Unit 2: Molecular and Ionic Compound Structure and Properties

Types of Chemical Bonds and Why Structure Controls Properties

Chemical substances behave the way they do—melting point, hardness, conductivity, and solubility—because of how their particles are held together and arranged. In this unit, the constant theme is connecting structure (what particles exist and how they’re organized) to properties (what you can observe and measure). Atoms engage in bonding because a bonded arrangement is typically a more stable, lower-potential-energy state than separated particles.

At the broadest level, bonding is driven by electrostatic attraction between positive and negative charges. The “types” of bonds (ionic, covalent, metallic) are models describing different ways electrons are involved in those attractions.

Ionic, covalent, and metallic bonding as a continuum

A common trap is to treat “ionic vs. covalent” as a strict either-or. In reality, many bonds have partial ionic character—electrons may be shared but unequally.

• Ionic bonding is the model used when electrons are transferred (or treated as transferred) and the compound is best described as a lattice of cations and anions (often metal + nonmetal).
• Covalent bonding is the model used when atoms (often nonmetals) share electrons to form discrete molecules (or sometimes extended networks).
• Metallic bonding is the model used when valence electrons are delocalized across many atoms, producing conductivity, malleability, and ductility.

A practical predictor for which model best describes a bond is electronegativity, the tendency of an atom to attract shared electrons. Large electronegativity differences often align with ionic behavior; small differences align more with covalent behavior.

Why bonding type predicts macroscopic properties

Different bonding models imply different particle arrangements and attraction strengths:

• Ionic compounds form repeating 3D lattices with strong attractions in many directions, so they are often strong solids at room temperature with high melting and boiling points, are brittle, and conduct electricity when molten or dissolved.
• Molecular covalent substances consist of separate molecules; their melting/boiling points depend heavily on intermolecular forces (Unit 3), and they are typically lower than ionic or network solids.
• Metals have mobile electrons, so they conduct and can deform without shattering.
• Covalent network solids (like diamond) are covalently bonded throughout the entire solid, giving extremely high melting points and hardness.

Exam Focus

• Typical question patterns:
  • Given elements (or electronegativities), predict whether a substance is ionic, metallic, or covalent and justify.
  • Compare properties (melting point, conductivity, brittleness) based on bonding/structure.
  • Identify bonding type from observations (e.g., conducts when molten, brittle crystal).
• Common mistakes:
  • Treating all polar covalent bonds as “ionic.” Polar covalent bonds still involve sharing.
  • Explaining ionic properties using “molecules” (ionic compounds are not made of molecules).
  • Forgetting that structure (lattice vs. molecules vs. network) is what drives properties.

Intramolecular Forces and Potential Energy (Why Bonds Form)

Bonds form because the system can reach a lower potential energy at an equilibrium distance where attractions outweigh repulsions. This energy perspective is the backbone for trends such as “higher charges give stronger ionic attractions” and “multiple bonds are shorter and stronger.”

Electrostatic attraction and Coulomb’s law

Opposite charges attract and like charges repel. The force magnitude is described by Coulomb’s law:

F = k\frac{q_1 q_2}{r^2}

Here, F is electrostatic force, k is a proportionality constant, q_1 and q_2 are charges, and r is the distance between charges.

Chemistry trend logic from this equation:

• Larger magnitude charges give stronger attractions (for example, ions with charges 2+ and 2- generally attract more strongly than 1+ and 1-).
• Smaller distances dramatically increase attraction because of the squared distance term.

Potential energy vs. internuclear distance: the “bond well” idea

As two atoms (or ions) approach, attraction lowers potential energy, but if they get too close, repulsions (electron-electron and nucleus-nucleus) rise sharply. The result is a potential energy curve with a minimum.

Interpreting a typical potential-energy diagram:

• The dashed line is often used to represent zero potential energy (separated particles). Values can be negative, indicating a more stable (lower-energy) bonded state.
• At very short distances (often labeled A on diagrams), atoms are too close; repulsion dominates and potential energy is high.
• At very large distances (often labeled D), atoms are too far to significantly attract, so potential energy approaches zero.
• At the minimum (often labeled C), attractive and repulsive effects balance; this is the equilibrium bond length where bonding is most stable.

