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5.4d MO row 2 and BO | General Chemistry | YouTubeToText
YouTube Transcript: 5.4d MO row 2 and BO
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so what can we do with these molecular
orbital diagrams in terms of predicting
our properties
so once we have this middle part of the
mo diagram
we can actually start to fill them out
so if we have something like
beryllium two we know that beryllium
each one of these beryllium um has
two electrons so two electrons
and then there's two beryllium so we
should end up getting
four electrons so we can go ahead and
take our mo diagrams and start filling
up from the bottom
up and we follow all the same rules that
we filled up with our atomic orbitals
we follow the same rules that we fill up
with the molecular orbitals
so we go ahead and fill this up we end
up getting one
the next orbital has to go in the same
molecular orbital but it has to have
opposite spin
and then we end up getting uh those are
two electrons
here's number three and here is number four
four
and this is now the mo diagram and the
electron configuration
for beryllium two so we can do this mo
diagram for the entire row
two periodic of the periodic table
so it will look something like this so
the lithium beryllium
um uh orbitals all will all look the same
same
and then what you have to watch out for
is right at oxygen
we have a switching of the two orbitals
so we end up getting a swap of these two
uh this uh sigma 2px
actually goes down in energy and your
book describes this as
there's a there's a some orbital mixing
between the s
um the the sigma and the the 2p
of the pi orbital so there's a little
bit of orbital mixing in there
that swaps those two but for our class
what we have to do is know that there's
a transition
in terms of the order and you should
have this molecular orbital diagram
available to you
for anything that you should need to do
with this
so we can start to fill up any of these
these orbitals
and look at this so we can go ahead and
look at
for example nitrogen we know that nitrogen
nitrogen
has five electrons in each one
so we should end up getting 10 electrons
uh on there so we end up getting 10
electrons so we're gonna go ahead and
fill up at the bottom
first one two three
four then we're gonna follow our rules
for filling up when they're degenerate
they're gonna go in parallel spins
first so this is five
six 7 8
9 10. so we can fill up the orbitals
just like we do with atomic orbitals to
get the molecular orbital configuration
now what we can also do is calculate
what we call
the bond order so your bond order
is going to be the number of bonding
electrons subtracted by the number
electrons and we're going to divide that
value by two
so if we go back and look at the
nitrogen we know that this is a sigma
over here this is a sigma star and then
these are
our pi and this is also over here
this is our sigma on here so if we look
at this
we look at the number of bonding uh
electrons we have
one two three
four five six seven eight
eight minus two in the antebonding
and then we're gonna go ahead and divide
this by two
equals three so we get a bond order of three
three
that matches what we draw in our lewis structure
structure
it also matches what we think of
in our valence bond theory kind of picture
picture
so that means all of our models are matching
matching
up with each other so it's a great place
to kind of think
about our bonding and the higher the
bond order
the stronger the interactions are
between the two atoms
and then if we end up going back to our
beryllium example remember we had four
electrons uh in here we end up getting
one two three four our bond order
equals zero so that means there is no
net bonding in there
so we would expect that something like
beryllium is going to be
not likely to form as a compound
so let's go ahead and think about this
in terms of
uh stability in terms of molecular hydrogen
hydrogen
and why is hydrogen in in h2 and helium
which is also a first row is also only
just in helium so if we end up going so
remember our
only bonding orbitals that are available
are our valence electrons uh and then
those are the 1s so that is why we are
going to work with the s
orbitals right now we end up getting a bonding
bonding
and an antibonding sigma sigma star
h2 we should anticipate two electrons
our bond order
is going to be equal to one on here
and if we end up looking at um
helium so we end up getting a hydrogen uh
uh
bond order one our helium
has four electrons for helium two
so we end up getting two more electrons
and our bond order
equals zero so that explains why
helium 2 doesn't actually exist however
we can go ahead and look at something like
like helium-2
positive so an ion of helium-2
that's going to have three electrons
because we start off with four electrons
to make it a positive charge it's a
minus one
and then what we can do when we fill
this out
we end up getting one two three
our bond order equals one half that
means there is a slight bonding interaction
interaction
and so in reality there is helium plus two
two
helium two plus that exists as a stable species
species
and that happens quite a bit in our sun
so that is one of the interactions that
is quite
um prominent in the sun itself
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