由元素在周期表中位置可知,A为氢、B为Na、F为Mg、M为Ca、L为Fe、G为Al、H为硅、C为碳、I为磷、D为氮、J为硫、E为氧、N为Br、K为Cl、O为氟;
(1)由表可知,L在第四周期第8列,即处于第四周期第Ⅷ族;A
2E为H
2O,Fe与水蒸气在高温条件下反应生成四氧化三铁与氢气,反应方程式为:3Fe+4H
2O(g)
Fe
3O
4+4H
2,
故答案为:第四周期第Ⅷ族;3Fe+4H
2O(g)
Fe
3O
4+4H
2;
(2)电子层结构相同,核电荷数越大离子半径越小,电子层越多离子半径越大,故离子半径:P
3->Cl
->Ca
2+>Mg
2+>Al
3+,故答案为:P
3->Cl
->Ca
2+>Mg
2+>Al
3+;
(3)MO
2为CaF
2,由钙离子与氟离子构成,其电子式为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/faedab64034f78f0d6777d427a310a55b3191c06?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;CJ
2为CS
2,分子中C原子与S原子之间形成2对共用电子对,用C原子、S原子电子式表示其形成过程为:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/aec379310a55b319cd8565c440a98226cffc1706?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,
故答案为:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/faedab64034f78f0d6777d427a310a55b3191c06?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/aec379310a55b319cd8565c440a98226cffc1706?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;
(4)同周期自左而右非金属性增强,同主族电子层越多非金属性越弱,故非金属性C弱于S,可以利用最高价含氧酸强酸制备弱酸进行验证,反应方程式为:Na
2CO
3+H
2SO
4=Na
2SO
4+CO
2↑+H
2O,
故答案为:弱;Na
2CO
3+H
2SO
4=Na
2SO
4+CO
2↑+H
2O;
(5)A与E形成的18个电子的化合物为H
2O
2,A与J形成18个电子的化合物为H
2S,过氧化氢具有强氧化性,可以氧化硫化氢,反应生成S与水,反应方程式为:H
2O
2+H
2S=2H
2O+S↓,
故答案为:H
2O
2+H
2S=2H
2O+S↓.