00:01
This problem, when done right, can take a considerable amount of time because you need to draw the lewis structures for both of these molecules, recognizing that the first one can be written as an ionic compound.
00:12
To do this, we need to calculate the total number of valence electrons that are available to us.
00:18
For the first compound that ends up being an ionic compound, we have five valence electrons coming from nitrogen.
00:29
We have four valence electrons from the three carbons.
00:35
And because one of the fluorides is going to be the anion, then for the cation portion, we only have eight fluorines with a total of seven valence electrons for each of them.
00:48
And then because it's a cation, we'll subtract one off.
00:51
Then we can take our 72 valence electrons and they will end up surrounding all the fluorides so that we have an octet, and then we will end up needing to share, we'll end up needing to share a pair of electrons between carbon and nitrogen because we will run out.
01:12
Therefore, now we have three electron groups surrounding the central nitrogen, thus this is a planar molecule if you exclude the fact that that last fluoride, that fluorine is now a fluoride, and outside of the chemical structure of this cation.
01:29
For the second one, we don't need to write it as an ionic compound...