Look at the electron-dot structures of the following molecules and ions: (b) Which can behave as a Lewis acid? Which can behave as a Lewis base?

Video transcript

Hello everyone. So in this video I want to classify the molecules or ions as either being lewis acids or louis spaces. So you can recall for lewis acids, these are just electron except ear's. And then we also have Lewis bases. And this is defined to be electron donors. All right. So these two are the definitions of what lewis acids and lewis bases are. So starting off with the first one we have H three C plus. So we see that on this ion we have a positive charge. Whenever we have a positive charge, it indicates that we have an electron deficient adam this being the carbon and this carbon wants to accept electrons because it wants to accept electrons. It's a lewis acid. Then next up we have carbon monoxide. So we have C. O. So this contains lone pair with negative formal charges. So we draw the lowest structure for this. We have a carbon here with a triple bond connected to the auction and then each atom here will have a lone pair. So this gives the carbon then a negative one formal charge in the oxygen a plus one formal charge. So we can see here for both of the atoms in this compounds that it contains a long pair with a negative formal charge. And because we have this carbon here with the negative one formal charge and go ahead and donate these electrons making it a louis space. Alright next we have a LCL three. So we know that this has an open valence shell. And because of this it can go ahead and accept electrons into the shell, making it a lewis acid. And lastly here we have this molecule here we see a double bond, so a pi bond and this consists of course high electrons which can be donated, making this a Lewis base. Alright, so just do some of our answers here. We have the lewis acids Being H three, C plus and a l c L three. And then for our louis spaces We have carbon monoxide and CH two, Though bounded by Ch two. So these two is going to be on my final answer for this problem.