problem set 1 key - Chemistry 312 Please turn in on...

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Please turn in on separate sheets, stapled together Problem Set 1 (Due beginning of class, Friday, July 1, 2011) 1. a. Based on periodic trends, would you expect Cl 2 or I 2 to be a better oxidizing agent ? Explain. By “better”, we mean “stronger”, and certainly Cl 2 is the stronger oxidant. Chlorine is nearer the top of the halogens column in the periodic table, and would thus more readily accept and stabilize an electron than iodine would. b. Based on periodic trends, would you expect K or Li to be a better reducing agent ? Explain. Again, by “better”, we mean “stronger”, and potassium is the stronger reductant. Potassium is nearer the bottom of the alkali metals column in the periodic table, and would thus more readily donate an electron than lithium would. c. Based on periodic trends, would you expect B or C to have a higher ionization potential ? Explain. By “ionization potential”, we mean “ionization energy”, and carbon should have a higher one than B, if only because C will have a higher Z eff than B due to its larger nuclear charge and same number of core electrons. 2. Rationalize the following values for successive ionization potentials of boron , where IP(n) corresponds to the energy required to remove the nth electron. n 1 2 3 4 5 IP(n) (MJ mol –1 ) 0.807 2.433 3.666 25.033 32.834 Definitely a jump in ionization energy from n = 3 to n = 4, as might be expected for boron, which has an electron configuration of (1s) 2 (2s) 2 (2p) 1 . By removing the three electrons in the incomplete second shell, boron becomes isoelectronic with helium, and thus gains the stability of having only complete electronic shells. Two subtle points: the first electron removed is from the 2p orbital, being unpaired, and thus is a lower energy cost to remove. The last electron removed (the last 1s electron) will be a very high energy cost to remove because it is attracted by all 5 protons in the nucleus with no screening by any other electrons. 3. Explain the decrease in first ionization energy (equivalent to the ionization potential) for oxygen versus nitrogen in the table of electronic properties found in Resource Section 2 of Atkins (p. 785). The electron configuration of oxygen is (1s)
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This note was uploaded on 01/19/2012 for the course CHEM 312 taught by Professor Staff during the Summer '08 term at University of Washington.

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problem set 1 key - Chemistry 312 Please turn in on...

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