00:01
Starting with part a, we're asked to draw the lewis structure of n4o, and that would look like this.
00:06
We'll have an oxygen, double bonded to a nitrogen, and we will have four nitrogens in a row.
00:16
There's double bonds here and here, and then we can just draw in the lone pairs of electrons on nitrogen's and oxygens to complete their octets and use all of our, all of the valence electrons that we have.
00:31
In total, there's 26 bounce electrons, since each nitrogen has five, adding to 20 plus the six from the oxygen.
00:42
For part b, we are asked the hybridization of the terminal nitrogen.
00:48
That would be this nitrogen here, and it is sp2 hybridized since it has three regions of electron density, so two loan pairs and one bond.
01:06
For part c, we are asked for the hybridization of the central nitrogen.
01:11
Would be this one here and it is also sp2 hybridized since it has three regions of electron density or two bonds and one lone pair...