AP Chemistry Unit 2 Practice Test & Exam | 15 Questions

1. 1. [Skill: 6A | Topic: 2.1] A student compares electronegativity values for several representative elements. Electronegativity (Pauling scale): - Li: 0.98, Be: 1.57, B: 2.04, C: 2.55, N: 3.04, O: 3.44, F: 3.98 - Na: 0.93, Mg: 1.31, Al: 1.61, Si: 1.90, P: 2.19, S: 2.58, Cl: 3.16 Which explanation best accounts for the trend in electronegativity across each period?

(A)Electronegativity increases because the effective nuclear charge experienced by valence electrons increases, increasing Coulombic attraction for shared electrons.(B)Electronegativity increases because additional electron shells are added, increasing the size of the atom and making it more able to attract shared electrons.(C)Electronegativity increases because metals on the left have more filled orbitals, which causes them to attract shared electrons more strongly.(D)Electronegativity increases because atoms on the right have higher first ionization energies, which decreases their attraction for shared electrons.

2. 2. [Skill: 1B | Topic: 2.1] A student is comparing bond polarity in two bonds found in organic molecules: C—H and C—F. The electronegativity values are C = 2.55, H = 2.20, and F = 3.98. Which statement best describes the bonding and charge distribution in these bonds?

(A)C—H is nonpolar covalent because C and H have similar electronegativities; C—F is polar covalent with δ− on F because F is more electronegative.(B)C—H is ionic because H is a nonmetal and C is a nonmetal; C—F is nonpolar covalent because both atoms are nonmetals.(C)C—H is polar covalent with δ− on H; C—F is ionic because the electronegativity difference is greater than 1.0.(D)C—H is metallic because carbon can delocalize electrons; C—F is polar covalent with δ− on C because carbon is less electronegative.

3. 3. [Skill: 4A | Topic: 2.1] Three solids, X, Y, and Z, are tested for melting point and electrical conductivity. Observed properties: - Solid X: Melting point is very high; does not conduct electricity as a solid; conducts electricity when molten. - Solid Y: Melting point is low; does not conduct electricity as a solid or when melted. - Solid Z: Melting point is moderate to high; conducts electricity as a solid. Which statement correctly identifies the dominant type of bonding in X, Y, and Z based on the properties?

(A)X is ionic, Y is molecular (covalent), and Z is metallic.(B)X is metallic, Y is ionic, and Z is molecular (covalent).(C)X is molecular (covalent), Y is metallic, and Z is ionic.(D)X is ionic, Y is metallic, and Z is molecular (covalent).

Refer to the figure below.

4. [Skill: 1.B | Topic: 2.2] A student compares the potential energy curves for two diatomic molecules, Molecule 1 and Molecule 2. [Image Cue]: Graph, "Potential Energy vs. Internuclear Distance", x-axis: internuclear distance (pm), y-axis: potential energy (kJ/mol). Two smooth curves labeled Molecule 1 and Molecule 2. Molecule 1 has a minimum at 110 pm with potential energy -400 kJ/mol. Molecule 2 has a minimum at 160 pm with potential energy -250 kJ/mol. Both curves approach 0 kJ/mol as distance becomes very large. Based on the graph, which statement is correct?

(A)Molecule 1 has a longer equilibrium bond length and a larger bond energy than Molecule 2.(B)Molecule 1 has a shorter equilibrium bond length and a larger bond energy than Molecule 2.(C)Molecule 1 has a shorter equilibrium bond length and a smaller bond energy than Molecule 2.(D)Molecule 1 and Molecule 2 have the same bond energy because both curves approach 0 kJ/mol at large distance.

5. [Skill: 6.A | Topic: 2.2] A student is comparing the relative strength of ionic interactions in several ionic compounds. The student is told to use Coulomb’s law reasoning: ionic interaction strength increases as the product of ionic charges increases and as the distance between ion centers decreases. The table provides approximate ionic radii. - $Na^+$: 102 pm, $Cl^-$: 181 pm - $Mg^{2+}$: 72 pm, $O^{2-}$: 140 pm - $K^+$: 138 pm, $Br^-$: 196 pm - $Ca^{2+}$: 100 pm, $F^-$: 133 pm Assume the distance between ion centers is approximately the sum of the ionic radii. Which compound is expected to have the strongest electrostatic attraction between its ions?

(A)NaCl, because it has relatively small ions compared with KBr.(B)KBr, because it contains larger ions, which increases polarizability and therefore attraction.(C)CaF2, because it contains a $2+$ cation, which doubles the attraction compared with $1+$ cations.(D)MgO, because it has ions with larger charges and a smaller ion-center distance than the other choices.

