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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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All right, so the bonding order, you're correct, should be 2, if we subtract the number of bonding minus anti-bonding electrons and take that in 1/2.
好，你们是对的，键序为，如果我们用成键数,减去反键数除以2。
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So each respectively, this is sorted 1 and 2 respectively 4 and 5, this is sorted but clearly they are not in the right order.
分别地，1磅和2磅的是有序的，同样4,4磅和5磅的也是有序的,但显然它们的顺序并不正确。
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It's going to weigh too much, we don't need to look at it. But it'll still be order 2 to the n.
我们没必要去看它,但是它还是要遵循2到n的顺序。
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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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PROFESSOR: And I didn't write up there but it is one, 1 and we can see that it's 1, 2 because it's 1/2 of 2, 4 minus 2, so 1/2 of 2, the bonding order is going to be equal to one.
教授：我没有写出来，但是是1，我们可以看到为什么是，因为它等于1/2的2，4减去，所以1/2的2，键序等于1。
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So what I want to point out is 3d2 what we said now is that the 3 d 2 is actually lower in energy, so if we were to rewrite this in terms of what the actual energy order is, 3d2 4s2 we should instead write it 3 d 2, 4 s 2.
所以我们想指出的是,我们现在所说的是,实际上能量比较低，所以如果我们重新的,写出实际的能量顺序，我们应该写出。
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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 what's the bonding order for c 2?
它的键序是多少？》
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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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So let's talk about what the bonding order is going to be for c 2.
让我们讨论一下C2的键序是多少。
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If you substitute it all in, you get basically order 2 to the n.
得到的是2的n次方个基本问题，指数级的，这是个问题，好。
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