Write acceptable Lewis structures for the following molecules: (a) H2 NNH2; (b) HOClO; (c) (HO)

Write acceptable Lewis structures for the following...

Key Concepts

Expanded Octet

Certain atoms, particularly those in the third period or beyond like sulfur, can possess more than eight electrons in their valence shell. This concept of expanded octet explains bonding in molecules where these atoms form more than four bonds, as seen in polyatomic ions and complex molecules.

Formal Charge

Formal charge is a bookkeeping tool used to assess the electron distribution in a molecule. It is calculated for each atom to help determine the most stable Lewis structure by minimizing the separation of charge and ensuring that the overall charge of the molecule is correctly represented.

Resonance Structures

Resonance involves the depiction of a molecule by two or more valid Lewis structures that differ only in the arrangement of electrons, not the arrangement of atoms. These structures contribute to the overall hybrid which better represents the actual electron distribution in the molecule.

Valence Electron Counting

This concept involves calculating the total number of valence electrons available from all atoms in a molecule. This count is crucial as it determines how electrons can be arranged around atoms in accordance with bonding and non-bonding pairs.

Lewis Structures

Lewis structures are diagrams that depict the bonding between atoms within a molecule and the lone pairs of electrons that may exist. They provide a simplified representation of a molecule's electron distribution, showing how valence electrons are allocated into bonds and non-bonding pairs.

Octet Rule and Its Exceptions

The octet rule states that atoms tend to form bonds until they are surrounded by eight valence electrons, achieving a noble gas configuration. However, there are exceptions, including molecules with an odd number of electrons, incomplete octets, or expanded octets, which occur especially with elements in the third period or beyond.

Transcript

00:01 Alright, so we're going to draw different loose structures for given compounds.

00:03 The first given compound is h2 and h2.

00:09 So this means that we have two nitrogen attached to each other, hydrogen is on the outside.

00:14 Now, i have the formal charge equation here in case i need it, and this is most likely going to be if we run out of electrons, or if we have something in 3p, 4p, 5p, or 6p orbital, so then we'll use this.

00:29 But nitrogen and hydrogen are not in any of those orbitals, so we're just going to use formal charge for this one.

00:36 You can to practice, but this is not going to need it.

00:40 We're going to do total number of valence elections.

00:45 So we have total four hydrants, two and two, plus two nitrogen's.

00:50 So four times one valence plus two times five.

00:55 That'll give me 10, that'll give me four.

00:57 That gives you 14.

00:58 We connect.

00:59 Every line has two electrons, so we use 246, 8, 10, 2, 4, 6, 8, 10.

01:09 We have 4 electrons left.

01:12 So i have 2, 4, 6, i need 2 here, 2 4x, i need 2 here, and that will give me 4, and that's it.

01:18 We check that hydrogen has 2 electrons, 2, 2, 2, 2, 2, and nitrogen has 8, 2468, and this molecule is stable.

01:26 You can't make double bonds because nitrogen can have more than 8 electrons around it, and hydrogen can't have double bonds around there.

01:34 So this will be the lose -out structure for a.

01:37 Let's go on to b.

01:46 So for b, we have h -o -c -l -o, and they're giving in the order, so it would be h -o -c -l -o.

01:54 We're going to do total number of valence electrons, so it will be one hydrogen, plus oxygen, 6, plus chlorine is 7, plus oxygen is 6, valence electrons, 6 plus 6 is 12, and then 1 plus 7 is 8 that will give you 20 connect 2 4 6 that will give you 14 so fill from outside in 2 4 6 i like to fill oxygens 7 8 9 10 11 12 13 and 14 all right so we have used all our valence electrons because we have a central atom that is in the 3p we should probably check for more charges right because this coin can have an expanded octet.

02:40 It means that it can have 8, 10, or 12 electrons around it.

02:46 So let's do formal charges.

02:49 So formal charge for hydrogen.

