College Chemistry • Atomic Structure & Periodicity

Valence Electrons and Ionic Compounds

How outermost electrons govern the formation, structure, and stability of ionic solids.

Laboratory test tubes with an atom and science notebook.
Inside this lesson 10 sections 5 questions 8 figures

SECTION 1

Historical Context & Motivation

The modern understanding of chemical bonding rests on a surprisingly recent insight: that only the outermost electrons of an atom participate meaningfully in bond formation. For much of the nineteenth century, chemists could describe the stoichiometry of salts such as NaCl and CaF2 without understanding why particular ratios occurred. Dalton's atomic theory provided combinatorial rules, but it offered no mechanism explaining why sodium invariably loses one unit of charge while chlorine gains one. The answer would emerge only after a series of discoveries linking atomic structure to electron configuration and, ultimately, to the concept of valence electrons.

1897 • Discovery of the Electron

J. J. Thomson's cathode-ray experiments reveal the electron as a universal subatomic particle, establishing that atoms have internal structure and that negative charges can be removed.

1913 • Bohr's Quantized Orbits

Niels Bohr proposes that electrons occupy discrete energy levels around the nucleus, providing the first model that explains why certain electron configurations are more stable than others.

1916 • Lewis's Electron-Pair Model

Gilbert N. Lewis introduces the cubical-atom model and the concept of shared electron pairs, articulating the octet rule and distinguishing ionic from covalent bonding for the first time.

1918 • Kossel's Ionic Bond Theory

Walther Kossel independently proposes that ionic compounds form when atoms transfer electrons to achieve noble-gas configurations, formalizing the electrostatic model of ionic bonding.

1926–1927 • Quantum Mechanical Framework

Schrödinger and Heisenberg develop wave mechanics and matrix mechanics, respectively, placing the concept of electron shells and subshells on rigorous mathematical footing and refining the notion of valence.

The central question that drove these developments can be stated succinctly: What determines the charge of an ion, and why do certain combinations of ions yield stable crystalline solids? Answering this question requires a thorough understanding of electron configurations, the periodic law, and the energetics of electron transfer—topics we now address in sequence.

SECTION 2

Core Principles & Definitions

Before examining ionic compounds in detail, it is essential to establish a precise vocabulary and a set of foundational ideas. Valence electrons are defined as the electrons in the outermost principal energy level (highest n quantum number) of an atom in its ground state. For main-group elements, valence electrons reside in ns and np orbitals, whereas for transition metals the picture is more nuanced because (n − 1)d electrons can also participate in bonding. The following grid summarizes the core concepts underlying this lesson.

1

Valence Electrons

Electrons in the highest principal energy level of a ground-state atom. These are the electrons that participate in chemical bonding and determine an element's chemical reactivity and group number placement.
2

Octet Rule

Main-group atoms tend to gain, lose, or share electrons until they possess eight electrons in their valence shell, achieving a noble-gas electron configuration (ns²np⁶). Hydrogen and lithium follow a duet analog.
3

Electron Transfer & Ion Formation

When atoms with vastly different electronegativities react, the more electropositive atom donates its valence electrons to the more electronegative atom, producing cations and anions held together by Coulombic attraction.
4

Crystal Lattice Energy

Ionic compounds do not exist as discrete molecules but as three-dimensional crystal lattices. The stability of the lattice is quantified by lattice energy—the enthalpy required to separate one mole of an ionic solid into gaseous ions.
5

Periodicity of Valence Count

The number of valence electrons for main-group elements equals the group number in the IUPAC system (Groups 1–2 and 13–18). This periodicity directly predicts common ion charges and compound stoichiometries.
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KEY TAKEAWAY

Think of an atom's valence electrons as the items in a person's hands at a crowded market: core electrons are safely packed in a backpack and rarely participate in transactions, but the items you are holding—your valence electrons—are readily exchanged, shared, or dropped. In ionic bonding, one trader hands items directly to another, and the resulting electrostatic attraction (positive seller, negative buyer) keeps them together as a stable partnership.

SECTION 3

Electron Transfer & Ion Formation — A Visual Explanation

The formation of an ionic compound can be visualized as a sequence of electron-transfer events in which a metal atom donates its valence electrons to a nonmetal atom. The diagram below illustrates this process for sodium chloride (NaCl), the archetypal ionic compound. Sodium ([Ne]3s1) possesses a single valence electron in its 3s orbital, whereas chlorine ([Ne]3s²3p5) has seven valence electrons and requires one more to complete its octet. Transfer of sodium's 3s electron to chlorine produces Na+ and Cl−, both isoelectronic with neon and argon, respectively.

