题目
M2LA F25 Expt 4 PreLab Quiz
多重下拉选择题
Part A. An iodine atom (I) has a Lewis structure with [ Select ] 4 6 8 1 2 3 5 7 dots (electrons) in the outer shell. Therefore, the atom has [ Select ] 3 2 4 1 0 unpaired electron(s) and [ Select ] -1 +1 0 charge. An iodide anion (I-) has a Lewis structure with [ Select ] 6 2 5 8 4 1 3 7 dots (electrons) in the outer shell. Therefore, the anion has [ Select ] 0 4 1 2 unpaired electron(s) and [ Select ] 0 +1 -1 charge. Part B. An iodide anion (I-) has [ Select ] 1 2 3 4 electrons in the 5s orbitals and [ Select ] 4 1 5 3 2 6 electrons in the 5p orbitals. The highest occupied molecular orbital (HOMO) is [ Select ] 6s 5p 4d 5s .
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标准答案
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思路分析
Part A involves filling in three sequential blanks about iodine (I) and its iodide (I-).
First blank (outer-shell dots for neutral I): The correct count is 7. In neutral iodine, the valence shell is 5s2 and 5p5, giving a total of 7 electrons in the outermost shell. The options provided include numbers like 4, 6, 8, 1, 2, 3, 5, and 7; 7 matches the known valence electron count for iodine, which determines its bonding behavior. Other choices would contradict the established electron configuration (for iodine, you never have only 4 or 6 valence electrons in this shell). For example, 8 would imply a full octet in the outer shell, which occurs for noble gases, not for iodine in its ground state.
Second blank (unpaired ......Login to view full explanation登录即可查看完整答案
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