AP Chemistry • Atomic Structure and Properties

Valence Electrons and Ionic Compounds

How outermost electrons drive the formation of ionic bonds and determine compound stoichiometry.

AP Chemistry illustration with a laboratory flask, atom, and microscope
Inside this lesson 10 sections 5 questions 8 figures

SECTION 1

Historical Context & Motivation

The question of why certain elements combine in fixed ratios while others remain inert puzzled chemists for well over a century. Early observations by Antoine Lavoisier and John Dalton established the law of definite proportions—compounds always contain elements in the same mass ratios—yet neither scientist could explain why those ratios existed. The answer ultimately lay in the electronic structure of atoms, specifically in the behavior of the outermost electrons. Understanding how valence electrons are gained, lost, or shared between atoms provided the conceptual bridge between macroscopic stoichiometry and atomic-level interactions, a bridge that remains central to modern chemistry.

1869 • Mendeleev's Periodic Table

Dmitri Mendeleev arranged elements by atomic mass and recurring chemical properties, revealing periodic trends that hinted at an underlying electronic basis for chemical behavior.

1897 • Discovery of the Electron

J.J. Thomson's cathode-ray experiments identified the electron as a subatomic particle, laying the groundwork for electron-transfer models of bonding.

1913 • Bohr Model of the Atom

Niels Bohr proposed quantized energy levels for electrons, giving chemists a framework for distinguishing inner-shell electrons from those in the outermost shell.

1916 • Lewis & Kossel's Electron-Transfer Theory

Gilbert N. Lewis and Walther Kossel independently proposed that atoms achieve noble-gas configurations by transferring or sharing electrons, establishing the modern concept of ionic and covalent bonding.

1927 • Quantum Mechanical Model

The Schrödinger equation replaced Bohr's orbits with orbitals, refining our understanding of electron distributions and enabling rigorous predictions of valence behavior.

The central question this lesson addresses is deceptively simple: What determines how many electrons an atom will gain or lose, and how does this electron transfer produce stable ionic compounds with predictable formulas? By connecting electron configurations to periodic trends, we can predict the charges of monatomic ions, write correct empirical formulas for ionic compounds, and rationalize the extraordinary stability conferred by lattice energy. These skills are foundational for Unit 2 of the AP Chemistry curriculum and recur in virtually every subsequent topic.

SECTION 2

Core Principles & Definitions

Before analyzing ionic compounds in detail, it is essential to establish the foundational ideas that govern how atoms interact through electron transfer. The principles below connect electron configuration—a topic covered extensively in the quantum model—to the observable chemistry of salt formation, crystal structures, and solubility behavior.

1

Valence Electrons

Electrons in the outermost principal energy level (highest value of n) that participate in chemical bonding. For main-group elements, the number of valence electrons equals the group number (using the 1–18 convention for Groups 1, 2, and 13–18).
2

Octet Rule & Exceptions

Atoms tend to gain, lose, or share electrons to achieve eight electrons in their valence shell, mimicking a noble-gas configuration. Important exceptions include hydrogen (duet), elements in Period 3+ (expanded octets), and certain transition metals.
3

Ionic Bonding via Electron Transfer

When a metal with low ionization energy encounters a nonmetal with high electron affinity, electron transfer produces oppositely charged ions. The resulting electrostatic attraction constitutes the ionic bond.
4

Charge Balance & Empirical Formulas

Ionic compounds are electrically neutral. The ratio of cations to anions in the empirical formula is determined by balancing the total positive and negative charges, yielding the lowest whole-number subscript ratio.
5

Lattice Energy & Stability

The energy released when gaseous ions assemble into a crystalline lattice quantifies the thermodynamic stability of an ionic compound. Lattice energy increases with higher ion charges and smaller ionic radii, consistent with Coulomb's law.
✦

KEY TAKEAWAY

Think of valence electrons as the currency atoms use in chemical transactions. A sodium atom is like a person carrying exactly one extra coin they are eager to spend, while a chlorine atom is like a merchant who needs exactly one coin to complete a transaction. When they meet, the exchange is immediate, both parties reach a stable state, and the "receipt" of that trade is the ionic bond holding NaCl together. The empirical formula is simply the bookkeeping that ensures every coin spent is accounted for by a coin received.

SECTION 3

Visual Explanation — Electron Transfer in NaCl

The formation of sodium chloride provides the archetypal illustration of ionic bonding. Sodium (Z = 11) has the electron configuration [Ne] 3s1, possessing a single valence electron in its third energy level. Chlorine (Z = 17) has the configuration [Ne] 3s2 3p5, meaning it requires one additional electron to complete its octet. The following diagram traces the electron transfer process, showing how each atom attains a noble-gas configuration upon forming an ion.

