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Unformatted text preview: Garg, Pallavi Homework 3 Due: Oct 23 2007, 11:00 pm Inst: Donna C Lyon 1 This print-out should have 46 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. This HW assignment is due Tuesday, Oc- tober 23, by 11PM. DO NOT WAIT UN- TIL THE LAST HOUR TO TURN IN YOUR HOMEWORK ANSWERS. KEEP UP WITH THE MATERIAL BY ANSWER- ING QUESTIONS AS THEY ARE COV- ERED IN CLASS. 001 (part 1 of 1) 10 points Bonding orbitals 1. are filled after anti-bonding orbitals. 2. hold all valence electrons. 3. are less stable than anti-bonding or- bitals. 4. are lower in energy than anti-bonding orbitals correct 5. hold non-bonding electrons. Explanation: Bonding orbitals are lower in energy and more stable than anti-bonding orbitals. 002 (part 1 of 1) 10 points Antibonding orbitals 1. lend instability to a molecule when popu- lated with electrons. correct 2. are higher in energy than bonding orbitals and are therefore populated with electrons prior to bonding orbitals. 3. are responsible for dipole moments in molecules. 4. are responsible for high ionization ener- gies in atoms. 5. are lower in energy than bonding orbitals and are therefore populated with electrons prior to bonding orbitals. Explanation: Antibonding orbitals have nothing to do with dipole moments or ionization energies. They form from the overlap of atomic or- bitals, are higher in energy than bonding or- bitals, are populated with electrons after the corresponding bonding orbitals are populated andlendinstabilitytothemolecule whenpop- ulated with electrons. 003 (part 1 of 1) 10 points A node on an antibonding molecular orbital 1. is the region in which the probability of finding an electron is zero. correct 2. always passes through the atomic nu- cleus. 3. only exists for a ? 1 s . 4. is the region in which the probability of finding an electron is the highest. Explanation: The elctron density is zero in the nodal plane of antibonding molecular orbitals. 004 (part 1 of 1) 10 points Consider the diatomic molecules CO, which has a bond order of 3, and O 2 , which has a bond order of 2. One can state that O 2 would be (easier, harder) to break apart than CO, and would also have (longer, more, shorter) bonds than CO. 1. easier; more 2. harder; longer 3. easier; shorter 4. harder; more 5. easier; longer correct 6. harder; shorter Explanation: Garg, Pallavi Homework 3 Due: Oct 23 2007, 11:00 pm Inst: Donna C Lyon 2 The larger the bond order, the more stable the molecule. This corresponds to stronger and shorter bonds. 005 (part 1 of 1) 10 points Which species has the highest bond order? 1. CO + 2. NO + correct 3. CO- 4. NO- 5. CO 3+ Explanation: The corresponding bond orders are CO 3+ : 1.5; CO + : 2.5; CO- : 2.5; NO + : 3; NO- : 2 006 (part 1 of 1) 10 points Based on MO theory, we would expect the order of increasing bond length for O 2 , O- 2 , and O + 2 to be 1. O- 2 , O 2 , O + 2 ....
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- Spring '07