问题 问答题

五种短周期元素A、B、C、D、E的原子序数依次增大,A和C同族,B和D 同族,C离子和B离子具有相同的电子层结构.A和B、D、E均能形成共价型化合物.A和B形成的化合物在水中呈碱性,C和E形成的化合物在水中呈中性.回答下列问题:

(1)五种元素中,原子半径最大的是______,非金属性最强的是______(填元素符号);

(2)由A和B、D、E所形成的共价型化合物中,热稳定性最差的是______(用化学式表示);

(3)A和E形成的化合物与A和B形成的化合物反应,产物的化学式为______,其中存在的化学键类型为______;

(4)D最高价氧化物的水化物的化学式为______;

(5)单质D在充足的单质E中燃烧,反应的化学方程式为______;D在不充足的E中燃烧,生成的主要产物的化学式为______;

(6)单质E与水反应的离子方程式为______.

答案

五种短周期元素A、B、C、D、E的原子序数依次增大.A和B形成的共价化合物在水中呈碱性,该化合物为NH3,则A为氢元素、B为氮元素;A和C同族,C的原子序数大于氮元素,故C为Na元素;B和D 同族,则D为磷元素;C和E形成的化合物在水中呈中性,则E为Cl元素,

(1)同周期自左而右原子半径增大,同主族自上而下原子半径增大,故Na元素的原子半径最大;

最高价含氧酸的酸性越强,中心元素的非金属性越强,高氯酸是最强的含氧酸,故Cl非金属性最强;

故答案为:Na;Cl;

(2)由A和B、D、E所形成的共价型化合物分别为NH3、PH3、HCl,非金属性越强氢化物越稳定,故热稳定性最差的是 PH3

故答案为:PH3

(3)A和E形成的化合物HCl,A和B形成的化合物NH3,二者反应生成NH4Cl,NH4Cl晶体中含有:离子键、共价键,

故答案为:NH4Cl;离子键、共价键;

(4)P元素的最高价氧化物的水化物的化学式为H3PO4

故答案为:H3PO4

(5)P在充足的氯气中燃烧生成五氯化磷,反应的化学方程式为2P+5Cl2

 点燃 
.
 
2PCl5;P在不充足的氯气中燃烧,生成的主要产物的化学式为PCl3

故答案为:2P+5Cl2

 点燃 
.
 
2PCl5;PCl3

(6)氯气与水反应生成盐酸与次氯酸,反应离子方程式为:Cl2+H2O=H++Cl-+HClO,

故答案为:Cl2+H2O=H++Cl-+HClO.

判断题
阅读理解

阅读理解。

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