Valence electrons and bonding: Worked examples — Hardware deep dive

Before we start, one promise: every symbol is earned before use. When we write $r_{0}$ , $U (r)$ , $E_{g}$ , or a group number, we say in plain words what it means and point at a picture.

The scenario matrix

Think of "bonding questions" as a grid. One axis is what the atom wants to do with electrons (give, take, share, or nothing). The other axis is what kind of question (classify, count, or compute with the energy curve). Here is every cell we must cover.

Cell	Case class	The trap / edge it tests
A	Metal + non-metal (big want-gap)	Sign of transfer → ionic
B	Identical atoms, 4 valence	Equal sharing → covalent (Si)
C	Two metals (loose electrons)	Delocalized → metallic
D	Degenerate: full outer shell (0 "want")	Noble gas — no bond
E	Degenerate: 1 valence vs 7 valence	Extreme want-gap, still ionic
F	Energy-curve calculus, general $m, n$	Find $r_{0}$ from $d U / d r = 0$
G	Energy-curve limiting behaviour	Deep well vs shallow well → band gap
H	Real-world word problem	Why doped silicon, not diamond
I	Exam twist (misleading count)	"7 valence ⇒ good conductor?" — NO

We build a "want" number for each atom: the number of electrons it must gain ( $+$ ) or lose ( $-$ ) to reach the stable count (8, or 2 for the first shell). This single signed number decides the whole story — so let's define it carefully.

Look at the number line above: valence $v$ runs $0 \to 8$ . On the left ( $v$ small, e.g. Na with 1) losing is cheap. On the right ( $v$ large, e.g. Cl with 7) gaining is cheap. Silicon sits dead centre at $v = 4$ : losing 4 costs the same as gaining 4, so it does neither — it shares. That is the geometric reason silicon is a semiconductor and not a metal or a salt.

Example A — Metal + non-metal (ionic)

Worked example Cell A: classify Na (group 1) + Cl (group 7)

Q: What bond forms between sodium and chlorine?

Forecast: Guess now — does someone give, take, or share? Which way do the electrons flow?

Step 1. Count valence: Na has $v = 1$ , Cl has $v = 7$ . Why this step? The outer-shell count is the only input that decides bonding; everything follows from it.

Step 2. Cheaper route for Na: lose $ℓ = 1$ vs gain $w = 7$ . Losing 1 wins. For Cl: lose $ℓ = 7$ vs gain $w = 1$ . Gaining 1 wins. Why this step? Nature takes the smaller of the two costs (our $w$ vs $ℓ$ rule).

Step 3. Na hands its 1 electron to Cl. Now Na $^{+}$ and Cl $^{-}$ attract by Coulomb force ⇒ ionic bond. Why this step? One atom donates, the other accepts — the definition of ionic.

Verify: After transfer, Na has 8 in its new outer shell (the shell below), Cl has $7 + 1 = 8$ . Both octets complete ✔. Charges: $+ 1$ and $- 1$ sum to $0$ (neutral compound) ✔.

Example B — Identical 4-valence atoms (covalent, Si)

Worked example Cell B: two silicon atoms

Q: What bond forms between two Si atoms, and how many electrons does one Si "see"?

Forecast: With equal atoms, can either one steal from the other? Guess the mechanism.

Step 1. Each Si has $v = 4$ , so $w = 8 - 4 = 4$ and $ℓ = 4$ . They are equal. Why this step? Equal costs ⇒ no atom is greedier, so no net transfer is possible.

Step 2. Since neither gives nor takes, each Si shares one electron with each of 4 neighbours. Why this step? Sharing lets both atoms count the pair — the only way to reach 8 with no theft.

Step 3. Si counts its own 4 + 4 shared = $4 + 4 = 8$ electrons ⇒ octet ✔ ⇒ covalent bond, forming the silicon lattice. Why this step? Each shared pair counts fully for both partners.