Electron Transfer: Formation of NaClNa[Ne]3s¹e⁻transferCl[Ne]3s²3p⁵After electron transferNa⁺[Ne]Cl⁻[Ar]Coulombic attraction
The sodium atom donates its single 3s valence electron (gold dot) to the chlorine atom, yielding Na+ (isoelectronic with Ne) and Cl− (isoelectronic with Ar). The resulting electrostatic attraction binds the ions together.

Several features of this diagram merit attention. First, notice that the sodium cation (Na+) is drawn smaller than neutral sodium because the loss of the outermost electron reduces electron–electron repulsion and contracts the remaining electron cloud, while the chloride anion (Cl−) is drawn larger because the added electron increases repulsion and expands the electron cloud. Second, both ions achieve noble-gas configurations—a recurring theme in ionic bond formation. Third, the dashed line between the ions symbolizes the non-directional Coulombic attraction that characterizes ionic bonding, in contrast to the directional electron sharing found in covalent bonds.

SECTION 4

Mathematical Framework — Energetics of Ionic Bond Formation

The spontaneous formation of ionic compounds from their constituent elements can be understood quantitatively through the interplay of several energy terms. The key quantities are ionization energy (IE), electron affinity (EA), and lattice energy (U). While the electron-transfer step itself is endothermic for most pairs (IE > |EA|), the large exothermic lattice energy that accompanies crystal formation drives the overall process to be thermodynamically favorable.

COULOMB'S LAW FOR ION PAIRS

E=k×(q+×q−)/dE = k × (q₊ × q₋) / d

where E is the electrostatic potential energy, k is the Coulomb constant (8.99 × 10⁹ N·m²/C²), q₊ and q₋ are ion charges, and d is the interionic distance. More negative E → stronger attraction.

BORN–LANDÉ EQUATION (LATTICE ENERGY)

U=−(NA×M×q+×q−)/(4πε0×r0)×(1−1/n)U = −(N_A × M × q₊ × q₋) / (4πε₀ × r₀) × (1 − 1/n)

where NA is Avogadro's number, M is the Madelung constant (geometry-dependent), r₀ is the nearest-neighbor interionic distance, ε₀ is the permittivity of free space, and n is the Born exponent (reflecting the compressibility of the ions).

BORN–HABER CYCLE (ENTHALPY OF FORMATION)

ΔH°f=ΔH°sub+½D0+IE−EA−UΔH°_f = ΔH°_sub + ½D₀ + IE − EA − U

The Born–Haber cycle dissects the overall enthalpy of formation (ΔH°f) into sublimation of the metal (ΔH°sub), dissociation of the diatomic nonmetal (½D₀), ionization energy (IE), electron affinity (EA), and lattice energy (U). This Hess's-law application confirms that the large lattice energy is the dominant stabilizing term.

The Born–Landé equation reveals two critical trends: lattice energy increases with higher ion charges and smaller interionic distances. This explains why MgO (with 2+ and 2− charges and small ionic radii) has a much higher melting point (2852 °C) than NaCl (801 °C), despite both adopting the rock-salt structure. The Madelung constant depends on the crystal geometry—1.7476 for the NaCl structure, 1.7627 for CsCl—and accounts for the net electrostatic effect of all ion–ion interactions in the lattice, not just nearest neighbors.

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Common Pitfall

Students often assume that a positive electron affinity means the process is exothermic. Be careful with sign conventions: in many general chemistry texts, EA is reported as a positive number when energy is released (thermodynamic convention uses a negative ΔH for exothermic processes). Always check whether your source defines EA as −ΔH or as ΔH for X(g) + e⁻ → X⁻(g). In this lesson, we follow the convention where EA is positive when energy is released upon electron attachment.

SECTION 5

Periodicity of Valence Electrons & Predicting Ionic Charges

One of the most powerful consequences of the periodic law is that the number of valence electrons for main-group elements can be read directly from the periodic table. Group 1 elements have one valence electron, Group 2 have two, Group 13 have three, and so on through Group 18 with eight (except helium, which has two). Transition metals complicate this pattern because their (n − 1)d orbitals are close in energy to the ns orbital, leading to variable oxidation states. The table below maps the main-group families to their valence counts and common ionic charges.

