00:01
So for this question, we are given several different lewis structures and we're asked to find what is wrong with each structure and then to write the correct lewis structure.
00:15
So let's start with a.
00:18
So when we're looking at a, the octet rule is satisfied for each atom and hydrogen is only bonded once.
00:28
So let's look at the formal charge and see if that will.
00:32
Help us.
00:35
So if we calculate the formal charge for each atom, hydrogen has two valence electrons minus, or hydrogen has one valance electron minus the one associated electron from the one bond.
00:53
So one minus one is zero.
00:58
And then we'll do the formal charge for carbon.
00:59
So carbon has four valence electrons minus how many associated electrons.
01:08
So we have the two lone pairs, which would be two, and then three bonds.
01:14
So we'll have one associated electron for each bond.
01:18
So two plus three is five associated electrons.
01:21
So four minus five means carbon has a negative one formal charge.
01:27
And then nitrogen has five valence electrons, and then it has four electrons in the lone pairs and two electrons and bonds.
01:38
So four plus two means it has six associated electrons.
01:44
So five minus six is also negative one.
01:50
So that's going to be what's wrong with this lewis structure.
01:56
The sum of the formal charges is negative two, and this is a neutral molecule.
02:01
So we want that formal charge to be zero.
02:07
So let's rewrite this.
02:11
Now we want it to be zero, but we also want the octet rule to still be satisfied.
02:19
So let's turn these loan pairs into a triple bond.
02:26
So now we have the octet rule for carbon satisfied, and then we're going to add a loan pair to nitrogen, which if we're looking at the structure, the total valence electrons for the structure was going to be 10 valence electrons.
02:43
So we have four bonds, which is eight electrons, and then two from the loan pair.
02:48
So we have 10 valence electrons.
02:50
So that is satisfied.
02:52
Each atom, the octet rule is satisfied.
02:55
Now let's calculate the formal charge.
02:57
So hydrogen, one valence electron minus one associate electrons with zero.
03:03
Carbon has four valence electrons minus four associate electrons, minus four associated.
03:07
So it's formal charge of zero.
03:11
And then nitrogen has five valence electrons minus its five associated electrons, right? two from the loan pairs and then three associated electrons from the three bonds.
03:22
So five valence minus five associated electrons means nitrogen has zero as its formal charge.
03:30
So this is going to be your correct lewis structure because the formal charge.
03:38
For each atom is zero.
03:41
Next we'll move to b.
03:44
So what is wrong with this lewis structure? so you should notice that hydrogen is double bonded to carbon and hydrogen can only have two valence electrons.
03:58
So it means it can only have a single bond.
04:01
It cannot have a double bond.
04:04
So let's go ahead and rewrite this lewis structure with hydrogen having a single bond to carbon on each side.
04:14
Now, with this structure, carbon's octet rule is not satisfied.
04:23
So we could either add a loan pair to each carbon, or we could add a triple bond in the middle.
04:32
Now, in order to figure out if there's a triple bond or two loan pairs, we know that this structure has a total of valence electrons of, so two valance electrons from the hydrogens, four valance electrons for each carbon.
04:49
So it has a total of 10 valence electrons that we have to include in the structure.
04:53
So if we write a triple bond, we have five bonds with two electrons each, and that gets us to our 10 valence electrons.
05:03
If we had written, say we instead of that triple bond, we had given each carbon alone pair, then we would have 12 valence electrons, which is too, too many.
05:22
So that's how we know instead of the lone pairs, we have a triple bond between the carbons.
05:29
So we have 10 valence electrons, the octet rule for the carbons is satisfied, and each hydrogen has a single bond.
05:39
So this is going to be our correct lewis structure.
05:42
We'll move on to c.
05:47
So what we can immediately notice is that the octet rule is not satisfied for any of the atoms, right? the oxygen only has six balance electrons.
06:00
The 10 only has four, right, two bonds.
06:07
So let's write the correct lewis structure with satisfying the octet rule for each atom.
06:17
So we'll write the skeletal structure as is.
06:20
So first let's figure out how many valence electrons that we have for the total structure.
06:25
So oxygen has six valence electrons, and we have two oxygens.
06:30
So we have 12 valence electrons from the oxygen.
06:34
10 has four valence electrons.
06:36
So we're going to have 16 total valence electrons for this structure.
06:44
So let's start adding more electrons.
06:48
So we have two bonds.
06:49
So that's four.
06:50
So 16 minus the four electrons we've already included, means we still have to add 12 electrons.
06:58
So if we add those 12 electrons as lone pairs for the oxygens, right? we add six electrons for each oxygen.
07:07
The octet rule for oxygen would be satisfied, but the octet rule for 10 would not be satisfied.
07:14
So let's do a double bond between 10 and oxygen.
07:23
So right now, tin's octet rule is satisfied.
07:29
Then we add two lone pairs to east oxygen to satisfy their octet rule.
07:35
So now the octet rule for each atom is satisfied.
07:41
And then we have four bonds, so that's eight electrons, and four lone pairs.
07:48
So that's eight electrons from the lone pairs, eight electrons from the bonds, gives us our 16 balance electrons for the structure.
07:58
So we know that that's going to be our correct lewis structure.
08:03
Next, we'll move on to d.
08:07
So it looks like for d, the octet rule is satisfied for each atom.
08:14
So let's look at its formal charge next.
08:18
So all the fluorines are the same.
08:22
They each have the same number of balance electrons.
08:24
They have the same number of lone pairs.
08:27
And bonds, so they have the same number associated electrons.
08:30
So if we figure out the formal charge for one fluorine, the formal charge for the other fluorines is going to be the same.
08:37
So fluorine has seven valence electrons and it has seven associate electrons.
08:48
We have six from the loan pairs and one from the bond.
08:52
So seven minus seven meet zero.
08:55
So fluorine is going to have a zero formal charge, which means each flore and he's going to have a zero formal charge.
09:01
Let's figure out the formal charge for boron.
09:04
Boron has three valence electrons.
09:07
And then as it's written in the structure, it has two electrons in a loan pair and three bonds.
09:15
So two plus three means that's five associated electrons.
09:20
So the three valence minus the five associated means that boron is going to have a negative two charge.
09:30
Now, we want this formal charge to be zero, which means we want the formal charge for each atom to be zero, which this is not zero.
09:40
So that's why this is incorrect.
09:43
Also, if there is going to be a negative charge, it's going to be towards the most electronegative atom, which would be fluorine, right? because boron is less electronegative than fluorine.
09:56
So let's try to rewrite the structure where boron has, as a zero formal charge.
10:05
So the fluorines already had a zero formal charge.
10:10
So we're going to keep them the exact same because we want that formal charge to say zero...