Verify: 4 bonds × 1 shared electron each = 4 shared; $4 + 4 = 8$ ✔. At $T = 0$ K all 4 are locked → insulator-like; a modest band gap $E_{g} \approx 1.1$ eV can free one → semiconductor ✔.

Example C — Two metals (metallic)

Example D — Degenerate: full shell (no bond)

Example E — Extreme want-gap (still ionic)

Worked example Cell E: aluminium (group 13,

$v = 3$ ) + fluorine (group 17, $v = 7$ ) Q: Bond type and formula for the Al–F compound?

Forecast: Al wants to lose 3, F wants to gain 1. How do the numbers balance?

Step 1. Al: $ℓ = 3$ (lose 3) vs $w = 5$ ; losing 3 is cheaper. F: $w = 1$ (gain 1) vs $ℓ = 7$ ; gaining 1 is cheaper. Why this step? Same cheaper-route rule; here the numbers are lopsided (3 vs 1).

Step 2. One Al gives 3 electrons, but each F only takes 1. So we need 3 F atoms per Al ⇒ formula $AlF_{3}$ . Why this step? Electrons donated must equal electrons accepted — conservation of charge.

Step 3. Al $^{3 +}$ + three F $^{-}$ ⇒ ionic (big electronegativity gap ⇒ transfer). Why this step? Metal + very electron-hungry non-metal always transfers.

Verify: Charge balance: $(+ 3) + 3 \times (- 1) = 0$ ✔. Al donated 3 = F accepted $3 \times 1$ ✔.

Example F — Energy curve, general exponents

Worked example Cell F: find

$r_{0}$ for general $U (r) = - A r^{- m} + B r^{- n}$ Q: With $m = 4, n = 8$ , find the equilibrium bond length $r_{0}$ in terms of $A, B$ .

Forecast: The parent's boxed formula is $r_{0} = (\frac{n B}{m A})^{1/ (n - m)}$ . Guess the answer before plugging in.

First, what do the symbols mean physically? $r$ is the distance between two atomic centres (see figure). $U (r)$ is the stored energy at that distance. The $- A / r^{m}$ term is long-range attraction (pulls atoms together); $+ B / r^{n}$ with $n > m$ is short-range repulsion (pushes them apart when too close). The bottom of the valley is where they settle: $r_{0}$ .

Step 1. Differentiate: $\frac{d U}{d r} = m A r^{- (m + 1)} - n B r^{- (n + 1)}$ . Why this step? The bottom of the valley has zero slope; $d U / d r$ is the slope. (This is why we use a derivative and not, say, an average: only the derivative pinpoints the flat minimum.)

Step 2. Set $= 0$ : $m A r^{- (m + 1)} = n B r^{- (n + 1)} \Rightarrow r^{n - m} = \frac{n B}{m A}$ . Why this step? Zero slope ⇒ attraction and repulsion forces exactly cancel — the definition of equilibrium.

Step 3. Plug $m = 4, n = 8$ : $r^{4} = \frac{8 B}{4 A} = \frac{2 B}{A} \Rightarrow r_{0} = (\frac{2 B}{A})^{1/4}$ . Why this step? $n - m = 4$ , so take the fourth root.

Verify: Matches boxed formula: $(\frac{n B}{m A})^{1/ (n - m)} = (\frac{8 B}{4 A})^{1/4} = (\frac{2 B}{A})^{1/4}$ ✔.

Example G — Limiting behaviour of the well depth

Worked example Cell G: deep well vs shallow well → band gap

Q: Two materials share the same functional form $U (r) = - A r^{- 6} + B r^{- 12}$ but material P has well depth $U_{m i n} = - 2$ eV and material Q has $U_{m i n} = - 8$ eV. Which is more likely a semiconductor, and which a hard insulator?

Forecast: Deeper well = ? Guess: does deep mean easy-to-break or hard-to-break?

