Metallic bonding — electron sea, band theory (intro): Worked examples

Before we compute anything, let us pin down the few symbols we will lean on, in plain words.

Definition The symbols we will use (each earned before use)

$E_{g}$ = band gap, the width in energy of the forbidden zone between the filled valence band and the empty conduction band. Measured in electron-volts (eV). One eV is the energy one electron gains crossing a 1-volt gap: $1 eV = 1.602 \times 1 0^{- 19} joules (J)$ .
$k_{B}$ = Boltzmann constant $= 1.381 \times 1 0^{- 23} J/K$ — the "exchange rate" between temperature and energy. Multiply it by a temperature $T$ (in kelvin, K) and you get a typical thermal energy $k_{B} T$ that jiggling atoms hand to electrons.
$T$ = absolute temperature in kelvin: $T (K) = T (° C) + 273$ . At room temperature $T \approx 300 K$ .
$n \propto e^{- E_{g} / (2 k_{B} T)}$ = how the number of thermally freed charge carriers depends on gap and temperature. The symbol $\propto$ means "grows in proportion to"; $e^{x}$ is the exponential function (a number $\approx 2.718$ raised to the power $x$ ).

The scenario matrix

Every case this topic can present falls into one of these cells. Each worked example below is tagged with the cell it covers.

Cell	Case class	What is being tested
A	Bond-strength comparison, more electrons donated	Na vs Mg vs Al ordering
B	Bond-strength comparison, same electrons, different size	radius controls strength
C	Zero-gap input ( $E_{g} = 0$ )	why a metal conducts at all $T$
D	Small-gap input (semiconductor)	numeric carrier ratio
E	Large-gap input (insulator)	why almost none jump
F	Limiting behaviour: $T \to 0$ and $T \to \infty$	what $e^{- E_{g} /2 k_{B} T}$ does at extremes
G	Sign-of-slope twist: does conductivity rise or fall with heat?	metal vs semiconductor opposite signs
H	Real-world word problem	pick a wire material
I	Exam twist / trap	"full band = insulator?" fallacy
J	Degenerate/overlap case	full VB but overlapping CB (Mg)

We now walk one example per cell (some cells share a figure).

Worked example

Example 1 — Cell A: more electrons donated (Na < Mg < Al) Q: Order Na, Mg, Al by metallic bond strength and hence melting point.

Forecast: Guess before reading: which do you think melts highest — the one that gives 1 electron or 3?

Step 1. Write valence: Na = [Ne] $3 s^{1}$ → donates 1 e⁻; Mg = [Ne] $3 s^{2}$ → 2 e⁻; Al = [Ne] $3 s^{2} 3 p^{1}$ → 3 e⁻. Why this step? The parent formula says strength $\propto$ (electrons donated)/(cation radius). Electrons donated is the first knob.
Step 2. More electrons in the sea → more electrostatic "glue" between the $+$ cations and the $-$ cloud, and the cation left behind carries more charge (Na⁺, Mg²⁺, Al³⁺). Why this step? Both the numerator (electrons) and the cation charge climb together, reinforcing the trend.
Step 3. Predict strength Na < Mg < Al, so m.p. Na < Mg < Al.

Verify: Real melting points: Na $98 ° C$ , Mg $650 ° C$ , Al $660 ° C$ . The order $98 < 650 < 660$ ✓ matches. (Uses Ionization energy & atomic radius.)

Worked example

Example 2 — Cell B: same electron count, size decides (Li vs Cs) Q: Both Li and Cs are group-1 (donate 1 e⁻). Which forms the stronger metallic bond?

Forecast: With the electron count tied, what's left to break the tie?

Step 1. Numerator (electrons donated) = 1 for both, so it cancels. Look at the denominator: cation radius. Why this step? When one knob is fixed, isolate the other — that is the whole point of a controlled comparison.
Step 2. Down group 1, atoms gain whole extra shells → radius grows. Li⁺ radius $\approx 76 pm$ , Cs⁺ $\approx 167 pm$ . Why this step? Strength $\propto 1/ (radius)$ ; a bigger radius spreads the $+$ charge out and weakens its pull on the sea.
Step 3. $\frac{1}{76} > \frac{1}{167}$ , so Li has the stronger bond.

Verify: m.p. Li $= 181 ° C$ , Cs $= 28 ° C$ . Indeed $181 > 28$ ✓ — the smaller cation melts higher.

Worked example

Example 3 — Cell C & J: zero gap / overlapping bands (why metals conduct) Q: Sodium's 3s band is only half full; magnesium's 3s band is full but overlaps its empty 3p band. Explain, with $E_{g}$ , why both conduct.

