英语周报2020~2021。八年级。新目标。答案

8.(15分)(1)哑铃形(或纺锤形)(1分)N原子的2p轨道电子处于半充满状态,比较稳定(2分)tt团世分As(2)sp°(1分)制炸药(或高能燃料)(1分)(3)①CO2(或N2O)(1分,其他合理答案也给分)②DE(2分)(4)1:224(2分)185610(2分)(5)Fe3C(2分)【解析】(1)基态N原子的核外电子排布式是1s2s2p3,最高能级2p能级的电子云轮廓图形状为哑铃形或纺锤形;N原子的2p轨道电子处于半充满状态,比较稳定,故N原子的第一电离能比O原子大;基态铁原子的价电子排布图为3d4s(2)晶体结构可知,氮原子的杂化轨道类型为sp这种高聚氮N一N键的键能为160kJ·mo-1,而N2中N=N键的键能为942k·mol-,高聚氮可能潜在的应用是制炸药或高能燃料。(3)①与N互为等电子体的分子有CO2N2O等。②生成HN的化学方程式为N2H4 HNO2=HN3 2H2O,A项正确;Na 半径小于K ,NaN的晶格能大于KN3,B项正确;氢叠氮酸(HN3)和水能形成分子间氢键,C项正确;HN3是极性分子,N2H4是非极性分子,D项错误;HN2分子中四个原子不在同一条直线上,E项错误。(4)A中含有1.5个Fe 、4个O,B中含有0.5个Fe 4个o-、4个Fe3 ,则该氧化物中Fe2 、Fe 、OP-的个数比为1:2:4。晶胞含有Fe 、Pe、O的个数分别是为816,32,一个品胞质量为g晶胞体积为8am,则品胞参数为x0ma(5)每个铁原子为两个八面体共用,故铁原子与碳原子的个数比是(6×b):1=3:1,则该化合物的化学式为Fe3C。

第二节One possible versionniceDear overseas students from America,Here comes the news that Chinese Bridge series activities will be held in our school. So all of participants arewarmly welcome in the pep rally to get fully prepared for thefolloThemed at promoting Chinese culture around theworld, Chinese Bridge acts as a platform to help narrow thegap of mutual understanding about China. Whoever isactively involved in it will develop a positive awareness ofChinese and its culture behind. Based on it, the pep rallis scheduled in our school auditorium from 14: 00 p.m15: 00 P. m. next Friday afternoonPlease make sure you are available at that time andThe student’ s Union

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