Lewis Dot Structures: Ions

Jules Bruno
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now recall that in a cat ion we have the loss of electrons by an element and then an ion. We have the gaining of an electron now, when lewis dot structures were going to say cat ions as a result, have less valence electrons, whereas an ions have mawr valence electrons here in this example question, it says, determine the formal charge off the nitrogen atom in the following ion. So here we have our nitrite ion. Now it doesn't matter if we deal with a neutral compound or if we deal with a one with a charge. We follow more or less the same rules in order to draw the lewis dot structure. So here we follow Step one, it says we need to determine the total number of Valence electrons off the structure. It's N o to minus, and we need to remember that Valence electrons equals group number of the element. Nitrogen is in Group five. A oxygen is in Group six A. We have one nitrogen, which is five valence electrons. We have to oxygen's. Each one is six Valence electron, so that's 12 combined. And then we realize here that minus one charge means that we have one extra electron. So what we have here is we have 18 total valence electrons for the nitrite ion. Step two, we're gonna place the least electro negative element in the center and connect all elements with single bonds here. Exceptions to this hydrogen never goes in the center. We don't have to worry about that. There's no hydrogen in this. Ion Callejon's the elements In group seven, A Onley makes single bonds as a surrounding element. We don't have any Hodgins either, so we don't worry about that. Nitrogen is less electoral negative than oxygen. So nitrogen goes in the center. Then we're gonna have our oxygen's here. We're going to connect our central element to the surrounding elements by single bonds. Realize here. That means that we have used a total of four valence electrons because each single bond has to valence electrons involved now here at electrons toe all the surrounding elements until they have eight electrons because we want to follow the octet rule exception again. Hydrogen on. Lee wants to electrons around it. So we have We have here two electrons in this bond. 345678 We have to valence electrons here. So 34567 and eight. So now we have eight total electrons around both oxygen's. So so far that's given us 16 total electrons used, meaning we have two electrons remaining that are unused. What do we do with them? Well, step four says we place any remaining electrons on the central atom, so we're gonna place them on nitrogen. When we do that, we have zero valence electrons left. Here's our issue, though. If any elements don't have eight octet electrons add double and triple bonds between them. So here the nitrogen has around it. 246 electrons. It needs two more to get to the octet rule. So what we're gonna do here is we're going to take from either oxygen. Doesn't matter. We're gonna take two electrons and use them to make a double bond here. So now oxygen has eight electrons around it, and the oxygen also has eight electrons around it. It's still using or sharing those electrons that were used to make that double bond. Now here, the formal charge could be used to determine if the Lewis dot structure is correct. Here. We need to determine the formal charge of the nitrogen within the structure. And one last thing we need to realize here is that because we're dealing with an eye on, we have to place the ion in brackets and it's charged in the top right corner. So here we're going to say, nitrogen. It's in Group five A. Because, remember, it's group number minus bonds at the Element is making its making three bonds, plus the number of non bonding electrons that it has. It has won two non bonding electrons, so that would come out to five minus five. So this equals zero for its formal charge. So the nitrogen Adam here would have a formal charge of zero because this is an ion, we place it in brackets with the charge on the outside. So here this would be the lewis dot structure for our nitrite ion