Ch 301 Exam 2 - Version Exam 2 Sutcliffe 1 This print-out...

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Unformatted text preview: Version Exam 2 Sutcliffe 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Note: The middle atom is C. OCS has what molecular geometry? 1. tetrahedral 2. linear correct 3. angular 4. T-shaped 5. trigonal Explanation: O b b b b b b C b b b b S b b b b b b For the central atom C, HED = 2, lone pairs = 0, the electronic geometry is linear and the molecular geometry is also linear. 002 10.0 points Which pair of elements is listed in order of increasing electronegativity? 1. N, O correct 2. N, C 3. F, Cl 4. S, As 5. S, Se Explanation: Electronegativity generally increases from left to right and from bottom to top of the Periodic Table. Thus N < O. 003 10.0 points In most Lewis dot structures of neutral com- pounds, carbon as the central atom will be bound to a maximum of how many atoms, ni- trogen will be bound to a maximum of how many atoms, and oxygen will be bound to a maximum of how many atoms? 1. 3; 3; 2 2. 4; 3; 2 correct 3. 5; 3; 2 4. 2; 2; 2 5. 4; 2; 2 6. 2; 3; 3 7. 3; 2; 2 8. 3; 2; 1 9. 4; 3; 3 Explanation: Carbon has four valence electrons; by bond- ing to four other atoms it can obtain an octet. Nitrogen has 5 valence electrons; it can form bonds to three other atoms to obtain an octet. Oxygen has six valence electrons; by bond- ing to two other atoms, it will obtain an octet. None of these atoms can have an expanded valence shell because the d-orbitals are not available to them. 004 10.0 points In the molecular orbital representation of ben- zene, the bonding between each C C pair is best thought of as 1. one sigma and one pi bond. 2. one sigma and half a delocalized pi bond. correct 3. one sigma bond. 4. one pi bond and half a delocalized pi bond. 5. one sigma and two pi bonds. Explanation: Version Exam 2 Sutcliffe 2 Six sigma bonding orbitals are formed from overlap of the 2 s atomic orbitals on the six carbons, and three pi bonding orbitals form a delocalized bonding system. 005 10.0 points The bond angle in the molecule H 2 O is less than the tetrahedral bond angle of 109.5 be- cause 1. it is a polar molecule. 2. the hydrogen atoms attract each other. 3. the repulsion between lone pairs of elec- trons is greater than the repulsion between bonding pairs. correct 4. the oxygen is sp 2 hybridized. Explanation: 006 10.0 points Write the ground-state electron configuration of Sb 3+ . 1. [Kr] 4 d 5 5 s 1 5 p 6 2. [Kr] 4 d 10 5 s 2 correct 3. [Kr] 4 d 10 5 p 2 4. [Kr] 4 d 10 5 s 1 5 p 1 5. [Kr] 4 d 8 5 s 1 5 p 3 Explanation: Write the electron configuration for the atom and then adjust for the charge of the ion. The Aufbau order of electron filling is 1 s , 2 s , 2 p , 3 s , 3 p , 4 s , 3 d , 4 p , 5 s , 4 d , 5 p , 6 s , 4 f , 5 d , 6 p , etc ....
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This note was uploaded on 11/15/2010 for the course CH 301 taught by Professor Fakhreddine/lyon during the Spring '07 term at University of Texas at Austin.

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Ch 301 Exam 2 - Version Exam 2 Sutcliffe 1 This print-out...

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