Valence Bond Theory with Examples Video Lecture | Chemistry Class 12 - NEET

okay this will be a short lecture on
valence bond theory if you remember from
chapter 10 1170 we try to find the
hybrid orbitals for a for the central
atom of different molecules so we had SP
sp2 sp3 sp3 d SP 3 D - 4 valence for the
hybrid orbitals for metals that are part
of a complex ion is a little trickier
same basic idea though so we have
chromium here they want to know the
hybrid orbital for chromium in here so
in this complex so we need to know the
charge on the chromium it's +3 we can
see that here is chromium but chromium
plus 3 we're going to get rid of this
and remember this is that exception so
we that's why we don't have two
electrons here so we get rid of one two
three so just make sure that it get rid
of them
now we need six um empty orbitals
because though each water and molecule
is going to come in with a lone pair
each ligand will so the ligand comes in
with the two electrons so what the
central atom has to have is an empty
orbital so we need six empty orbitals
now these ones I've scratched out those
are empty so we just we can't touch
these even though they're half filled
and that's what we were looking for back
in chapter 10 now we need completely
empty orbitals to allow the two
electrons from the ligand to attach so
we start here 1 2 3 4 5 6 so this would
be D 2 sp3 because we're going order
there's two DS 1s three P's let me
rewrite that so d2 sp3 those are the
hybrid orbitals now when they hybridize
they turn into six that are exactly the
same kind of shape though they go off in
six different directions and this is the
hybrid over the hybridization is D 2 SP
3 so what you do is you find the central
atom you write out the elect the orbital
diagram and you can use the noble gas
abbreviation
you are right it out with the full
electrons then you take away the
electrons the s electrons go first
that's always the case we learned that
back in chapter chapter 9 actually the
the for the transition metals we've
removed the 4s electrons first and so
let's do it again down here we have iron
it's in now we the cyanide has a minus 1
there's six of them so minus six so this
must be plus three again here's the
orbital diagram for iron so since we got
here three of electrons we get rid of
these three of these two and sorry I
can't erase them but just know that
that's now empty and then we gotta get
rid of a third and we get rid of the one
that's paired up so now we can't touch
any of these so we get we have to use
this one these next three and then we go
into the four DS so that's why I didn't
draw the four DS up here because I
didn't need to we we had six but here we
needed them so now this is sp3d2 that
should be very familiar and we know that
sp3d2 s like four sf6 a molecule that
we'd see back in chapter 10 created an
octahedral structure and that's the
that's what we have here for each of
these this one just turns out doesn't
have the right orbitals to do something
that we're familiar with but you just
fill in it so if this one was missing we
would have started here this would been
d SP P 3d so they can get a little wild
looking but just follow that procedure
right off the orbital diagram remove the
electrons starting with the 4s the the
guys that I give you these uh now this
is plus 3d the metals that I give you
will be from that first row so you'll be
able to use this this orbital diagram
straight straight out I just have to put
in the different number of electrons and
take away the number it doesn't have to
be plus three they just happen to be
here to be plus two plus one plus four
so just make sure you remove the right
number of electrons and then go from
there