The depth of the well relates to bond strength (bond energy): deeper well means more energy required to separate the particles.

Bond energy (bond enthalpy) as an average measure

A bond energy is the energy required to break one mole of a specific bond type in the gas phase (reported as an average across many molecules).

Key structural trends:

• Single bonds are generally longer and weaker than double bonds.
• Double bonds are generally longer and weaker than triple bonds.
• Double and triple bonds are shorter and stronger than single bonds, but they are not double or triple the strength.

This happens because multiple bonds add pi bonding in addition to a sigma bond, increasing electron density between nuclei.

Worked example: reasoning about ionic attraction

Compare the relative strength of attraction between:

• Pair A: Na^+ and Cl^-
• Pair B: Mg^{2+} and O^{2-}

Step-by-step reasoning:

1. Charges: Pair B has charges of magnitude 2 instead of 1.
2. Coulombic attraction scales with q_1 q_2. Magnitudes: Pair A gives 1 \times 1 = 1, Pair B gives 2 \times 2 = 4.
3. If distances were similar (often they are not), Pair B would be much stronger; even if distance differs, higher charges strongly increase attraction.

So compounds with 2+ and 2- ions often have higher melting points and larger lattice energies than 1+ and 1- compounds.

Exam Focus

• Typical question patterns:
  • Explain why bond strength changes with atomic/ionic size or charge.
  • Rank compounds by lattice energy or melting point using charge and size reasoning.
  • Interpret a potential energy diagram (bond length at minimum, bond energy as well depth).
• Common mistakes:
  • Saying “bonds form to reach octets” as the energy explanation (octets are a model; energy minimization is the driving idea).
  • Forgetting distance matters: smaller ions bring charges closer, strengthening attraction.
  • Confusing bond energy (breaking a bond costs energy) with bond formation (releases energy).

Ionic Solids: Lattices, Formulas, and Properties

Ionic bonds are often described as forming between a metal and a nonmetal, and the solid is held together by electrostatic attraction between cations and anions arranged in a lattice. Salts are held together by ionic bonding.

Why ionic compounds form lattices (not molecules)

Ionic substances do not exist as separate molecules like “NaCl molecules.” Each ion attracts many oppositely charged neighbors in all directions, so the stable structure is a repeating 3D pattern.

That’s why an ionic chemical formula is an empirical formula: it gives the simplest whole-number ratio of ions that creates neutrality. The appropriate particle-language term is formula unit, not molecule.

Determining formulas from charges (charge balance)

The rule is electrical neutrality: total positive charge equals total negative charge.

Example: aluminum oxide.

• Aluminum forms Al^{3+}.
• Oxygen forms O^{2-}.
  To balance charge, use the least common multiple of 3 and 2, which is 6:
• 2 aluminum ions give total +6.
• 3 oxide ions give total -6.
  Formula: Al_2O_3.

The criss-cross method can work, but charge-balance reasoning is safer and helps catch errors.

Lattice energy (qualitative) and trends

Ionic substances are typically strong with high melting and boiling points. Melting point varies with the strength of ionic attraction:

• Greater ionic charge generally gives greater attraction and higher lattice energy (for example, MgO with 2+ and 2- tends to have stronger ionic attraction than NaCl with 1+ and 1-).
• If charges are the same, smaller ions bring charges closer and generally increase attraction.

Lattice energy is the energy released when gaseous ions form an ionic solid (or equivalently, the energy required to separate the solid into gaseous ions). AP typically expects qualitative trend reasoning.

Key properties of ionic solids (and structural reasons)

1. High melting and boiling points: Many strong attractions throughout the lattice must be overcome.
2. Brittleness: If layers shift, like charges can align, causing strong repulsion and fracture.
3. Electrical conductivity:
   • Solid ionic compounds conduct poorly because ions are locked in place.
   • Molten ionic compounds conduct because ions can move.
   • Aqueous ionic solutions conduct because ions are solvated and mobile.

Worked example: predicting properties from structure

A solid is crystalline, has a high melting point, is brittle, and does not conduct as a solid but does conduct when molten.

Reasoning:

1. High melting point suggests strong attractions (ionic or network).
2. Brittleness strongly points to ionic crystals (metals are malleable; network solids are very hard).
3. Conducts only when molten indicates mobile charged particles appear upon melting.