Electron Transfer: Na + Cl → Na⁺ + Cl⁻11+valence e⁻Na atom[Ne] 3s¹1 valence e⁻e⁻ transfer17+Cl atom[Ne] 3s²3p⁵7 valence e⁻After Electron TransferNa⁺[Ne] — 10 e⁻Cl⁻[Ar] — 18 e⁻Electrostatic Attraction (Ionic Bond)
The diagram shows sodium's single valence electron (gold dot, top shell) transferring to chlorine. After transfer, Na+ achieves the [Ne] configuration and Cl− achieves the [Ar] configuration. The electrostatic attraction between the resulting oppositely charged ions constitutes the ionic bond.

Several features of this diagram merit close attention. First, note that sodium's valence electron (highlighted in gold) sits in the third principal energy level—far from the nucleus—making it relatively easy to remove. The first ionization energy of sodium is only 496 kJ/mol, reflecting the weak hold the nucleus exerts on this lone 3s electron. Second, chlorine's seven valence electrons (pink dots on the outer shell) leave precisely one vacancy in the 3p subshell. Adding one electron to chlorine releases 349 kJ/mol of energy, its electron affinity. The combination of sodium's low ionization energy and chlorine's highly exothermic electron affinity makes the electron-transfer process energetically favorable even before accounting for lattice energy, the enormous stabilization gained when ions assemble into the crystal lattice.

SECTION 4

Mathematical Framework — Lattice Energy & Coulomb's Law

While the AP Chemistry exam does not require calculating lattice energies from first principles, understanding the mathematical relationships governing ionic interactions enables powerful qualitative predictions. Coulomb's law provides the foundation: the electrostatic potential energy between two point charges is proportional to the product of their charges and inversely proportional to the distance between them.

COULOMB'S LAW (ELECTROSTATIC ENERGY)

E=k×(q1×q2)/rE = k × (q₁ × q₂) / r

Where E is the electrostatic potential energy, k is Coulomb's constant (8.99 × 10⁹ N·m²/C²), q₁ and q₂ are the ion charges, and r is the interionic distance. For oppositely charged ions, E is negative (attractive).

BORN-LANDÉ APPROXIMATION (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 equilibrium interionic distance, and n is the Born exponent related to compressibility. This equation is beyond the AP exam scope but illustrates why lattice energy increases with charge and decreases with ionic radius.

CHARGE BALANCE FOR EMPIRICAL FORMULAS

(cationcharge)×(numberofcations)+(anioncharge)×(numberofanions)=0(cation charge) × (number of cations) + (anion charge) × (number of anions) = 0

This fundamental constraint determines the subscripts in every ionic formula. For example, if the cation is Al3+ and the anion is O2−, the formula must be Al₂O₃ so that (+3)(2) + (−2)(3) = 0.

For the AP exam, the critical qualitative takeaway is that lattice energy increases as ionic charges increase and as ionic radii decrease. This explains why MgO (with +2 and −2 ions) has a much higher melting point (2852 °C) than NaCl (with +1 and −1 ions, melting at 801 °C), even though both adopt the same rock-salt crystal structure. The charge ratio matters more than the size ratio in most comparisons, because the Coulombic energy depends on the product of the charges, which quadruples when each charge doubles.

SECTION 5

Valence Electrons Across the Periodic Table

The periodic table is, in essence, organized by valence electron count. For main-group (representative) elements, the group number directly encodes the number of valence electrons: Group 1 elements have one, Group 2 have two, and Groups 13–18 have three through eight, respectively. Transition metals present a more complex picture because their (n−1)d electrons can participate in bonding to varying degrees, leading to multiple common oxidation states. The diagram below maps the relationship between periodic table position, valence electron count, and the most common ion charges formed by main-group elements.

