短周期主族元素X、Y、Z、W、M、N 的原子序数依次增大,且原子最外层电子数之和为23.X的原子半径比Y的小

2025-03-16 01:15:19
推荐回答(1个)
回答1:

短周期主族元素X、Y、Z、W、M、N 的原子序数依次增大,Z的最外层电子数是电子层数的3倍,原子只能有2个电子层,最外层电子数为6,则Z为O元素;X的原子半径比Y的小,结合原子序数可知,X只能处于第一周期,故X为H元素;X与W同主族,W原子序数大于O,故W为Na;M是地壳中含量最高的金属元素,则M为Al;所以原子最外层电子数之和为23,则Y、N最外层电子数之和=23-1-6-1-3=12,Z的原子序数小于O,可知Y最外层电子数为5、N最外层电子数为7,故Y为N元素、N为Cl元素,
(1)M为Al,其离子结构示意图为:,X、Y形成的含18电子的化合物为N2H4,电子式为,W、Z形成的只含离子键的化合物为Na2O,用电子式表示其形成过程为:
故答案为:
(2)同周期自左而右原子半径减小,同主族自上而下原子半径增大,故原子半径:Al>N>O;
非金属性N<Cl,Y的氢化物稳定性<N的氢化物稳定性,
故答案为:Al>N>O;<;
(3)H和O组成的化合物中,既含有极性共价键又含有非极性共价键的是H2O2,此化合物可将碱性工业废水中的CN-氧化为碳酸盐和氨气,相应的离子方程式为:H2O2+CN-+OH-=CO32-+NH3↑,
故答案为:H2O2;H2O2+CN-+OH-=CO32-+NH3↑;
(4)由以上元素中两种或多种组成的化合物A、B、C、D有如图转化关系,已知A溶液显碱性,可促进水的电离,可推知A为偏铝酸钠、B为HCl、C为氢氧化铝、D为氯化铝,用离子方程式表示氯化铝溶液显碱性的原因:AlO2-+2H2O Al(OH)3+OH-,A与D反应的离子方程式为:Al3++3AlO2-+6H2O=4 Al(OH)3↓,
故答案为:AlO2-+2H2O Al(OH)3+OH-;Al3++3AlO2-+6H2O=4 Al(OH)3↓.

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