前四周期原子序数依次增大的六种元素A、B、C、D、E、F中,A、B属于同一短周期元素且相邻,A元素所形成的

2025-03-16 12:17:45
推荐回答(1个)
回答1:

前四周期原子序数依次增大的六种元素A、B、C、D、E、F中,A、B属于同一短周期元素且相邻,A元素所形成的化合物种类最多,则A为碳元素、B为N元素;C、D、E、F是位于同一周期的金属元素,只能处于第四周期,基态C、F原子的价电子层中未成对电子均为1个,且C、F原子的电子数相差为10,可推知C为K、F为Cu,基态D、E原子的价电子层中未成对电子数分别为4、2,且原子序数相差为2,D、E价电子排布分别为3d64s2,3d84s2,故D为Fe、E为Ni,
(1)六种元素中K的金属性最强,其第一电离能最小,非金属性越强电负性越大,故电负性最大的是N,
故答案为:K;N;
(2)黄血盐的化学式K4[Fe(CN)6],CN-与氮气互为等电子体,CN-中存在C≡N三键,故1mol CN-中含有π键的数目为2NA,黄血盐晶体中含有离子键、配位键、共价键,没有金属键、氢键、分子间的作用力,
故答案为:K4[Fe(CN)6];2NA;def;
(3)Ni2+的价层电子排布式为3d8,故价电子排布图为
①CH2=CH2、③、④HCHO中C原子价层电子对数都是3,没有孤电子对,C原子采取sp2杂化,②HC≡CH为C原子价层电子对数是2,没有孤电子对,C原子采取sp杂化,HCHO分子的立体结构为 平面三角形,它加成产物为甲醇,甲醇分子之间形成氢键,其熔、沸点比CH4的熔、沸点高,
故答案为:;①③④;平面三角;加成产物CH3OH分子之间能形成氢键;
(4)金属K晶体为体心立方堆积,晶胞结构为图Ⅰ,晶胞中K原子配位数为8,金属Cu晶体为面心立方最密堆积,晶胞结构为图Ⅱ,以顶点Cu原子研究与之最近的原子位于面心,每个顶点Cu原子为12个面共用,晶胞中Cu原子配位数为12,K、Cu两种晶体晶胞中金属原子的配位数之比为 8:12=2:3;
金属K的晶胞中,K原子数目=1+8×
1
8
=2,若K原子的摩尔质量为M,阿伏加得罗常数为NA,则晶胞质量=
2×M
NA
g,设该晶胞的密度为a g/cm3,则晶胞体积=
2M
NA
g
a g/cm3
=
2M
aNA
cm3,晶胞棱长=
3
2M
aNA
cm,设K原子半径为r,则晶胞体对角线长度=4r,故3(
3
2M
aNA
cm)2=(4r)2,故r=
3
4
×
3
2M
aNA

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