Molecular and Ionic Compound Structure and Properties — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: The full scope of AP Chemistry Unit 2, including all 7 core sub-topics: types of chemical bonds, intramolecular force and potential energy, ionic solid structure, metal and alloy structure, Lewis diagrams, resonance, formal charge, and VSEPR and bond hybridization.

You should already know: Basic atomic structure and valence electron counting. Periodic trends in electronegativity and ionization energy. The difference between atoms, ions, and compounds.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. Why This Matters

This unit is the foundational link between the atomic structure you learned in Unit 1 and every subsequent topic about compound properties, reactions, and intermolecular forces in the rest of the AP Chemistry course. Almost all observable properties of compounds—from the conductivity of copper wire to the hardness of table salt to the shape of drug molecules that bind to enzymes—can be traced back to the bonding and structure concepts you learn here. This unit teaches you to move beyond just writing chemical formulas to explaining why a compound has the properties it does, which is the core of what AP Chemistry tests: connecting microscopic structure to macroscopic behavior. Because this content appears across nearly every unit on the exam, a weak foundation here will make topics like intermolecular forces, thermodynamics, and acid-base reactivity much harder than they need to be. Per the AP CED, this unit accounts for 7–9% of your total exam score, and concepts from this unit are integrated into roughly 15–20% of all exam questions as supporting reasoning.

2. Unit Concept Map: How Sub-Topics Build On Each Other

The 7 sub-topics of this unit follow a logical, sequential progression, starting from the most basic classification of bonds and ending with full 3D structure of covalent molecules:

1. Types of chemical bonds: The starting point, where you classify bonds as ionic (electron transfer, large electronegativity difference), covalent (electron sharing, small electronegativity difference), or metallic (delocalized electron sea, metal atoms only). This sets the framework for all further analysis: your classification tells you which structural model to use.
2. Intramolecular force and potential energy: Next, you connect bond type to bond strength via potential energy curves, learning how bond length relates to bond dissociation energy. This gives you the energetic foundation to explain stability and reactivity later.
3. Structure of ionic solids: Next, you explore the bulk structure of ionic compounds, how ionic lattices form, and how lattice energy relates to ionic charge and radius. This connects ionic bonding to bulk properties like melting point and conductivity.
4. Structure of metals and alloys: Next, you cover bulk structure of metallic compounds, the electron sea model, and how alloying changes metal properties. This completes your coverage of non-molecular bulk structures.
5. Lewis diagrams: You then shift to covalent molecular compounds, starting with 2D representations of molecules that show valence electron distribution. This is the required precursor to all further analysis of molecular geometry.
6. Resonance and formal charge: This refines Lewis diagrams, giving you a tool to select the most stable valid Lewis structure when multiple arrangements exist.
7. VSEPR and bond hybridization: The final capstone of the unit, where you extend 2D Lewis diagrams to predict 3D molecular geometry and bond hybridization, connecting structure to molecular shape and reactivity.

A Guided Tour: Solving A Multi-Concept Unit Problem

Let's work through a common multi-part exam-style question step-by-step to see how each sub-topic connects: Question: Explain why sodium chloride (NaCl) has a melting point of 801°C, while chlorine gas (Cl₂) has a boiling point of -34°C. Then draw the Lewis structure for Cl₂, calculate formal charge for each atom, and predict its molecular geometry.