Conclusion: it is an ionic solid.

Exam Focus

• Typical question patterns:
  • Determine an ionic formula from ion charges or from a model/diagram.
  • Explain conductivity differences (solid vs. molten vs. aqueous) using particle motion.
  • Rank lattice energy or melting point using ionic charge and radius reasoning.
• Common mistakes:
  • Calling an ionic formula a “molecule.” Use “formula unit” for ionic compounds.
  • Saying ionic solids conduct because they “have charges.” Charges must be mobile to conduct.
  • Assuming “higher molar mass” means “higher melting point” for ionic compounds (charge and ionic size dominate).

Metallic Bonding and Alloys: The Electron-Sea Model

Metallic bonding is well described by the electron-sea model: a lattice of metal cations surrounded by delocalized valence electrons. Unlike ionic solids, the key mobile charge carriers are electrons.

Why metals conduct electricity and heat

Delocalized electrons move through the solid when an electric potential is applied, explaining:

• Electrical conductivity in the solid phase.
• Thermal conductivity, because mobile electrons transfer kinetic energy efficiently.

Why metals are malleable and ductile

If metal-ion layers shift, the delocalized electron sea continues to provide attraction, so the structure can deform without shattering. This explains malleability and ductility.

Alloys: tuning properties by mixing atoms

An alloy is a mixture of elements with metallic properties. Alloys often become stronger/harder because disrupting the regular lattice makes it harder for layers to slide.

Two common structural categories:

• Substitutional alloys form when atoms of similar size replace host metal atoms (example: brass, mainly Cu and Zn).
• Interstitial alloys form when much smaller atoms fit into gaps between metal atoms (example: steel, where C occupies interstitial spaces in Fe).

Exam Focus

• Typical question patterns:
  • Explain metal conductivity and malleability using the electron-sea model.
  • Compare metals vs. ionic solids in conductivity and mechanical properties.
  • Predict how an alloy changes properties compared to the pure metal.
• Common mistakes:
  • Explaining metallic bonding as “shared electrons between two atoms” (that’s covalent; metallic electrons are delocalized across many atoms).
  • Saying metals are brittle like ionic solids.
  • Forgetting that solid metals conduct (a key contrast with ionic solids).

Lewis Diagrams: A Model for Covalent Bonding and Valence Electrons

A Lewis diagram (Lewis structure) represents valence electrons in molecules and polyatomic ions. It’s a model for predicting connectivity, bond order, resonance, and (with VSEPR) shape and polarity.

Why valence electrons matter

Bonding is primarily about valence electrons, since core electrons are not typically involved. Many atoms reach lower energy by achieving a noble-gas-like configuration:

• Octet rule: many atoms tend to form bonds until surrounded by 8 valence electrons.
• Duet rule: hydrogen tends to have 2 valence electrons.

Also, covalent molecules can vary widely in size; there is no strict “size limit” on how many atoms can be covalently connected.

Step-by-step method for drawing Lewis structures

1. Count total valence electrons using the periodic table.
   • Add electrons for a negative charge.
   • Subtract electrons for a positive charge.
2. Choose a skeletal structure.
   • The least electronegative atom is often central (never H).
3. Connect atoms with single bonds (each bond is 2 electrons).
4. Complete octets on terminal atoms first.
5. Place remaining electrons on the central atom.
6. If the central atom lacks an octet, form multiple bonds by converting lone pairs on terminal atoms into bonding pairs.

Exceptions to the octet rule you must recognize

1. Incomplete octets (electron-deficient):

   • Hydrogen and helium follow a duet; hydrogen bonds, while helium essentially does not form typical bonds.
   • Boron is commonly stable with 6 electrons (example: BF_3).
   • Beryllium can also form electron-deficient compounds.

2. Odd-electron species (radicals):

   • Total valence electron count is odd (example: NO).

3. Expanded octets (period 3 or below):

   • Elements such as Si, P, S, and Cl can have more than 8 electrons around the central atom in some valid structures (example: SF_6).
   • Period 2 elements (C, N, O, F) do not expand octets in typical Lewis structures.
   • Noble gases can sometimes form compounds under the right conditions.

