Valence electrons and bonding: Exercises — Hardware deep dive

Definition Two words this page leans on — pinned here so nothing is used before it is defined

Octet target (why 8?). Atoms are most stable when their outermost shell holds 8 electrons (a filled $s^{2} p^{6}$ set). Why 8 and not some other number? A filled outer $s$ subshell (2 slots) plus a filled outer $p$ subshell (6 slots) makes $2 + 6 = 8$ — a symmetric, low-energy, closed arrangement, the same one the chemically inert noble gases have. Nature slides toward low energy, so atoms rearrange their outer electrons to reach this count. Whenever this page says "needs 4 more" or "reach 8," this is the rule it invokes — the parent Valence electrons and bonding builds it in full. Electronegativity ( $χ$ ). A single number, on the Pauling scale (dimensionless, running from about $0.7$ for cesium to $4.0$ for fluorine), that says how strongly one atom pulls shared electrons toward itself inside a bond. Why we need it: it lets us predict, from two numbers, whether a shared pair stays roughly in the middle (both pull about equally ⇒ covalent) or gets dragged almost entirely onto one atom (a big pull difference ⇒ transfer ⇒ ionic). A large gap in $χ$ between two atoms signals ionic; a small gap signals covalent.

Level 1 — Recognition

Goal: read a fact straight off the definition.

Recall Solution 1.1

What we need: the outermost occupied shell count. Silicon's electron configuration is $[Ne] 3 s^{2} 3 p^{2}$ . The neon core $[Ne]$ is the filled inner shells (10 electrons) — those are spectators. The highest shell is $n = 3$ , holding $3 s^{2}$ (2 electrons) plus $3 p^{2}$ (2 electrons). $2 + 2 = 4 valence electrons$ See Atomic structure and electron shells for why the $[Ne]$ core is inert.

Recall Solution 1.2

Ionic → transfer (one atom donates, the other accepts).
Covalent → share (a pair belongs to both atoms at once).
Metallic → delocalize (electrons pool into a shared "sea"). Mnemonic from the parent: "I Steal, We Share, They Swim."

Level 2 — Application

Goal: apply one rule to a fresh case.

Recall Solution 2.1

Step 1 — count valence. Outer shell $n = 3$ : $3 s^{2} + 3 p^{3} = 2 + 3 = 5$ valence electrons. Step 2 — distance to octet. The octet target is 8 (see the pinned definition — a full $s^{2} p^{6}$ outer set), so it needs $8 - 5 = 3$ more. Why this matters: when phosphorus sits in a silicon lattice it forms 4 bonds like its neighbours but arrives with 5 electrons — one left over. That spare electron is the whole idea behind Doping and carriers.

Recall Solution 2.2

Step 1 — count valence. Mg has 2, O has 6. Step 2 — name the deciding energies. Two measurable quantities settle whether electrons transfer (ionic) or stay shared (covalent):

Ionization energy = energy to remove an electron from an atom. Metals like Mg have a low ionization energy (its first two ionizations, $\approx 7.6$ eV and $\approx 15$ eV, are modest for a metal), so it gives electrons up cheaply.
Electron affinity = energy released when an atom gains an electron. Oxygen has a large, favourable electron affinity (first electron affinity $\approx 1.5$ eV released), so it grabs electrons and lowers its energy. Step 3 — compare the routes with actual electron counts. To reach the octet target of 8, each atom has a shortest path and a longer path:
O has 6, so it can gain 2 (short path) — and gaining electrons releases energy for oxygen. To instead reach 8 by losing, O would have to shed all 6 outer electrons, paying six successive ionization energies (each larger than the last as the atom grows more positive) — an enormous energy cost. So O's cheap route is to gain 2, not lose 6.
Mg has 2, so it can lose 2 (short path) at its modest ionization energies. To instead reach 8 by gaining, Mg would have to accept 6 electrons, cramming them onto an atom that has little pull for them (Mg's electron affinity is essentially zero/unfavourable) — again a huge cost with no energy payback.
The quantitative point: the two short paths dovetail perfectly — Mg loses exactly the 2 that O wants to gain — and each short path is downhill or cheap, whereas both long (gain-6 / lose-6) paths pile up many ionization energies with no compensating release. Losing/gaining 6 is therefore energetically far worse than the 2-for-2 exchange. Step 4 — decide. Mg donates 2 → $Mg^{2 +}$ ; O accepts 2 → $O^{2 -}$ . Opposite charges attract by Coulomb force ⇒ ionic bond. Why not covalent? Sharing wins only when the two atoms pull on the shared electrons comparably — that is, when their electronegativities $χ$ (defined at the top: how hard an atom tugs shared electrons) are close. Here the gap is large: O has $χ \approx 3.4$ , Mg has $χ \approx 1.3$ on the Pauling scale, a difference of about $2.1$ . A gap that big ( $≳ 1.7$ is the usual ionic threshold) means the shared electrons would sit almost entirely on oxygen — and "electrons sitting entirely on one atom" is transfer, i.e. ionic.

