推荐回答(2个)
对于ABn型号分子,先算中心原子A周围的电子对数,
电子对数=(A的最外层电子数
+B的化合价的绝对值
*n
)/2
(B是O,S,就不算电子)、
电子对数=2,
A原子
采取
sp杂化
2个杂化轨道
电子对数=3,
A原子
采取
sp2杂化,3个
电子对数=4,
A原子
采取
sp3杂化,4个
CO2
电子对数=
4
/2=2
sp
杂化
CH4
电子对=
(4+1*4)/2=4
sp3杂化,
BF3
(3+1*3)/2=3
sp2
BeCl2
(2+2*1)/2=2
sp
烃不属于ABn型
,不能用直接用上述方法判断
C2H4
这个
分子中含有C=C, C采取sp2杂化
C6H6 凯库勒式中也含有C=C,C采取sp2杂化
记住物质的
空间构型
基本
直线型的
SP1
平面性的
SP2
空间三角锥
SP3
其实
这个问题理解起来很难
基本就是
蛋鸡问题
问你
为什么
SP1杂化
回答说因为空间构型直线型
问你
为什么
直线型
回答
SP1
杂化
至于杂化数目
大约记点
最外层电子数
还有
基本的常识
就行了
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