Refer to the figure below.

6. [Skill: 1.A | Topic: 2.3] A student is asked to draw a particulate model (2D slice) of an ionic solid formed from equal numbers of $M^{2+}$ and $X^{2-}$ ions. Two proposed 2D repeating patterns are shown. Pattern 1 is a checkerboard arrangement in which each $M^{2+}$ is surrounded (up, down, left, right) by $X^{2-}$ ions. Pattern 2 is a striped arrangement in which ions alternate by rows: one full row of $M^{2+}$ ions next to a full row of $X^{2-}$ ions. Which pattern best represents how ions are arranged in a stable ionic crystal lattice, based on Coulomb’s law?

(A)Pattern 1, because it maximizes attractions between oppositely charged ions while minimizing repulsions between like-charged ions.(B)Pattern 1, because it maximizes repulsions between like-charged ions, which increases the stability of the solid.(C)Pattern 2, because keeping like-charged ions together reduces the number of ion–ion interactions and therefore lowers the total energy.(D)Pattern 2, because alternating by rows increases the distance between opposite charges and strengthens the attractive force.

7. [Skill: 1.B | Topic: 2.3] Two different particulate models are proposed for an ionic solid composed of $Na^+$ and $Cl^-$. In Model A, each $Na^+$ has $Cl^-$ ions as its nearest neighbors in all directions. In Model B, some nearest-neighbor positions of $Na^+$ are occupied by other $Na^+$ ions (and similarly, some nearest neighbors of $Cl^-$ are other $Cl^-$ ions). A student claims Model B could still represent a stable ionic crystal because overall the numbers of positive and negative ions are equal. Which of the following best explains why the student’s claim is incorrect?

(A)Equal numbers of ions do not matter; ionic solids form only when all ions have identical sizes.(B)Model B increases the number of like-charge nearest-neighbor interactions, increasing electrostatic repulsion and raising the potential energy of the crystal.(C)Model B decreases the number of opposite-charge interactions, which makes the crystal neutral and therefore more stable.(D)Model B places ions farther apart on average, which increases the magnitude of the electrostatic attraction between opposite charges.

8. [Skill: 1.B | Topic: 2.4] A student compares the properties of a metal wire with those of an ionic solid crystal. Observations: - A copper wire conducts electricity as a solid. - Solid sodium chloride does not conduct electricity, but molten sodium chloride does. - Copper can be hammered into thin sheets without shattering. Which model best represents the bonding in solid copper and explains these observations?

(A)Copper atoms share electrons in localized directional bonds, creating neutral molecules that can slide past one another.(B)Copper consists of positive metal ions arranged in a lattice surrounded by a sea of delocalized valence electrons that can move throughout the solid.(C)Copper consists of alternating positive and negative ions held together by electrostatic attractions; electrons are fixed in place in the solid.(D)Copper consists of atoms held together primarily by intermolecular forces between neutral atoms; mobile ions carry charge through the solid.

9. [Skill: 1.B | Topic: 2.4] An engineer is developing two alloys and provides the atomic radii shown. Atomic radii (pm): - Fe: 126 - C: 70 - Cu: 128 - Zn: 134 Alloy 1 is produced by adding a small amount of carbon to iron. Alloy 2 is produced by mixing zinc with copper. Which statement best describes the structures of Alloy 1 and Alloy 2 in terms of where the added atoms are located in the metal lattice?

(A)Alloy 1 is substitutional because carbon replaces iron atoms; Alloy 2 is interstitial because zinc fits into the spaces between copper atoms.(B)Alloy 1 is interstitial because carbon occupies spaces between iron atoms; Alloy 2 is substitutional because zinc atoms replace some copper atoms in the lattice.(C)Both alloys are interstitial because the added atoms are smaller than the host metal atoms.(D)Both alloys are substitutional because all metal lattices can only accommodate substitution of atoms, not interstitial atoms.

10. [Skill: 1.A | Topic: 2.5] A student is asked to draw a Lewis diagram for phosgene, $COCl_2$. The student correctly determines that the total number of valence electrons is 24 and proposes the following skeleton arrangement of atoms: Cl–C–O with the second Cl bonded to C. Which of the following Lewis diagrams is most consistent with the established principles for constructing Lewis diagrams (correct total valence electrons and octets where applicable)?

