CHEM 548: Materials Chemistry

Lecture 5: Electron Counting and Stoichiometry

Wed 09/09/2026

Electron counting from the octet rule to the generalized 8−N rule and Zintl phases, and the crystal chemical formula

Worked examples: $\mathrm{SiO_2}$, $\mathrm{NaTl}$, $\mathrm{SnCl_2}$, $\mathrm{CaSb}$, $\mathrm{Al_2O_3}$, $\mathrm{Na_3N}$

Reading: Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry
Ch. 1 §1.3, §1.5.6

Learning Objectives

Octet ruleexplain how directional covalent bonding satisfies it

Bond countspredict them from Lewis structures and the trend from group 7A to 4A

Bonding classesdistinguish covalent, metallic, and ionic bonding by electron availability

VEC per anioncount the valence electrons an anion receives

Generalized 8−N rulepredict anion–anion bonds, cation–cation bonds, and cation lone pairs

Zintl phasesrecognize an electropositive metal donating electrons into a covalent network or dimer

The octet rule

Periodic table with Lewis dot symbols, colored by metal, metalloid, and nonmetal.

Electron count and covalent bonds

Across the main groups: Al with 12 nearest neighbors through Si, P4, and S8 to Cl2 with one bond per atom; metallic bonding gives way to covalent bonding.

Valence Electron Count per Anion

Cation Cthe more electropositive element, m per formula unit

Anion Athe more electronegative element, n per formula unit

Valence electrons per anion, against the octet

$\mathrm{VEC}_A = \dfrac{\textcolor{#ff5a5f}{e_C} \cdot \textcolor{#e0b060}{m} + \textcolor{#ff5a5f}{e_A} \cdot \textcolor{#e0b060}{n}}{\textcolor{#e0b060}{n}}$

valence electrons per cation valence electrons per anion

Octet obeyed$\mathrm{VEC}_A = 8$

The generalized octet rule

SiO₂

$\mathrm{VEC}_A = \dfrac{e_C \cdot m + e_A \cdot n}{n}$

SiO₂

$\mathrm{VEC}_A = \dfrac{e_C \cdot m + e_A \cdot n}{n}$

$\mathrm{VEC}_A = \dfrac{4 \cdot 1 + 6 \cdot 2}{2} = \dfrac{16}{2} = 8$

SiO₂

$\mathrm{VEC}_A = \dfrac{e_C \cdot m + e_A \cdot n}{n}$

The octet rule is obeyed$\mathrm{VEC}_A = \dfrac{4 \cdot 1 + 6 \cdot 2}{2} = \dfrac{16}{2} = 8$

NaTl

$\mathrm{VEC}_A = \dfrac{e_C \cdot m + e_A \cdot n}{n}$

$\mathrm{VEC}_A = \dfrac{1 \cdot 1 + 3 \cdot 1}{1} = 4$

Silicon dioxide is the better test, because the bonding in it is highly covalent.

NaCl was the ionic case — SiO₂ tests the count on covalent bonding

Let us put silicon and its two oxygens into the count.

[advance: work the silicon dioxide electron count]

The oxygens are the anions here, so they go on the bottom of the fraction as well as the top.

SiO₂ inputs: Si 4 electrons ×1, O 6 electrons ×2
Numerator 4 + 12 = 16 over two oxygens — 8 per oxygen, octet obeyed
Running tally of electrons — ionic or covalent alike

Now try the same count on sodium thallide.

[advance: now count sodium thallide]

Sodium thallide is where the count gets interesting.

NaTl: Na 1 + Tl 3, one each — 4 per thallium
Four is not eight — octet unsatisfied, yet NaTl is a real compound

The generalized 8 minus N rule tells us what a structure does when the count misses eight.