Level 3 — Analysis

Goal: combine rules and explain the "why".

Recall Solution 3.1

Step 1: Si has 4 valence electrons and needs 4 more to hit the octet target of 8 (the pinned "why 8?" rule). Step 2: It cannot easily donate 4 (too many to strip) or accept 4 (too many to grab) — it is balanced (its electronegativity $χ \approx 1.9$ sits right in the middle, so it neither dumps nor snatches). So it shares: it forms one covalent bond per missing electron ⇒ 4 bonds, to 4 neighbours. Step 3: Each bond is a shared pair counted fully by both atoms. So one Si contributes 4 of its own electrons and counts 4 more (one from each neighbour): $4_{own} + 4_{shared-in} = 8 ✓$ Walk the figure below, piece by piece. The single violet disc at the centre labelled "Si" is our chosen atom. The four magenta discs at the surrounding corners are its four nearest neighbours — drawn at the corners of a tetrahedron (three splayed out, one tucked toward us) because that is the real 3-D shape silicon adopts to keep the four bonds as far apart as possible. Each orange line joining the centre to a corner is one covalent bond, and the two small navy dots sitting on each orange line are the shared electron pair for that bond. Now do the counting on the picture: pick any orange line and notice both discs it touches lay equal claim to its navy dots — the central Si counts all four lines' near-side electrons plus its own four, and each corner atom likewise counts the same shared pair from its side. That deliberate double-counting of every shared pair is exactly the arithmetic $4 + 4 = 8$ from Step 3 made visible. This tetrahedral unit is the repeating seed of the Silicon crystal lattice.

Figure: central Si (violet) covalently bonded to 4 neighbours (magenta); orange lines = shared pairs, navy dots = shared electrons, tetrahedral geometry.

Recall Solution 3.2

Step 1 — the cold state (bond picture). At absolute zero every valence electron is paired up in a covalent bond. No electron is free to drift ⇒ no current ⇒ it acts like an insulator. Step 2 — the band picture (what is really going on). When many Si atoms sit in a lattice, their shared electrons are not trapped in one bond — quantum mechanics spreads them into delocalized states across the whole crystal. These states clump into two allowed energy ranges: a full lower band (the valence band) and an empty upper band (the conduction band), separated by a forbidden gap of width $E_{g} \approx 1.1$ eV. See Energy bands and band gap. "Breaking a bond" is the everyday shorthand for promoting one electron across this gap, from the valence band into the conduction band — it is a change of collective electronic state, not the snapping of one isolated stick. Step 3 — supply energy. Heat, light, or a field can hand an electron $\geq E_{g}$ . Promoted to the conduction band, the electron is now free to roam. Step 4 — two carriers appear. The promoted electron is a negative carrier; the vacancy it left in the valence band behaves like a mobile hole (a positive carrier). Both conduct — see Doping and carriers. Step 5 — conclusion. Higher temperature ⇒ more electrons promoted across $E_{g}$ ⇒ more carriers ⇒ conductivity rises with temperature. That temperature-controllable, mid-sized gap is exactly what makes silicon a semiconductor — see Conductors insulators and semiconductors.

Level 4 — Synthesis

Goal: build and solve a small model.

