This is the H2SO4 Lewis structure: sulfuric
acid.
The key to understanding this Lewis structure
is that you have these H's in front, and then
you have this polyatomic ion.
That means you're going to have an acid and
that these H's are going to be attached to
the outside of the Oxygens.
So we'll put the Sulfur in the center; it's
the least electronegative.
We have 4 Oxygens.
And then the Hydrogens, as we said, they'll
go on the outside of the Oxygens.
We have 2 Hydrogens.
For H2SO4 we have a total of 32 valence electrons.
Put a pair between the atoms.
We're forming chemical bonds here.
We've used 8, 10, 12, and then around the
outside to fill the octets on the Oxygens,
14, and 32.
So this looks like a pretty good Lewis structure
for H2SO4.
We've used all 32 valence electrons and each
of the atoms has a full outer shell.
We should check the formal charges, though.
Sulfur is in period 3 on the periodic table,
and it can hold more than 8 valence electrons.
It's a good idea just to see how the formal
charges pan out.
Turns out that Sulfur has a +2 formal charge
and each of these Oxygens here has a -1.
We'd like our formal charges to be as close
to 0 as possible.
So what we can do is form double bonds with
the Oxygens--the Oxygens that don't have a
Hydrogen on them-- and see how that works
out with the formal charges.
So let's take these electrons here to form
a double bond with Sulfur, and we'll do the
same thing over here.
Since we had a +2, we're going to need to
move both of those in there to get rid of
that.
When we form those two double bonds and recalculate
our formal charges, we'll find that the formal
charge on each atom in the H2SO4 Lewis structure
is now 0.
That makes this a much better structure for
the sufuric acid molecule.
So when you see something like Sulfur there,
make sure you check the formal charges.
I'm Dr. B., and thanks for watching.
