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【题目】A、B、C、D 4种元素的核电荷数依次增加,他们的离子电子层数相同且最外层电子数均为8。A原子的L层电子数与K、M层电子数之和相等,D原子的K、L层电子数之和等于电子总数的一半。回答下列问题:

1这4种元素的符号依次是A________,B________,C________,D________。它们的原子半径由大到小的顺序是_________________。

2分别写出这4种元素的最高价氧化物对应水化物的化学式:_________________,

分别比较酸性和碱性的强弱:_________________。

3写出气态氢化物的化学式:_______________。比较其稳定性:______________,理由是_________。

【答案】1S Cl K Ca

rK>rCa>rS>rCl

2H2SO4HClO4、KOH、CaOH2

酸性:HClO4H2SO4、碱性:KOH>CaOH2

3HCl、H2S HCl>H2S 氯的原子半径小于硫,氯的非金属性强

【解析】由A原子的L层电子数与K、M层电子数之和相等,所以A的核电荷数为2×8=16,A为硫元素。D原子的K、L层电子数之和等于电子总数的一半,所以D原子的核电荷数是2+8×2=20,D为钙元素。根据核电荷数依次增大,并都形成离子,B为氯元素,C为钾元素。

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