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    【题目】A、B、C、D、E五种元素的原子序数依次递增，且均属于短周期元素，已知A是非金属元素，其电子层数与最外层电子数相等。Bn＋和C（n＋1）＋与氖原子具有相同的电子层结构，且C是两性元素。Dn－、E（n－1）－与氩原子具有相同的电子层结构。

    （1）写出这五种元素的元素符号

    A．__________；B.__________；C.__________；D.__________；E.__________。

    （2）写出E单质与NaOH溶液反应的化学方程式：____________________。

    （3）比较A、B、C、D、E的半径大小_____________________。

    【答案】（1）H Mg Al S Cl

    （2）Cl2＋2NaOH===NaCl＋NaClO＋H2O

    （3）r（A）＜r（E）＜r（D）＜r（C）＜r（B）

    【解析】短周期元素中电子层数与最外层电子数相等的有H、Be、Al。其中H为非金属元素，所以A为H。由于C为两性元素，则Bn＋、C（n＋1）＋分别为Mg2＋、Al3＋。由于n＝2，则Dn－、E（n－1）－分别为S2－、Cl－。因此A、B、C、D、E分别为H、Mg、Al、S、Cl。A元素原子半径最小，B、C、D、E同周期，序数增大，其半径减小。

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