The following three Lewis structures can be drawn for N2O:

(a) Using formal charges, which of these three resonance forms is likely to be the most important?

Video transcript

Hello everyone today, we are being presented with the following problem and asked to solve for it. So we have three possible lewis structures that can be drawn as shown below for oxygen cyanide based on formal charge alone, determine the most important resident structure for oxygen cyanide. So by most important this is implicating that the negative charge. So the negative charge is on the most electro negative atom. So it's the most electro negative atom. If we recall our periodic trend electrode negativity increases as we go from as we go up the periodic table and to the right meaning that in this case oxygen is the most electoral negative atom. So it's most electro negative in this example. And so we want to look and see where this negative charge outside the brackets here plays onto each of these atoms. And to see if it's on the oxygen atom. And so we're gonna go one by one. So for the first one we're going to calculate the formal charge for oxygen. And that equation is going to be simply the number of valence electrons, oxygen is in group six has six valence electrons where they're going to subtract the number of lines we see attached to that oxygen. So three and they're going to subtract the number of electrons or dots that we see around that oxygen. And so that's two. And so that leaves us with a formal charge of positive one, we then move on to the carbon. It's the same procedure. Carbon is in group four. So it has four valence electrons minus how many lines are bond lines we see around it and we see four individual bond lines and we see zero electrons. So that has a formal charge of zero. We then move on to the nitrogen. Lastly nitrogen is in group five. So it of course has five valence electrons. It has one bond line around it. It has six individual electrons surrounding it, giving us a formal charge of -2. And we said that the most electronic of atom here is oxygen and oxygen, oxygen's formal charge here is greater than that of the nitrogen. Therefore it can't be this first structure here. Moving on to the second structure, we're going to calculate the formal charge once again. So we're gonna take our six valence electrons subtract our to bond lines by our four electrons. We're gonna get zero for the carbon, we're going to get our four valence electrons minus the four bond lines minus zero electrons. Another zero. And then for nitrogen we're going to have our five valence electrons minus r two bond lines subtracted by four electrons, giving us a formal charge of negative one. Once again, nitrogen is not the most electro negative atom here. Therefore it cannot be structure. To lastly we're gonna calculate the formal charge for the different atoms in our last structure. So for oxygen we have six valence electrons minus our one bond line minus our six individual electrons here giving us a formal charge of negative one. For our carbon it's going to be the four valence electrons minus the four bond lines minus the number of electrons we see around it, which is zero. And then for nitrogen, lastly, we're going to have our five valence electrons minus r three bond lines minus our two individual electrons that we see giving us zero. And notice here that the most electro negative atom oxygen has the most negative charge. And so therefore our answer is going to be structure number three, I hope this helped, and until next time.