Ap Chemistry Unit 2 Practice Test

The journey through AP Chemistry Unit 2, focusing on chemical and molecular structure, demands a solid grasp of fundamental concepts. Successfully navigating this unit requires consistent practice and a strategic approach to problem-solving. This article provides a comprehensive practice test, complete with detailed explanations, to solidify your understanding and boost your confidence for the AP exam.

AP Chemistry Unit 2: Practice Test

This practice test covers essential topics from Unit 2, including:

• Lewis Structures: Drawing and understanding the limitations of Lewis structures.
• VSEPR Theory: Predicting molecular geometry and bond angles.
• Hybridization: Determining the hybridization of atoms in molecules.
• Bonding Theories: Understanding sigma and pi bonds, and molecular orbital theory.
• Polarity: Determining molecular polarity based on bond polarity and molecular geometry.
• Isomerism: Identifying structural and geometric isomers.
• Resonance: Drawing and understanding resonance structures.
• Formal Charge: Calculating formal charge to determine the most stable resonance structure.

Instructions:

• Answer all questions to the best of your ability.
• Show your work for calculation-based questions.
• Refer to the periodic table and relevant constants as needed.
• Compare your answers with the provided solutions and explanations.

Let's begin!

Multiple Choice: (1 point each)

1. Which of the following molecules is nonpolar? (A) H₂O (B) NH₃ (C) CO₂ (D) SO₂ (E) CH₃Cl

2. The central atom in XeF₄ has how many lone pairs of electrons? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4

3. What is the hybridization of the central atom in BF₃? (A) sp (B) sp² (C) sp³ (D) sp³d (E) sp³d²

4. Which of the following molecules has a tetrahedral electron geometry but a bent molecular geometry? (A) CH₄ (B) NH₃ (C) H₂O (D) BF₃ (E) BeCl₂

5. Which of the following statements is true regarding sigma (σ) and pi (π) bonds? (A) Sigma bonds are weaker than pi bonds. (B) Sigma bonds are formed by end-to-end overlap of atomic orbitals. (C) Pi bonds are formed by end-to-end overlap of atomic orbitals. (D) Sigma bonds are only found in single bonds. (E) Pi bonds allow for free rotation around the bond axis.

6. Which of the following molecules exhibits resonance? (A) CH₄ (B) H₂O (C) O₃ (D) NH₃ (E) CO₂

7. What is the bond order of N₂? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5

8. Which of the following molecules is polar? (A) CCl₄ (B) BF₃ (C) CS₂ (D) NH₃ (E) XeF₄

9. Which of the following has the largest bond angle? (A) CH₄ (B) NH₃ (C) H₂O (D) BF₃ (E) CO₂

10. What is the formal charge on the central nitrogen atom in the most stable resonance structure of N₂O (N-N-O arrangement)? (A) -2 (B) -1 (C) 0 (D) +1 (E) +2

Free Response: (Points as indicated)

1. (10 points) Draw Lewis structures for the following molecules/ions. Include all resonance structures if applicable and indicate formal charges on all atoms in each structure. (a) SO₂ (b) NO₃⁻ (c) OCN⁻

2. (10 points) For the following molecules: (a) H₂S (b) PCl₅ (c) XeF₂ Determine the following for each molecule: * Lewis structure * Electron geometry * Molecular geometry * Bond angles (approximate) * Hybridization of the central atom * Polarity (polar or nonpolar)

3. (5 points) Using molecular orbital theory, determine the bond order of O₂⁺. Is it paramagnetic or diamagnetic? Explain.

4. (5 points) Explain the difference between structural isomers and geometric isomers. Give an example of each.

5. (5 points) Explain why the bond angle in H₂O is smaller than the bond angle in CH₄, even though both have a tetrahedral electron geometry.

Answer Key and Explanations

Multiple Choice:

1. (C) CO₂

   • Explanation: CO₂ has a linear molecular geometry with two polar C=O bonds. Due to the symmetrical arrangement, the bond dipoles cancel each other out, resulting in a nonpolar molecule. H₂O is bent and polar. NH₃ is pyramidal and polar. SO₂ is bent and polar. CH₃Cl is tetrahedral but the C-Cl bond makes it polar overall.

2. (C) 2

   • Explanation: XeF₄ has a square planar molecular geometry. Xenon has 8 valence electrons. Four are used to form bonds with fluorine, leaving 4 electrons or 2 lone pairs.

3. (B) sp²

   • Explanation: Boron has 3 valence electrons, all of which are involved in sigma bonding with the three fluorine atoms. There are no lone pairs on boron. Therefore, the steric number is 3, corresponding to sp² hybridization and a trigonal planar geometry.

4. (C) H₂O

   • Explanation: Oxygen has 6 valence electrons. In H₂O, two are used to bond with hydrogen, leaving two lone pairs. The electron geometry is tetrahedral (4 electron domains), but the molecular geometry is bent due to the repulsion of the lone pairs. CH₄ is tetrahedral. NH₃ is tetrahedral electron geometry and trigonal pyramidal molecular geometry. BF₃ is trigonal planar. BeCl₂ is linear.

