0:02 so what can we do with these molecular
0:04 orbital diagrams in terms of predicting
0:05 our properties
0:08 so once we have this middle part of the
0:10 mo diagram
0:13 we can actually start to fill them out
0:15 so if we have something like
0:18 beryllium two we know that beryllium
0:22 each one of these beryllium um has
0:25 two electrons so two electrons
0:28 and then there's two beryllium so we
0:29 should end up getting
0:31 four electrons so we can go ahead and
0:34 take our mo diagrams and start filling
0:35 up from the bottom
0:38 up and we follow all the same rules that
0:41 we filled up with our atomic orbitals
0:43 we follow the same rules that we fill up
0:44 with the molecular orbitals
0:46 so we go ahead and fill this up we end
0:48 up getting one
0:50 the next orbital has to go in the same
0:52 molecular orbital but it has to have
0:53 opposite spin
0:56 and then we end up getting uh those are
0:57 two electrons
1:00 here's number three and here is number four
1:01 four
1:04 and this is now the mo diagram and the
1:05 electron configuration
1:09 for beryllium two so we can do this mo
1:12 diagram for the entire row
1:15 two periodic of the periodic table
1:18 so it will look something like this so
1:20 the lithium beryllium
1:23 um uh orbitals all will all look the same
1:24 same
1:26 and then what you have to watch out for
1:29 is right at oxygen
1:32 we have a switching of the two orbitals
1:36 so we end up getting a swap of these two
1:39 uh this uh sigma 2px
1:42 actually goes down in energy and your
1:43 book describes this as
1:46 there's a there's a some orbital mixing
1:47 between the s
1:50 um the the sigma and the the 2p
1:52 of the pi orbital so there's a little
1:54 bit of orbital mixing in there
1:57 that swaps those two but for our class
1:58 what we have to do is know that there's
1:59 a transition
2:01 in terms of the order and you should
2:03 have this molecular orbital diagram
2:04 available to you
2:07 for anything that you should need to do
2:08 with this
2:11 so we can start to fill up any of these
2:12 these orbitals
2:15 and look at this so we can go ahead and
2:16 look at
2:18 for example nitrogen we know that nitrogen
2:19 nitrogen
2:22 has five electrons in each one
2:26 so we should end up getting 10 electrons
2:28 uh on there so we end up getting 10
2:30 electrons so we're gonna go ahead and
2:31 fill up at the bottom
2:35 first one two three
2:38 four then we're gonna follow our rules
2:41 for filling up when they're degenerate
2:43 they're gonna go in parallel spins
2:47 first so this is five
2:50 six 7 8
2:54 9 10. so we can fill up the orbitals
2:57 just like we do with atomic orbitals to
3:00 get the molecular orbital configuration
3:03 now what we can also do is calculate
3:04 what we call
3:07 the bond order so your bond order
3:11 is going to be the number of bonding
3:15 electrons subtracted by the number
3:22 electrons and we're going to divide that
3:24 value by two
3:25 so if we go back and look at the
3:28 nitrogen we know that this is a sigma
3:30 over here this is a sigma star and then
3:32 these are
3:36 our pi and this is also over here
3:39 this is our sigma on here so if we look
3:40 at this
3:42 we look at the number of bonding uh
3:43 electrons we have
3:47 one two three
3:50 four five six seven eight
3:54 eight minus two in the antebonding
3:57 and then we're gonna go ahead and divide
3:58 this by two
4:01 equals three so we get a bond order of three
4:02 three
4:09 that matches what we draw in our lewis structure
4:10 structure
4:13 it also matches what we think of
4:16 in our valence bond theory kind of picture
4:17 picture
4:20 so that means all of our models are matching
4:20 matching
4:23 up with each other so it's a great place
4:24 to kind of think
4:28 about our bonding and the higher the
4:29 bond order
4:31 the stronger the interactions are
4:34 between the two atoms
4:36 and then if we end up going back to our
4:38 beryllium example remember we had four
4:42 electrons uh in here we end up getting
4:46 one two three four our bond order
4:49 equals zero so that means there is no
4:50 net bonding in there
4:52 so we would expect that something like
4:54 beryllium is going to be
4:59 not likely to form as a compound
5:01 so let's go ahead and think about this
5:02 in terms of
5:04 uh stability in terms of molecular hydrogen
5:05 hydrogen
5:10 and why is hydrogen in in h2 and helium
5:13 which is also a first row is also only
5:16 just in helium so if we end up going so
5:17 remember our
5:20 only bonding orbitals that are available
5:22 are our valence electrons uh and then
5:25 those are the 1s so that is why we are
5:26 going to work with the s
5:28 orbitals right now we end up getting a bonding
5:29 bonding
5:33 and an antibonding sigma sigma star
5:37 h2 we should anticipate two electrons
5:39 our bond order
5:43 is going to be equal to one on here
5:47 and if we end up looking at um
5:50 helium so we end up getting a hydrogen uh
5:51 uh
5:54 bond order one our helium
5:59 has four electrons for helium two
6:01 so we end up getting two more electrons
6:02 and our bond order
6:06 equals zero so that explains why
6:09 helium 2 doesn't actually exist however
6:11 we can go ahead and look at something like
6:12 like helium-2
6:19 positive so an ion of helium-2
6:21 that's going to have three electrons
6:23 because we start off with four electrons
6:25 to make it a positive charge it's a
6:26 minus one
6:28 and then what we can do when we fill
6:30 this out
6:34 we end up getting one two three
6:38 our bond order equals one half that
6:40 means there is a slight bonding interaction
6:41 interaction
6:44 and so in reality there is helium plus two
6:44 two
6:48 helium two plus that exists as a stable species
6:49 species
6:52 and that happens quite a bit in our sun
6:55 so that is one of the interactions that
6:56 is quite
6:59 um prominent in the sun itself
Chrome Extension