Example 1: Lewis structure of CO_2

1. Valence electrons: C has 4, each O has 6, total 16.
2. Skeleton: O–C–O.
3. Single bonds use 4 electrons.
4. Complete octets on O atoms (12 more electrons).
5. Carbon still lacks an octet, so convert one lone pair from each O into a bonding pair.

Final: O=C=O, each O has two lone pairs.

Example 2: Lewis structure of NH_4^+

1. Valence electrons: N has 5; 4 H contribute 4; total 9.
2. Positive charge means subtract 1 electron: total 8.
3. Skeleton: N single-bonded to four H.
4. Four bonds use all 8 electrons, leaving no lone pairs.

Result: N has an octet via four bonding pairs; the ion has a +1 charge.

Exam Focus

• Typical question patterns:
  • Draw the correct Lewis structure for a molecule or polyatomic ion, including brackets and charge.
  • Identify the number of bonding pairs and lone pairs on a given atom from a Lewis diagram.
  • Decide whether multiple bonds are needed to satisfy octets.
• Common mistakes:
  • Forgetting to adjust electron count for ionic charge.
  • Putting hydrogen in the center or giving hydrogen an octet.
  • Drawing expanded octets for period 2 atoms (C, N, O, F).

Resonance and Formal Charge: Choosing the Best Lewis Representation

Some species cannot be represented accurately by a single Lewis structure. Resonance and formal charge help create a better electron-delocalization model.

Resonance: when one Lewis structure is not enough

Resonance structures differ only in electron placement, not atom placement. The real molecule/ion is a resonance hybrid with delocalized electron density; it is not rapidly “flipping” between drawings.

Resonance helps explain:

• equal bond lengths where you might expect different lengths
• added stability from spreading charge over multiple atoms

Formal charge: an electron-accounting tool

Formal charge is a bookkeeping method (not the same as oxidation number). A reliable formula is:

FC = V - N - \frac{B}{2}

where V is valence electrons of the free atom, N is nonbonding electrons on that atom, and B is bonding electrons in bonds to that atom.

An equivalent way to say this is: formal charge equals valence electrons minus “assigned” electrons (lone-pair electrons plus half of bonding electrons).

Good Lewis structures generally:

• minimize the magnitude of formal charges
• place negative formal charges on more electronegative atoms
• avoid unnecessary charge separation

Example: nitrate ion, NO_3^-

A valid structure has N central with three O atoms. To satisfy octets, you typically draw one N=O and two N–O single bonds with an overall −1 charge. Any oxygen can be the double-bonded one, creating three equivalent resonance structures.

Consequence: all three N–O bonds are equivalent in the resonance hybrid with bond order between a single and a double bond.

Example: carbonate resonance and bond order

The carbonate ion (commonly written as CO_3^{2-}) can be drawn with two single bonds and one double bond, where the double bond can be on any oxygen. Even though one drawing shows a double bond, the resonance hybrid has three equal C–O bonds with lengths and strengths between a single and double bond.

A simple way to quantify “in between” is average bond order:

• single bond has bond order 1
• double bond has bond order 2

Add the bond orders shown across the resonance set and divide by the number of resonance forms. For carbonate, one way to compute the average is:

\frac{1+2+1}{3} = 1.33

Example: using formal charge to choose between structures

If one candidate structure has formal charges of +1 and −1 on adjacent atoms while another has smaller magnitudes (often 0), the second is usually preferred because it has less charge separation. When deciding between structures with similar charge magnitudes, place negative formal charge on the more electronegative atom (for example, negative on O is typically more reasonable than negative on N).

Exam Focus

• Typical question patterns:
  • Draw resonance structures and indicate resonance with a double-headed arrow.
  • Calculate formal charges and use them to justify the best Lewis structure.
  • Explain equal bond lengths or unexpected stability using resonance (delocalization).
• Common mistakes:
  • Moving atoms instead of only electrons when drawing resonance.
  • Treating the resonance hybrid as “one structure is correct and the others are wrong.”
  • Confusing formal charge with actual partial charges (formal charge is model-based bookkeeping).

VSEPR Theory: Predicting Molecular Shape from Electron Domains

Once you have a Lewis structure, you can predict 3D shape using VSEPR theory (Valence Shell Electron Pair Repulsion). Electron regions arrange to minimize repulsions; when there are more than two electron regions, shape depends on the number of bonding regions and lone pairs on the central atom.

