CH101 Exam Study Notes

PRACTIC EXAM 3 - CH101 Study Notes

This document provides an exhaustive collection of questions and relevant concepts from Practice Exam 3 for CH101, encompassing a range of topics including chemical bonding, electron configuration, molecular geometry, and polarity.

Exam Structure

The exam consists of 25 questions, each worth 4 points.
Total points for the exam: 100.
Answers are to be written on a Scantron sheet, with all work shown on provided paper to be turned in.

Questions and Concepts

Bonding

Identify the Weakest Bond.
- Options:
  - A) single covalent bond
  - B) double covalent bond
  - C) triple covalent bond
  - D) All of the above bonds are the same strength.
Electron Groups in sp3d Hybridization.
- Number of electron groups:
  - A) 2
  - B) 3
  - C) 4
  - D) 5
  - E) 6
  - Correct Answer: D) 5
Sigma and Pi Bonds in a Double Bond.
- Options:
  - A) 1 sigma, 1 pi
  - B) 2 sigma, 1 pi
  - C) 2 sigma, 2 pi
  - D) 1 sigma, 2 pi
  - Correct Answer: A) 1 sigma, 1 pi
  - Explanation: A double bond consists of one sigma bond and one pi bond.
Lewis Structure for CH2CHCH3.
- Options:
  - A) 8 sigma, 1 pi
  - B) 9 sigma, 0 pi
  - C) 9 sigma, 1 pi
  - D) 7 sigma, 2 pi
  - E) 8 sigma, 2 pi
  - Correct Answer: C) 9 sigma, 1 pi
  - Analysis: Drawn structure must confirm the number of sigma and pi bonds present.
Electron Groups in Trigonal Planar Shape.
- Options:
  - A) 1
  - B) 2
  - C) 3
  - D) 4
  - E) 5
  - Correct Answer: C) 3
Electron Geometry (eg), Molecular Geometry (mg), and Hybridization for SF4.
- Options:
  - A) eg = tetrahedral, mg = tetrahedral, sp3
  - B) eg = trigonal pyramidal, mg = trigonal pyramidal, sp3
  - C) eg = octahedral, mg = square planar, sp3d2
  - D) eg = octahedral, mg = octahedral, sp3d2
  - E) eg = trigonal bipyramidal, mg = seesaw, sp3d
  - Correct Answer: E) eg = trigonal bipyramidal, mg = seesaw, sp3d

Bond Angles and Structures

Bond Angle for Octahedral Shape.
- Options:
  - A) 109.5°
  - B) 180°
  - C) 120°
  - D) 105°
  - E) 90°
  - Correct Answer: E) 90°
Best Lewis Structure for PO43⁻.
- Options: (Visual structures not provided in the transcript)
Valence Electrons for ClO4⁻.
- Options:
  - A) 30
  - B) 32
  - C) 39
  - D) 40
  - E) 41
  - Correct Answer: B) 32

Resonance and Formal Charges

Best Lewis Structure for BeF2.
- Options: (Visual structures not provided in the transcript)
Resonance Structures for OCN⁻.
- Contributions:
  - A) O(2 lone pairs) C N (2 lone pairs)
  - B) O(1 lone pair) C-N(3 lone pairs)
  - C) O(1 lone pair) C(2 lone pairs) N(1 lone pair)
  - D) O(3 lone pairs)—C N(with 1 lone pair)
  - E) They all contribute equally to the correct structure of OCN⁻.
  - Correct Answer: C) O(1 lone pair) C(2 lone pairs) N(1 lone pair)
Formal Charge on Central N in NO2⁺ Lewis Structure.
- Options:
  - A) -1
  - B) 0
  - C) +1
  - D) +2
  - E) -2
  - Correct Answer: C) +1
Electron Geometry and Molecular Geometry for ICl2⁻.
- Options:
  - A) eg = tetrahedral, mg = bent
  - B) eg = tetrahedral, mg = trigonal pyramidal
  - C) eg = trigonal bipyramidal, mg = linear
  - D) eg = trigonal bipyramidal, mg = trigonal planar
  - E) eg = octahedral, mg = linear
  - Correct Answer: C) eg = trigonal bipyramidal, mg = linear

