OK, this is Dr. B. Let's do the Lewis structure
for CH2O, methanal or formaldehyde.
Start with the valence electrons.
Looking at the periodic table, Carbon has
4.
Hydrogen, in group 1, has 1 and Oxygen, in
group 6 or 16, has 6; but we have two Hydrogens,
so let's multiply that by 2.
Let's add them up: 4 plus 2 plus 6 equals
12 total valence electrons to work with.
All right: so we know Hydrogens always go
on the outside, and in this one, Carbon's
the least electronegative, so that should
be in the center.
So let's put our Carbon right here, put our
Hydrogens on the outside, and then let's put
our Oxygen up here.
We have 12 valence electrons to work with,
so let's spread those out and form some chemical
bonds.
We'll put the bonds between these atoms so
now we have chemical bonds, they're all joined
together, and we've used 2, 4, 6, and we want
to have 12.
So 2, 4, 6, 8, 10, 12.
We can check now and see if we have octets.
Hydrogen only needs two valence electrons,
so each of those Hydrogens, they're OK.
Oxygen needs 8, it's fine with 8.
Carbon needs 8, as well.
It only has 6.
What we'll do is take these two up here, and
let's move them between the Oxygen and Carbon;
we're going to share them.
And so now we can check and see if we've fulfilled
our octets.
We have 2, 4, 6, 8, 10, 12, so we have the
same number of valence electrons; but now
Oxygen has 2, 4, 6, 8 still.
Hydrogens are good with two each.
And the Carbon has 2, 4, 6, 8 as well.
So we've used the 12 valence electrons and
we've fulfilled the octets.
We can also write this in a structural formula
that'll look like this right here, where the
double bond here is represented by this--these
two lines here.
And we'll spread our valence electrons a little
nicer there, so they're spread out.
That's the Lewis structure for CH2O.
This is Dr. B., and thanks for watching.
