.已知分子E和分子G反应,生成两种分子L和M(组成E、G、L、M分子的元

.已知分子E和分子G反应,生成两种分子L和M

(组成E、G、L、M分子的元素原子序数小于10 “○”代表原子“____”表示化学键 ),如下图,则下列判断错误的是(      )(未配平)

A. G是最活泼的非金属单质               B. L中的化学键是极性键

C. E能使紫色的石蕊试液变蓝             D. M的化学性质活泼

 

答案

D

解析:组成E、G、L、M分子的元素原子序数小于10,所以E是氨气,M是氮气,G是氟气,L是HF。氮气中含有氮氮三键,键能大,性质稳定,D不正确。答案是D。

 

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