Add electrons to complete the following Lewis structures. In each case, the charge is placed as

Add electrons to complete the following Lewis structures. In...

Key Concepts

Octet Rule

The octet rule is a guideline that atoms tend to bond in such a way that they have eight electrons in their outer (valence) shell, mimicking the electron configuration of noble gases. This rule helps in determining the placement of electrons but also has exceptions in species with an odd number of electrons or electron-deficient atoms.

Electronegativity and Charge Distribution

Electronegativity—an atom's ability to attract electrons—plays a key role in determining where charges are best placed within a molecule. More electronegative atoms are preferred sites for carrying a negative formal charge, while less electronegative atoms are better suited for carrying a positive charge. This concept ensures that the atomic charges in the Lewis structure are placed in locations that facilitate maximum molecular stability.

Formal Charge

Formal charge is calculated by comparing the number of valence electrons an atom has in its free state with the number calculated from its assignment in the Lewis structure (counting each bond as splitting electrons equally). Minimizing formal charges in a structure generally leads to a more stable molecule.

Lewis Structures

Lewis structures are diagrams that represent the bonding between atoms of a molecule and illustrate the placement of both bonding and nonbonding (lone pair) electrons. These diagrams are a basic tool for visualizing the molecular structure and the distribution of valence electrons among atoms.

Valence Electron Count

The valence electron count involves summing up the valence electrons from all atoms present in a molecule. Adjustments are made for ions by adding electrons for negative charges and subtracting electrons for positive charges. This total is used as a basis for distributing electrons in the Lewis structure.

Transcript

00:01 Okay, so for this question, they want us to draw the lewis structure and then assign the formal negative charge to the correct atom.

00:08 For part a, we have h3co minus.

00:12 So, h3co minus.

00:15 Now, the first thing that we have to do is calculate how many electrons we're going to need to draw.

00:20 So we can do that by calculating the valance electron for every single atom present in the compound.

00:26 So we have three hydrogen atoms.

00:28 Hydrogen belongs to group 1a.

00:30 This means that they have one balanced electron.

00:33 Carbon below, we have one carbon.

00:35 Carbon belongs to group 4a.

00:36 This means that they have four valance electrons.

00:39 And then we have one oxygen atom.

00:41 An oxygen belongs to group 6a, so we have six valance electrons.

00:46 And then we got to be careful here.

00:47 There's a minus charge here.

00:48 This means that we have to add one more electron, right? so if it's a negative charge, that means you add an electron.

00:55 If it's a positive charge, that means you remove an electron.

00:58 Because an electron is a negative charge.

01:02 So let's calculate how many uses this.

01:03 So this is 10 .14.

01:05 So there is 14 electrons.

01:08 So let's just draw a carbon here.

01:11 So and then hydrogen coming off.

01:15 So usually carbon is most likely always decentralized.

01:18 So you can always draw carbon and then there's branches coming out of carbon.

01:23 And then we have oxygen coming off.

01:25 So carbon has four bonds in order to have a complete octet.

01:28 That's one of the rules.

01:30 Drawing a lewis structure.

01:31 There are obviously exceptions like hydrogen where they only need two electrons instead of eight and there's like elements like sulfur phosphorus xenon which can exceed the octet but for this part we don't have to worry about that so there's four bonds each bond has two electrons right so that is eight electrons that we've used so we have six left now we have to put one two three four five six now we have no electrons left and they want us to determine the formal charge.

02:09 So in order to determine the formal charge, all we got to do is take the balanced electrons and subtract it from the number of electrons around the atom and subtract the number of bonds around the atom.

02:23 So carbon has four valance electrons and we see that there's four bonds around carbon, so the charge is zero.

02:30 Now oxygen has six valence electrons and we see that there is six electrons around the oxygen.

02:35 And then there is a bond around oxygen, so we minus one, so we get negative one, so it's negative charge on oxygen, and hydrogen are zero charged.

02:48 So we know that there is a negative charge on the oxygen, so that is the answer to part a.

02:56 Now for part b, the compound is nh2 minus, so nitrogen, we have one of them.

03:05 It's group 5a, so it's five valance electrons.

03:08 We have two hydrogens and hydrogen is 1a, so it's one balance electron plus one, remember, because it's a negative charge.

03:17 So we get 8.

03:24 Now we draw nitrogen in the middle, and then we have two hydrogens coming up.

03:31 So so far there's two bonds, so we've used up four of them, and we have four left.

03:36 Now we add one, two, three, four onto the nitrogen.

03:40 We have zero left.

03:41 Now let's calculate the formal charge.

03:43 So nitrogen is five valance electrons.

03:49 Right now we have four electrons around the nitrogen, so minus four, and then we have two bonds.

03:55 So minus two, and we get negative one, and the nitrogen has a negative one charge.

04:02 Now for part c, it's cn, so cn minus, so carbon has four valance electrons, nitrogen has five valance electrons, plus one, because of the negative charge and we get 10.

04:21 Now we have carbon, nitrogen, right, one bond.

04:27 So we use one.

04:28 So there's eight left.

04:33 We have eight left.

04:34 So let's three more bonds on here.

04:40 All right.

04:40 So now we have four left.

04:42 So when you draw lewis structures, it's all about drawing a structure, seeing if it works.

04:46 If it doesn't, then you change something about it.

04:49 So i'm just going to put three bonds here for now.

04:52 And now we have four electrons left.

04:54 Let's put two on nitrogen and two on carbon.