IUPAC GroupValence e⁻Common Ion ChargeExample IonIsoelectronic Noble Gas
1 (Alkali metals)1+1Na⁺Ne
2 (Alkaline earth metals)2+2Mg²⁺Ne
13 (Boron group)3+3Al³⁺Ne
15 (Pnictogens)5−3N³⁻Ne
16 (Chalcogens)6−2O²⁻Ne
17 (Halogens)7−1Cl⁻Ar
18 (Noble gases)8 (2 for He)0 (rarely form ions)——

Valence electron count and typical ionic charges for main-group elements.

Simplified Periodic Table — Valence Electrons & Ion ChargesMETALS — tend to LOSE valence e⁻ → CATIONSNONMETALS — tend to GAIN e⁻ → ANIONSGrp 1Grp 2Grp 13Grp 14Grp 15Grp 16Grp 17Grp 181 e⁻2 e⁻3 e⁻4 e⁻5 e⁻6 e⁻7 e⁻8 e⁻Na→ Na⁺Mg→ Mg²⁺Al→ Al³⁺N→ N³⁻O→ O²⁻Cl→ Cl⁻Arstable+1+2+3−3−2−10Example Ionic Compounds FormedNaClNa⁺ + Cl⁻MgOMg²⁺ + O²⁻CaF₂Ca²⁺ + 2 F⁻Charge balance: total cation charge must equal total anion charge for a neutral compound.Lose electrons (metals)Gain electrons (nonmetals)Increasing electronegativity →
A simplified representation of main-group elements showing the number of valence electrons, common ion charges, and representative ionic compounds. Metals (violet borders) lose electrons to form cations, while nonmetals (green borders) gain electrons to form anions. Charge balance determines the formula of each compound.

The diagram above reveals a key pattern: metals on the left side of the periodic table form cations by losing their few valence electrons, while nonmetals on the right side form anions by gaining enough electrons to complete their octets. Group 14 elements (C, Si, Ge) occupy an intermediate position and generally prefer covalent bonding because losing or gaining four electrons is energetically prohibitive. The formula of any ionic compound is then determined by the constraint of electrical neutrality: the total positive charge contributed by cations must exactly balance the total negative charge of anions.

SECTION 6

Worked Example — Predicting the Formula and Lattice Energy Trend for an Ionic Compound

Consider the reaction between calcium (Group 2) and chlorine (Group 17). We will predict the formula of the resulting ionic compound and explain why its lattice energy is significantly higher than that of NaCl.

Predicting the Formula of Calcium Chloride and Comparing Lattice Energies

  1. 1

    Step 1 — Determine Valence Electrons

    Calcium is in Group 2 with the electron configuration [Ar]4s². It has 2 valence electrons. Chlorine is in Group 17 with the configuration [Ne]3s²3p⁵, giving it 7 valence electrons. Each chlorine atom needs 1 more electron to achieve an octet.
    Ca: 2 valence e⁻; Cl: 7 valence e⁻ (needs 1 more)
  2. 2

    Step 2 — Predict Ion Charges

    Calcium will lose its two 4s electrons to form Ca²⁺, achieving the electron configuration of argon [Ar]. Each chlorine atom will gain one electron to form Cl⁻, achieving the electron configuration of argon [Ar] as well. Notice that both ions become isoelectronic with argon, satisfying the octet rule.
    Ca → Ca²⁺ + 2e⁻; each Cl + e⁻ → Cl⁻
  3. 3

    Step 3 — Apply Charge Balance

    Electrical neutrality requires the total positive charge to equal the total negative charge. One Ca²⁺ contributes a +2 charge. Each Cl⁻ contributes a −1 charge. Therefore, two Cl⁻ ions are needed: (+2) + 2(−1) = 0. The formula is CaCl₂.
    CaCl₂
  4. 4

    Step 4 — Write the Net Ionic Equation

    The overall electron-transfer process can be written as: Ca(g) → Ca²⁺(g) + 2e⁻ and 2Cl(g) + 2e⁻ → 2Cl⁻(g). Combining these: Ca(g) + 2Cl(g) → Ca²⁺(g) + 2Cl⁻(g). The gaseous ions then assemble into the crystal lattice, releasing the lattice energy.
    Ca(g) + 2Cl(g) → CaCl₂(s)
  5. 5