Recall Solution 4.1

Why $d U / d r = 0$ ? The atoms settle where the net force vanishes — the bottom of the energy well. Force is the negative slope of $U$ , so a flat slope ( $d U / d r = 0$ ) means no push either way. Read the figure before the algebra. The horizontal axis is the separation $r$ between the two atoms; the vertical axis is their shared potential energy $U (r)$ . Follow the violet curve from the right: far apart (large $r$ ) the energy is near zero (the dotted navy line marks $U = 0$ , "atoms not interacting"). As they approach, the attraction pulls the curve down into a valley, then at very small $r$ the sharp repulsion shoots it steeply up. The single lowest point of that valley is the magenta dot — that is where the slope is momentarily flat, i.e. $d U / d r = 0$ . The dashed orange vertical line dropped from the dot to the $r$ -axis reads off the equilibrium separation $r_{0}$ , and the vertical drop from the dotted $U = 0$ line down to the dot is the well depth, the energy you must pay to pull the atoms back apart.

Figure: violet curve $U (r)$ ; magenta dot = minimum at $r_{0} = (2 B / A)^{1/6}$ ; the vertical distance from the dot up to the dotted $U = 0$ line is the well depth (dissociation energy).

Step 1 — differentiate. Using $\frac{d}{d r} r^{- k} = - k r^{- k - 1}$ : $\frac{d U}{d r} = 6 A r^{- 7} - 12 B r^{- 13}$ Step 2 — set to zero. At the minimum (the magenta dot) the slope is flat: $6 A r^{- 7} - 12 B r^{- 13} = 0 \Rightarrow 6 A r^{- 7} = 12 B r^{- 13}$ Step 3 — solve, one algebraic move at a time.

Move 1 — clear the negative powers so we can isolate $r$ . Multiply both sides by $r^{13}$ . On the left, $r^{- 7} \cdot r^{13} = r^{6}$ ; on the right, $r^{- 13} \cdot r^{13} = r^{0} = 1$ : $6 A r^{6} = 12 B .$
Move 2 — get $r^{6}$ alone. Divide both sides by $6 A$ (allowed since $A > 0$ , so we never divide by zero), isolating the power of $r$ : $r^{6} = \frac{12 B}{6 A} = \frac{2 B}{A} .$
Move 3 — undo the 6th power. Take the positive real 6th root of both sides (we keep only the positive root because a separation cannot be negative): $r_{0} = (\frac{2 B}{A})^{1/6}$ This matches the general formula $r_{0} = (\frac{n B}{m A})^{1/ (n - m)}$ with $m = 6, n = 12$ .

Recall Solution 4.2

Step 1 — separation. $r_{0} = (\frac{2 B}{A})^{1/6} = (\frac{2 ( 1 )}{2})^{1/6} = 1^{1/6} = 1.$ Step 2 — evaluate $U$ at $r_{0} = 1$ . $U (1) = - \frac{2}{1 ^{6}} + \frac{1}{1 ^{12}} = - 2 + 1 = - 1 .$ The negative sign means the bonded state sits below the "atoms far apart" energy (which is $0$ as $r \to \infty$ — the dotted line in the figure). The dissociation energy is the depth $∣ U_{m i n} ∣ = 1$ unit: that much must be supplied to pull the atoms apart. A deeper well ⇒ a stronger bond — but remember (see the caveat above) this dissociation energy is a mechanical bond-breaking energy, not the electronic band gap $E_{g}$ .

Level 5 — Mastery

Goal: the general, edge-case-proof version.

Recall Solution 5.1

Step 1 — first derivative. $U^{'} (r) = \frac{m A}{r ^{m + 1}} - \frac{n B}{r ^{n + 1}}, U^{'} (r_{0}) = 0 \Rightarrow \frac{m A}{r _{0}^{m + 1}} = \frac{n B}{r _{0}^{n + 1}} .$ Step 2 — second derivative (a minimum needs $U^{''} (r_{0}) > 0$ , i.e. the curve bends upward like a bowl): $U^{''} (r) = - \frac{m ( m + 1 ) A}{r ^{m + 2}} + \frac{n ( n + 1 ) B}{r ^{n + 2}} .$ Step 3 — substitute the balance condition. From Step 1, $\frac{n B}{r _{0}^{n + 1}} = \frac{m A}{r _{0}^{m + 1}}$ ; divide both sides by $r_{0}$ to match the $U^{''}$ powers: $\frac{n B}{r _{0}^{n + 2}} = \frac{m A}{r _{0}^{m + 2}} .$ Insert into $U^{''}$ :

