端位O原子参与大π键的电子如何判断?是2个还是1个啊???

2025-03-17 02:07:33
推荐回答(1个)
回答1:

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先写出相应的结构式,只画单键,再在每个原子旁标注剩余的成键电子数(需判断杂化类型),比如NO2,N和O周围有4个形成π键的电子,O原子参与大π键的电子各为1,共有3个原子,故NO2为π3,4
同理ClO2中为π3,5,O原子参与大π键的电子各为1,Cl为3
这两种物质中,N和Cl均为sp2杂化,均有2个未成键电子,其余的价电子有2个形成sigma键,其余形成大π键。氧原子均为sp杂化,1个O有4个未成键电子,1个形成sigma键,一个参与形成大π键。 
大多数情况O作为端原子时只有1个参与大π键的电子(sp杂化)如CO2,N2O3,N2O5。但在N2O中则有3个,该分子中有2个π3,4;NO有3个。
O作为中心原子时,如O3中有π3,4,中间的O有2个电子参与形成大π键。
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