Generalized 8−N Rule

VEC_A < 8anion–anion bonds

VEC_A > 8cation–cation bonds or cation lone pairs

Electrons missing or left over, per anion

$\mathrm{VEC}_A = 8 + \textcolor{#ff5a5f}{CC}(\textcolor{#e0b060}{m}/\textcolor{#e0b060}{n}) - \textcolor{#ff5a5f}{AA}$

CC: electrons left on each cation AA: anion–anion bonds per anion

When the count misses eight, the compound has to make up the difference somewhere, and the generalized 8 minus N rule says where.

8 + CC(m/n) − AA: AA anion–anion bonds, CC electrons left on each cation
Below eight: anion bonds to other anions, as P, S, Si do as elements
Above eight: electrons stay on the cation — cation–cation bonds or lone pairs

Next we count sodium thallide, where thallium is the anion even though it is a metal.

[advance: count the electrons in sodium thallide]

Sodium thallide is the compound to try it on, and its count came out below eight.

Count: Na 1 + Tl 3 over the one thallium = 4 per thallium

Next we put that four into the rule and solve for the number of thallium-thallium bonds.

[advance: put the four into the rule, solve for the anion-anion bonds, and show the NaTl network]

Now put four in on the left.

8 > 4, so cation–cation bonds and lone pairs are no help — middle term drops out
AA = 4: each thallium bonds to four other thalliums, as the structure shows
Thallium builds a diamond network, sodium cations in its voids
Tl plus sodium's electron = carbon or silicon's count, so it builds their structure

Sodium thallide is the first member of a family we call the Zintl phases.

Zintl phases

Apply the generalized 8−N rule

SnCl₂ and CaSb

SnCl₂

$\mathrm{VEC}_A = \dfrac{4 \cdot 1 + 7 \cdot 2}{2} = \dfrac{18}{2} = 9$

$\mathrm{VEC}_A = 8 + CC(m/n) - AA$

$9 = 8 + CC(1/2) -$ $AA$

There are 2 electrons localized on Sn²⁺ that form a lone pair$CC = 2$

Start with tin dichloride.

SnCl₂: 4 + 14 = 18 over two chlorines — 9 per chloride
9 > 8: chlorides cannot take all of tin's electrons — AA out, CC = 2
Lone pair on Sn²⁺: 5p goes first, pair sits in 5s, nonbonding — common for Sn(II)

Now the second one, calcium antimonide.

[advance: now the calcium antimonide count]

Calcium antimonide is CaSb, one calcium and one antimony.

CaSb: 2 + 5 = 7 per antimony — below 8, CC out, one anion–anion bond each
Dimer from one bond each — Sb halogen-like, I₂
[Sb₂]⁴⁻: covalent polyatomic anion, charge balanced by the Ca²⁺ around it

Those two examples close out electron counting, so let me pull the story together.

Summary

Valence electrons per anion (VEC/A) counts what the cations donate and the anions accept. The octet then fixes the bond count across the p block: 7A (7e⁻) → 1 bond, 6A → 2, 5A → 3, 4A → 4 (diamond, Si, Ge); further left, bonding goes metallic through delocalized electrons.

VEC/A	What the generalized 8−N rule says	Example
= 8	The octet closes on the anion as drawn, even in a largely covalent solid	SiO₂
< 8	Too few electrons: anion–anion bonds form	Tl–Tl in NaTl Sb₂⁴⁻ in CaSb
> 8	Too many: the excess stays on the cation as lone pairs or cation–cation bonds	Sn²⁺ in SnCl₂

Zintl phases: an electropositive metal donates its electrons and the electronegative element builds the covalent structure — NaTl diamond-like, SrGa₂ graphite-like sheets.

Learning Objectives

Stoichiometry constraintexplain how a formula constrains the crystal structure

Crystal chemical formuladefine it and write it for a compound

Three balancesbalance electroneutrality, connectivity, and bond valence

Bond graphsconstruct one and read coordination numbers and bond valences back for SiO₂, TiO₂, CaF₂, Al₂O₃, and Na₃N

Parsimony and symmetryapply Pauling’s and Brown’s rules to a candidate structure

Structures follow stoichiometry

Four oxide structures: La2Zr2O7, TiO2, Al2O3, and ReO3.