    Step 5 — Compare Lattice Energies (CaCl₂ vs. NaCl)

    According to Coulomb's law, E ∝ (q₊ × q₋)/d. In NaCl, q₊ = +1 and q₋ = −1, while in CaCl₂ the cation charge doubles to +2 (q₋ remains −1 for each Cl⁻). Additionally, Ca²⁺ has a smaller ionic radius (100 pm) than Na⁺ (102 pm), meaning the interionic distance in CaCl₂ is slightly shorter. Both factors contribute to a larger lattice energy. Experimentally, the lattice energy of CaCl₂ is approximately 2258 kJ/mol compared to 786 kJ/mol for NaCl.
    U(CaCl₂) ≈ 2258 kJ/mol ≫ U(NaCl) ≈ 786 kJ/mol
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Key Insight

Lattice energies are not simply proportional to the charge product for a single ion pair; the crystal structure and the number of formula units per lattice cell also matter. CaCl₂ adopts a rutile-type structure with a Madelung constant of approximately 4.71, significantly different from the NaCl rock-salt value of 1.748. Always consider both the Coulombic contribution per ion pair and the crystal geometry.

SECTION 7

Ionic Bonding in Context — Strengths, Limitations, and Comparisons

Ionic bonding is only one of several bonding paradigms. Understanding when the ionic model applies—and when it breaks down—is crucial for predicting material properties. The table below compares ionic bonding with covalent bonding across several dimensions.

PropertyIonic CompoundsCovalent Compounds
Bond formationElectron transfer between atoms of very different electronegativities (ΔEN > 1.7 as a rough guideline)Electron sharing between atoms of similar electronegativities (ΔEN < 1.7)
Physical state (room temp.)Crystalline solids with high melting and boiling pointsGases, liquids, or low-melting solids (molecular); network covalent solids can have very high melting points
Electrical conductivityConductive when dissolved in water or molten (mobile ions); non-conductive as solidsGenerally non-conductive in any phase (exceptions: graphite, conductive polymers)
SolubilityOften soluble in polar solvents ("like dissolves like"); guided by lattice energy vs. hydration energyOften soluble in nonpolar solvents; polar covalent molecules dissolve in polar solvents
Structural unitExtended 3D crystal lattice—no discrete moleculesDiscrete molecules (or extended networks for SiO₂, diamond)
BrittlenessBrittle—displacement of ion layers brings like charges together, causing repulsion and fractureVariable—molecular crystals are soft; network solids are hard but brittle

Comparison of ionic and covalent bonding paradigms.

It is important to recognize that the distinction between ionic and covalent bonding is not binary. Fajans' rules predict that small, highly charged cations will polarize the electron cloud of nearby anions, introducing significant covalent character into nominally ionic bonds. For example, LiI is classified as ionic but exhibits substantial covalent character because the small Li⁺ ion strongly distorts the large, easily polarized I⁻ ion. This continuum perspective becomes essential in advanced inorganic chemistry.

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KEY TAKEAWAY

The ionic-versus-covalent dichotomy is an idealization. Real bonds lie on a spectrum parameterized by electronegativity difference and ion polarizability. Treat the ionic model as a powerful first approximation for compounds of alkali/alkaline-earth metals with halogens or chalcogens, but be prepared to invoke partial covalent character—especially for transition-metal compounds, post-transition metals like Pb²⁺ and Sn²⁺, and small cations paired with large anions.

SECTION 8

Connection to Advanced Theory — Band Theory and Solid-State Chemistry

The localized electron-transfer picture of ionic bonding provides a solid foundation, but more advanced treatments reframe the discussion in terms of electronic structure of extended solids. In band theory, the discrete energy levels of individual ions broaden into continuous bands when 10²³ ions assemble into a crystal. Ionic compounds typically exhibit a large band gap between the filled valence band (derived from anion orbitals) and the empty conduction band (derived from cation orbitals), which explains their electrical insulating behavior. The magnitude of this band gap correlates with lattice energy and electronegativity difference.

