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So let's figure out the bond order for our two molecules here that we figured out the electron configuration for.
让我们看看这里,两个分子的键序是多少。
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/2 So the bond order is going to be equal to 1/2, and then it will be 2 minus 2.
它的键序等于，然后乘以2减去2。
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And you can go ahead and tell me what you think the bond order is going to be for this molecule.
你们告诉我你觉得,这个分子的键序应该是怎样的。
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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 what we see is a bond order of 0, and again, the bond is very, very weak.
我们看到键序是0，同样的，键非常非常的弱。
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So we know that it's 1, because we have 1, 2, 3, 4 bonding, minus 2 anti-bonding, and 1/2 of that is a bond order of 1.
我们知道它是，因为我们有1，2，3，4个成键，减去2个反键，它的一半就是键序为1。
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For hydrogen our bond order is going to equal 1/2, 2 minus 0.
对于氢原子键序等于1/2，2减去。
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And the bond order you get out will either be, for example, zero, which would mean that you have no bond, 5 or you could have 1, a single bond, 1 . 5, a 1 and 1/2 bond, 2, a double bond, and so on.
你得到的键序要么是比如说是零，这意味着没有键，或者你会得到1，单键1。,1又二分之一键，2，一个双键，等等等等。
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So what I want to tell you is we also always get the same bond order if we instead only deal with the valence electrons.
我想要说的是我们如果,只考虑价电子也可以得到相同的键序。
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So let's just prove that to ourselves and figure out the bond order just using valence electrons.
让我们证明一下这一点,来看看只用价电子算出键序。
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/2 So this would mean the bond order is equal to 1/2, and in terms of valence electrons, how many bonding valence electrons do we have?
这意味着键序等于，对于价电子，有多少个成键价电子？
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So for the bond order we want to take 1/2 of the total number of bonding electrons, so that's going to be 4 minus anti-bonding is 4, so we end up getting a bond order that's equal to 0.
键序等于1/2乘以,总的成键电子数，也就是4，减去反键电子数，也就是4。,所以最后得到键序为0。
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What kind of a bond is a bond order of 1?
什么样的键键序是1？
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So we would predict a bond order of 1.
所以我们预测键序是。
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What would the bond order be for this bond?
这个键的键序是多少？
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So our bond order for h e 2 is going to be equal to 0.
所以He2的键序等于0。
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So what I want to point out with this case in beryllium is that you don't have to use all of the electrons to figure out the bond order, and in fact, once you get to molecules 10 that are from atoms with atomic numbers of 8 or 10, you're not going to want to maybe draw out the full molecular orbital diagram.
我要指出的是,在Be这种情况下,你不需要利用,所有的电子来指导键序，实际上，一旦分子中,原子序数到达了8或者，你也许不想画出,整个分子轨道图。
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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 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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