00:01
Okay, this is problem 118 from chapter 16, and we have this equation at the top.
00:08
And say the acetone is that potassium iodine is soluble in, but potassium bromide isn't soluble in.
00:17
So when the iodine ion approaches carbon, it is opposite to the bromine, and bromide is replaced with iodine.
00:24
And this precipitates as potassium bromine and the other iodine reacts with ethyde.
00:31
So a is asking if the carbon is bonded to the halogen as carbon 1, demonstrated by c -1 right here, what is the shape in hybridization? okay, so i am going to draw iodine ethane.
00:49
So we're going to have our carbon in the middle, and this is bonded to two hydrogens.
00:57
And so one right here and then there's one attached to a dashed line and it's also attached to one iodine and with the wedge it is attached to a methyl group so h3c okay so as you can tell by this structure there are no loan pairs and there are four bonding groups so this makes a shape of the structure, tetrahedral, and the hybridization as sp3.
01:51
And hybridization is a mixture of an orbital forming new orbitals from an element.
01:58
Okay.
02:00
So b is asking in a transition state, one of an unhybridized 2p orbital of c -dash -1, so carbon one overlaps an orbital of iodine and the other overlaps bromide.
02:16
And the transitions to state, what is the shape and hybridization? okay...