ConceptIntroductory (This Lesson)Advanced Treatment
Bonding descriptionFull electron transfer → cation + anion → Coulombic attractionBand theory: overlap of atomic orbitals across the lattice produces valence and conduction bands separated by a band gap
Lattice energyBorn–Landé equation with point-charge modelKapustinskii equation, Born–Mayer equation, and DFT calculations that include electron correlation
Ionic vs. covalentΔEN threshold (≈1.7) as heuristicBader charge analysis, electron localization function (ELF), and quantum theory of atoms in molecules (QTAIM)
Crystal structureRadius-ratio rules predict coordination geometriesStructure prediction using Pauling's rules, Goldschmidt tolerance factor, and computational crystal-structure prediction (e.g., AIRSS, CALYPSO)
PropertiesQualitative: high m.p., brittle, conductive when moltenQuantitative property prediction via Born–Haber cycles, Hess's law thermodynamic cycles, and computational thermodynamics

Progression from introductory ionic-bonding concepts to advanced solid-state theory.

As you advance in your studies, you will encounter topics such as defect chemistry (vacancies, interstitials, and Schottky/Frenkel defects in ionic crystals), superionic conductors (materials like AgI where ions become mobile even in the solid state), and perovskite structures (ABO₃) that underpin modern photovoltaics and superconductors. Each of these fields builds directly upon the valence-electron and lattice-energy concepts introduced here, so mastering the fundamentals will pay dividends in upper-division coursework and research.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Strontium (Sr) is in Group 2 and Period 5 of the periodic table. Without looking up its electron configuration, determine how many valence electrons strontium has, predict the charge of the strontium ion, and identify the noble gas whose electron configuration Sr²⁺ would match. Explain the reasoning behind each prediction.
Show answer
Strontium is in Group 2, so it has 2 valence electrons in its outermost shell (5s²). To achieve a noble-gas configuration, strontium loses both 5s electrons, forming Sr²⁺ with a charge of +2. The resulting electron configuration is [Kr], isoelectronic with krypton (Z = 36). This reasoning follows directly from the periodic law: Group 2 elements always form 2+ cations because removing two electrons empties the outermost s subshell and exposes the fully filled shell below. The pattern is consistent across Be²⁺ → [He], Mg²⁺ → [Ne], Ca²⁺ → [Ar], and Sr²⁺ → [Kr].

Question 2: BASIC CALCULATION

Predict the formula of the ionic compound formed between aluminum (Al) and oxygen (O). Show how charge balance determines the subscripts. Write the electron configurations of Al³⁺ and O²⁻ to verify that both achieve noble-gas configurations.
Show answer
Aluminum (Group 13) has 3 valence electrons and forms Al³⁺ by losing all three: [Ne]3s²3p¹ → [Ne]. Oxygen (Group 16) has 6 valence electrons and forms O²⁻ by gaining two: [He]2s²2p⁴ → [He]2s²2p⁶ = [Ne]. For charge balance, the total positive charge must equal the total negative charge: if we use 2 Al³⁺ ions, total positive charge = +6; if we use 3 O²⁻ ions, total negative charge = −6. The formula is therefore Al₂O₃ (aluminum oxide, also known as alumina). Both ions are isoelectronic with neon, confirming satisfaction of the octet rule. Al₂O₃ has an extremely high lattice energy (~15,916 kJ/mol) owing to the high charges and small ionic radii.

Question 3: INTERMEDIATE

Rank the following ionic compounds in order of increasing lattice energy and justify your ranking using Coulomb's law: NaF, CaO, KBr. You may consult ionic radii: Na⁺ = 102 pm, F⁻ = 133 pm, Ca²⁺ = 100 pm, O²⁻ = 140 pm, K⁺ = 138 pm, Br⁻ = 196 pm.
Show answer
Lattice energy is proportional to (q₊ × q₋)/d, where d ≈ r₊ + r₋. For KBr: charges are +1 and −1, d = 138 + 196 = 334 pm, so the Coulombic factor ∝ 1/334. For NaF: charges are +1 and −1, d = 102 + 133 = 235 pm, factor ∝ 1/235. For CaO: charges are +2 and −2, d = 100 + 140 = 240 pm, factor ∝ 4/240. Computing relative values: KBr ∝ 0.00299, NaF ∝ 0.00426, CaO ∝ 0.01667. Thus, the ranking in order of increasing lattice energy is KBr < NaF < CaO. This matches experimental values: KBr ≈ 672 kJ/mol, NaF ≈ 923 kJ/mol, CaO ≈ 3401 kJ/mol. The dramatic increase for CaO illustrates how doubling both ion charges quadruples the charge product, making ion charge the dominant factor.