5. (B) Sigma bonds are formed by end-to-end overlap of atomic orbitals.

   • Explanation: Sigma (σ) bonds are formed by the direct, head-on overlap of atomic orbitals. Pi (π) bonds are formed by the side-by-side overlap of p orbitals. Sigma bonds are generally stronger than pi bonds. Sigma bonds are present in all single, double, and triple bonds. Pi bonds restrict rotation around the bond axis.

6. (C) O₃

   • Explanation: Ozone (O₃) has two resonance structures where the double bond can be located between either of the two oxygen atoms. CH₄, H₂O, and NH₃ do not have multiple ways to draw their Lewis structures with equivalent bonding. While CO₂ has resonance structures, they are not equally contributing (one with a formal charge of -1 and +1).

7. (C) 3

   • Explanation: N₂ has a triple bond (N≡N). Molecular orbital configuration is (σ2s)² (σ2s*)² (σ2p)² (π2p)⁴. Bond order = (bonding electrons - antibonding electrons)/2 = (8-2)/2 = 3

8. (D) NH₃

   • Explanation: NH₃ has a trigonal pyramidal shape. The N-H bonds are polar and, due to the shape, the bond dipoles do not cancel out, resulting in a net dipole moment. CCl₄ is tetrahedral and symmetrical, so the bond dipoles cancel. BF₃ is trigonal planar and symmetrical, so the bond dipoles cancel. CS₂ is linear and symmetrical, so the bond dipoles cancel. XeF₄ is square planar and symmetrical, so the bond dipoles cancel.

9. (E) CO₂

   • Explanation: CO₂ is linear, with a bond angle of 180°. BF₃ is trigonal planar, with bond angles of 120°. CH₄ is tetrahedral, with bond angles of 109.5°. NH₃ is trigonal pyramidal, with bond angles slightly less than 109.5° due to lone pair repulsion. H₂O is bent, with bond angles even smaller than NH₃ due to the stronger repulsion of two lone pairs.

10. (D) +1

    • Explanation: The most stable resonance structure of N₂O is N=N=O. Nitrogen (left) has 5 valence electrons. It is surrounded by 4 electrons (2 bonds). Formal Charge = 5-4 = +1.

Free Response:

1. (a) SO₂ (10 points)

   • Lewis Structure: O=S-O ↔ O-S=O (Sulfur has one lone pair)
   • Formal Charges: In each structure, one O has a formal charge of -1, the S has a formal charge of +1, and the other O has a formal charge of 0.

   (b) NO₃⁻ (10 points)

   • Lewis Structure:
     • O=N-O⁻ ↔ O⁻-N=O ↔ O-N=O

     •      |      |      |
       

     •      O⁻     O     O⁻
       

   • Formal Charges: In each structure, one O has a formal charge of -1, the N has a formal charge of +1, and the other two O have a formal charge of 0.

   (c) OCN⁻ (10 points)

   • Lewis Structure:
     • ⁻O-C≡N: ↔ O=C=N⁻ ↔ :O≡C-N²⁻
   • Formal Charges:
     • ⁻O-C≡N: (-1, 0, 0)
     • O=C=N⁻ (0, 0, -1)
     • :O≡C-N²⁻ (+1, 0, -2)
   • The most stable structure is O=C=N⁻ because it has the lowest magnitude of formal charges on each atom and the negative formal charge is on the more electronegative Nitrogen atom.

2. (a) H₂S (10 points)

   • Lewis Structure: H-S-H (Sulfur has two lone pairs)
   • Electron Geometry: Tetrahedral
   • Molecular Geometry: Bent
   • Bond Angles: Approximately 109.5° (slightly smaller due to lone pair repulsion)
   • Hybridization of the Central Atom: sp³
   • Polarity: Polar

   (b) PCl₅ (10 points)

   • Lewis Structure: Phosphorus bonded to five chlorine atoms (no lone pairs on P)
   • Electron Geometry: Trigonal Bipyramidal
   • Molecular Geometry: Trigonal Bipyramidal
   • Bond Angles: 90°, 120°, 180°
   • Hybridization of the Central Atom: sp³d
   • Polarity: Nonpolar

   (c) XeF₂ (10 points)

   • Lewis Structure: F-Xe-F (Xenon has three lone pairs)
   • Electron Geometry: Trigonal Bipyramidal
   • Molecular Geometry: Linear
   • Bond Angles: 180°
   • Hybridization of the Central Atom: sp³d
   • Polarity: Nonpolar

3. (5 points)

   • O₂ has the molecular orbital configuration (σ2s)² (σ2s*)² (σ2p)² (π2p)⁴ (π2p*)². O₂⁺ has one less electron, so the configuration is (σ2s)² (σ2s*)² (σ2p)² (π2p)⁴ (π2p*)¹.
   • Bond order = (bonding electrons - antibonding electrons) / 2 = (8 - 3) / 2 = 2.5
   • O₂⁺ is paramagnetic because it has unpaired electrons in the π2p* antibonding molecular orbitals.

