AP Chemistry, AP exams, 7–9%

Unit 2: Compound Structure and Properties

Unit 2 moves from atoms to bonded structures: ionic, covalent, and metallic bonding, and the geometry that follows from electron domains. Lewis structures, resonance, formal charge, and VSEPR shapes are the working tools. The unit connects bonding directly to physical properties such as melting point and conductivity.

Position: 2 of 9 units
Published weighting: 7–9% in the official framework
Topics covered: 6 topics in this guide

What is covered

Topics inside this unit.

These topics describe the scope students most often review for this part of AP Chemistry. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

Types of chemical bonds: Ionic, covalent, and metallic bonding placed on a continuum by electronegativity difference.
Intramolecular force and potential energy: Reading potential energy curves for bond length and bond energy, and relating them to bond order.
Structure of ionic solids, metals, and alloys: Lattice arrangement, the origin of high melting points, and how interstitial and substitutional alloys differ.
Lewis diagrams: Building valid structures, tracking formal charge, and handling expanded and incomplete octets.
Resonance and formal charge: Choosing among structures using formal charge, and what delocalization implies about bond length and strength.
VSEPR and hybridization: Predicting electron-domain and molecular geometry, bond angles, polarity, and the matching hybridization.

Study tips

How to practice Compound Structure.

Section-specific guidance that follows from how this material is usually tested.

Draw Lewis structures with a fixed routine (count electrons, connect, complete octets, check formal charge) so the process never varies.
Pair each geometry with its bond angles and an example molecule on a single card set.
Practice going from structure to a physical property prediction; that direction is what free-response questions ask for.

Where this fits

Unit 2: Compound Structure and Properties is 2 of 9 units in the AP Chemistry guide, and it carries a published weighting of 7–9% in the official framework. Use the full guide to balance it against everything else you have to review.

Back to the AP Chemistry guide Official exam information

Sample Compound Structure questions

Practice questions, not yet expert-reviewed

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

Question 1Foundational
What is the molecular geometry of a molecule whose central atom has four bonding pairs and no lone pairs, such as CH4?
1. ATrigonal planar
2. BTrigonal pyramidal
3. CBent
4. DTetrahedral
Show answer
Answer: D. Tetrahedral
Four electron domains repel each other into the arrangement that maximizes their separation, which is a tetrahedron with bond angles of about 109.5 degrees. Trigonal pyramidal is the tempting distractor, but that shape requires one of the four domains to be a lone pair, as in NH3. Trigonal planar corresponds to only three electron domains.
Key insight: Count electron domains first, then subtract lone pairs to name the shape the atoms actually make.
Question 2Foundational
What type of bond forms between two atoms with a very large electronegativity difference, such as sodium and chlorine?
1. AIonic bond
2. BNonpolar covalent bond
3. CPolar covalent bond
4. DMetallic bond
Show answer
Answer: A. Ionic bond
When the electronegativity difference is large, one atom effectively transfers an electron to the other and the resulting oppositely charged ions attract in a lattice. A polar covalent bond is the tempting distractor because it also involves unequal sharing, but there the electrons remain shared rather than transferred. Metallic bonding requires two metals sharing delocalized electrons.
Key insight: Bond type is a continuum set by electronegativity difference: equal sharing, unequal sharing, then transfer.
Question 3Intermediate
In the ammonium ion NH4+, nitrogen forms four single bonds and has no lone pairs. What is the formal charge on that nitrogen atom?
1. A-1
2. B0
3. C+1
4. D+4
Show answer
Answer: C. +1
Formal charge equals valence electrons minus nonbonding electrons minus half of the bonding electrons, so nitrogen gives 5 minus 0 minus 4, which is +1. Answering 0 forgets that nitrogen normally holds a lone pair and only three bonds. The +1 formal charge on nitrogen accounts for the entire charge of the ion, since each hydrogen has a formal charge of 0.
Key insight: Formal charges in a Lewis structure must sum to the overall charge of the species.
Question 4Intermediate
The nitrate ion NO3- is represented by three equivalent resonance structures. What does that predict about its three nitrogen-oxygen bonds?
1. AOne is shorter than the other two because it is a double bond
2. BAll three are full double bonds and therefore very short
3. CAll three are identical and intermediate between a single and a double bond
4. DTwo are single bonds and one alternates rapidly between positions
Show answer
Answer: C. All three are identical and intermediate between a single and a double bond
Resonance means the real ion is a single delocalized structure rather than a mixture that flips between forms, so all three bonds are equivalent with a bond order of about 1.33. Experimental bond lengths confirm that they are equal and lie between typical single and double bond lengths. The idea that one bond is shorter comes from reading a single Lewis structure too literally.
Key insight: Resonance structures are drawing limitations, not real states the molecule flips between.
Question 5Advanced
MgO and NaF have similar distances between the centers of neighboring ions, yet MgO melts at a far higher temperature. What best explains the difference?
1. AMgO forms hydrogen bonds between its ions while NaF does not
2. BMgO contains 2+ and 2- ions, so the Coulombic attraction is much stronger
3. CMgO has the larger molar mass, so its particles move more slowly
4. DMgO is held together by covalent bonds while NaF is ionic
Show answer
Answer: B. MgO contains 2+ and 2- ions, so the Coulombic attraction is much stronger
Coulomb's law makes the force of attraction proportional to the product of the charges and inversely proportional to the distance between the ion centers. Going from 1+ and 1- ions to 2+ and 2- ions multiplies that product by four while the distance stays about the same, so far more energy is needed to pull the lattice apart. Molar mass is the tempting distractor, but it does not determine how strongly ions are held in a lattice.
Key insight: When interionic distances are similar, ionic charge magnitude decides lattice strength and melting point.

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Unit 2: Compound Structure and Properties