Lewis Dot Structures: Ions

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

So here it says an ionic compound contains a cat ion, which is a positive ion connected to an anti on, which is a negative ion here. We need to draw the lewis dot structure for the Ionic compound of sodium nitrate. So here, what we're gonna do is step zero. We're going to break up the Ionic compound into its two ionic forms. In this case would break up into an A plus and n 03 minus. Now, for the Poly Atomic Ion, which is the nitrate ion, we're gonna follow steps 127 to draw its lewis dot structure. So here, if we want to draw this, we're gonna say that nitrogen group five days. So it has five violence. Oxygen's in group six A. And there's three of them and then minus one means we gained an outside electron. So we're gonna add one more electron to this giving us 24 valence electrons. We're gonna place the least electro negative element in the center, which is nitrogen. It's gonna form single bonds with our three oxygen's. Remember, the surrounding elements have to fulfill the octet rule, so make sure you add enough electron, so that they have eight. At this present moment we have, we've used all 24 of our electrons, so we have none left. But we run into an issue because nitrogen on Lee has six electrons around it. Remember, if an element is not fulfilling the octet rule, just take one of the lone pairs and use it to make either a double bond or triple bond. Nitrogen requires two more to get to the octet rule. So we're just gonna take one of these lone pairs and draw it here so that we have now eight electrons around nitrogen. Now, step eight, we're gonna place both ions near one another because opposite charges attract. Okay, so here we put this in brackets because ions are placed in brackets and then you would just place the sodium ion nearby. You could put it right here to the left of it. So this illustration will be the way to depict the ionic compound of sodium nitrate