[题目]根据元素周期表1~20号元素的性质和递变规律,回答下列问题:(1)属于金属元素的有种,其中金属性最强的元素的单质与氧气反应生成的化合物的化学式为 (填两种).(2)最高正化合价最高的元素是 (填元素符号),其最高正化合价为 .(3)单质既能与HCl反应,又能与NaOH反应的是 ,其最高价氧化物对应的水化物具有 (填“酸性 “碱性或“两性 ).( 题目和参考答案——青夏教育精英家教网—

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【题目】根据元素周期表1~20号元素的性质和递变规律,回答下列问题:

（1）属于金属元素的有种,其中金属性最强的元素的单质与氧气反应生成的化合物的化学式为（填两种）。

（2）最高正化合价最高的元素是（填元素符号）,其最高正化合价为。

（3）单质既能与HCl反应,又能与NaOH反应的是 ,其最高价氧化物对应的水化物具有（填“酸性”“碱性”或“两性”）。

（4）第三周期中,原子半径最大的元素是（稀有气体元素除外）。

（5）比较Si、N最简单氢化物的稳定性: （用化学式表示）。

【答案】（1）7;K2O、K2O2（KO2也可以）

（2）Cl;+7价

（3）Al;两性

（4）Na

（5）NH3>SiH4

【解析】前20号元素中,（1）属于金属元素的共有7种,其中金属性最强的元素是钾,钾单质与氧气反应生成的化合物有K2O、K2O2、KO2;（2）最高正化合价最高的元素是Cl,其最高正化合价为+7价;（3）既能与HCl反应又能与NaOH反应的单质是铝,其最高价氧化物对应的水化物是Al（OH）3,具有两性;（4）同一周期从左往右,原子半径逐渐减小,故第三周期中,原子半径最大的是Na;（5）非金属性:N>P>Si,故NH3的稳定性大于SiH4的稳定性。

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