Exam 2-solutions

Exam 2-solutions - Version 331 – Exam 2 – mccord –...

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Unformatted text preview: Version 331 – Exam 2 – mccord – (50970) 1 This print-out should have 34 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. McCord CH301 001 10.0 points What is the molecular geometry of SbCl 2 − 5 ? Antimony (Sb) can be an exception to the octet rule, accepting up to twelve electrons. 1. square planar 2. trigonal bipyramidal 3. square pyramidal correct 4. octahedral 5. seesaw Explanation: 002 10.0 points Which of the following has bond angles of exactly 120 ◦ ? 1. HO − 2 2. CS 2 − 3 correct 3. S 2 − 3 4. NO − 2 5. O 3 Explanation: While CS 2 − 3 and NO − 2 both have three re- gions of electron density around the central atom, in NO − 2 one of these regions is a lone pair which will repel the other regions more strongly, making the bond angle less than 120 ◦ . Only in CS 2 − 3 would the bond angle be exactly 120 ◦ . 003 10.0 points Which of the following has the largest radius? 1. S 2 − correct 2. K + 3. S 4. Cl 5. Cl − Explanation: Although size decreases from left to right across the periodic table due to increasing effective nuclear charge, the negative ions will be the largest ions, and the negative ion with the largest charge will have the largest radius since the electrons will repel each other and the effective nuclear charge will be insufficient to overcome this repulsion. 004 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. 150 pm 3. 110 pm 4. 120 pm correct 5. 55 pm 6. 75 pm 7. 135 pm Explanation: The bond is a double bond so you add 60 pm to 60 pm and get 120 pm. Version 331 – Exam 2 – mccord – (50970) 2 005 10.0 points In the Lewis dot structure of the molecule ClF 3 , how many unbonded electron pairs are found around the central atom? 1. 2 correct 2. 4 3. 4. 1 5. 3 Explanation: Draw the Lewis dot structure. ClF 3 has 3 Cl F single bonds and 2 lone pairs on Cl. 006 10.0 points Draw the Lewis structure for CO. 1. b b b b b b O b b b b b b C 2. b b b b b b O b b b b b b O C 3. b b b b b b O b b b b b b C 4. b b b b b b O b b b b C 5. O b b b b b b b b b b C 6. O b b C 7. b b b b b b b b b b O C correct 8. b b b b b b b b O b b b b b b b b C 9. O b b b b b b b b b b C 10. O b b b b b b b b C Explanation: C has 4 valence electrons; O has 6: b b b b C b b b b b b O Total number of valence electrons: C 1 × 4 e − = 4 e − O 1 × 6 e − = 6 e − 10 e − b b b b b b b b b b O C or b b b b O C This configuration gives each atom a com- plete octet....
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This note was uploaded on 12/06/2010 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-solutions - Version 331 – Exam 2 – mccord –...

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