(A)Cl–C(=O)–Cl, with 2 lone pairs on O and 3 lone pairs on each Cl(B)Cl–C(–O)–Cl, with 3 lone pairs on O and 3 lone pairs on each Cl(C)Cl=C–O–Cl, with 2 lone pairs on O and 2 lone pairs on each Cl(D)Cl–C≡O–Cl, with 1 lone pair on O and 3 lone pairs on each Cl

11. [Skill: 1.B | Topic: 2.5] A student is drawing a Lewis diagram for formaldehyde, $CH_2O$. The student records the following information: - Total valence electrons: $2(1) + 4 + 6 = 12$ - Skeleton: H–C–H with O attached to C Which Lewis diagram correctly completes the structure using 12 valence electrons and satisfying the octet rule for C and O?

(A)H–C(=O)–H, with 2 lone pairs on O(B)H–C(–O)–H, with 3 lone pairs on O(C)H–C(–O)–H, with 1 lone pair on O(D)H=C=O, with 1 lone pair on O

12. [Skill: 4.A | Topic: 2.6] A researcher measures the three N–O bond lengths in the nitrate ion, $NO_3^-$. The data show that all three N–O bonds have the same length (within experimental uncertainty). Which of the following Lewis-diagram descriptions best accounts for the experimental bond-length data?

(A)Three equivalent resonance structures can be drawn, each with one N=O double bond and two N–O single bonds; the resonance hybrid has three identical N–O bonds (average bond order greater than 1 but less than 2).(B)A single Lewis structure with three N–O single bonds best accounts for the data because it gives all three N–O bonds the same bond order of 1.(C)Two equivalent resonance structures can be drawn; the resonance hybrid therefore contains two identical N–O bonds and one distinct N–O bond.(D)A single Lewis structure with three N=O double bonds best accounts for the data because it gives all three N–O bonds the same bond order of 2.

13. [Skill: 1.A | Topic: 2.6] A student attempts to draw a Lewis structure for nitric oxide, NO. The student counts 11 valence electrons and is concerned that the octet rule cannot be satisfied for both atoms. Which of the following statements best explains the limitation of the Lewis structure model for NO and the most appropriate way to represent its valence electrons?

(A)Add an extra electron to the structure so that both atoms can achieve octets; the correct Lewis structure for NO therefore has 12 valence electrons.(B)Because NO has an odd number of valence electrons, any Lewis diagram must include one unpaired electron; therefore, not all atoms can have complete octets in a single Lewis structure.(C)Nitrogen can expand its octet, so the best Lewis structure has a triple bond and enough lone pairs to give both atoms at least eight electrons.(D)Resonance between two Lewis structures can eliminate the odd electron, allowing both atoms to satisfy the octet rule in the resonance hybrid.

Refer to the figure below.

14. **1.** [Skill: 2.A | Topic: 2.7] Two different molecules have the same molecular formula, $C_2H_2Cl_2$. In molecule I, the two Cl atoms are on the same side of the C=C bond. In molecule II, the two Cl atoms are on opposite sides of the C=C bond. Which of the following best predicts a difference in a measurable molecular property of I and II?

(A)Molecules I and II have the same dipole moment because rotation around the C=C bond averages the bond dipoles over time.(B)Molecule II has a dipole moment, while molecule I does not, because placing identical atoms on the same side causes dipoles to cancel.(C)Molecule I has a dipole moment, while molecule II does not, because the $π$ bond prevents rotation and the bond dipoles cancel only in molecule II.(D)Neither molecule has a dipole moment because both contain a nonpolar C=C bond and the molecule is therefore nonpolar overall.

15. **2.** [Skill: 4.A | Topic: 2.7] A student collects bond-length data for two carbon–oxygen containing species. - In $CO_2(g)$, the measured C–O bond length is 116 pm. - In the carbonate ion, $CO_3^{2-}(aq)$, the measured average C–O bond length is 128 pm. Which statement best explains these observations using Lewis structures, bond order, and electron-domain geometry?

(A)In $CO_3^{2-}$ the carbon is $sp^3$ hybridized, so increased lone-pair repulsion on carbon lengthens the C–O bonds compared with $CO_2$.(B)In $CO_3^{2-}$ there is one C=O double bond and two C–O single bonds, so the average C–O bond order is greater than 2, making the average bond length longer than in $CO_2$.(C)In $CO_3^{2-}$ resonance leads to three equivalent C–O bonds with bond order $\frac{4}{3}$, so the average bond length is between a single and double bond; in $CO_2$ each C–O bond has bond order 2 and carbon is $sp$ hybridized, consistent with a shorter bond length.(D)In $CO_2$ the linear geometry causes less effective orbital overlap, lowering bond energy and increasing bond length compared with $CO_3^{2-}$.

AP Chemistry Unit 2: Compound Structure and Properties Practice Exam

Exam Details

A. Multiple-Choice Questions (15 Questions)

AP Chemistry Unit 2: Compound Structure and Properties Practice Exam

Exam Details

A. Multiple-Choice Questions (15 Questions)