Main-Group Elements: Valence Electrons → Common Ion ChargesMetals lose electrons (cations) • Nonmetals gain electrons (anions)Group 1Group 2Group 13Group 14Group 15Group 16Group 17Group 18Val. e⁻12345678Ion1+2+3+varies3−2−1−0Representative Elements & Their IonsNa → Na⁺loses 1 e⁻[Ne]Mg → Mg²⁺loses 2 e⁻[Ne]Al → Al³⁺loses 3 e⁻[Ne]N → N³⁻gains 3 e⁻[Ne]O → O²⁻gains 2 e⁻[Ne]F → F⁻gains 1 e⁻[Ne]Cations (metals lose e⁻)Anions (nonmetals gain e⁻)Group 14 elements (C, Si) typically form covalent bonds rather than monatomic ions.Group 18 noble gases have 8 valence e⁻ and are generally unreactive.
Main-group elements form predictable ions by either losing all valence electrons (metals, red borders) or gaining enough to complete an octet (nonmetals, cyan borders). Each resulting ion has a noble-gas electron configuration. Group 14 elements are a borderline case—they typically form covalent rather than ionic bonds because gaining or losing four electrons would be energetically prohibitive.
GroupValence e⁻Electron Config. PatternCommon IonIsoelectronic With
1 (Li, Na, K)1ns¹1+Previous noble gas
2 (Be, Mg, Ca)2ns²2+Previous noble gas
13 (Al, Ga)3ns²np¹3+Previous noble gas
15 (N, P)5ns²np³3−Next noble gas
16 (O, S)6ns²np⁴2−Next noble gas
17 (F, Cl, Br)7ns²np⁵1−Next noble gas
18 (Ne, Ar, Kr)8ns²np⁶0 (no ion)Already a noble gas

Main-group valence electrons, common ions, and isoelectronic noble gases

SECTION 6

Worked Example — Predicting the Formula of Calcium Phosphide

Let us walk through the complete process of predicting the empirical formula of the ionic compound formed between calcium and phosphorus. This example illustrates how valence electron counts dictate ion charges and how charge balance determines stoichiometry.

Determining the Formula of Calcium Phosphide

  1. 1

    Step 1 — Identify Element Types and Valence Electrons

    Calcium (Ca) is in Group 2 of the periodic table, so it is a metal with 2 valence electrons (electron configuration [Ar] 4s²). Phosphorus (P) is in Group 15, a nonmetal with 5 valence electrons (electron configuration [Ne] 3s²3p³). Since one is a metal and the other a nonmetal, we expect ionic bonding.
  2. 2

    Step 2 — Determine Expected Ion Charges

    Calcium will lose its 2 valence electrons to achieve the argon noble-gas configuration, forming Ca²⁺. Phosphorus will gain 3 electrons to complete its octet and achieve the argon configuration, forming P³⁻.
    Ca → Ca²⁺ (loses 2 e⁻) | P → P³⁻ (gains 3 e⁻)
  3. 3

    Step 3 — Apply Charge Balance

    The compound must be electrically neutral. We need the total positive charge to equal the total negative charge in magnitude. The least common multiple of 2 and 3 is 6. Therefore, we need 3 calcium ions (total charge = +6) and 2 phosphide ions (total charge = −6).
    3 × (+2) + 2 × (−3) = +6 − 6 = 0 ✓
  4. 4

    Step 4 — Write the Empirical Formula

    The cation is written first, followed by the anion. The subscripts represent the number of each ion required for charge neutrality.
    Ca₃P₂ — calcium phosphide
  5. 5

    Step 5 — Verify

    Confirm that the formula is in lowest whole-number ratio: 3 and 2 share no common factor greater than 1, so Ca₃P₂ is already the empirical formula. The total number of electrons lost (3 × 2 = 6) equals the total number of electrons gained (2 × 3 = 6), confirming electron conservation.
    Final answer: Ca₃P₂

SECTION 7

Ionic vs. Covalent Bonding — Strengths & Limitations

While the ionic model provides an elegant and often accurate description of compounds formed between metals and nonmetals, it is important to recognize its boundaries. Real chemical bonds exist on a continuum between purely ionic and purely covalent character, and the simple electron-transfer model breaks down for certain compounds. The following comparison highlights where the ionic model excels and where more nuanced approaches are necessary.

PropertyIonic CompoundsCovalent Compounds
Bond formationElectron transfer between atoms with large electronegativity differences (ΔEN > 1.7 as a rough guideline)Electron sharing between atoms with small electronegativity differences (ΔEN < 1.7)
Typical participantsMetal + nonmetal (e.g., NaCl, MgO, CaF₂)Nonmetal + nonmetal (e.g., H₂O, CO₂, CH₄)
Melting pointGenerally high (strong Coulombic forces in lattice)Generally low (weak intermolecular forces between discrete molecules)
Electrical conductivityConducts when molten or dissolved (mobile ions); insulator as solidGenerally nonconducting (no mobile charged particles)
StructureExtended 3D crystal lattice; no discrete moleculesDiscrete molecules (exceptions: network covalent solids like SiO₂)
SolubilityOften soluble in polar solvents like waterOften soluble in nonpolar organic solvents