Electron domains: what counts as a region of electron density

Count electron domains around the central atom:

• each single bond counts as 1 domain
• each double bond counts as 1 domain
• each triple bond counts as 1 domain
• each lone pair counts as 1 domain

Double and triple bonds are treated the same as single bonds for the purpose of predicting overall geometry because VSEPR counts regions, not individual shared pairs.

Electron geometry vs. molecular geometry

• Electron geometry: arrangement of all electron domains (bonds + lone pairs).
• Molecular geometry: arrangement of atoms only (ignore lone pairs when naming shape).

Common electron geometries and molecular geometries

Bond angles and lone-pair effects

Ideal bond angles:

• linear: 180°
• trigonal planar: 120°
• tetrahedral: 109.5°

Lone pairs repel more strongly than bonding pairs, compressing bond angles:

• CH_4: 109.5°
• NH_3: about 107°
• H_2O: about 104.5°

Worked example: shape of SO_2

1. Draw Lewis structure (S central, two O atoms; resonance is common).
2. Electron domains on S: two bonding regions + one lone pair = 3 domains.
3. Electron geometry: trigonal planar.
4. Molecular geometry: bent.
5. Bond angle: less than 120° due to lone-pair compression.

Exam Focus

• Typical question patterns:
  • Determine electron geometry and molecular geometry from a Lewis structure.
  • Predict relative bond angles when lone pairs are added (compare related molecules).
  • Use VSEPR to explain polarity or physical properties through shape.
• Common mistakes:
  • Counting multiple bonds as multiple electron domains.
  • Naming electron geometry when the question asks for molecular geometry (or vice versa).
  • Assuming bond angles are always the ideal values even with lone pairs.

Bond Hybridization and Sigma/Pi Bonding (Connecting Geometry to Orbitals)

Hybridization is a model that connects electron-domain geometry to orbital orientation, and it helps distinguish sigma and pi bonding in single vs. multiple bonds.

Sigma bonds and pi bonds

• A sigma bond forms from end-to-end overlap along the internuclear axis. Every single bond is a sigma bond, and the first bond between two atoms is sigma.
• A pi bond forms from side-by-side overlap of unhybridized p orbitals. In a double bond, the second bond is pi; in a triple bond, the second and third bonds are pi.

Multiple bonds contain:

• double bond: 1 sigma + 1 pi
• triple bond: 1 sigma + 2 pi

Pi bonding restricts rotation, which is why double bonds can create rigid geometry.

How to determine hybridization from electron domains

Hybridization aligns with the number of electron domains (same counting used in VSEPR):

This “domain counting” approach is highly effective for typical AP Chemistry molecules.

Worked example: hybridization and bonding in C_2H_4 (ethene)

1. Each carbon has three electron domains: two C–H single bonds and one C–C bonding region (the double bond counts as one domain).
2. Three domains means trigonal planar electron geometry and sp2 hybridization.
3. The C=C double bond consists of one sigma bond (sp2–sp2 overlap) and one pi bond (p–p overlap).

Result: planar geometry around each carbon and restricted rotation about the double bond.

Worked example: hybridization in CO_2

Central carbon has two electron domains (two double bonds, each counts as one domain). Two domains gives linear geometry and sp hybridization.

Exam Focus

• Typical question patterns:
  • Determine hybridization of a central atom from a Lewis structure (via electron domains).
  • Identify sigma vs. pi bonds and count how many of each are present.
  • Relate multiple bonds to shorter bond length and restricted rotation.
• Common mistakes:
  • Counting a double bond as two electron domains when assigning hybridization.
  • Saying a double bond is “two sigma bonds” (it is one sigma and one pi).
  • Forgetting that lone pairs count as domains for hybridization assignment.

Molecular Polarity: From Bond Dipoles to Net Dipole

Polarity is a major structure-to-property bridge because it depends on electronegativity and geometry.

Bond polarity: unequal sharing of electrons

A polar covalent bond results from unequal sharing due to electronegativity differences. The bond has a dipole pointing toward the more electronegative atom. Polarity is a spectrum, not just “polar vs. nonpolar.”

