CrazCo Posted January 16, 2009 Share Posted January 16, 2009 1. you know methanol is CH3OH, which turns out to be tetrahedral in shape, correct? but it doesn't necessarily look like its exclusively tetrahedral, so can you have more than one shape in a bond? 2. BBr3 and NI3 are different in shape because of what? - would saying that since the central atom has no free electrons in the NI3 bond but there are in the BBr3 bond, making it so the extra, free electrons push the bonded electrons away for more moving space be an adequate response? Link to comment Share on other sites More sharing options...
UC Posted January 16, 2009 Share Posted January 16, 2009 I take it that you are learning VESPR theory? There are two "shapes" in a methanol molecule, since the hydrogens don't have any p-orbitals to hybridize. One of the two is tetrahedral. Keep in mind that it isn't a single bond that is shaped, but the cluser of bonds attached to a central atom. You should look into lone pairs and how they situate themselves in hybridized orbitals There is a key difference between boron and nitrogen and it has everything to do with valence electrons. good luck Link to comment Share on other sites More sharing options...
hermanntrude Posted January 16, 2009 Share Posted January 16, 2009 the answer you gave contains some signs you know what you want to say. However, you'll need to re-word it carefully, and also switch it backwards... you said boron has "free electrons" ( by which i think you mean "a lone pair") and that nitrogen doesn't. It's the other way around, of course. Link to comment Share on other sites More sharing options...
CrazCo Posted January 16, 2009 Author Share Posted January 16, 2009 Yes, sorry, I thought since the outer shell of nitrogen was full it had no free pairs, brain fart guess. I wonder if that would have anything to do with it considering the boron still wouldn't have his valence shell filled. Link to comment Share on other sites More sharing options...
hermanntrude Posted January 16, 2009 Share Posted January 16, 2009 you need to approach these questions systematically. First, draw a Lewis diagram of the molecule in question, using the rules for that. Then you can simply sum the numbers of lone pairs and bonding pairs on the central atom Link to comment Share on other sites More sharing options...
CrazCo Posted January 16, 2009 Author Share Posted January 16, 2009 I understand now, thanks . Link to comment Share on other sites More sharing options...
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