PracticeExamI Solutions Test 2

PracticeExamI Solutions Test 2 - Version 392 Exam 2 McCord...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

Page1 / 10

PracticeExamI Solutions Test 2 - Version 392 Exam 2 McCord...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online