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Unformatted text preview: Version 392 Exam 2 McCord (53130) 1 This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. McCord CH301tt This exam is only for McCords Tues/Thur CH301 class. If Quest works properly, you should be able to see your score for this exam by 11:30 PM tonight. Do realize however that Quest has been taking up to 8 hours to get through the grading, so your score might appear bit by bit. I will post an announcement on our web page and/or Quest when I feel like all bubblesheets have been graded. PLEASE carefully bubble in your UTEID and Version Number! We av- erage about 20-30 students per exam that misbubble this information. Get it right! 001 10.0 points Consider the species H 2 CCH + . C H H C H a b + What hybridization is at a and b, respec- tively? 1. sp ; sp 2. sp 3 ; sp 3. sp 2 ; sp 3 4. sp 2 ; sp 2 5. sp 3 ; sp 3 6. sp 2 ; sp correct Explanation: C H H C H sp 2 sp + 002 10.0 points A double bond consists of 1. two pi bonds. 2. one sigma and one pi bond. correct 3. two sigma bonds. 4. one sigma bond containing four elec- trons. 5. two beta bonds. 6. an alpha and a beta bond. Explanation: The sigma bonds overlap along a line drawn between the two nuclei, and the pi bonds overlap perpendicular to this line. 003 10.0 points Identify the antibonding orbital corre- sponding to the -bond in H 2 . 1. 2. None of these 3. 4. correct 5. Explanation: The term sigma implies the entity has symmetry of rotation of any angle around a central axis; like viewing a cylinder down the long axis from one of the circular ends. A sigma-antibonding molecular orbital is Version 392 Exam 2 McCord (53130) 2 formed when two atomic orbitals overlap in such a way that they are out of phase. In H 2 the atomic orbitals which combine to form the antibonding sigma molecular orbital are both 1 s : H 1 s + H 1 s H 2 004 10.0 points The compound ClF 5 (where Cl is the central atom) has an electronic geometry of octahe- dral. What would be the quantities for the Lewis dot formula? N (needed electrons), A (available electrons) and S (shared electrons) 1. N = 48; A = 42; S = 10 2. N = 50; A = 42; S = 10 3. N = 48; A = 42; S = 12 4. N = 50; A = 42; S = 8 5. N = 50; A = 42; S = 12 6. N = 52; A = 42; S = 12 7. N = 52; A = 42; S = 10 correct 8. N = 48; A = 42; S = 8 9. N = 52; A = 42; S = 8 Explanation: The number of electrons needed in this case must be adjusted for an expanded octet for the central atom, chlorine, which will have 6 pairs of electrons (12 total) for its valence shell. Each of the fluorines will have a true octet. Thus N = 12 + 5 (8) = 52 A = 7 + 5 (7) = 42 S = N- A = 10 ....
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