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1·A coordinate covalent bond is one where both bonding electrons are from one of the atoms involved in the bond.

配位共价键是其中两个键合电子均来自该键所涉及的原子之一的键。

2·Step six is thinking about do we have any bonding electrons left?

第六步是想一想我们有多少个成键电子剩下？

3·And step six asks us, well, do we have any bonding electrons left?

而第六步问我们，好，我们还有成键电子剩下吗？

4·So what we need to do is fill in those extra bonding electrons into our bonds.

因此我们需要，把这些电子也填到键中。

5·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。

6·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。

7·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。

8·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。

9·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。

10·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。