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If these bonds were all completely of equal distance apart, whether is was a lone pair or bonding electrons, 5° the angles would be 109 . 5 degrees.
如果不管它是孤对，还是成键，它们等距分开的话，键角是109。
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But because there's this lone pair here, it's pushing down on the other bonds, 5° so we end up with an angle of less than 109 . 5 degrees.
但因为这里有孤对，它会把其它键向下压，所以键角会小于109。
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They want to get as far away from each other possible, the ideal angle is 120. But what we have here is a four-membered ring, so what angle does 90° that have to be, that bond? 90 degrees.
它们想要尽量远离彼此，最理想的是形成120°键角,但现在是个四元环,所以这键角应该多大？
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This is the easiest question all day, what is the bond angle between all of these?
这个问题很简单，键角是多少？
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The bond, it turns out, 5° is 104 . 5 degrees, that h o h bond.
这个键即，ＨＯＨ键角，是104。
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So again, if we think about that shape of that carbon atom, it's going to be trigonal planar, 120° it's going to have bond angles of 120 degrees, because we have this set up of having three hybrid orbitals.
如果我们考虑碳原子的形状，它是平面三角形，键角是，因为我们有这三个杂化轨道。
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So, in a settling, what is the bond angle here?
这里的键角是多少？
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What do you know the bond angle should be?
你们觉得键角是多少？
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Ok So, according to this model, 90° what we're seeing is a bond angle of 90 degrees.
我听到不同的答案，OK，，I，hear，a，mix。,根据这个模型，我们看到键角是。
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So if we still have an angle of a 109 . 5 degrees, and again, we still have four unpaired electrons available for bonding, we can make one of those bonds with another s p 3 hybridized carbon, so we're going to make up one pair here.
如果键角仍然是109。,同样，我们还有4个未配对的电子可以用来成键，我们可以用其中的一个,和另外一个sp3杂化碳原子成键，这样我们可以组成一对。
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So if we do this and we form the molecule ammonia, let's switch to a clicker question, and have you tell me what the bond angle - is going to be in ammonia -- HNH Actually, let me draw it on the board as you look -- actually, can you put the class notes on, since you don't actually have your notes to refer to.
如果这样做的话，我们就可以形成氨分子，让我们来做一个课堂练习,你们告诉我氨分子中的键角是多少-,键角，the，h，n，h，bond，angle。,实际上让我在黑板上画出它来，这样你们可以看到实际上-,你能把课堂讲义放出来吗，因为你们没有讲义可以参考。
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