So, what we end up having is three of these pi -- 2 p y 2 p y bonds, we can have one between these two carbons here.

我们剩下的有三个π键-,2py2py键，在这两个碳原子之间会有一个。

麻省理工公开课 - 化学原理课程节选

So what we see is that those six pi electrons are actually going to be de-localized around all six of those atoms.

我们看到,六个π电子会,巡游在所有六个原子周围。

麻省理工公开课 - 化学原理课程节选

That's -Pi times a vector.

就等于用 -π 乘以那个矢量

耶鲁公开课 - 基础物理课程节选

I know the energy in this first pair would equal -e^2 That is just going to equal minus e squared over 4 pi epsilon zero r naught.

我们明白第一对的能量将会等于,等于,/4πε0，R圈。

麻省理工公开课 - 固态化学导论课程节选

So z equals 7 -- this is the cut-off where, in fact, the sigma orbital is going to be higher in energy than the pi 2 p orbitals.

所以z等于7-这是分界点，实际上，sigma轨道能量,要比π2p轨道高。

麻省理工公开课 - 化学原理课程节选

And what we see here is now when we're combining the p, we have our 2 p x and our 2 p y orbitals that are lower in energy, and then our pi anti-bonding orbitals that are higher in energy.

这里我们看到,当我们结合p轨道时，在低能处我们有,2px和2py轨道，π反键轨道在更高的能级处。

麻省理工公开课 - 化学原理课程节选

z And what you need to remember is if the z 8 is equal to eight or greater, such as oxygen being the cut-off point, this sigma 2 p orbital is actually lower in energy than the pi 2 p orbitals, the molecular orbitals.

你们要记住如果,等于或者大于，就像O是分界点，这时sigma2p轨道,比π2p轨道能量更低。

麻省理工公开课 - 化学原理课程节选

So, the number of nuclei, 119 if we were to sit and count these as well, is 119. So, we'll multiply that by just pi, r squared, to get that cross-section, and divide all of that by 1 . 39 meters squared.

如果你们数的话，原子核的数是，我们用它乘以πr的平方，得到横截面积，除以1。39平方米。

麻省理工公开课 - 化学原理课程节选