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Unformatted text preview: Version 261 – Exam 2 – Laude – (53755) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 6.0 points Which of the following statements regarding intermolecular forces (IMF) is/are true? I) Intermolecular forces result from attrac- tive forces between regions of positive and negative charge density in neighbor- ing molecules. II) The stronger the bonds within a molecule are, the stronger the intermolecular forces will be. III) Only non-polar molecules have London forces. 1. II only 2. I and III 3. III only 4. II and III 5. I only correct 6. I, II, and III 7. I and II Explanation: Statement I is true - all IMF result from Coulombic attraction. Statements II and III are both false; the strength of the bonds within a molecule have no bearing on the strength of the bonds between molecules; all molecules have London forces. 002 6.0 points What atomic orbitals comprise the molecular orbitals in SiBr 4 ? 1. 3 p, 4 p 2. 3 s, sp 3 3. sp 3 , sp 3 correct 4. 3 p, sp 3 5. 3 s, 4 p Explanation: The silicon atom is sp 3 hybridized and the bromine atoms are as well. 003 6.0 points Assign the strongest relevant intermolecular force to the following species, respectively: RbCl 2 , C 6 H 6 (benzene) , HI , Fe 2 O 3 , CH 2 NH . I) ionic forces II) hydrogen bonding III) dipole-dipole IV) van der Waals’ forces 1. III, II, IV, III, III 2. I, IV, III, I, II correct 3. II, IV, III, IV, IV 4. III, IV, I, I, II 5. I, II, III, II, I 6. I, II, IV, I, III 7. I, III, III, IV, II Explanation: Rubidium Chloride and Iron(III) oxide are both ion-ion. Benzene is non-polar and thus has only van der Waal’s forces. Hydroiodic acid is polar and has dipole-dipole interac- tions. Methylimine has H-bonding. 004 6.0 points Which of the following statements concerning valence bond theory is/are true? I) Hybridizing one 2 s orbital with three 2 p orbitals would produce four sp 3 orbitals. II) When a 2 s orbital is hybridized with a single 2 p orbital, there will be a single unhybridized 2 p orbital left to form a π bond. III) Among the types of hybrid orbitals we have learned about, the minimum amount of s-character for a hybrid or- Version 261 – Exam 2 – Laude – (53755) 2 bital is 1/6 th . 1. II only 2. I only 3. II, III 4. III only 5. I, II 6. I, III correct 7. I, II, III Explanation: Statement II is false because there will be a pair of unhybridized p orbitals available for π bonding. Statements I is true because hy- bridizing any number of atomic orbitals al- ways results in an equal number of hybrid orbitals. Statement III is true because sp 3 d 2 hybrid orbitals have 1/6 th s-character....
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This test prep was uploaded on 06/06/2010 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.
- Fall '07