Exam 2 - Version 319 Exam 2 McCord (53110) This print-out...

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Version 319 – Exam 2 – McCord – (53110) 1 This print-out should have 36 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. McCord CH301mwf This exam is only For McCord’s MW± 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 aver- age about 20-30 students per exam that mis- bubble this inFormation. Get it right! There are NO spaces in UTEIDs. 001 10.0 points How many σ - and π -bonds, respectively, are there in acrolein (CH 2 CHCHO)? 1. 5 and 4 2. 7 and 1 3. 5 and 2 4. 7 and 2 correct 5. 4 and 2 Explanation: C H H C H C H O 002 10.0 points Use VSEPR theory to predict the molecular geometry oF the molecule HCN. 1. octahedral 2. trigonal-planar 3. None oF these 4. linear correct 5. trigonal-pyramidal 6. bent or angular 7. tetrahedral 8. trigonal-bipyramidal Explanation: molecular geometry oF HCN = ? The Lewis structure For HCN is H C N or C N H Although there is an octet oF electrons around C, six electrons are combined to Form a triple bond. The two areas oF high elec- tron density around the central atom make the molecular geometry linear. 003 10.0 points The central atom in a compound with the Formula Sb± 2 H 3 is sp 3 d hybridized. 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 = 12 2. N = 32; A = 22; S = 10 correct 3. N = 30; A = 20; S = 12 4. N = 48; A = 42; S = 10 5. N = 32; A = 20; S = 10 6. N = 32; A = 20; S = 12
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Version 319 – Exam 2 – McCord – (53110) 2 7. N = 30; A = 22; S =8 8. N = 32; A = 24; S 9. N = 30; A = 22; S = 10 Explanation: The number of electrons needed in this case must be adjusted for an expanded octet for the central atom antimony, which will have 5 pairs of electrons (10 total) for its valence shell. Each of the Fuorines will have a true octet. Thus N = 10 + 2 (8) + 3 (2) = 32 A = 5 + 2 (7) + 3 (1) = 22 S = N - A = 10 . 10 electrons are shared in the 3 single bonds to the hydrogens and 2 single bonds to the Fuorines. 004 10.0 points Which of the following statements concerning molecular orbital theory is/are true? I) Bonding orbitals are equal in energy to their corresponding anti-bonding orbitals II) Adding electrons to anti-bonding orbitals destabilizes molecules. III) Unlike when we ±ll atomic orbitals, we don’t use Hund’s Rule to ±ll molecular orbitals. 1. II only correct 2. I only 3. I, II, III 4. III only 5. I, III 6. II, III 7. I, II Explanation: Statement I is false because bonding or- bitals are lower in energy than their corre- sponding anti-bonding orbitals. Statement III is false, because degenerate molecular or- bitals are half-±lled before being completely ±lled, according to Hund’s rule, just like atomic orbitals.
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This note was uploaded on 04/30/2011 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.

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Exam 2 - Version 319 Exam 2 McCord (53110) This print-out...

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