Exam 02 - Su Yung Exam 2 Due 3:00 pm Inst FAKHREDDINE This print-out should have 23 questions Multiple-choice questions may continue on the next

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Su, Yung – Exam 2 – Due: Oct 19 2007, 3:00 pm – Inst: FAKHREDDINE 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. Each question is worth as assigned. 001 (part 1 o± 1) 10 points N 2 has a bond order o± 3 and O 2 has a bond order o± 2. Based on this in±ormation, choose the response that best completes the ±ollowing sentence, in the order listed. N 2 is (less, more) stable than O 2 , and has a (larger, shorter) bond length and a (higher, lower) bond en- ergy. 1. more; shorter; lower 2. less; longer; lower 3. more; shorter; higher correct 4. more; longer; higher 5. less; shorter; lower Explanation: The higher the bond order, the more sta- ble the molecule. Stable bonds have a high bond energy since they are hard to break apart. As bond order increases, the bond length gets shorter. Triple bonds have shorter bond length than double or single bonds. 002 (part 1 o± 1) 8 points Assuming these can all be made, which would have the largest diameter? 1. Cl 2+ 2. Cl 1 - 3. Cl 1+ 4. Cl 2 - correct 5. Cl 0 Explanation: All o± the species have the same number o± protons. As thenumber o±electrons increases, the attractive ±orce per electron decreases and they are ±ound ±arther ±rom the nucleus, re- sulting in larger particles. 003 (part 1 o± 1) 8 points The correct ±ormula ±or iron(III) carbonate is 1. Fe 2 CO 3 . 2. Fe 3 (CO 3 ) 2 . 3. Fe 2 (CO 3 ) 3 . correct 4. Fe 3 (CO 3 ) 3 . 5. FeCO 3 . 6. Fe 3 CO 3 . 7. Fe 2 (CO 3 ) 2 . 8. Fe(CO 3 ) 3 . Explanation: The iron(III) ion is Fe 3+ ; the carbonate ion is CO 2 3 - . Three CO 2 3 - are needed to balance the charge on every two Fe 3+ . (This gives a total anion charge o± - 6 and a total cation charge o± +6.) The ±ormula is Fe 2 (CO 3 ) 3 . 004 (part 1 o± 1) 8 points Which o± the ±ollowing elements has the high- est ionization energy? 1. P correct 2. Si 3. S 4. Allwouldhaveequalionizationenergies. 5. Cannot determine without more in±orma- tion. Explanation: P=[Ne]3 s 2 3 p 3 neutralatomhasadditional stability due to the ±act that its outermost or- bital is hal±-flled (3 p 3 ). This is re±erred to
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Su, Yung – Exam 2 – Due: Oct 19 2007, 3:00 pm – Inst: FAKHREDDINE 2 as half-±lled stability. In such a case the ion- ization energy is higher than what would be expected for a simple left-to-righttrend across the periodic table. If such a simple trend were followed, we would expect the ionization en- ergy of P to be between Si and S. In reality, since P has added stability due to a half-±lled orbital, it takes more energy to remove an electron from P than from Si or S. 005
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This note was uploaded on 04/15/2009 for the course CH 54740 taught by Professor Fakhreddine during the Spring '09 term at University of Texas at Austin.

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Exam 02 - Su Yung Exam 2 Due 3:00 pm Inst FAKHREDDINE This print-out should have 23 questions Multiple-choice questions may continue on the next

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