Comparison of ionic and covalent bonding models

✦

KEY TAKEAWAY

The ionic model is like a simplified circuit diagram in electrical engineering—it captures the essential behavior (current flow, voltage drops) with high accuracy for most practical purposes, even though it ignores quantum tunneling and relativistic effects at the component level. Similarly, the electron-transfer model correctly predicts formulas, charges, and physical properties for the vast majority of metal-nonmetal compounds, but it oversimplifies the partial covalent character present in highly polarizing ions like Al³⁺ or Li⁺. For the AP exam, use the ionic model confidently for main-group metal–nonmetal combinations and be aware of Fajans' rules as the bridge to understanding covalent character in "ionic" compounds.

Conductivity probe in saltwaterDissolved ions carry charge through saltwater between immersed electrodes.

SECTION 8

Connection to Advanced Theory — Beyond the Octet

The valence electron framework developed in this lesson serves as the launching pad for several advanced topics you will encounter throughout the AP Chemistry curriculum and in university-level courses. Understanding how the simple ionic model connects to these more sophisticated treatments will deepen your ability to reason about chemical behavior.

This Lesson's ConceptAdvanced Extension
Octet rule drives ion chargeTransition metals form multiple oxidation states because d-orbital electrons have variable participation in bonding (e.g., Fe²⁺ and Fe³⁺)
Lattice energy ∝ q₁q₂/rBorn-Haber cycles use Hess's law to calculate lattice energies indirectly from measurable thermodynamic quantities (ionization energy, electron affinity, enthalpy of formation)
Ionic compounds dissolve in waterDissolution involves ion-dipole interactions and hydration enthalpy; solubility rules emerge from the competition between lattice energy and hydration energy
Charge balance → empirical formulaPolyatomic ions (SO₄²⁻, NO₃⁻, PO₄³⁻) introduce complex charge-balance scenarios with internal covalent bonding and external ionic bonding
Electronegativity difference → bond typePauling's electronegativity scale and percent ionic character calculations provide a quantitative measure of bond polarity beyond the binary ionic/covalent classification

How foundational concepts connect to advanced AP and university topics

As you progress through AP Chemistry, you will revisit valence electrons repeatedly—in Lewis structures (Unit 2), in VSEPR theory for molecular geometry (Unit 2), in intermolecular forces (Unit 3), and in redox chemistry (Unit 4). The concept of electron transfer that produces ionic compounds is, in a sense, the simplest case of a broader theme: all chemistry is fundamentally about the rearrangement of valence electrons. Master the principles here, and every subsequent topic builds naturally on this foundation.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Sulfur is in Group 16 of the periodic table. When sulfur forms a monatomic ion in an ionic compound, what is the most likely charge on that ion, and which noble gas is it isoelectronic with?
  • A. S²⁺, isoelectronic with silicon
  • B. S⁶⁺, isoelectronic with neon
  • C. S²⁻, isoelectronic with argon
  • D. S⁴⁻, isoelectronic with calcium
Show answer (C)
The correct answer is C. Sulfur has 6 valence electrons (Group 16) and needs 2 more to complete its octet. By gaining 2 electrons, sulfur forms the S²⁻ ion with 18 electrons total, which is isoelectronic with argon (18 electrons). Option A is incorrect because sulfur is a nonmetal that gains electrons rather than losing them in ionic compounds. Option B is incorrect because while S⁶⁺ could theoretically exist (losing all valence electrons), this would require enormous ionization energies and does not occur in simple ionic compounds; furthermore, S with 16 − 6 = 10 electrons would be isoelectronic with neon, but this is not the most likely monatomic ion. Option D is incorrect because gaining 4 electrons would give sulfur 20 electrons with a [Ar]3s² configuration, which is not a noble-gas configuration and is energetically unfavorable.

Question 2: BASIC CALCULATION

What is the correct empirical formula for the ionic compound formed between barium (Ba) and nitrogen (N)?
  • A. BaN
  • B. Ba₂N₃
  • C. Ba₃N₂
  • D. Ba₃N
Show answer (C)
The correct answer is C. Barium is in Group 2 and forms Ba²⁺ by losing its 2 valence electrons. Nitrogen is in Group 15 and forms N³⁻ by gaining 3 electrons. To achieve charge neutrality, we need the least common multiple of 2 and 3, which is 6. This requires 3 barium ions (total charge = +6) and 2 nitride ions (total charge = −6), giving Ba₃N₂. Option A is incorrect because Ba²⁺ and N³⁻ do not balance as a 1:1 ratio (net charge = +2 − 3 = −1). Option B is incorrect because 2(+2) + 3(−3) = +4 − 9 = −5 ≠ 0; this reverses the correct subscripts. Option D is incorrect because 3(+2) + 1(−3) = +6 − 3 = +3 ≠ 0.

