Compound Structure and Properties
Unit 2 of AP Chemistry, worth 7–9% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Types of bonds, intramolecular force and potential energy, structure of ionic solids and metals, Lewis diagrams, resonance and formal charge, VSEPR and bond hybridisation.
How this unit is tested
Start every bonding question by comparing electronegativities: a large difference between a metal and nonmetal signals an ionic bond, a small or zero difference between nonmetals signals covalent, and identical metal atoms signal metallic bonding with delocalized electrons. For any potential-energy or lattice-energy comparison, go straight to Coulomb's law: energy scales with the product of the charges and inversely with the distance between centers, so compare charge magnitude first, then ionic radius.
For Lewis structures, always count total valence electrons first (adding one electron per negative charge, subtracting one per positive charge), place the least electronegative atom in the center, connect atoms with single bonds, distribute remaining electrons to complete octets on outer atoms, then convert lone pairs to multiple bonds on the central atom if it is electron-deficient. When more than one valid structure exists, that's resonance — draw every reasonable structure and use formal charge to rank them.
Formal charge is a bookkeeping tool, not a real charge: calculate it for every atom, and prefer structures where formal charges are closest to zero, and where any negative formal charge sits on the more electronegative atom. Once the structure is settled, apply VSEPR: count the total electron domains (bonding regions plus lone pairs) around the central atom to get the electron-domain geometry, then note which domains are lone pairs to get the molecular geometry and the hybridization of the central atom's orbitals.
Solids questions (ionic crystals and metals) are really about explaining macroscopic properties from microscopic structure: rigid, directional ionic bonds in a lattice explain brittleness and high melting points, while nondirectional delocalized electrons in a metal explain malleability and conductivity. Practice translating structure to property in a sentence, since that is how the exam usually asks it.
What you have to know
Coulomb's law for lattice/ionic interactions
The potential energy of interaction between two ions is $E \propto \dfrac{q_1 q_2}{r}$, where $q_1$ and $q_2$ are the ion charges and $r$ is the distance between their centers. Lattice energy increases in magnitude with larger charges and smaller ionic radii.
Formal charge
$FC = (\text{valence electrons of free atom}) - (\text{nonbonding electrons}) - \dfrac{1}{2}(\text{bonding electrons})$. The sum of formal charges in a structure equals the overall charge on the species.
VSEPR electron-domain counting
Count the number of bonding domains (single, double, or triple bonds each count as one domain) plus lone pairs on the central atom. This total determines electron-domain geometry (2=linear, 3=trigonal planar, 4=tetrahedral, 5=trigonal bipyramidal, 6=octahedral); removing lone-pair positions from the description gives the molecular geometry.
Hybridization from electron domains
2 domains = sp, 3 domains = sp2, 4 domains = sp3, 5 domains = sp3d, 6 domains = sp3d2. Each sigma bond uses one hybrid orbital; remaining unhybridized p orbitals form pi bonds.
Resonance and stability preference
When multiple valid Lewis structures differ only in electron placement (not atom position), the true structure is a resonance hybrid; the most representative structures minimize formal charges and place negative formal charge on the more electronegative atom.
14 practice questions
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Which type of bonding is characterized by valence electrons delocalized in an 'electron sea' surrounding a lattice of cations?
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Answer. Metallic bonding
In metals, valence electrons are not localized between specific atoms but move freely throughout the lattice, allowing metal atoms to slide past each other without breaking bonds. This explains conductivity and malleability. -
Rank NaCl, MgO, KBr, and CaS from largest to smallest lattice energy based on Coulomb's law.
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Answer. MgO > CaS > NaCl > KBr
Lattice energy scales with the product of ionic charges divided by the distance between ion centers. MgO and CaS both have 2+/2- ion pairs (larger charge product) versus 1+/1- for NaCl and KBr, so they dominate; MgO beats CaS because Mg2+ and O2- are smaller ions than Ca2+ and S2-, giving a shorter distance. Among the 1+/1- pairs, NaCl has smaller ions than KBr, so its lattice energy is larger. -
Rank the carbon-carbon bonds in ethane (C-C), ethene (C=C), and ethyne (C≡C) from shortest to longest.
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Answer. C≡C < C=C < C-C
Higher bond order pulls the bonded atoms closer together because more shared electron density increases attraction, so the triple bond is shortest and strongest, while the single bond is longest and weakest. -
On a plot of potential energy versus internuclear distance for two approaching atoms, where does the potential energy reach its minimum?
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Answer. At the equilibrium bond length, where attractive and repulsive forces balance
As atoms approach, potential energy decreases due to attraction between nucleus and electrons; at very short distances, nucleus-nucleus repulsion dominates and energy rises sharply. The minimum point on the curve corresponds to the most stable internuclear distance, the bond length. -
Draw a Lewis structure of the sulfate ion, SO4^2-, in which sulfur forms only single bonds to each of the four oxygen atoms. What is the formal charge on sulfur in this structure?
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Answer. +2
Sulfur has 6 valence electrons, 0 nonbonding electrons, and 4 bonding pairs (8 bonding electrons) in this structure, so FC = 6 − 0 − (8/2) = 6 − 4 = +2. This high formal charge is one reason chemists often draw sulfate with some S=O double bonds to lower sulfur's formal charge toward zero, using sulfur's expanded octet capability. -
The thiocyanate ion, SCN⁻, has resonance structures with the negative formal charge on either sulfur or nitrogen depending on double bond placement. Which structure is favored, and why?