= \frac{mA}{r_0^{m+2}}\Big[(n+1) - (m+1)\Big] = \frac{mA}{r_0^{m+2}}\,(n-m).$$ **Step 4 — sign check.** Since $A>0$, $m>0$, $r_0>0$, and $n>m$ gives $n-m>0$, every factor is positive: $$U''(r_0) > 0 \;\Rightarrow\; \text{genuine minimum} \;\Rightarrow\; \text{a stable bond.} \checkmark$$ **Edge case $n = m$ (equal powers).** If the two exponents are equal the potential collapses to $$U(r) = -\frac{A}{r^{m}} + \frac{B}{r^{m}} = \frac{B - A}{r^{m}},$$ a single power law with no competition between the terms. Its derivative $U'(r) = -m(B-A)\,r^{-m-1}$ is **never zero** for finite $r>0$ (unless $A=B$, which makes $U\equiv 0$). So the curve is monotone — it slides one way with no valley — and there is **no equilibrium separation and no bond**. **Edge case $n < m$ (repulsion falls off *slower* than attraction).** Now look at the two terms at small $r$: the term with the *larger* exponent blows up fastest as $r\to0$. Here that larger exponent belongs to the **attraction** $-A/r^{m}$, so at short range attraction *dominates* and $U\to-\infty$ — the atoms are pulled ever closer with no repulsive wall to stop them, so they **collapse together**. And the stationary point our formula would spit out is actually a **maximum**: repeating Step 4, the curvature factor $(n-m)$ is now *negative*, giving $U''(r_0)<0$, a hilltop the atoms roll off of, i.e. an *unstable* equilibrium — not a bond. **The moral.** The single inequality $n>m$ — "repulsion is the *sharper*, faster-rising term at short range" — is exactly the condition that (i) makes the curvature $(n-m)$ positive, carving out a true bowl-shaped well, and (ii) supplies the hard inner wall that keeps the atoms from collapsing. Drop it and the bond disappears.

Recall Solution 5.2

Step 1 — same geometry, different gap. Both have the same tetrahedral, 4-bond arrangement, so both build a filled valence band and an empty conduction band. What differs is the width of the forbidden gap $E_{g}$ between them. Step 2 — why the gap differs. Carbon's atoms are smaller and its shared electrons are held more tightly, so its bonding and antibonding states split further apart — a wider band gap. Silicon's larger atoms bind less tightly, giving a narrower gap. (This is a band-structure statement, not a claim that one covalent stick is stronger than another.) Step 3 — what the gap does. Carbon: $E_{g} \approx 5.5$ eV — far more than room-temperature thermal energy ( $\sim 0.025$ eV) can supply, so almost no electrons are promoted across it ⇒ essentially no free carriers ⇒ insulator. Step 4 — silicon's middle ground. Silicon's $E_{g} \approx 1.1$ eV is small enough that a tiny but non-zero fraction of electrons cross at room temperature, and doping can add carriers on demand. Not zero (insulator) and not overflowing (metal) — the controllable in-between that defines a semiconductor. Full band picture: Energy bands and band gap and Conductors insulators and semiconductors.

Recall Self-check: did you climb the whole ladder?

L1 count valence off a configuration ::: Silicon $3 s^{2} 3 p^{2} \Rightarrow 4$ . L2 pick a bond type from ionization energy vs electron affinity ::: Mg (low ionization energy) donates, O (high electron affinity) accepts $\Rightarrow$ ionic. L3 explain conduction rising with temperature ::: electrons promoted across $E_{g} \approx 1.1$ eV from valence to conduction band, leaving mobile holes. L4 derive $r_{0}$ for $- A r^{- 6} + B r^{- 12}$ ::: $r_{0} = (2 B / A)^{1/6}$ . L5 prove it's a minimum ::: $U^{''} (r_{0}) = \frac{m A}{r _{0}^{m + 2}} (n - m) > 0$ when $n > m$ .