Question 4: APPLIED

A materials scientist wants to develop a ceramic insulator with the highest possible melting point. She is considering three candidate materials: NaCl, MgO, and BaS. Using your knowledge of lattice energies and the factors that influence them, rank these three materials by expected melting point (lowest to highest) and explain which candidate best meets her requirements. Ionic radii: Mg²⁺ = 72 pm, O²⁻ = 140 pm, Ba²⁺ = 135 pm, S²⁻ = 184 pm.
Show answer
Melting point in ionic compounds correlates strongly with lattice energy. For NaCl (+1/−1, d = 102 + 181 = 283 pm): the low charge product of 1 and moderate d give the lowest lattice energy. For BaS (+2/−2, d = 135 + 184 = 319 pm): the charge product is 4, but the interionic distance is large because both ions are sizable. For MgO (+2/−2, d = 72 + 140 = 212 pm): the charge product is also 4, but both ions are much smaller, yielding a significantly shorter interionic distance. The Coulombic factors are: NaCl ∝ 1/283 = 0.0035, BaS ∝ 4/319 = 0.0125, MgO ∝ 4/212 = 0.0189. Thus, the ranking by increasing melting point is NaCl (801 °C) < BaS (~2227 °C) < MgO (2852 °C). The scientist should choose MgO, which is indeed widely used as a refractory ceramic in furnace linings and crucibles.

Question 5: CRITICAL THINKING

Silver chloride (AgCl) has a lattice energy of 905 kJ/mol, which is comparable to NaF (923 kJ/mol), yet AgCl is famously insoluble in water (Ksp = 1.77 × 10⁻¹⁰) while NaF is quite soluble (≈ 42 g/L at 25 °C). If both compounds have similar lattice energies, what other thermodynamic quantity must differ to explain this solubility discrepancy? Develop a thermodynamic argument using the enthalpy of solution framework: ΔH°soln ≈ U − ΔH°hydration.
Show answer
Dissolution of an ionic solid requires overcoming the lattice energy (endothermic) and is compensated by ion hydration enthalpies (exothermic). For NaF, the sum of hydration enthalpies of Na⁺ (−406 kJ/mol) and F⁻ (−506 kJ/mol) gives about −912 kJ/mol, which nearly matches the lattice energy of 923 kJ/mol, yielding a small positive ΔH°soln that is offset by the favorable entropy of dissolution. For AgCl, although the lattice energy is similar (905 kJ/mol), Ag⁺ is a d¹⁰ ion with poor hydration enthalpy (−473 kJ/mol) and Cl⁻ has a hydration enthalpy of −363 kJ/mol, totaling only about −836 kJ/mol. The resulting ΔH°soln is more positive (+69 kJ/mol vs. +11 kJ/mol for NaF), and the entropy gain is insufficient to overcome this larger endothermic barrier. Furthermore, the bonding in AgCl has significant covalent character (Ag⁺ is small and polarizing, Cl⁻ is polarizable), meaning the true binding energy exceeds what a purely ionic lattice-energy model predicts. This problem highlights that solubility depends on the balance between lattice energy and hydration energy, not on lattice energy alone.

SUMMARY

Lesson Summary

Summary

Valence electrons are the electrons in the outermost principal energy level of an atom, and their number for main-group elements is determined directly by the group number in the periodic table. When atoms with low ionization energies (metals) interact with atoms possessing high electron affinities (nonmetals), electron transfer occurs: the metal loses valence electrons to form a cation while the nonmetal gains electrons to form an anion. Both ions typically achieve noble-gas electron configurations, satisfying the octet rule.

The resulting oppositely charged ions are held together by Coulombic attraction in a three-dimensional crystal lattice, not as discrete molecules. The stability of this lattice is quantified by lattice energy, which increases with higher ion charges and smaller ionic radii, as described by the Born–Landé equation. The formula of any ionic compound is determined by charge balance—the total cation charge must exactly equal the total anion charge. These principles, together with the Born–Haber cycle, provide a complete thermodynamic framework for predicting whether ionic compounds will form, their formulas, and their physical properties such as melting point, solubility, and electrical conductivity.

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