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So for example I defined p, remember back over here, as a Cartesian point, but I can actually ask for its polar form.
我可以要求返回它的极坐标形式，这里对我是可访问的，好，这很棒，请再记另外一个为什么。
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OK, in fact if we look at this, if I say, print of p, it prints it out in that form.
好，实际上如果，我们来看看这个，如果我输入，p的显示。
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Remember, we didn't hybridize the 2 p y orbital, so that's what we have left over to form these pi bonds.
记住，我们并没有杂化2py轨道，这是我们剩下的那个行成了π键。
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The second problem is, 1 suppose actually I had p 1 and p 2 were in polar form, and I ran add points on them.
第二个点也就是，半径为3角度为1,也就是差不多在这，那么什么点，对不起。
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Having created c p 1 and c p 2, I had this weird looking form here.
我可以在类中建立一些属性，我可以给类增加一些特征。
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p But it now says that in any, in no matter what kind of form I made it from, I can get out that kind of information.
例如我定义了，记得这里吧，是一个笛卡尔卡。
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We can still hybridize all these orbitals, however, so we can still form four hybrid orbitals, which are again, 2 s p 3 hybrid orbitals.
但我们仍然可以杂化这些轨道，所以我们还是可以形成4个杂化轨道，同样的，是2sp3杂化轨道。
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So some p orbitals form pi molecular orbitals, and some form sigma p orbitals.
有些p轨道会形成π分子轨道，有些会形成p轨道。
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So it already should make sense why we have that p orbital there, in order to form a pi bond, we're going to need a p orbital.
这里有p轨道是很合理的,为什么我们在这里有P轨道，为了形成一个π键，我们需要一个p轨道。
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All right, so that's s p 3 hybridization, but those aren't the only type of hybrid orbitals that we can form. Let's take a look at what happens if instead of combining all four orbitals, we just combine three of those orbitals, and what we'll end up with is s p 2 hybridization.
好了，这是sp3杂化，但这并不是我们可以,形成的唯一类型的杂化轨道,让我们来看看，不是四个轨道结合，我们仅仅结合3个轨道。
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