Forecast: Mg's valence band is full — does "full band" mean "can't conduct"? Guess.

Step 1. For Na: the highest band is half-occupied, so filled levels sit directly beneath empty ones with no gap. Effective $E_{g} = 0$ . Why this step? Conduction needs an empty state an electron can slide into under a field; a half-filled band supplies that for free.
Step 2. For Mg: the 3s band is full, but the 3p band overlaps it (they share the same energy range), so empty states again touch filled ones. Effective $E_{g} = 0$ . Why this step? Overlap (left figure) is the degenerate case where "full band" is not an insulator — this kills the common trap (Cell I).
Step 3. With $E_{g} = 0$ , the carrier formula gives $n \propto e^{- 0/ (2 k_{B} T)} = e^{0} = 1$ for all $T$ — carriers exist at every temperature, no thermal "climb" required.

Verify: $e^{- 0/ (2 k_{B} T)} = 1$ independent of $T$ ✓. Both Na and Mg are conductors — see Electrical conductivity.

Worked example

Example 4 — Cell D & E: numeric carrier ratio, Si vs diamond Q: At $T = 300 K$ , compare the intrinsic carrier count of silicon ( $E_{g} = 1.1 eV$ ) with diamond ( $E_{g} = 5.5 eV$ ). By what factor does Si beat diamond?

Forecast: Diamond's gap is 5× Si's. Guess: is diamond 5× worse, or wildly worse?

Step 1. Compute $k_{B} T$ in eV at $300 K$ : $k_{B} T = (1.381 \times 1 0^{- 23}) (300) J = 4.14 \times 1 0^{- 21} J$ . Divide by $1.602 \times 1 0^{- 19} J/eV$ → $k_{B} T \approx 0.0259 eV$ . Why this step? The exponent $E_{g} / (2 k_{B} T)$ mixes gap (in eV) with thermal energy; both must be in the same unit or the ratio is meaningless.
Step 2. Exponent for Si: $\frac{1.1}{2 ( 0.0259 )} = 21.2$ . For diamond: $\frac{5.5}{2 ( 0.0259 )} = 106.2$ . Why this step? The carrier count is $e^{- exponent}$ ; a bigger exponent means exponentially fewer carriers.
Step 3. Ratio $\frac{n _{Si}}{n _{diamond}} = e^{- 21.2} / e^{- 106.2} = e^{(106.2 - 21.2)} = e^{85} \approx 8 \times 1 0^{36}$ . Why this step? Dividing exponentials subtracts exponents; the difference of gaps, not their ratio, drives the answer.

Verify: $e^{85}$ is astronomically large — Si has roughly $1 0^{37}$ × more carriers than diamond, so Si is a semiconductor and diamond an insulator ✓ (matches Semiconductors & doping). A "5× bigger gap" produced a $1 0^{36}$ × effect — the exponential's whole point.

Worked example

Example 5 — Cell F: limiting behaviour $T \to 0$ and $T \to \infty$ Q: For a semiconductor ( $E_{g} > 0$ ), what does $n \propto e^{- E_{g} / (2 k_{B} T)}$ do as (a) $T \to 0 K$ and (b) $T \to \infty$ ?

Forecast: At absolute zero, is a semiconductor a conductor or an insulator? Guess, then check.

Step 1. As $T \to 0$ : the denominator $2 k_{B} T \to 0^{+}$ , so the fraction $E_{g} / (2 k_{B} T) \to + \infty$ , so the exponent $- \infty$ , so $e^{- \infty} \to 0$ . Why this step? No thermal energy means no electron can pay the gap toll — carriers vanish, so a pure semiconductor becomes an insulator at 0 K.
Step 2. As $T \to \infty$ : the fraction $\to 0$ , so $e^{0} = 1$ — the factor saturates at 1, the maximum. Why this step? At huge temperature, thermal energy dwarfs the gap; the gap stops mattering and carriers plateau.
Step 3. Between these, $n$ rises smoothly (the right figure): steeply at first, then leveling off — a classic exponential-approach curve.

Verify: $lim_{T \to 0^{+}} e^{- E_{g} / (2 k_{B} T)} = 0$ and $lim_{T \to \infty} e^{- E_{g} / (2 k_{B} T)} = 1$ ✓ for any $E_{g} > 0$ .

Worked example

Example 6 — Cell G: the sign-of-slope twist (metal vs semiconductor) Q: Heat a copper wire and a silicon chip. In which does conductivity rise and in which does it fall with temperature? State why the signs are opposite.