Question 3: INTERMEDIATE

Arrange the following ionic compounds in order of increasing lattice energy: NaF, CaO, KBr.
  • A. CaO < NaF < KBr
  • B. KBr < NaF < CaO
  • C. NaF < KBr < CaO
  • D. KBr < CaO < NaF
Show answer (B)
The correct answer is B. Lattice energy depends on the product of the ion charges (numerator in Coulomb's law) and inversely on the interionic distance (denominator). CaO has the highest lattice energy because both ions are doubly charged (Ca²⁺ and O²⁻), so the charge product is 4, compared to 1 for NaF and KBr. Between NaF and KBr, both have singly charged ions, but NaF has smaller ions (Na⁺ = 116 pm, F⁻ = 119 pm) than KBr (K⁺ = 152 pm, Br⁻ = 182 pm), giving NaF a shorter interionic distance and higher lattice energy. Therefore the order is KBr < NaF < CaO. Option A reverses the ranking entirely. Option C incorrectly places NaF below KBr. Option D incorrectly ranks CaO below NaF, ignoring the dominant effect of the quadrupled charge product.

Question 4: APPLIED

5 pts
A student investigates two unknown white crystalline solids, Compound X and Compound Y. Both dissolve in water. When the student tests the solutions with a conductivity apparatus, Compound X produces a bright bulb glow while Compound Y produces no glow. The student also determines that Compound X has a melting point of 801 °C, while Compound Y melts at 186 °C. (a) Based on the conductivity data, classify Compound X and Compound Y as ionic or molecular (covalent). Justify your reasoning for each compound. (b) Compound X is determined to be NaCl. Write the electron configurations for the Na⁺ and Cl⁻ ions and identify the noble gas each is isoelectronic with. (c) Compound Y is determined to be glucose (C₆H₁₂O₆). Explain why glucose dissolves in water but does not conduct electricity, referencing the type of bonding present. (d) A third compound, MgO, has a melting point of 2852 °C. Using Coulomb's law principles, explain why MgO has a much higher melting point than NaCl, even though both are ionic compounds.
Show answer
Point-by-point rubric: (a) [1 point] Compound X is ionic because its aqueous solution conducts electricity, indicating the presence of mobile ions (Na⁺ and Cl⁻) that carry charge through solution. Compound Y is molecular (covalent) because its solution does not conduct, indicating no ions are produced upon dissolution—the compound remains as intact, neutral molecules. (b) [1 point] Na⁺: 1s²2s²2p⁶ (10 electrons), isoelectronic with neon. Cl⁻: 1s²2s²2p⁶3s²3p⁶ (18 electrons), isoelectronic with argon. Each ion has achieved a noble-gas electron configuration by transferring one electron from Na to Cl. (c) [1 point] Glucose contains only covalent (intramolecular) bonds between C, H, and O atoms. When dissolved, glucose molecules remain intact rather than dissociating into ions. Glucose dissolves because its numerous −OH groups form hydrogen bonds with water molecules (like dissolves like), but since no charged species are produced, the solution cannot conduct electricity. (d) [2 points — 1 for identifying charge difference and 1 for connecting to Coulomb's law and/or ionic radii] MgO contains Mg²⁺ and O²⁻ ions, each with a charge magnitude of 2, whereas NaCl contains Na⁺ and Cl⁻, each with a charge magnitude of 1. By Coulomb's law, the attractive force between ions is proportional to the product of the charges (|q₁ × q₂|). For MgO this product is 2 × 2 = 4, compared to 1 × 1 = 1 for NaCl—a fourfold increase. Additionally, Mg²⁺ (86 pm) and O²⁻ (126 pm) are smaller ions than Na⁺ (116 pm) and Cl⁻ (167 pm), resulting in a shorter interionic distance that further increases the lattice energy. The combination of higher charges and smaller radii gives MgO a dramatically higher lattice energy and therefore a much higher melting point.