Stoichiometry Constraints

Cation Cthe less electronegative element, charge c+

Anion Athe more electronegative element, charge a−

Assumptionevery atom of one element sits in identical surroundings

Three balances must hold

crystal chemical formula

$\mathrm{C}^{\textcolor{#ff5a5f}{c+}}_{\textcolor{#e0b060}{m}}\textcolor{#8d90e8}{[N]}\,\mathrm{A}^{\textcolor{#ff5a5f}{a-}}_{\textcolor{#e0b060}{n}}\textcolor{#8d90e8}{[M]}$

charge atoms in the formula coordination number

Cation Cthe less electronegative element, charge c+

Anion Athe more electronegative element, charge a−

Assumptionevery atom of one element sits in identical surroundings

Three balances must hold

crystal chemical formula

$\mathrm{C}^{\textcolor{#ff5a5f}{c+}}_{\textcolor{#e0b060}{m}}\textcolor{#8d90e8}{[N]}\,\mathrm{A}^{\textcolor{#ff5a5f}{a-}}_{\textcolor{#e0b060}{n}}\textcolor{#8d90e8}{[M]}$

charge atoms in the formula coordination number

Electroneutrality$\textcolor{#e0b060}{m} \times \textcolor{#ff5a5f}{c} = \textcolor{#e0b060}{n} \times \textcolor{#ff5a5f}{a}$

Connectivity$\textcolor{#e0b060}{m} \times \textcolor{#8d90e8}{N} = \textcolor{#e0b060}{n} \times \textcolor{#8d90e8}{M}$

Bond valence$\textcolor{#ff5a5f}{c}/\textcolor{#8d90e8}{N} = \textcolor{#ff5a5f}{a}/\textcolor{#8d90e8}{M}$

Givensilicon and oxygen, Si tetrahedral (CN 4), Si 4+, O 2−

Findempirical formula, coordination number of oxygen, Si–O bond valence

Givensilicon and oxygen, Si tetrahedral (CN 4), Si 4+, O 2−

Findempirical formula, coordination number of oxygen, Si–O bond valence

crystal chemical formula

$\mathrm{Si}^{\textcolor{#ff5a5f}{4+}}_{\textcolor{#e0b060}{m}}\textcolor{#8d90e8}{[4]}\,\mathrm{O}^{\textcolor{#ff5a5f}{2-}}_{\textcolor{#e0b060}{n}}\textcolor{#8d90e8}{[M]}$

charge atoms in the formula coordination number

Electroneutrality$\textcolor{#e0b060}{m} \times \textcolor{#ff5a5f}{4} = \textcolor{#e0b060}{n} \times \textcolor{#ff5a5f}{(2)}$

Connectivity$\textcolor{#e0b060}{m} \times \textcolor{#8d90e8}{N} = \textcolor{#e0b060}{n} \times \textcolor{#8d90e8}{M}$

Bond valence$\textcolor{#ff5a5f}{c}/\textcolor{#8d90e8}{N} = \textcolor{#ff5a5f}{a}/\textcolor{#8d90e8}{M}$

Givensilicon and oxygen, Si tetrahedral (CN 4), Si 4+, O 2−

Findempirical formula, coordination number of oxygen, Si–O bond valence

crystal chemical formula

$\mathrm{Si}^{\textcolor{#ff5a5f}{4+}}_{\textcolor{#e0b060}{m}}\textcolor{#8d90e8}{[4]}\,\mathrm{O}^{\textcolor{#ff5a5f}{2-}}_{\textcolor{#e0b060}{n}}\textcolor{#8d90e8}{[M]}$

charge atoms in the formula coordination number

Electroneutrality$\textcolor{#e0b060}{m} \times \textcolor{#ff5a5f}{4} = \textcolor{#e0b060}{n} \times \textcolor{#ff5a5f}{(2)}$$\mathrm{SiO}_2$

Connectivity$\textcolor{#e0b060}{1} \times \textcolor{#8d90e8}{4} = \textcolor{#e0b060}{2} \times \textcolor{#8d90e8}{M}$$M = 2$

Bond valence$\textcolor{#ff5a5f}{4}/\textcolor{#8d90e8}{4} = \textcolor{#ff5a5f}{2}/\textcolor{#8d90e8}{2}$$= 1$

Take a binary compound of two elements, the more electronegative one the anion of charge a minus, the more electropositive one the cation of charge c plus, and assume every atom of a given element sits in identical surroundings.

