a. Draw a Lewis structure for each of the following:
b. ...

Question

a. Draw a Lewis structure for each of the following:

b. Draw a structure for each of the species that shows approximate bond angles.

c. Which species have no dipole moment?

1. CH₃N₂⁺

Accepted Answer

Hey, everyone, let's do this problem and says for the following species, one draw an appropriate Lewis structure to draw structure that shows approximate bond angles. And three does it have a zero dipole moment? So first, we're drawing a Lewis structure. So let's review how to do that. So first, when drawing a Lewis structure, we want to count the total number of valence electrons for each atom. And if the compound has a positive or negative charge, we need to add or subtract a value to account for those charges. So if we have a negative charge, we add one, if we have a positive charge, we subtract one, right. Alright, then we're going to draw our Sigma bond framework that basically just shows all the connectivity of the atoms using only single bonds, Sigma bonds, then we might have some electrons left over and some atoms might not have their octet filled. So we are going to use the lone pairs to complete the octet. And if we run out of electrons to add, we can form some double bonds or triple bonds if needed. Alright. And lastly, we need to figure out if any of our atoms have a formal charge. This will be a hint based on what the charge of the overall compound is, we basically just have to find which Adam that charge belongs to. So for formal charge, we calculate this using the number of valence electrons for that atom minus the number of non bonding electrons or loan pairs minus the number of bonding electrons Divided by two. Or that would also be the number of bonds, right, since there are two electrons in a bond and we are dividing the number the bonding electrons divided by two. So that would also be this whole term would be the same as the number of bonds. I also like to lump these two together and call them touching electrons because we're counting the number of electrons that are touching that specific atom. So let's apply this to our compound. So counting the number of valence electrons, we have four for carbon two, for two hydrogen, for for another carbon one, for another hydrogen, 10, for two nitrogen atoms. And we have a positive charge. So we're going to subtract one. So adding that up, we get a value of 20 valence electrons. So drawing our Lewis structure from left to right, we have CH two C H in two. So in, in and that is it. So how many electrons did we use for this? 2468, 10, 12, so 20 minus 12 valence electrons means we have eight left over or that will be four lone pairs, right, two electrons per lone pair. So let's add our lone pair to some of our atoms that need some more electrons. So this carbon over here needs some more electrons. It only had 246. So we'll add a lone pair there this in our nitrogen czar lacking electrons. So we'll add our other lone pairs, the three lone pairs to those nitrogen, but it's still not satisfied, right? This second carbon that's attached to the nitrogen, its octet is not complete. Our end nitrogen definitely does not have enough electrons. So let's form some double bonds or maybe triple bonds. So we can form a double bond between these two carbons with this lone pair on the carbon. And these nitrogen can actually form a triple bond with those two lone pairs coming from the inner nitrogen to give us our final Louis structure. Okay. So now all atoms have eight total electrons, but we know that the compound has a positive charge. So where does that positive charge go? Well, let's calculate the formal charge on this nitrogen right here our inner nitrogen. So nitrogen likes to have five valence electrons. But how many are touching it? One from each bond? Right? 1234. So five minus four is plus one. So there is a positive formal charge on that nitrogen. So this is our Lewis structure for part one. So part two says draw a structure that shows approximate bond angles right. This structure, it's all pretty linear and 90 degree angles, 101 180 degree angles. But that's not necessarily true. It depends on the hybridization of the central atoms. So lets redraw our compound looking at the hybridization. So starting from left to right are end carbon has charge centers. So that's going to be S P two hybridized. So that will make the hydrogen carbon hydrogen bond angle 120°. the hydrogen carbon, high carbon hydrogen carbon carbon Bond angle also 120°. So let's draw that. So carbon hydrogen, hydrogen carbon and now let's make our next carbon in different color. We'll do blue. So these angles are 1 20 right? That's what we drew over here. So now our next carbon also has three charge centers, 123. So that will also be sp two hybridized. So those bond angles will also be 120°. So this is a double bond and then we have our hydrogen in our nitrogen. So that means our nitrogen, carbon, hydrogen bond, nitrogen, carbon hydrogen, that is 1 20 our nitrogen carbon carbon bond right here is 1 20 and our hydrogen carbon carbon bond is 1 oops 1 20. Alright. And then lastly, we have our nitrogen and that has to charge centers around it. So that's going to be sp hybridized, which means it'll be linear, right? When we have a triple bond, The bond angle is 180°. So this blue was SP two and our nitrogen with the positive charge or the nitrogen, nitrogen carbon bond right Is going to be 180. So that is all the bond angles For this compound. And the last question says, does it have a zero Die Paul moment? So where are their die poles in the structure? Let's redraw it really quick, technical difficulties here. So we know that this as a group or the end 2-plus group is an electron withdrawing group. So that is going to create a di poll pulling down, right? It's going to pull those electrons down this way, but there are no other die poles throughout the compound, right? And we would have to have one that is equal and in the opposite direction in order to create an overall zero dipole moment for this compound. But since this is the only die pole, that means the compound itself does not have a zero dipole moment. So the answer for part three is no. So those are the correct answers for this problem which makes d the correct choice. That's it for this one. I hope this video is helpful and thanks for watching.

a. Draw a Lewis structure for each of the following:
b. Draw a structure for each of the species that shows approximate bond angles.
c. Which species have no dipole moment?
1. CH₃N₂⁺

Key Concepts

Lewis Structures

Molecular Geometry and Bond Angles

Dipole Moment

Watch next