4. (5 points)

   • Structural isomers have the same molecular formula but different connectivity of atoms. For example, butane (CH₃CH₂CH₂CH₃) and isobutane (CH₃CH(CH₃)CH₃).
   • Geometric isomers (also called cis-trans isomers) have the same connectivity of atoms but different spatial arrangements of atoms around a rigid bond (e.g., a double bond or a ring). For example, cis-2-butene and trans-2-butene. In cis-2-butene, the two methyl groups are on the same side of the double bond, while in trans-2-butene, they are on opposite sides.

5. (5 points)

   • Both H₂O and CH₄ have a tetrahedral electron geometry. However, H₂O has two lone pairs on the central oxygen atom, while CH₄ has no lone pairs on the central carbon atom. Lone pairs exert a greater repulsive force than bonding pairs. The two lone pairs in H₂O push the bonding pairs closer together, resulting in a smaller bond angle (approximately 104.5°) compared to the bond angle in CH₄ (109.5°). The repulsion between the bonding pairs in CH₄ is more uniform, leading to the ideal tetrahedral angle.

Strategies for Success

Mastering AP Chemistry Unit 2 requires more than just memorizing facts. Here are some effective strategies:

• Practice, Practice, Practice: The more you practice drawing Lewis structures, predicting molecular geometries, and applying bonding theories, the better you'll become. Work through numerous practice problems from textbooks, online resources, and past AP exams.
• Understand the Underlying Concepts: Don't just memorize rules. Understand why molecules have certain shapes or why some bonds are polar while others aren't. This deeper understanding will help you apply the concepts to new and unfamiliar situations.
• Visualize Molecules in 3D: Use molecular modeling kits or online simulations to visualize molecules in three dimensions. This will help you better understand the relationships between electron geometry, molecular geometry, and bond angles.
• Master VSEPR Theory: VSEPR theory is the foundation for predicting molecular shapes. Make sure you understand the different electron geometries and how lone pairs affect molecular geometry.
• Draw Lewis Structures Accurately: Pay close attention to valence electrons, formal charges, and resonance structures. A correctly drawn Lewis structure is essential for predicting molecular properties.
• Review Molecular Orbital Theory: While often challenging, understanding the basics of molecular orbital theory can help you predict bond order and magnetic properties.
• Focus on Polarity: Understand how bond polarity and molecular geometry contribute to overall molecular polarity. Be able to predict whether a molecule is polar or nonpolar based on its structure.
• Create Flashcards: Use flashcards to memorize key definitions, rules, and exceptions.
• Work in Groups: Collaborate with classmates to solve problems and discuss challenging concepts. Explaining concepts to others can solidify your own understanding.
• Seek Help When Needed: Don't hesitate to ask your teacher or tutor for help if you're struggling with a particular topic.
• Review Regularly: Consistent review is key to retaining information. Set aside time each week to review Unit 2 concepts.
• Take Practice Tests Under Timed Conditions: Simulate the actual AP exam by taking practice tests under timed conditions. This will help you improve your pacing and test-taking skills.
• Analyze Your Mistakes: Carefully analyze your mistakes on practice tests to identify areas where you need to improve. Focus your studying on those areas.
• Stay Organized: Keep your notes, practice problems, and other study materials organized so you can easily find what you need.
• Stay Positive: Believe in yourself and your ability to succeed. A positive attitude can make a big difference in your performance.

Common Mistakes to Avoid

• Incorrectly Drawing Lewis Structures: Not counting valence electrons correctly, violating the octet rule (especially for elements in the third period and beyond), and not showing formal charges when necessary.
• Confusing Electron Geometry and Molecular Geometry: Failing to account for lone pairs when determining molecular geometry.
• Misunderstanding VSEPR Theory: Not correctly applying the rules of VSEPR theory to predict molecular shapes.
• Ignoring Lone Pair Repulsion: Forgetting that lone pairs exert a greater repulsive force than bonding pairs, which can affect bond angles.
• Incorrectly Determining Hybridization: Not correctly counting the number of sigma bonds and lone pairs around the central atom.
• Misunderstanding Polarity: Not recognizing the relationship between bond polarity, molecular geometry, and overall molecular polarity.
• Not Considering Resonance: Failing to draw all possible resonance structures and not understanding their relative contributions.
• Incorrectly Calculating Formal Charge: Making errors in calculating formal charges, which can lead to incorrect Lewis structures.
• Failing to Apply Molecular Orbital Theory Correctly: Making errors in constructing molecular orbital diagrams and calculating bond order.
• Poor Time Management: Spending too much time on difficult questions and not enough time on easier ones.

By diligently practicing and understanding the concepts presented in this article, you will be well-prepared to tackle AP Chemistry Unit 2 with confidence. Remember to review your work, seek help when needed, and stay focused on your goals. Good luck!