Molecular polarity: vector sum of bond dipoles

To decide whether a molecule is polar:

1. Identify polar bonds.
2. Determine 3D geometry (VSEPR).
3. Check whether bond dipoles cancel by symmetry.

Examples highlighting geometry’s role

• CO_2: polar C=O bonds but linear symmetry cancels dipoles, so nonpolar.
• H_2O: bent shape prevents cancellation, so polar.
• BF_3: polar B–F bonds but trigonal planar symmetry cancels, so nonpolar.
• NH_3: trigonal pyramidal; lone pair breaks symmetry, so polar.

Why polarity matters for properties

Polarity influences solubility and intermolecular attractions. Polar substances tend to dissolve in polar solvents; ionic compounds dissolve in polar solvents through ion-dipole interactions. Polarity also affects boiling/melting trends via intermolecular forces (expanded in Unit 3) and how molecules align in electric fields.

Exam Focus

• Typical question patterns:
  • Determine whether a molecule is polar based on Lewis structure and VSEPR geometry.
  • Compare physical properties (like boiling points) using polarity as part of the reasoning.
  • Identify which bonds are polar and indicate dipole directions.
• Common mistakes:
  • Declaring “has polar bonds, therefore polar molecule” without considering geometry.
  • Ignoring lone pairs that destroy symmetry and create polarity.
  • Mixing up formal charge (Lewis bookkeeping) with polarity (electron distribution in space).

Types of Solids: Molecular, Ionic, Metallic, and Covalent Network

A frequent AP task is classifying a solid based on properties and then justifying those properties using the correct particle model.

Four major categories and particle-level structures

1. Molecular solids

   • Particles: neutral molecules
   • Attractions: intermolecular forces (dispersion, dipole-dipole, hydrogen bonding)
   • Properties: generally lower melting points; poor electrical conductivity

2. Ionic solids

   • Particles: cations and anions in a lattice
   • Attractions: strong Coulombic attractions
   • Properties: high melting points, brittle, conduct when molten/aqueous

3. Metallic solids

   • Particles: metal cations in a lattice with delocalized electrons
   • Properties: conduct as solids; malleable/ductile

4. Covalent network solids

   • Particles: atoms covalently bonded in an extended lattice (a network of covalent bonds)
   • Properties: very hard; very high melting and boiling points; usually poor electrical conductors because electrons cannot move freely through the lattice
   • Notable exception: graphite conducts due to delocalized electrons within layers

Network solids are especially common for carbon (diamond, graphite) and silicon-containing solids (such as quartz) because having four valence electrons supports extensive covalent bonding.

Real-world examples (and what to notice)

• Diamond: carbon network, each C bonded to four others; extremely hard, very high melting point, electrical insulator.
• Graphite: planar carbon sheets with delocalized electrons; conducts along sheets; layers slide, making it softer.
• Silicon dioxide (quartz): network solid; high melting point.
• Solid I_2: molecular solid held by dispersion forces; relatively low melting point compared to ionic/network solids.

Example classification question

A solid has a very high melting point and does not conduct electricity in solid or molten form.

• If ionic, it would conduct when molten.
• If metallic, it would conduct as a solid.
• If molecular, melting point would likely be much lower.

Best match: covalent network solid.

Exam Focus

• Typical question patterns:
  • Identify solid type from properties (melting point, conductivity, hardness).
  • Explain melting-point differences using attraction type/strength.
  • Compare two solids (e.g., graphite vs. diamond) based on structure.
• Common mistakes:
  • Confusing molecular solids with covalent network solids (both involve covalent bonds, but structures are fundamentally different).
  • Saying “ionic solids conduct electricity” without specifying molten/aqueous.
  • Forgetting graphite’s conductivity exception and why it happens.

Conductivity as a Structure Clue (Phase Matters)

Conductivity depends on whether charged particles are mobile. A useful way to summarize typical AP expectations is:

Important clarifications:

• Covalent substances generally do not conduct electricity. Pure water conducts very poorly; tap water conducts better because it contains dissolved ions.
• Ionic substances do not conduct as solids because ions cannot move in the rigid lattice. When molten or aqueous, ions are mobile and conductivity becomes significant.
• For ionic solutions, conductivity depends on how many ions are present:
  • Higher concentration means more charge carriers. For example, 1.0 M NaCl conducts better than 0.1 M NaCl.
  • Substances that dissociate into more ions generally conduct better at the same molar concentration. For example, 1.0 M CaCl_2 (three ions per formula unit) typically conducts better than 1.0 M NaCl (two ions per formula unit).