Hybridization and Molecular Geometry

Determine Hybridization at Labeled Atoms.
- Options:
  - A) 1 = sp2, 2 = sp3, 3 = sp2
  - B) 1 = sp2, 2 = sp3, 3 = sp3
  - C) 1 = sp3, 2 = sp3, 3 = sp3
  - D) 1 = sp3, 2 = sp3, 3 = sp2
  - E) 1 = sp, 2 = sp, 3 = sp2
Order of Increasing F-A-F Bond Angle for PF3, OF2, PF4⁺.
- Options:
  - A) PF3 < OF2 < PF4⁺
  - B) OF2 < PF3 < PF4⁺
  - C) OF2 < PF4⁺ < PF3
  - D) PF4⁺ < OF2 < PF3
  - E) PF4⁺ < PF3 < OF2
Determine Polar Molecules from List: PCl5, COS, XeO3, SeBr2
- Options:
  - A) 2
  - B) 0
  - C) 1
  - D) 3
  - E) 4

Molecular Orbitals

Statements on Molecular Orbitals.
- Options:
  - A) The total number of molecular orbitals formed doesn't always equal the number of atomic orbitals in the set.
  - B) A bond order of 0 represents a stable chemical bond.
  - C) When two atomic orbitals come together to form two molecular orbitals, one molecular orbital will be lower in energy than the two separate atomic orbitals, and one molecular orbital will be higher in energy than the separate atomic orbitals.
  - D) Electrons placed in antibonding orbitals stabilize the ion/molecule.
  - E) All of the above are true.
  - Correct Answer: C)
Molecular Orbital Diagram for Paramagnetism.
- Options:
  - A) O22⁻
  - B) Ne22⁺
  - C) O22⁺
  - D) F22⁺
  - E) None of the above are paramagnetic.
Lone Pairs in H3P and Molecular Geometry.
- Options:
  - a. 0, linear
  - b. 1, trigonal (or triangular) planar
  - c. 1, trigonal (or triangular) pyramid
  - d. 2, bent (or angular)
  - e. 2, tetrahedral
  - Correct Answer: c) 1, trigonal (or triangular) pyramid

VSEPR Theory and Molecular Polarity

Conditions for Different Electron-Pair Geometry vs. Molecular Geometry in VSEPR Theory.
- Options:
  - a. in molecules with at least one lone pair of electrons on the central atom
  - b. in molecules with no lone pairs of electrons on the central atom
  - c. in molecules with steric bulk
  - d. in molecules that exceed an octet around the central atom
  - e. in molecules that have less than an octet around the central atom
Polar or Nonpolar Character of Molecules.
- List:
  - CH2Cl2
  - SCl2
  - SiCl4
- Options:
  - a. nonpolar, nonpolar, nonpolar
  - b. nonpolar, nonpolar, polar
  - c. polar, nonpolar, polar
  - d. polar, polar, polar
  - e. polar, polar, nonpolar
Statements about Central Atoms with Six Shared Electrons.
- Options:
  - a. The central must have an expanded octet.
  - b. The central atom must be a nonmetal.
  - c. The central atom must be from the third or subsequent periods.
  - d. The central atom must have an octahedral molecular geometry.
  - e. The central atom is small and has low electronegativity.
Identifying Ionic Compounds.
- Options:
  - a. CaSO4
  - b. H2S
  - c. CH4
  - d. NH3
  - e. He
Correct Lewis Dot Structure for CH2O.
- Options: (Visual structures not provided in the transcript)
Formal Charges in [SO4]2-.
- Options:
  - a) +2 on sulfur, -2 on oxygen
  - b) +3 on sulfur, -1 on oxygen
  - c) +1 on sulfur, -1 on oxygen
  - d) -1 on sulfur, +2 on oxygen
  - e) +2 on sulfur, -1 on oxygen

Conclusion

This study guide encompasses key topics that are essential for the CH101 exam preparation, covering bonds, molecular structure, hybridization, polarity, and VSEPR theory. Students should familiarize themselves with these concepts and practice drawing Lewis structures and molecular orbital diagrams to enhance understanding.