The crystal chemical formula records all of that at once
Charge as a superscript (red), number of atoms as a subscript (gold), coordination number in brackets (blue)
N is the cation's coordination number, M the anion's

Three balances have to hold.

[advance: the three balances]

Take the three balances one at a time.

Electroneutrality: m × c = n × a — the cation charge in the formula unit cancels the anion charge
Oxidation numbers stand in for true charges
Connectivity: m × N = n × M — a bond leaving a cation arrives at an anion, so both sides count the same bonds
Bond valence: c/N = a/M — oxidation number spread over the atom's bonds, close to bond order
Same electrons in the bond, so it comes out the same from either side

Now a worked case, silicon and oxygen.

[advance: pose the silicon-oxygen question]

So take the binary compound of silicon and oxygen.

Silicon strongly prefers tetrahedral coordination and nearly always adopts it with oxygen
From that one fact: the empirical formula, the coordination number of oxygen, the Si–O bond valence

Start with the charges and the electroneutrality balance.

[advance: write the electroneutrality balance]

Silicon is in the plus four oxidation state, oxygen is minus two, and we are given silicon's coordination number as four.

Oxidation numbers, not the electron count: O is −2 in an oxide, Si +4 to balance it
Si gives up all four valence electrons, so the two agree here — the balances always take the oxidation number
Electroneutrality: m × 4 = n × 2 — twice as much oxygen as silicon
Empirical formula SiO₂

The other two balances follow from the same numbers.

[advance: complete the connectivity and bond-valence balances]

Take connectivity first.

Connectivity: 1 × 4 = 2 × M — oxygen is two coordinate, linear or bent
Bond valence: 4/4 from silicon, 2/2 from oxygen — one per Si–O bond
Valence 1 = one unit of the +4 and one of the −2 per bond = a single bond, one shared pair
Charge per bond, not electrons per bond — TiO₂ will give 2/3, a number no molecule has

The same bookkeeping is easier to see as a drawing. That drawing is the bond graph.

Bond graphs

$\mathrm{Si}^{\textcolor{#ff5a5f}{4+}}\textcolor{#8d90e8}{[4]}\mathrm{O}^{\textcolor{#ff5a5f}{2-}}_2\textcolor{#8d90e8}{[2]}$

Every atom in the empirical formula goes into the bond graph, and then we draw the bonds as lines.

SiO₂: Si four coordinate, four lines out, two to each oxygen
Read back: O two coordinate, Si⁴⁺/4 and O²⁻/2 — valence one per bond

Now the same drawing for titanium dioxide.

[advance: add the titanium dioxide bond graph]

Titanium dioxide is the next step.

TiO₂: Ti larger, CN 6 octahedron — three lines per oxygen, O three coordinate
Not a triple bond — lines are coordination, not bond order
Ti⁴⁺/6 and O²⁻/3 = two thirds — noninteger valence, unlike carbon's four
TiO₂ is our commonest white pigment — paint and some sunscreens

Now calcium fluoride.

[advance: add the calcium fluoride bond graph]

Calcium fluoride is bigger again.

CaF₂: Ca commonly 8; connectivity makes each fluoride four coordinate
The fluorite structure from last lecture: tetrahedral fluoride, eight fluorides on a cube around Ca
Bond valence 2/8, a quarter per bond

Behind these assignments sit two rules about symmetry.

