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

Everything here builds on the parent topic. We will reuse only ideas already earned there; anything numerical is re-derived in place.

Before we start, two tools appear repeatedly — let us pin them down in plain words.

L1 — Recognition

Recall Solution

What: We match a description to a named model. Why: Recognition first — you must own the vocabulary before you reason with it. This is the electron-sea (Drude/Lorentz) model. The roaming particles are the delocalised valence electrons. The positive kernels are the cations (atoms that gave up those valence electrons). See Giant structures / lattices for the lattice geometry.

Recall Solution

What: Sort three gaps into conductor / semiconductor / insulator. Why: The size of the forbidden gap alone decides the electrical class (see Electrical conductivity). (a) $E_{g} = 0$ → conductor (metal): filled and empty states touch, electrons move with no climb. (b) $E_{g} = 1.1$ eV → semiconductor (this is silicon): small enough that heat promotes some carriers. (c) $E_{g} = 5.5$ eV → insulator (this is diamond): the gap is a wall almost nothing crosses.

L2 — Application

Recall Solution

Step 1 — count donated electrons. Na is [Ne]3s¹ → 1 e⁻; Mg is [Ne]3s² → 2 e⁻; Al is [Ne]3s²3p¹ → 3 e⁻. Why: more electrons = more glue. Step 2 — compare cation size. Across the period nuclear charge rises and radius shrinks: Na⁺ > Mg²⁺ > Al³⁺. Why: stronger pull from the nucleus contracts the ion (Ionization energy & atomic radius). Step 3 — combine both knobs. Both effects push the same way: strength Na < Mg < Al. Predicted m.p. order: Na < Mg < Al. (Real data: 98 °C, 650 °C, 660 °C — confirms it.)

Recall Solution

What: Take a ratio of the two Boltzmann factors. Why the ratio, not each alone? The unknown proportionality constant cancels in a ratio, so we can answer cleanly. $\frac{n _{Si}}{n _{dia}} = \frac{e ^{- E_{g}^{Si} / (2 k_{B} T)}}{e ^{- E_{g}^{dia} / (2 k_{B} T)}} = e^{- (E_{g}^{Si} - E_{g}^{dia}) / (2 k_{B} T)} = e^{(5.5 - 1.1) / (2 \times 0.02585)}$ The exponent is $4.4/0.0517 = 85.1$ , so the ratio is $e^{85.1} \approx 9 \times 1 0^{36}$ . What it means: silicon has astronomically more carriers than diamond at the same $T$ — that single exponential is why one is a useful semiconductor and the other an insulator (Semiconductors & doping).

L3 — Analysis

Recall Solution

Step 1 — identify what's the same. Both are metals; both use an electron sea. True but incomplete. Step 2 — find the hidden extra bonding. Na donates 1 s-electron. Fe is [Ar]3d⁶4s²; its partially filled 3d orbitals also overlap and delocalise (Transition metals & d-orbitals), so Fe pours ~8 electrons into the sea, not 1. Step 3 — apply the strength knob. Vastly more sea electrons + a small, highly-charged kernel → far stronger glue. Conclusion: Fe ≫ Na in bond strength → Fe melts far hotter (Fe ≈ 1538 °C vs Na ≈ 98 °C). The flaw was treating "metal" as one uniform mechanism; d-orbital participation changes everything.

Recall Solution

Step 1 — the naive worry. A full band has no empty states, so on its own it cannot conduct. Step 2 — the resolution: band overlap. Mg's filled 3s band overlaps the empty 3p band in energy. Look at Figure s01: the two shaded bands share an energy range instead of leaving a gap. Step 3 — the consequence. Because they overlap, there are empty states (from 3p) sitting at the same energy as the topmost filled ones — so electrons can accelerate. $E_{g}$ is effectively $0$ . Takeaway: "full band" ≠ "insulator". Only full band + a gap above it insulates.

Figure — Metallic bonding — electron sea, band theory (intro)

L4 — Synthesis

Recall Solution

Step 1 — write the ratio. Same material, two temperatures: $\frac{n ( 400 )}{n ( 300 )} = exp [- \frac{E _{g}}{2 k _{B}} (\frac{1}{400} - \frac{1}{300})] .$ Why a ratio again: the proportionality constant cancels. Step 2 — the temperature bracket. $\frac{1}{400} - \frac{1}{300} = 0.0025 - 0.003333 = - 0.0008333 K^{- 1}$ . It is negative, so the minus sign flips the exponent positive — carriers rise with heat, as expected for a semiconductor. Step 3 — plug in. $\frac{E _{g}}{2 k _{B}} = \frac{0.7}{2 \times 8.617 \times 1 0 ^{- 5}} = 4062 K$ . Exponent $= - 4062 \times (- 0.0008333) = 3.385$ . Result: factor $= e^{3.385} \approx 29.5$ . Heating 100 K multiplies carriers ~30-fold. Contrast this with a metal, whose conductivity drops on heating — see the trap below.

Recall Solution

Step 1 — form the index $S = \frac{electrons}{radius} \times 100$ (the ×100 just gives friendly numbers). Step 2 — compute.

Na: $1/102 \times 100 = 0.98$
Mg: $2/72 \times 100 = 2.78$
Al: $3/54 \times 100 = 5.56$ Step 3 — rank. $S_{Al} > S_{Mg} > S_{Na}$ , matching the melting-point order Na < Mg < Al. Both knobs (more electrons and smaller radius) reinforce each other down this list.

L5 — Mastery

Recall Solution

Part (a) — read the sign of the trend.

X's conductivity falls with heat → this is the metal / electron-sea fingerprint (electron–phonon scattering dominates; no gap to open). X is a conductor.
Y's conductivity rises steeply with heat → carriers are being unlocked across a gap → Y is an intrinsic semiconductor (band picture, Semiconductors & doping).

Part (b) — invert the carrier law to get $E_{g}$ . Using $\frac{n ( 600 )}{n ( 300 )} = exp [- \frac{E _{g}}{2 k _{B}} (\frac{1}{600} - \frac{1}{300})] = 1000$ :

Temperature bracket: $\frac{1}{600} - \frac{1}{300} = 0.0016667 - 0.0033333 = - 0.0016667 K^{- 1}$ .
Take logs: $6.908 = - \frac{E _{g}}{2 k _{B}} \times (- 0.0016667) = \frac{E _{g}}{2 k _{B}} \times 0.0016667$ .
So $\frac{E _{g}}{2 k _{B}} = \frac{6.908}{0.0016667} = 4145 K$ .
Then $E_{g} = 4145 \times 2 \times 8.617 \times 1 0^{- 5} = 0.714 eV$ . Result: $E_{g} \approx 0.71$ eV — a small-gap semiconductor, consistent with the strong thermal response. The opposite signs of the two trends, not their magnitudes, were the key clue.

Recall Solution

Step 1 — grant the true part. Both do involve electrostatic attraction between positive and negative charge. Fair. Step 2 — locate the decisive difference: localisation. In Ionic bonding electrons are transferred and pinned onto specific anions at fixed lattice sites — the bonding is directional. In a metal the electrons are delocalised and shared by every cation. Step 3 — a property that separates them.

Mechanical: slide a layer in an ionic crystal → like-charges align → repulsion → it shatters (brittle). Slide a layer in a metal → the directionless electron sea re-glues → it bends (malleable). One property, opposite outcome.
Electrical: ionic solids do not conduct when solid (charges fixed); metals do (electrons roam). See Figure s02. Conclusion: shared electrostatics ≠ same bond. Delocalisation is the dividing line, and it shows up plainly as malleability and conductivity.