Step 1. Well depth $∣ U_{m i n} ∣$ = energy to pull atoms apart = bond strength. Why this step? Depth measures how far you must climb out of the valley (figure s02, amber depth).

Step 2. Deeper well ⇒ stronger bond ⇒ larger band gap $E_{g}$ . Q ( $8$ eV) is much deeper than P ( $2$ eV). Why this step? Harder-to-break bonds need more energy to free an electron ⇒ wider gap.

Step 3. Limiting reasoning: as $∣ U_{m i n} ∣ \to$ large, $E_{g}$ grows → few electrons ever free → insulator (that's Q). As $∣ U_{m i n} ∣ \to$ moderate, thermal energy can occasionally break a bond → semiconductor (that's P). Why this step? This is exactly the limit that separates diamond ( $E_{g} \approx 5.5$ eV, insulator) from silicon ( $E_{g} \approx 1.1$ eV, semiconductor).

Verify: Ratio of depths $8/2 = 4$ , so Q's bond is 4× stronger ✔. Deeper ⇒ insulator (Q), shallower ⇒ semiconductor (P) ✔.

Example H — Real-world word problem

Worked example Cell H: "Why dope silicon instead of using pure diamond?"

Q: A student wants a controllable conductor for a transistor. Diamond (carbon, $v = 4$ ) and silicon (also $v = 4$ ) both form 4 covalent bonds. Why is doped silicon used, not diamond?

Forecast: Both have 4 valence electrons and the same lattice idea. What single number decides?

Step 1. Both C and Si have $v = 4$ ⇒ both form 4 covalent bonds. Bonding pattern is identical. Why this step? Rules out valence count as the difference.

Step 2. The difference is bond strength / well depth ⇒ band gap: diamond $E_{g} \approx 5.5$ eV, silicon $E_{g} \approx 1.1$ eV. Why this step? From Example G, deeper well = larger gap; carbon's smaller atoms bond more tightly.

Step 3. Ratio $5.5/1.1 = 5$ : diamond's gap is $5 \times$ silicon's. Thermal/impurity energy easily bridges $1.1$ eV but not $5.5$ eV. So silicon's carriers can be switched on by doping; diamond stays an insulator. Why this step? Controllability needs a gap thermal + dopant energies can cross.

Verify: $5.5/1.1 = 5.0$ exactly ✔ — diamond's gap is 5× larger, confirming silicon as the practical semiconductor ✔.

Example I — Exam twist (misleading count)

Worked example Cell I: "Chlorine has 7 valence electrons, almost a full shell — so it's an excellent conductor, right?"

Q: True or false, with reasoning.

Forecast: More valence electrons feels like more conduction. Trust it or not?

Step 1. Cl has $v = 7$ , needs $w = 1$ to reach 8 ⇒ it grabs one electron tightly. Why this step? A small $w$ means an atom that hoards, not one that frees electrons.

Step 2. Grabbed electrons are locked into completing octets, not delocalized. Conduction needs free/mobile electrons, not merely many. Why this step? From the parent note: it's mobility, not count, that conducts.

Step 3. Therefore FALSE — chlorine (as $Cl_{2}$ / chloride) is an insulator despite 7 valence electrons. Why this step? Ties the count-vs-mobility distinction to the classic exam trap.

Verify: $w = 8 - 7 = 1$ (grabs, doesn't free) ✔. High valence ≠ conduction ✔ — matches the parent's third [!mistake].

Recall Self-test: match each example to its matrix cell

Na+Cl ::: Cell A (ionic transfer) Si+Si ::: Cell B (covalent sharing) Neon ::: Cell D (zero-want, no bond) $U = - A r^{- 4} + B r^{- 8}$ , find $r_{0}$ ::: Cell F, answer $(2 B / A)^{1/4}$ Deep vs shallow well ::: Cell G (band-gap limit) Diamond vs silicon ::: Cell H (band gap 5.5 vs 1.1 eV) Chlorine "conductor?" ::: Cell I (FALSE — count ≠ mobility)