1. Step 1 (Types of chemical bonds): First, classify the bonds in each compound. NaCl has an electronegativity difference of ~2.2 (Na = 0.9, Cl = 3.2), so it forms an ionic compound with ionic bonding. Cl₂ is two identical atoms with 0 electronegativity difference, so it forms a nonpolar covalent molecular compound. Without this first classification step, you cannot correctly explain the melting/boiling point difference.
2. Step 2 (Structure of ionic solids vs covalent molecules): NaCl forms a crystalline ionic lattice held together by strong electrostatic attractions between Na⁺ and Cl⁻ ions. Large amounts of energy are required to break these strong lattice attractions, hence the high melting point. Cl₂ is a molecular compound, so only weak intermolecular forces between separate Cl₂ molecules need to be overcome to boil it, hence the very low boiling point.
3. Step 3 (Lewis diagrams, formal charge, VSEPR): Draw the Lewis diagram for Cl₂: total valence electrons = 7*2 = 14. A single bond between the two Cl atoms, 6 lone electrons on each atom, gives 14 electrons and satisfies the octet rule. Calculate formal charge for each Cl: $FC=\text{valence electrons}-(\text{lone pair electrons}+1/2\text{bonding electrons})=7-(6+1)=0$, which confirms the structure is stable. VSEPR: each Cl has 1 bonding domain, 3 lone pairs, so the molecular geometry is linear.

This tour shows how you must apply sub-topics in sequence: you cannot predict geometry before drawing a correct Lewis diagram, and you cannot explain bulk properties before classifying the bond type.

Exam tip: For any multi-part question that asks you to explain a property, always start by classifying the bond type—AP exam graders look for this first step as the foundation of your reasoning, and missing it will cost you points even if your final conclusion is correct.

3. Cross-Cutting Common Pitfalls (and how to avoid them)

• Wrong move: Confusing intramolecular bonds with intermolecular forces when explaining bulk properties of molecular compounds. Why: Students learn intramolecular bond strength first in this unit, so they incorrectly assume breaking covalent bonds is required to melt or boil a molecular compound. Correct move: When explaining melting/boiling point of any molecular compound, explicitly state that you are overcoming intermolecular forces between molecules, not breaking the intramolecular covalent bonds within molecules.
• Wrong move: Assuming all compounds with a metal are ionic. Why: The introductory rule of thumb "metal + nonmetal = ionic" is not a hard rule, so students misclassify compounds like AlCl₃ or metallic alloys as ionic. Correct move: Always calculate electronegativity difference to confirm bond classification, even if the compound contains a metal.
• Wrong move: Counting the number of lone pairs on the central atom incorrectly when doing VSEPR. Why: Students forget to subtract electrons used in bonding from the total valence electron count when drawing Lewis diagrams, leading to wrong number of electron domains. Correct move: After adding all valence electrons, count your drawn electrons and adjust the number of lone pairs or multiple bonds to match the total before doing VSEPR.
• Wrong move: Forgetting that formal charge is a bookkeeping tool, not an actual charge on atoms in a covalent molecule. Why: The formula for formal charge makes it look like actual ionic charge, so students incorrectly claim that a non-zero formal charge means the molecule is ionic. Correct move: Always state that formal charge is used to select the most stable Lewis structure among multiple valid options, and does not represent actual full charge separation in covalent molecules.
• Wrong move: Assuming all ionic solids have the same high melting point regardless of ion size and charge. Why: Students memorize that ionic solids have high melting points, so they incorrectly predict that NaF has a higher melting point than MgO, for example. Correct move: When comparing lattice energy of two ionic compounds, always check ionic charge first (higher charge = higher lattice energy) then ionic radius (smaller radius = higher lattice energy).
• Wrong move: Confusing electron domain geometry with molecular geometry in VSEPR. Why: Both are calculated from the number of domains, but students often mix them up when asked for molecular geometry. Correct move: When asked for molecular geometry, only count bonding domains around the central atom, not lone pairs, to get the correct shape.

4. Quick Check: When To Use Which Sub-Topic

For each of the following questions, which sub-topic from this unit would you use to answer it?

1. Which has a higher melting point: KBr or CaO?
2. What is the shape of the nitrate ion (NO₃⁻)?
3. Which Lewis structure of OCN⁻ is most stable?
4. Why is bronze harder than pure copper?
5. Why is a triple bond shorter and stronger than a single bond between the same two atoms?

5. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following correctly ranks the compounds below from lowest melting point to highest melting point? Compounds: Cl₂, CaCl₂, NaCl A) Cl₂ < NaCl < CaCl₂ B) NaCl < CaCl₂ < Cl₂ C) CaCl₂ < Cl₂ < NaCl D) Cl₂ < CaCl₂ < NaCl

Worked Solution: First, classify each compound by bond type. Cl₂ is a covalent molecular compound, so it has the lowest melting point because only weak intermolecular forces must be overcome to melt it. This eliminates options B and C, which do not list Cl₂ first. Next, compare the two ionic compounds NaCl and CaCl₂. NaCl is made of Na⁺ (+1) and Cl⁻ (-1), while CaCl₂ is made of Ca²⁺ (+2) and Cl⁻ (-1). Higher ionic charge leads to stronger electrostatic attractions in the ionic lattice and higher melting point. So the order from lowest to highest is Cl₂ < NaCl < CaCl₂. The correct answer is A.

Question 2 (Free Response)

The molecule dinitrogen monoxide (N₂O) has three possible valid Lewis structures with the N-N-O atomic skeleton. (a) Classify the bonding in N₂O as ionic, covalent, or metallic, and justify your answer. (b) Draw all three possible Lewis structures for N₂O, and calculate the formal charge of each atom in all three structures. (c) Identify the most stable Lewis structure and justify your choice using formal charge.

Worked Solution: (a) N₂O is composed of two nonmetals, nitrogen and oxygen, with an electronegativity difference of only 0.4 (N = 3.0, O = 3.5). A small electronegativity difference means electrons are shared, so bonding is covalent. (b) Total valence electrons = 5*2 + 6 = 16. The three valid structures are:

1. Structure 1 (Single N-N, triple N-O): Formal charges: End N = 0, Central N = +1, O = -1
2. Structure 2 (Double N-N, double N-O): Formal charges: End N = -1, Central N = +1, O = 0
3. Structure 3 (Triple N-N, single N-O): Formal charges: End N = 0, Central N = +1, O = -1 (c) The most stable structure places the negative formal charge on the most electronegative atom. Oxygen is more electronegative than nitrogen, so Structure 2 (negative charge on N) is eliminated. Both Structures 1 and 3 place the negative charge on O, but Structure 3 (triple N-N, single N-O) is most stable because it avoids placing a high formal charge on the electronegative O atom in a triple bond, matching experimental bond length data.

Question 3 (Application / Real-World Style)

Dental fillings are often made of amalgam, an alloy of mercury, silver, tin, and copper. Pure silver metal is soft and wears down quickly when used as a filling. Explain why adding tin and mercury to silver makes the alloy harder and more durable than pure silver, connecting your answer to the structure of metals and alloys.

Worked Solution: Pure silver has a regular, uniform crystalline lattice of silver atoms, all of the same size. The uniform lattice allows layers of silver atoms to slide past each other easily when stress is applied, making pure silver soft and malleable. When tin and mercury atoms are added to form the amalgam alloy, these atoms have different atomic radii than silver atoms. The different-sized atoms disrupt the regular crystalline lattice of pure silver. This prevents layers of silver atoms from sliding past each other easily when force is applied. As a result, the amalgam alloy is much harder and more resistant to wear than pure silver, making it ideal for long-lasting dental fillings.

6. Quick Reference Cheatsheet

7. What's Next (Individual Sub-Topics in This Unit)

This unit overview gives you the big picture of how all the sub-topics in Unit 2 connect to each other. Next, you will dive into each individual sub-topic to master the specific skills and concepts needed for the exam. Mastery of each sub-topic here is required to build the full foundation needed for later units: after completing all sub-topics in this unit, you will move on to Unit 3: Intermolecular Forces and Properties, where you will connect compound structure to intermolecular forces and bulk properties of liquids, solids, and gases. Without mastering the bonding and structure concepts from this unit, you will not be able to correctly explain intermolecular force strength or predict properties like boiling point and vapor pressure. Below are links to each individual sub-topic study guide in this unit:

• Types of chemical bonds
• Intramolecular force and potential energy
• Structure of ionic solids
• Structure of metals and alloys
• Lewis diagrams
• Resonance and formal charge
• VSEPR and bond hybridization