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。

麻省理工公开课 - 计算机科学及编程导论课程节选

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。

麻省理工公开课 - 计算机科学及编程导论课程节选

Because if we think about wave behavior of electrons and we're forming bonds, then what we have to do is have constructive interference of 2 different electrons, right, to form a bond, we want to and together those probabilities.

如果我们考虑,电子的波动行为,并且，我们要成键的话,我们要,把，这些概率，加在一起,如果。

麻省理工公开课 - 化学原理课程节选

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。

麻省理工公开课 - 化学原理课程节选

These 2 are going to come together like this, and the first bond that we're going to form is going to be a sigma bond, right, so we see that here. If we're looking head on, we see they form a sigma bond.

它们两个会靠近到一起，首先会形成的是,一个sigma键，对吧，我们在这里可以看出来，我们看到它们形成一个sigma键。

麻省理工公开课 - 化学原理课程节选

In contrast, if we have destructive interference, what we're going to form is a sigma 2 s star, and what does the star designate?

相反，如果我们是相消干涉，我们会形成sigma2s星，星代表什么？

麻省理工公开课 - 化学原理课程节选

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。

麻省理工公开课 - 化学原理课程节选

And if we picture those two coming together, we form the h 2 molecule.

如果我们想象它们两个靠近时，我们会形成H2分子。

麻省理工公开课 - 化学原理课程节选