Forecast: "Heat = energy," so surely both conduct better hot? Guess — one is a trap.

Step 1. Copper (metal, $E_{g} = 0$ ): all valence electrons are already free. Heating does not add carriers ( $n$ is fixed). What it does add is cation vibration. Why this step? If $n$ can't grow, the only thing left to change is how easily existing carriers move.
Step 2. Hotter cations vibrate harder and scatter the drifting electrons (electron–phonon scattering), so resistance rises → conductivity falls as $T$ rises. Negative slope. Why this step? This is the parent note's key correction to the naive "hot = better."
Step 3. Silicon (semiconductor, $E_{g} = 1.1 eV$ ): heating creates carriers via $n \propto e^{- E_{g} / (2 k_{B} T)}$ , and this exponential gain overwhelms the scattering loss → conductivity rises. Positive slope. Why this step? Opposite sign, opposite dominant mechanism — the whole distinction in one sentence.

Verify: Sign check — metal: $d n / d T = 0$ but scattering ↑ ⇒ $σ$ ↓; semiconductor: $\frac{d}{d T} e^{- E_{g} / (2 k_{B} T)} > 0$ ⇒ $n$ ↑ ⇒ $σ$ ↑ ✓. (Derivative is positive because raising $T$ shrinks the positive exponent, see VERIFY.)

Worked example

Example 7 — Cell H: real-world word problem (choosing a wire) Q: You must pick a metal for house wiring that stays conductive across cold winters and hot summers, and bends around corners without snapping. Between an ionic solid (NaCl) and a metal (Cu), which — and why using this chapter's models?

Forecast: Both contain charged particles. Why can't salt carry your current?

Step 1. NaCl electrons are localised & transferred to specific Cl⁻ sites (see Ionic bonding); no mobile carriers in the solid → it does not conduct as a solid. Why this step? Conduction demands delocalised charge; localisation rules NaCl out immediately.
Step 2. Cu has a delocalised electron sea ( $E_{g} = 0$ ) → conducts at all everyday temperatures (Cell C result). Why this step? We already proved $e^{- 0/ (2 k_{B} T)} = 1$ ⇒ carriers at every $T$ , so cold winters/hot summers both work.
Step 3. Bending: pushing Cu's cation layers just re-forms the non-directional sea (malleable), while NaCl lines up like charges → shatters (brittle). Why this step? This is the malleable-vs-brittle contrast from the parent note applied to the real spec.

Verify: Cu chosen ✓ — conducts at all $T$ (Cell C) and is malleable (electron-sea non-directionality). NaCl fails both.

Worked example

Example 8 — Cell I: exam trap ("a full valence band can never conduct") Q: True or false: "Because magnesium's 3s valence band is completely full, magnesium cannot conduct electricity." Give the corrected reasoning.

Forecast: Full band = no empty states = no conduction… surely true? Spot the trap.

Step 1. It is true that a full band, in isolation, offers no empty states to accelerate into. Why this step? Grant the part that's correct so the fix is precise, not a blanket denial.
Step 2. But Mg's empty 3p band overlaps the full 3s band in energy — empty states sit right at the same energy as filled ones (Example 3 figure, left). Why this step? Overlap re-introduces empty adjacent states, so the "isolated full band" premise doesn't hold for Mg.
Step 3. With empty states available at $E_{F}$ , effective $E_{g} = 0$ and Mg conducts. So the statement is False.

Verify: "Full band ≠ insulator when a band overlaps it." Mg is a known conductor ✓ — consistent with Example 3's $e^{0} = 1$ result.

Recall Quick self-test (cover the answers)

Which cell had $E_{g} = 0$ and why does that make $n$ constant in $T$ ? ::: Cell C — $e^{- 0/ (2 k_{B} T)} = 1$ for every temperature, so carriers exist always. A gap 5× larger gave a carrier ratio of roughly what? ::: About $e^{85} \approx 8 \times 1 0^{36}$ — the difference of gaps, not the ratio, sits in the exponent. Metal heated: conductivity up or down, and by what mechanism? ::: Down — electron–phonon scattering rises while carrier number stays fixed. Semiconductor at $T \to 0$ : conductor or insulator? ::: Insulator — $e^{- E_{g} / (2 k_{B} T)} \to 0$ . Why doesn't solid NaCl conduct though it has charges? ::: Its charges are localised/transferred, not delocalised — no mobile carriers.

Covalent bonding & MO theory · Transition metals & d-orbitals · Giant structures / lattices · Electrical conductivity · Semiconductors & doping

The scenario matrix

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