Question 5: CRITICAL THINKING

5 pts
The table below shows the lattice energies and melting points for four sodium halides: | Compound | Lattice Energy (kJ/mol) | Melting Point (°C) | Anion Radius (pm) | |----------|------------------------|--------------------|-----------------| | NaF | 923 | 993 | 119 | | NaCl | 786 | 801 | 167 | | NaBr | 747 | 747 | 182 | | NaI | 704 | 661 | 206 | (a) Describe the trend in lattice energy as you move from NaF to NaI. Identify the variable responsible for this trend, given that the cation and its charge remain constant. (b) Using Coulomb's law, explain quantitatively why NaF has a lattice energy approximately 1.31 times that of NaI. Reference specific data from the table. (c) A student predicts that NaAt (sodium astatide, where At is astatine in Group 17) would have a lattice energy less than 704 kJ/mol. Evaluate this prediction and justify whether it is reasonable. (d) The data show that the melting point of NaBr (747 °C) is numerically equal to its lattice energy (747 kJ/mol). A student claims this means melting point and lattice energy are always equal. Critique this claim using evidence from the table.
Show answer
Point-by-point rubric: (a) [1 point] Lattice energy decreases from NaF (923 kJ/mol) to NaI (704 kJ/mol). Since the cation (Na⁺) and all ion charges (+1 and −1) remain constant across the series, the variable responsible is the increasing anion radius: F⁻ (119 pm) < Cl⁻ (167 pm) < Br⁻ (182 pm) < I⁻ (206 pm). As the anion radius increases, the interionic distance r increases, and since lattice energy is inversely proportional to r, the lattice energy decreases. (b) [1 point] Coulomb's law states E ∝ q₁q₂/r. Since q₁ and q₂ are identical for all four compounds (+1 and −1), the ratio of lattice energies should approximate the inverse ratio of the interionic distances. The interionic distance is the sum of the cation and anion radii. For NaF: r ≈ 116 + 119 = 235 pm. For NaI: r ≈ 116 + 206 = 322 pm. The ratio r(NaI)/r(NaF) = 322/235 ≈ 1.37, so we expect the lattice energy of NaF to be about 1.37 times that of NaI. The observed ratio is 923/704 ≈ 1.31, which is in reasonable agreement—the slight discrepancy arises because the simple Coulomb model does not account for additional factors like electron-electron repulsion and the Born exponent. (c) [1 point] The prediction is reasonable. Astatine is below iodine in Group 17, so the At⁻ ion would have an even larger ionic radius than I⁻ (206 pm). Since lattice energy is inversely proportional to interionic distance, a larger anion radius means a greater Na–At distance and therefore a lower lattice energy than the 704 kJ/mol observed for NaI. The trend in the data consistently shows decreasing lattice energy with increasing halide size, and extrapolating this trend supports the student's prediction. (d) [2 points — 1 for identifying the claim as incorrect, 1 for providing specific counterexamples with reasoning] The student's claim is incorrect. The numerical equality for NaBr is coincidental because lattice energy (kJ/mol) and melting point (°C) are measured in entirely different units and are not expected to be numerically equal. The table itself provides counterexamples: NaF has a lattice energy of 923 kJ/mol but a melting point of 993 °C, NaCl has 786 kJ/mol but melts at 801 °C, and NaI has 704 kJ/mol but melts at 661 °C. While there is a positive correlation between lattice energy and melting point (higher lattice energy generally corresponds to higher melting point), they are not numerically equal. Melting point depends on the kinetic energy needed to disrupt the entire crystal lattice, which involves factors such as crystal structure and entropy of fusion—not just the pairwise Coulombic interaction.

SUMMARY

Lesson Summary

Summary

Valence electrons are the electrons in the outermost principal energy level of an atom, and for main-group elements their count equals the group number. These outermost electrons dictate chemical reactivity: metals with low ionization energies readily lose valence electrons to form cations, while nonmetals with high electron affinities readily gain electrons to form anions. The resulting oppositely charged ions attract through Coulombic forces, assembling into extended crystal lattices that characterize ionic compounds. The charge balance principle — total positive charge must equal total negative charge — determines the empirical formula of every ionic compound.

The stability of ionic compounds is quantified by lattice energy, which increases with higher ion charges and smaller ionic radii, consistent with Coulomb's law. Key physical properties — high melting points, brittleness, and electrical conductivity in the molten or dissolved state — follow directly from the extended lattice structure and the strength of interionic attractions. Understanding valence electrons and ionic bonding provides the essential foundation for Lewis structures, molecular geometry, intermolecular forces, and the redox reactions you will encounter throughout AP Chemistry.

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