00:01
For the problem we're going to consider next, we are given the information that some chemists believe that satisfying the octet rule is the primary objective.
00:11
This should be the top criterion for determining the dominant lewis structure.
00:15
Other chemists believe that finding the best formal charge combination in the lewis structure should be dominant.
00:23
We are going to be considering the dihydrogen phosphate ion, and we're told that the h bonds to the o.
00:33
We're going to draw the structures because we're asked to determine a, best for the octet rule.
00:49
B, we're asked to find which one would be best considering formal charges, and c, any other options.
01:04
I'll do these on the next page.
01:08
So i'm going to start by drawing the three structures i could think of.
01:15
Fyi got a little help on my third.
01:17
So i'm going to draw all three structures.
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First, i determined, i'm going to write this up here again.
01:26
Was it p .o4 or p .o .3? a p .o .4.
01:29
I determined that this has 32 valence electrons.
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I like to always write that down so i can make sure i've done it right.
01:41
So for my first structure, i had nothing but single bonds.
01:46
So i'm going to start with my p in the center.
01:59
I'll fill in my other electrons later.
02:02
For my second structure, i had one double bond.
02:09
Of course, this one would exhibit resonance.
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And for the third one, i had four double bonds.
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Now i'm going to fill in the rest of the electrons.
02:43
For this one, each of my oxygens bonded to a hydrogen had four unshared.
02:51
And these other two, of course, each had six.
02:56
You can count these if you'd like.
02:59
Here again, and this again has six.
03:04
Well, my double bonded only has two pair.
03:09
Here, i only needed one pair of electrons, and each of my double -bonded oxygen, i only need two.
03:18
Now i'm going to, since these have charges, i'm going to indicate the charges.
03:30
There we go.
03:32
Next, we're going to do formal charges.
03:37
I'm going to go back to the other page and make a quick notation if i have room.
03:40
I do.
03:41
So for formal charges, a double -bonded, what was my substance here, a p, just double -bonded, to the o will be the same no matter what.
04:02
A p single bonded to an o, that o will be the same no matter what.
04:07
Or that o will be the same for each of these.
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And my h will be the same.
04:14
You'll know what i mean in a moment.
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So let's start.
04:16
I like to start with p's and our p always has five valence electrons.
04:30
And we're going to count our unshared pairs of electrons.
04:34
Well you can look at all of these.
04:35
You can see that they all have zero unshared pairs of electrons.
04:40
On our first we have 246, 8, so 8 divided by 2.
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My first one is going to have 8 plus 1.
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My second one is 246, 8, 10, 10 divided by 2.
04:57
The second one we have a 0.
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And in the third we have 246, 8, 10, 12, 14, 16, which equals 8.
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Boy, i didn't even know if i figured that one out.
05:12
It's so big.
05:13
That would be a plus 3.
05:15
24, 6, 8, 10, 12, 14, 16, is 8...