Exam 2-solutions-quy - Version 153 Exam 2 quy (50970) 1...

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Unformatted text preview: Version 153 Exam 2 quy (50970) 1 This print-out should have 35 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Which of the following compounds is non- polar? 1. O 3 2. PH 3 3. PCl 3 4. NH 3 5. BH 3 correct Explanation: NH 3 , PH 3 and PCl 3 are tetrahedral with one lone pair and are therefore not symmet- ric; their polar bonds do not cancel, and the molecule is polar. O 3 is bent, with a lone pair on the central oxygen and is thus polar. BH 3 is trigonal planar. The polar B H bonds cancel and the molecule is not polar. 002 10.0 points In the Lewis dot structure of the molecule C 3 H 7 Cl , how many unbonded electron pairs are found in the molecule? 1. 1 2. 2 3. 4 4. 6 5. 3 correct Explanation: Draw the Lewis dot structure. Place the hydrogens and chlorine around the carbon atoms, C 3 H 7 Cl has single bonds and 3 lone pairs on Cl. 003 10.0 points Consider the following covalent bond radii: Single Double Triple C 77 pm 67 pm 60 pm N 75 pm 60 pm 55 pm O 74 pm 60 pm- S 102 pm-- What is the approximate length of the NN bond in nitrogen hydride (HNNH) using the table values? 1. 60 pm 2. 75 pm 3. 120 pm correct 4. 150 pm 5. 55 pm 6. 110 pm 7. 135 pm Explanation: The bond is a double bond so you add 60 pm to 60 pm and get 120 pm. 004 10.0 points Identify the compound with the most polar bond. 1. NH 3 correct 2. SbH 3 3. AsH 3 4. PH 3 Explanation: Calculate the difference in the electronega- tivities (EN): EN As H 2 . 2- 2 . 2 = 0 . N H 3 .- 2 . 2 = 0 . 8 P H 2 . 2- 2 . 2 = 0 . Sb H 2 . 2- 2 . 1 = 0 . 1 Version 153 Exam 2 quy (50970) 2 The N H bond is the most polar of these. 005 10.0 points A molecule has one lone pair of electrons on the central atom and three atoms bonded to the central atom. The central atom follows the octet rule. What is its electronic arrange- ment and its hybridization? 1. pyramidal; sp 2 2. tetrahedral; sp 3 correct 3. trigonal planar; sp 2 4. trigonal planar; sp 3 5. tetrahedral; sp 2 6. pyramidal; sp 3 7. angular; sp 3 Explanation: One lone pair plus 3 bonded pairs equals 4 electronic regions. This means that the central atom is tetrahedral and hybridization is sp 3 . 006 10.0 points Use VSEPR theory to predict the molec- ular geometry of the molecule whose Lewis structure is given as S b b b b F b b b b b b F b b b b b b 1. octahedral 2. tetrahedral 3. None of these 4. linear 5. bent or angular correct 6. trigonal-bipyramidal 7. trigonal-pyramidal 8. trigonal-planar Explanation: The central atom has 2 bonding pairs and 2 lone pairs of electrons. The four areas of high electron density make the electronic geometry tetrahedral and the molecular geometry bent or angular....
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Exam 2-solutions-quy - Version 153 Exam 2 quy (50970) 1...

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