The connections between atoms in the amino acid histidine are shown. Complete the electron-dot structure by adding multiple bonds and lone pairs of electrons. Give the hybridization on the numbered carbon and nitrogen atoms. (LO 8.4)

(a) N1 = sp2, N2 = sp, C1 = sp2, C2 = sp3 (b) N1 = sp2, N2 = sp2, C1 = sp3, C2 = sp2 (c) N1 = sp3, N2 = sp, C1 = sp2, C2 = sp3 (d) N1 = sp3, N2 = sp2, C1 = sp2, C2 = sp3

Video transcript

Hello Everyone in this video, we're given this incomplete structure here and we're trying to fill in the missing multiple bonds as well as lone pairs. And we have to determine the hybridization of are highlighted atoms, which is this carbon here and this auction in red. So first we have to know that carbon likes to have four bonds and we only have two carbon atoms here. So we go ahead and give it a double bond. I'll do this in green, nitrogen likes to have three bonds and one lone pair here in the structure we see we have two nitrogen and each have three bonds already. So we can go ahead and add those necessary lone pairs to fulfill its octet. So again, I'll do this in this green color. We'll just add one lone pair to each nitrogen atom. And lastly we are dealing with auctions. The auction likes to have two bonds and two lone pairs. And we can see each of these auctions here only have a single bond. So we go ahead and add in our double bond. Now that it has a double bond, of course it needs, it's necessary to lone pairs to fulfill its octet. So again, we go ahead and add those lone pairs in as well. And now taking a look at this carbon here, we need to go ahead and figure out what the hybridization is. So for carbon here, do this in red. So carbon we can see has three bonding groups and that's just one carbon one hydrogen and one nitrogen. So those are the bonding groups. And because it has three bonding groups, its hybridization is going to be S. P two. So carbons hybridization is sP two, that's one of their answers. Next one will do in purple and that's for our auction atom here. You can see here, the auction has one bonding group And two lone pairs. So the bonding group would be this carbon here and then we have two long pairs. So it has one bonding group and two lone pairs. That's three groups total, and that's still an sP two hybridization. So that's going to be our second answer for this problem.