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So let's have a clicker question here on how many total pi bonds do you expect to see in benzene?
让我们来做个课堂习题,你们觉得苯环里有一共多少个π键？
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Sorry, said that wrong, p1 radius 1 and angle 2, 2 radians is a little bit more than pi half.
而是半径和角度的表示，在这个例子中点，并不对应这个点,它实际上对应的是。
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There's absolutely no reason I couldn't have switched it around and said that instead the pi orbitals form between these atoms instead of those first atoms I showed.
我完全没有理由,不能把它转过来,现在π键在这些原子间,而不是我开始展示的那些原子间。
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And so both of these lobes together constitute a pi bond.
上下两片叶一起组成了一根π键。
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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。
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We can make some substitutions here using some of the derivation on the previous board which will give us the Planck constant divided by 2 pi mass of the electron times the Bohr radius.
在这里我们也可以,用我以前在黑板上写过的一些词来取代它,得到的是普朗克常数除以2π电子质量,再乘以波尔半径。
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So if we want to talk about the volume of that, we just talk about the surface area, which is 4 pi r squared, and we multiply that by the thickness d r.
如果我们要讨论它的体积，我们要用的是表面面积，也就是４πｒ的平方，乘以厚度ｄｒ
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So even though we see a nodal plane down the center, I just want to really point out that it's only when we have a nodal plane in the internuclear or the bond axis that we're calling that a pi orbital.
虽然在中间有个节面，我想要指出的是,只有节面在核间轴,或者键轴上时,我们才叫它π轨道。
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So in order to rotate a double bond, you have to actually break the pi bond, so essentially what you're doing is breaking the double bond.
为了能够旋转双键，你必须打破一个π键，本质上我们要做的就是打破双键。
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So if I try to rotate my 2 atoms, you see that I have to break that pi bond, because they need to be lined up so that the electron density can overlap.
如果我要试着转动两个原子，你会看到我必须要打破一个π键，因为他们需要连接起来,让那些电子能够重叠。
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Pi orbitals are a molecular orbital that have a nodal plane through the bond axis.
轨道是沿着键轴,有节面的分子轨道。
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So let's fill it out in this way, 2p keeping in mind that we're going to fill sigma out the pi 2 p's before the sigma.
让我们这样填上去，记住我们先填π,轨道再填。
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So when we go ahead and name these, we're going to call these pi orbitals.
我们继续来命名，我们叫它π轨道。
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px We'll call it either pi 2 p x, 2py if we're combining the x orbitals, or pi 2 p y.
我们叫它π,如果是x轨道组合的话，或者π
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r=nh/mv That will give us 2 pi r goes as nh over mv.
2π
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n*pi/L It turns out it is n pi over l.
结果是。
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q1*q2/ That's simply q1, q2 over 4 pi epsilon zero R.
那只是简单的。
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And now the force, in its most general term / is q1q2 over 4 pi epsilon zero, which is the conversion factor r squared.
库仑力的最基本形式,就是，其中r是一个变量。
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So our second bond is going to be a pi bond.
第二个键是π键。
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r So the circumference is 2 pi r.
周长=2π
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STUDENT: Pi.
学生：π
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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。
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If we have a double bond, we know we need to have only one sigma bond, and we're also going to have one pi bond.
如果我们有双键，我们知道我们需要一个sigma键，还需要一个π键。
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So, as they're very quietly handing out your class notes, let's think about what this bond is here, this boxed bond, is it a pi bond or a sigma bond?
在他们发讲义的同时，我们来看看这个键，方框里的这个键，它是sigma键还是π键？
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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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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。
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That's 4 pi epsilon zero 3 r naught and so on and so on.
也就是，4πε0，3R圈,以此类推。
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py So this will be pi carbon 2 p y, carbon 2 p y.
这是π碳2py碳。
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And we wrote something that looks, the energy is equal to minus the Madelung constant times Avogadro's number, 0R0 q1 q2 over 4 pi epsilon zero R zero.
我们写下了，晶格能等于负的马德隆常数,乘以阿伏伽德罗常数，乘以q1q2除以4πε
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