02:51 One, we're going to use the blue one.

02:52 So this is the one given to you.

02:54 This is the one that i like to use.

02:56 So one valence electron minus zero non -bonding electrons, minus one bond, zero.

03:02 Then i'm going to call this 0.

03:05 It's going to be six valence electrons, minus four non -bonding electrons, minus two bonds.

03:11 This is going to be zero.

03:13 And we're going to do the scoring right here.

03:16 This is 7 valence electrons, minus 4 non -bonding electrons, 1, 2, 4, minus 2 bonds, which would give you positive 1.

03:29 Not ideal.

03:30 Let's see if we can get it to 0 in a second.

03:32 It's going to be 02, and there's going to be 6 valence electrons minus 6 non -bonding electrons, minus 1 bond, which would give you negative 1.

03:43 Now, ideally, since this molecule doesn't have a charge, you want the whole thing to be zero, right? so as ideally what you want.

03:52 Since chlorine can have an expanded octet, what i'm going to do is i'm actually going to recalculate it.

03:59 And what i will do is i'm going to take and give chlorine more than 8.

04:05 And i'm going to do that because this one is not zero.

04:07 I'm going to take two from there and share it and see if that works.

04:11 And it does become kind of like a trial and error kind of thing.

04:14 All right, so now we're re -calculate cloying.

04:17 So we have seven valence electrons, minus four non -bonding electrons, minus three bonds.

04:24 That will give me formal charge of zero.

04:25 Looking good.

04:26 O2 will have six valence electrons, minus four non -bonding electrons, minus two bonds, and that will be zero.

04:34 So this is you lose that structure for b.

04:39 Okay? let's go on to c.

04:47 So for c, we have h -o -2 -s -o.

04:52 That means that the sulfur is in the metal, oxygen, and then oh, and then oh.

04:58 Right, so that too means that they belong to a sulfur, and this oxygen also belongs to a sulfur, but those two will come together.

05:05 So then we count a total number of valent electrons.

05:08 We have two hydrogens plus three oxygens plus one sulfur.

05:12 You can come from here.

05:13 So this is two times one valent electrons.

05:16 Oxygen has 6 valence electrons and sulfur has 6 so we have 24 electrons here plus 2 will give me 26 connect so because this is 0h it means that this hydrogen belongs to is oxygen and the same here so we have used 2 4 so each line represents 2 electrons 2 4 6 8 10 we have 16 electrons to use so i'm gonna start here 2 4 6 6 6, 8, 10, 12, 14, and then 16.

05:55 Because the software can have an expanded of z, it's in the 3p orbital, we should do formal charges to make sure that this is the most stable molecule.

06:05 So let's do formal charges.

06:07 So formal charge for the hydrogens, which look the same.

06:10 So it will be 1, valn's electron, minus 0, non -bonding electrons, minus 1 bond, it will be 0.

06:18 Zero, zero.

06:20 We're going to do 01.

06:21 We're going to call those two, since they're exactly the same, we're going to call them 1.

06:24 So 6 valence electron, minus 4 non -bonding electrons, minus 2 bonds, 4 more charge of 0.

06:32 So far, so good.

06:34 We have sulfur, 6 vaguelan electrons, minus 2 non -bonding electrons, minus 3 bonds, that will give you positive 1.

06:42 Not that good.

06:44 Let's see if we can bring it to 0 in a second.

06:45 Let's calculate oxygen 2.

06:46 So oxygen 2 will be 6 valence electrons, minus 6 not bonding electrons, minus 1 bond, so this is negative 1.

06:55 All right, so the idea, since this molecule doesn't have a charge, if we can get that to 0, then we should.

07:01 Since sulfur can have an expanded octet, it means that we can actually bring that to 0 by giving them more electrons.

07:10 Right.

07:10 Now you have to be careful of where you take the electrons from.

07:16 I'm going to take it from this oxygen, because this oxygen is by...

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