X为无色液体,与过氧化钠反应得到B与气体Y,则X是H
2O,Y是O
2,B是NaOH.H与Na
2O
2可发生化合反应,生成的盐与Ba
2+反应可生成不溶于稀G的白色沉淀,说明H应该是SO
2.SO
2具有还原性,被过氧化钠氧化生成硫酸钠,硫酸钠和钡离子反应生成硫酸钡白色沉淀.SO
2被氧气氧化生成三氧化硫,三氧化硫溶于水生成硫酸,即I是SO
3,K是H
2SO
4.一个D分子中含有10个电子,经过系列反应得到强酸G,则D是NH
3,则E是NO,F是NO
2,所以G是HNO
3.A是由短周期元素组成的酸式盐,所以A是NH
4HSO
3,则C是Na
2SO
3,
(1)D为NH
3,电子式为:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/eaf81a4c510fd9f9d9155ddb262dd42a2834a405?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/eaf81a4c510fd9f9d9155ddb262dd42a2834a405?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;
(2)D+H+X→A的化学方程式为:SO
2+NH
3+H
2O=NH
4HSO
3,故答案为:SO
2+NH
3+H
2O=NH
4HSO
3;
(3)C→H的离子方程式为:SO
32-+2H
+=SO
2↑+H
2O,故答案为:SO
32-+2H
+=SO
2↑+H
2O;
(4)组成单质Y元素为氧元素,其基态原子电子排布式为:1S
22S
22P
4,故答案为:1S
22S
22P
4.