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So we started with 10 valence electrons, we used up 8 of those electrons in terms of making bonds.
我们一开始有十个价电子，然后用掉了八个电子来成键。
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So, if we hybridize just these three orbitals, what we're going to end up with is our s p 2 hybrid orbitals.
我们会看到现在有3个未配对的电子，可以成键。
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Well, if this bond has completely identically equal sharing of electrons, then this bond will be nonpolar.
如果一根键连的两个原子,对键上的电子吸引程度是完全等价的，那么这根键是非极性的。
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So what we end up with in terms of our bonding electrons is going to be 6 bonding electrons.
因此最终我们需要六个成键电子,那么我们可以来把它们填上。
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And because of the way those antibonding orbitals are stacked, the two electrons go one each into those antibonding.
因为这样,反键轨道被堆积了，这两个电子都填到各自的反键轨道。
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So again, this is an anti-bonding orbital, and what you see is that there is now less electron density between the two nuclei than there was when you had non-bonding.
同样的，这是反键轨道，你们看到当你有反键轨道的时候,两个原子核中间的电子密度更小了。
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Yeah, so also 4. We started with 10 valence electrons, we used up 6 of those as bonding electrons, so we have 4 left, which will be lone pair electrons.
对，也是四个，我们从十个价电子开始，只用了六个来成键，因此我们还剩下四个，它们将成为孤对电子。
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Clearly, we put 2 for each bond, and now we end up having 2 remaining bonding electrons left.
显然，我们在每个键处放上两个电子，那么最后我们还剩下两个成键电子。
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We also have six spots available to form hydrogen bonds, so we can go ahead and fill in those electrons as well.
我们还有6个地方可以供氢原子成键，我们可以继续把这些电子填充进去。
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You can see that there is no value in studying inner shell electrons to ask questions about bonding.
你应该了解考虑成键时,去考虑内层电子,是毫无意义的。
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So let's see, we started with 8 bonding electrons, and we used up only 4, so the answer is yes, we have 4 bonding electrons left.
那么让我们来看看，我们一开始有八个成键电子，然后只用掉了四个，因此答案应该是还有剩余，我们还剩下了四个成键电子。
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The next thing that we want to do is figure out do we have any bonding electrons left.
下一步要做的是看看我们,还有没有剩下的成键电子。
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And we know that it's electron density between the nuclei that holds two atoms together in a bond.
我们知道是两个原子核之间的,电子密度保持两个原子在一起成键的。
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Then minus 1/2 of 2, because we only have one bond or 2 electrons in a bond.
然后减去二的二分之一，因为我们只有一个键，一个键就是两个电子。
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And then we end up having 6 shared electrons, 2 from each of the bonds, so we end up with a formal charge on sulfur of plus 1.
然后我们有六个共用电子，每个键两个，因此最终硫的形式电荷量为正一。
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Because step five is that we need to fill in our bonding electrons, and we start it with filling in two electrons per bond.
因为第五步要做的是将我们的成键电子填在这，所以我们开始在每个键处填上两个电子。
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For the sulfur, we start off with 6 valence electrons, minus 4 lone pair electrons, minus 2, taking in account our bonding electrons, so we end up with a formal charge of 0.
对于硫，我们从六个价电子开始，减去四个孤对电子，再减去二，算上我们的成键电子，因此最终我们有零个形式电荷。
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The first one will be above and below the bond axis is where we'll see the electron density, and the second will be perpendicular to that, so it will be a density in front of and behind the bond axis.
第一个是在键轴之上和之下,我们可以看到电子密度，另外一个垂直于它，所以在键轴之前和之后有电子密度。
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So what we'll have here is a trigonal planar case, and you can see that we only have three electrons that are set for bonding, so we'll add three hydrogens, and for b h 3, we'll get a stable structure here.
让电子劲量远离的时候，不用考虑它,这个例子是平面三角形，你可以看到，只有3个电子可以成键。
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We see three bonding pairs so this is a triple bond, indeed a multiple bond.
我们看到3对成键电子所以这是一个三重键，它确实是多重键。
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This energy level diagram helps us understand the relationship between electron filling and bond strength.
能级图能帮助我们,理解电子填充,和键强的关系。
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And we give different names, depending on what kind of electrons we're dealing with, so, for example, with h c l here, we can talk about having bonded versus lone pair electrons.
我们还起了不同的名字，给我们要处理的不同类型的电子，以氯化氢为例，我们来介绍一下成键电子与孤对电子。
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The way that we can figure this out is using something called bond order, and bond order is equal to 1/2 times the number of bonding electrons, minus the number of anti-bonding electrons.
我们可以用叫做,键序的概念来弄明白它，键序等于1/2乘以成键电子,数目减去反键电子数目。
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So why don't you tell me what the valence bond description would be of these carbon hydrogen bonds?
你们来告诉我,碳氢键的价,电子是怎样的？
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So we have 18 electrons, and the next thing that we need to figure out is how many bonding electrons we have.
那么我们有十八个电子，下一步要做的是判断,我们有多少个成键电子。
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And step four is going to have us figure out how many bonding electrons we have, so we have 16 minus 10, is going to be 6 bonding electrons.
而第四步需要判断,我们有多少个成键电子，那么我们有十六减十，也就是六个成键电子。
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That means that we need 12 minus 8, 8 or 4 bonding electrons in our structure.
这意味着在这个结构里，我们需要12减去,等于4个成键电子。
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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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So to figure out bonding electrons, -- what we take is that number 18, which is our total number of electrons we need to fill valence shells, and we subtract it from our number of valence electrons, which is 10.
那么为了找出成键电子，我们将十八，也就是填满所有价壳层,所需要电子的总个数，减去我们所有的价电子的个数，也就是十。
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Now we have 6 things around the nitrogen, and we have 8 around the carbon.
现在我们有六个成键电子在氮周围，有八个在碳周围。
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