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Answer. The structure with the negative formal charge on nitrogen (more electronegative than sulfur) is favored
When more than one resonance structure gives formal charges of the same magnitude, the more stable structure places negative formal charge on the more electronegative atom. Since nitrogen is more electronegative than sulfur, the structure S=C=N with negative charge concentrated toward nitrogen contributes more to the actual resonance hybrid. -
Draw the Lewis structure for BF3. Why does boron violate the octet rule, and what is boron's formal charge in the resulting structure?
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Answer. Boron ends up with only 6 electrons (an incomplete octet), and its formal charge is 0
Boron has only 3 valence electrons and forms 3 single bonds to fluorine, giving it 3 bonding pairs and no lone pairs, so it has 6 electrons total rather than 8. Because fluorine's strong electronegativity resists forming a double bond that would give boron more electrons, boron is commonly drawn with an incomplete octet, and FC = 3 − 0 − 3 = 0. -
Nitrogen dioxide, NO2, is an odd-electron molecule. How many valence electrons must be distributed in its Lewis structure, and what structural feature results from this odd number?
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Answer. 17 valence electrons total, resulting in one unpaired electron (a radical) on the nitrogen atom
N contributes 5 and each O contributes 6, for 5 + 12 = 17 electrons, an odd number that cannot be fully paired. After forming one N=O double bond and one N-O single bond with complete oxygen octets, nitrogen is left with a single unpaired electron rather than a lone pair, making NO2 a radical. -
What is the molecular geometry of ClF3, and how many lone pairs does the central chlorine atom have?
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Answer. T-shaped, with 2 lone pairs on chlorine
Chlorine has 7 valence electrons; forming 3 bonds to fluorine uses 3 electrons, leaving 2 lone pairs, for a total of 5 electron domains. Five domains give a trigonal bipyramidal electron-domain geometry, and placing the 2 lone pairs in equatorial positions (to minimize repulsion) leaves the 3 bonded fluorines in a T-shaped molecular geometry. -
What is the hybridization of the central beryllium atom in BeCl2, and what molecular geometry results?
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Answer. sp hybridization, linear geometry
Beryllium forms 2 single bonds to chlorine and has no lone pairs, giving 2 electron domains. Two domains correspond to sp hybridization and a linear electron-domain and molecular geometry (180 degree bond angle). -
What is the hybridization of sulfur in SF6, and what is the molecular geometry?
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Answer. sp3d2 hybridization, octahedral geometry
Sulfur forms 6 single bonds to fluorine with no lone pairs, giving 6 electron domains. Six domains require sp3d2 hybridization (using an expanded octet with d orbitals) and produce an octahedral molecular geometry with 90 degree bond angles. -
In the Lewis structure of formaldehyde, CH2O (C double-bonded to O, single-bonded to two H atoms), how many total sigma bonds and pi bonds are present?
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Answer. 3 sigma bonds and 1 pi bond
The two C-H bonds are each single (1 sigma each), and the C=O double bond consists of 1 sigma bond plus 1 pi bond. Adding these gives 2 + 1 = 3 sigma bonds total and 1 pi bond total, consistent with carbon's sp2 hybridization (3 sigma bonds from hybrid orbitals) and one unhybridized p orbital forming the pi bond. -
Explain, at the particle level, why ionic solids are brittle while metals are malleable.
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Answer. Ionic solids have fixed, directional electrostatic bonds in a rigid lattice that fracture when like-charged ions are forced into contact; metals have nondirectional delocalized electrons that let atoms slide past each other without breaking bonds
Applying stress to an ionic crystal shifts rows of ions so that ions of the same charge line up, and their mutual repulsion shatters the lattice along a cleavage plane. In a metal, the delocalized 'sea' of electrons continues to bond the metal cations together even as their positions shift, so the metal deforms rather than breaking. -
A student compares HF and NaF and must classify their bond types. What type of bond is present in each, and what electronegativity-difference reasoning justifies the classification?
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Answer. HF has a polar covalent bond; NaF has an ionic bond
H and F are both nonmetals sharing electrons, giving a covalent bond, but the fairly large electronegativity difference between H and F makes it strongly polar. Na is a metal and F is a nonmetal with a very large electronegativity difference, causing essentially complete electron transfer and forming an ionic bond rather than a shared pair.
What people get wrong
- Forgetting to add or subtract electrons for the ion's overall charge when counting valence electrons for a polyatomic ion; always adjust the total electron count before drawing the skeleton.
- Treating one resonance structure as 'the' correct structure and assigning fixed bond lengths to it; instead recognize the resonance hybrid has bond lengths and orders intermediate between the individual structures.
- Confusing electron-domain geometry with molecular geometry when lone pairs are present on the central atom; state both separately, since lone pairs change the molecular shape (e.g., bent vs. linear) but not the electron-domain arrangement.
- Assuming formal charge of zero everywhere is always achievable; some structures require formal charges, and the goal is minimizing magnitude and placing negative charge appropriately, not forcing all charges to zero.
- Assuming higher ionic charge always means higher lattice energy without checking ionic radius; Coulomb's law depends on both charge and distance, so a smaller ion pair with lower charge can still have comparable or greater lattice energy than expected.
- Miscounting bonding domains in multiple bonds for VSEPR by counting each pi bond separately; a double or triple bond between two atoms still counts as only one electron domain.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Chemistry. Free, and no account needed to start.