electrons — the electrons in the outermost shell of an atom; only these take part in bonding.
Covalent bond — a shared pair of electrons, one from each of two neighbouring atoms.
Band gap $E_{g}$ — the minimum energy needed to promote a bonded (valence) electron up into the conduction band, where it can move through the crystal. It is not torn out of the material — it stays inside the lattice but is no longer tied to one bond. Measured in electron-volts (eV).
Hole — the empty slot left in a bond after an electron leaves; it behaves like a mobile positive charge.
$n$ — the number of free (conduction) electrons per unit volume (units: $cm^{- 3}$ ).
$p$ — the number of holes per unit volume (units: $cm^{- 3}$ ).
$n_{i}$ — the intrinsic carrier concentration: in pure silicon, breaking a bond makes one electron and one hole together, so $n = p = n_{i}$ . It counts how many electron–hole pairs exist per $cm^{- 3}$ at a given temperature.
$k_{B}$ — the Boltzmann constant, the fixed number that converts temperature into energy. In two handy unit systems: $k_{B} \approx 8.617 \times 1 0^{- 5} eV/K$ or $1.381 \times 1 0^{- 23} J/K$ . $T$ is the absolute temperature in kelvin (K).

True or false — justify

Silicon at 0 K is a perfect insulator

True — every valence electron is locked in a covalent bond, so there are zero free carriers to move charge.

A single silicon atom in the crystal owns 8 electrons all by itself

False — it contributes 4 and shares 4 from neighbours; it only "sees" 8, it does not own 8.

There are 8 covalent bonds around each silicon atom

False — there are 4 bonds; each bond is a pair (2 electrons), so 4 bonds × 2 = 8 shared electrons, not 8 bonds.

Breaking one covalent bond creates exactly one free electron

False — it creates a pair: one free electron AND one hole, because the electron that leaves exposes an empty slot.

Diamond and silicon have the same lattice geometry

True — both are diamond cubic with 109.5° tetrahedral bonds; they differ in band-gap size, not shape.

Silicon conducts because it is a soft metal

False — it is a semiconductor; it has no free electrons at 0 K, unlike a metal which has a free-electron sea at all temperatures.

Raising temperature lowers the intrinsic carrier concentration

False — heat breaks more bonds, so

n_{i} \propto e^{- E_{g} / (2 k_{B} T)}

rises steeply with temperature.

A hole is empty space and therefore cannot carry current

False — when a neighbouring bonded electron hops into the slot, the slot moves the opposite way, acting as a mobile positive charge.

The tetrahedral picture and the flat 2D textbook picture describe different bonding

False — the flat grid is a cartoon of the same tetrahedral bonding; both say "4 bonds, 8 shared electrons per atom."

Silicon could reach an octet by donating its 4 valence electrons

True in principle but energetically absurd — stripping 4 electrons costs far too much energy, so sharing is chosen instead.

Why questions

Why does silicon share electrons instead of giving or grabbing them?

With exactly 4 valence electrons it sits halfway to an octet; donating 4 or stealing 4 both cost too much energy, so mutual sharing is the cheapest route to a full shell.

Why does a full outer shell (octet, 8 electrons) make silicon stable at 0 K?

Every electron is tied into a bond, so none are free to move — a stable, insulating configuration with no carriers.

Why is the "2" needed in the exponent of

n_{i} \propto e^{- E_{g} / (2 k_{B} T)}

One bond-break produces two carriers (

n = p = n_{i}

), and isolating

n_{i}

from the pair relation

n p = n_{i}^{2}

takes a square root, which halves the exponent to

E_{g} / (2 k_{B} T)

Why does the intrinsic carrier count rise so steeply with temperature?

It sits inside an exponential

e^{- E_{g} / (2 k_{B} T)}

; as

T

grows the magnitude of the negative exponent shrinks, and small changes in the exponent cause large multiplicative jumps in

n_{i}

Why is the band gap's size, not the bonding type, what decides semiconductor vs insulator?

Diamond and silicon bond identically, yet diamond (5.5 eV) is an insulator and silicon (1.12 eV) a semiconductor — only the energy needed to promote an electron across the gap differs. See Band theory of solids.

Why do textbooks draw the 3D tetrahedron as a flat grid?

To make "4 bonds, 8 shared electrons" easy to see on paper; the true 109.5° tetrahedral geometry is hard to draw but not needed to count bonds.

Why does the freed electron leave a positively-behaving hole rather than just a neutral gap?

The crystal was electrically neutral; removing a negative electron leaves a net local positive charge, and that charge appears to move as electrons shuffle to fill the slot.

Why must

E_{g}

and

k_{B} T

share the same energy unit in the exponent?

An exponent has to be a pure number; only if both are in eV (or both in joules) does

E_{g} / (k_{B} T)

cancel to dimensionless, which is why we quote

k_{B}

in eV/K when

E_{g}

is in eV.

Edge cases

At exactly 0 K, how many holes exist in pure silicon?

Zero — holes only appear when bonds break, and at 0 K no bond has the thermal energy to break, so

n_{i} = 0

If you cool doped silicon toward 0 K, does it still conduct?

Only barely — thermal pair generation freezes out, and even donor/acceptor carriers can "freeze" onto their atoms, so conduction collapses toward zero (carrier freeze-out).

What happens to

n_{i}

in the limit

E_{g} \to 0

The exponent

e^{- E_{g} / (2 k_{B} T)} \to 1

, so carriers are generated freely at any temperature — the material behaves like a conductor (a zero-gap or metallic case).

What happens to

n_{i}

in the limit

E_{g} \to

very large (like diamond)?

The exponent

\to 0

, so essentially no bonds break at room temperature — the material is an insulator.

Can a single covalent bond hold three electrons?

No — a covalent bond is defined as a shared pair (2 electrons); three electrons is not a stable single covalent bond in this lattice.

Is a silicon atom on the crystal surface fully bonded like an interior atom?

No — surface atoms lack some neighbours, leaving "dangling bonds" (unshared valence electrons), an edge case that matters for real-device surfaces though the ideal bulk model ignores it.

Does adding heat change silicon's bonding type from covalent to metallic?

No — heat only breaks some covalent bonds to free carriers; the remaining lattice is still covalently bonded. The bonding type never becomes metallic.

In pure silicon, does

n

equal

p

Yes — every broken bond makes one electron and one hole together, so

n = p = n_{i}

; this equality is exactly what defines the intrinsic case.

Recall One-line self-test

Cover every answer above. If for any item your reason was "because that's the rule," you have not understood it — go back and re-derive it from "4 valence electrons, 4 neighbours, small gap, and $n_{i} \propto e^{- E_{g} / (2 k_{B} T)}$ ."

Connections

1.3.03 Covalent bonding in silicon crystals (Hinglish) — the parent topic
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

Question bank — Covalent bonding in silicon crystals