Resonance Structures

Now, with residents structures, we're gonna say a set of two or more valid lewis dot structures for Polly Atomic ions possessing at least one pie bond. Now we're gonna say, in a resident structure, we have the movement of Onley electrons from either a pie bond or loan pair. Here, this depicts two different ways that we can show our nitride ion here. The pi bon could be either on the left oxygen or it could be on the right oxygen. Both are valid depictions of the Nitrite island now here to show resident structures. We utilize double sided arrows there used to show that the resident structures are equal or equivalent with each other. At this level of chemistry, we're going to say that their equivalent. But once you get the higher levels of chemistry such as organic chemistry, your later learn that not all residents structures are equivalent to one another. Now we're gonna say the rial structure is represented by the composite off the resident structures and it's called the Residents Hybrid. Now the residents hybrid is the composite of all major residents structures. When we say composite, we really need an average of the two resident structures to draw the residents hybrid. We place a dotted line anywhere a pie bond has been. So there was a pie bond on the left oxygen. So we do a dotted line there. There was a double bond on the right oxygen. So we show a dotted line there. So this would represent the residents hybrid off our nitride ion.

here, it says, determine the remaining resident structures possible for the carbonate ion, which is C 03 to minus. So here we have our carbonate ion. We have our pie. Are double bond on the oxygen on the top. Realize here that the oxygen that are single bonded have formal charges equal to minus one. So they're depicted that way. And we have to remember to carry over those formal charges of minus one for any other oxygen's that are single body. So remember in residents were just moving around the electrons from the pi bonds. So wait, another resident structure I could show is, what if it's not the top oxygen that's double bonded, but it's the one on the left. Then we notice here that the double bonded oxygen has two lone pairs. So here's two lone pairs and the single bonded oxygen's have three lone pairs. Remember the single bonnet Oxygen's have formal charges of minus one, and here we put it in brackets because it has a charge. But remember, they're connected with double bonds. But who says that the oxygen on the left side is double bonded? What if it's the oxygen on the right side that's double bonded. The ones that are single bonded, remember, have three lone pairs. The one that's double bonded has to remember the single body Oxygen's each. Have a formal charge of minus one. Put it in brackets with the charge on the outside. And remember, these double sided arrows are what connects these resonance structures to one another. So just remember, we're just showing different places. The Pi Bon could have been within each of these structures. Each of them represents a residence structure for the carbonate ion.

here, we say that the average charges the charge and element possesses from the overall charge off all its residents. Structures here, it says, determine the average charge of the oxygen atom within the phosphate ion. So phosphate ion basically has phosphate double bonded toe one oxygen, but then single, bonded, tow all the others. It gives us four resident structures Now, if only given the molecular formula, then draw one of the residents structures and have given multiple resident structures again. Just choose one here. We're going to calculate the formal charge for the elements and add them to determine their overall charge. So we're gonna focus on one of them. Let's focus on this one. Remember, formal charge equals group number, minus the bonds. The element is making, plus each non bonding electron calculated or added individually. So let's say that the double bonded oxygen, oxygen A and all these single bonded ones are oxygen. Be so Oxygen A is in Group six A. It's making to bonds, and it has four electrons that are not bonding, so its formal charges zero the oxygen be. They are in group six A as well. One of them is making one bond, and it has six non bonding electrons. So this one is zero, and these are all negative one. So now we're gonna add up all the formal charges together. So zero plus negative one plus negative one plus negative one equals negative three for the overall charge, Step four says, divide the overall charge by the total number of those elements. So we have a negative three overall charge and we're dealing with four oxygen's. So each of those oxygen's would have a charge of minus 3/4 so we'd say their average charges is minus 3/4. It's a simple is that draw the molecule correctly, Then you're going toe. Figure out the total overall charge of all of the elements in question and divide by the total number that exist. That gives us each other individual or average charges.