All things equal nature favors symmetric structures

Pauling, parsimonyfew chemically distinct coordination environments per ion, ideally one

Brown, maximum symmetrythe most symmetric structure the constraints allow

Bond graphs: Al₂O₃ and Na₃N

GivenAl₂O₃ and Na₃N, rule of parsimony obeyed

Findcoordination numbers and bond valences

Here are two compounds for you, aluminum oxide and sodium nitride.

Assume parsimony holds in both — give the coordination numbers and the bond valences

Next we pick a coordination number for aluminum and see what the oxygens end up with.

[advance: try tetrahedral aluminum]

Start with aluminum oxide.

Al₂O₃: Al +3, O −2 — electroneutrality gives two Al and three O
Assume a CN for aluminum: in oxides it is tetrahedral or octahedral
Try tetrahedral: four bonds from each aluminum

Now let us see whether that works.

[advance: cross out the tetrahedral attempt]

Count the lines arriving at each oxygen and the attempt fails.

Two oxygens at three bonds, the third at two
Violates parsimony — two chemically different oxygens
Not forbidden, and real structures do it, but avoid it where we can

So four-coordinate aluminum is out, and we go back to the bare atoms.

[advance: clear the graph and start again]

Clear the bonds and keep the same five atoms.

Two aluminums, three oxygens stay — only the bonds go

Now try octahedral aluminum.

[advance: the octahedral answer, with bond valence one half]

Now make the aluminum octahedral, six bonds from each aluminum, twelve bonds in all, and each of the three oxygens takes four of them.

Every oxygen the same environment — parsimony satisfied; formula Al₂[6]O₃[4]
Bond valence 3/6 and 2/4 — one half either way

Now the second compound, sodium nitride.

[advance: set up the sodium nitride graph]

Sodium nitride is the same exercise.

Na₃N: N −3, Na +1 — one nitrogen, three sodiums
N takes four or six neighbors, rarely more; large Na runs six to twelve
Take nitrogen at six, the top of its range

Now split those six bonds among the three sodiums.

[advance: the sodium nitride answer]

The formula and the bond valence come out of that graph.

Six bonds split three ways — Na two coordinate, formula Na₃[2]N[6]
Bond valence 1/2 and 3/6 — one half again
Two coordinate sodium is an extreme outlier — Na₃N first made 2002
Ordinary route gives sodium azide, not the nitride
Na₃N decomposes near 360 K even once made

That is the whole toolkit, so let me summarize.

Summary

Crystal structures are not random. The crystal chemical formula records the oxidation states, coordination numbers, and bonding, and three balances have to close on it.

Balance	What balances	Rule
Electroneutrality	Charges	Cation and anion charges sum to zero across the formula
Connectivity	Bond counts	Bonds leaving the cations equal the bonds arriving at the anions
Bond valence	Bond strength	Oxidation state ÷ coordination number; fractional values are allowed

A bond graph draws that bookkeeping out; Pauling’s parsimony minimizes distinct environments and Brown’s maximum symmetry takes the most symmetric structure the stoichiometry allows.

SiO₂, TiO₂, CaF₂, Al₂O₃, and Na₃N all fall out of those rules, the rare ones included: Na₃N was first synthesized in 2002 and is unstable at 360 K.

Practice questions

Work the Lecture 5 practice questions before the next class. They cover the valence electron count per anion, the generalized 8−N rule and Zintl phases, and the crystal chemical formula with its three balances.

Open the Lecture 5 practice questions

Every question carries a worked explanation, so you can check your reasoning as you go.

CHEM 548: Materials Chemistry

Lecture 5: Electron Counting and Stoichiometry

Learning Objectives

The octet rule

Electron count and covalent bonds

Valence Electron Count per Anion

The generalized octet rule

Generalized 8−N Rule

Zintl phases

Apply the generalized 8−N rule

SnCl2 and CaSb

Summary

Learning Objectives

Structures follow stoichiometry

Stoichiometry Constraints

Bond graphs

All things equal nature favors symmetric structures

Bond graphs: Al2O3 and Na3N

Summary

Practice questions

SnCl₂ and CaSb

Bond graphs: Al₂O₃ and Na₃N