Exam Focus

• Typical question patterns:
  • Use conductivity in different phases to identify bonding/solid type.
  • Compare ionic-solution conductivity using concentration and dissociation into ions.
• Common mistakes:
  • Saying ionic solutions conduct because “electrons are free”; in ionic conduction, ions carry charge.
  • Forgetting the phase dependence (solid ionic compounds do not conduct well).

Solid-State Structure: Unit Cells, Packing, and Density

Many solids (especially metals and ionic compounds) are crystalline with a repeating 3D pattern. A unit cell is the smallest repeating box that can be stacked to build the full crystal.

Counting particles in a unit cell

Particles on boundaries are shared:

• Corner: shared by 8 unit cells, counts as \frac{1}{8}
• Edge-centered: shared by 4, counts as \frac{1}{4}
• Face-centered: shared by 2, counts as \frac{1}{2}
• Completely inside: counts as 1

Common cubic metallic unit cells

• Simple cubic (SC): corners only, atoms per unit cell = 1
• Body-centered cubic (BCC): corners + 1 body atom, atoms per unit cell = 2
• Face-centered cubic (FCC): corners + face centers, atoms per unit cell = 4

These totals come from counting:

• SC: 8 \times \frac{1}{8} = 1
• BCC: corners contribute 1 plus 1 body atom = 2
• FCC: corners contribute 1 and faces contribute 6 \times \frac{1}{2} = 3, total 4

Packing efficiency

Modeling atoms as hard spheres:

• SC: about 52%
• BCC: about 68%
• FCC: about 74%

Higher packing efficiency often (not always) correlates with higher density.

Relating atomic radius to unit cell edge length

Geometric relationships used in density problems:

a = 2r

(for SC, atoms touch along an edge)

a = \frac{4r}{\sqrt{3}}

(for BCC, atoms touch along the body diagonal)

a = 2\sqrt{2}r

(for FCC, atoms touch along the face diagonal)

Density of a crystalline metal from its unit cell

For a cubic unit cell:

\rho = \frac{Z M}{N_A a^3}

where Z is atoms per unit cell, M is molar mass (g/mol), N_A is Avogadro’s number, and a is edge length (cm).

Worked example: density from an FCC unit cell

A metal is FCC with M = 63.55 g/mol and a = 3.61 \times 10^{-8} cm.

1. For FCC, Z = 4.
2. Use:

\rho = \frac{Z M}{N_A a^3}

3. Substitute:

\rho = \frac{4 \times 63.55}{(6.022 \times 10^{23})(3.61 \times 10^{-8})^3}

4. Compute volume:

• a^3 = (3.61^3) \times 10^{-24} cm³
• 3.61^3 \approx 47.0
  So a^3 \approx 4.70 \times 10^{-23} cm³.

4. Finish:

• Denominator: (6.022 \times 10^{23})(4.70 \times 10^{-23}) \approx 28.3
• Numerator: 4 \times 63.55 = 254.2

\rho \approx \frac{254.2}{28.3} \approx 8.98 \text{ g/cm}^3

Ionic solids and unit cells (conceptual)

Ionic crystals can also be described using unit cells. AP questions often provide a diagram and ask you to count ions (using corner/edge/face sharing rules) and reduce to get the empirical formula. Memorizing named ionic structures is less reliable than counting from the diagram.

Exam Focus

• Typical question patterns:
  • Count atoms/ions in a unit cell diagram and determine the formula.
  • Identify SC, BCC, or FCC from a diagram and state atoms per unit cell.
  • Calculate density using \rho = \frac{Z M}{N_A a^3}, sometimes with radius-to-edge relationships.
• Common mistakes:
  • Forgetting sharing fractions for corners/edges/faces.
  • Mixing up BCC and FCC (BCC has a body atom; FCC has face atoms).
  • Unit errors in density problems (especially converting edge length to cm before cubing).