Exam 2 Quy - Version 247 Exam 2 quy(50970 This print-out...

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Version 247 – 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 10.0points The octet rule indicates that 1. atoms gain or lose electrons to form bond- ing orbitals. 2. most atoms are stabilized by a filled outer s and p shell. correct 3. atoms are most stable when involved in formation of eight bonds. 4. atoms prefer the octahedral electronic ge- ometry. 5. all of the other responses are false. Explanation: The octet rule states that atoms in com- pounds are generally most stable when they achieve noble gas configurations. Noble gases have filled outer s and p orbitals. Since an s orbital can hold two electrons and a p orbital is filled with 6 electrons, this is a total of 8 valence electrons. 002 10.0points 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. 150 pm 2. 120 pm correct 3. 60 pm 4. 110 pm 5. 75 pm 6. 135 pm 7. 55 pm Explanation: The bond is a double bond so you add 60 pm to 60 pm and get 120 pm. 003 10.0points What is the electronic geometry around ni- trogen in the molecule CH 3 CH 2 NH 2 ? 1. trigonal planar 2. trigonal pyramidal 3. tetrahedral correct 4. linear 5. square planar 6. bent Explanation: 004 10.0points Identify the compound with the most polar bond. 1. SbH 3 2. PH 3 3. NH 3 correct 4. AsH 3 Explanation: Calculate the difference in the electronega- tivities (ΔEN): ΔEN As H 2 . 2 - 2 . 2 = 0 . 0 N H 3 . 0 - 2 . 2 = 0 . 8 P H 2 . 2 - 2 . 2 = 0 . 0 Sb H 2 . 2 - 2 . 1 = 0 . 1
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Version 247 – Exam 2 – quy – (50970) 2 The N H bond is the most polar of these. 005 10.0points The electronic geometry of NH 3 (ammonia) is 1. tetrahedral. correct 2. trigonal pyramidal. 3. trigonal planar. 4. linear. 5. bent. Explanation: N H H H Note: the H-N-H bond angle is just under 109.5 due to the lone pair. 006 10.0points The electronic arrangment is the same as the molecular shape when 1. there are more shared electrons than non- shared electrons. 2. the atoms are joined by sigma bonds. 3. the number of bonding orbitals equals the number of anti-bonding orbitals. 4. there are no lone pairs of electrons on the central atom. correct 5. the molecule is not polar. Explanation: The only time that the molecular shape and electronic arrangement are the same is when there are no lone pairs of electrons on the central atom. 007 10.0points What is the molcular shape of COCl 2 ?. 1. octahedral 2. trigonal planar correct 3. tetrahedral 4. square planar 5. seesaw 6. trigonal pyramidal Explanation: 008 10.0points Use VSEPR theory to predict the molecular geometry of the molecule SiBr 4 . 1. trigonal-bipyramidal 2. tetrahedral correct 3. octahedral 4. trigonal-pyramidal 5. bent or angular 6. trigonal-planar 7. None of these 8. linear Explanation: The Lewis structure for SiBr 4 is Si Br Br Br Br The central atom Si has 4 bonding pairs and no lone pairs of electrons. The four areas of high electron density make the electronic geometry tetrahedral and the molecular ge- ometry tetrahedral.
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