Hi, this is Dr. B. Let's do the SO4 2- Lewis
structure, for the sulfate ion.
On the periodic table: Sulfur, 6 valence electrons;
Oxygen also has 6, we have 4 Oxygens, multiply
by 4; and these 2 valence electrons up here,
we need to add those, as well.
That gives us a total of 32 valence electrons.
We'll put the Sulfur in the center, and then
the four Oxygens will go on the outside.
Next, we'll draw bonds between the Sulfur
and the Oxygens, so there we have four bonds
and we've used eight valence electrons.
Let's go around the outer atoms and make sure
they have octets.
So we've used 8, 10, 12, and 32.
Looking at the structure here, we see that
each of the Oxygens has 8 valence electrons;
2, 4, 6, 8; as does the Sulfur here, 2, 4,
6, 8.
But we're not quite done yet.
Sulfur is in the third period of the periodic
table.
That means it can hold more than 8 valence
electrons.
So we really do need to check our formal charges.
So to calculate the formal charge on the Sulfur:
we see that Sulfur, on the periodic table,
group 16 or 6, has 6 valence electrons.
Up here, all of the electrons, all of them
are involved in bonds, so that's going to
be zero.
And the bonding electrons, 2, 4, 6, 8; we've
used 8 of those, and we'll divide that by
2.
Six minus zero minus 4 gives us a +2 formal
charge for the Sulfur.
For the Oxygen, it's also in group 6 or 16,
so it has 6 valence electrons.
Up here, nonbonding, we have 6; and then bonding,
we have 2.
And all these Oxygens are the same, so we
only need to do one.
Six minus 6 minus 2 gives us a minus 1 formal
charge for each Oxygen.
If we add up all the formal charges, the -1,
-1, -1, -1 and +2, we do get a total charge
of negative 2.
That does make sense, but with formal charges,
we want them to be as close to zero as possible
for the atoms.
So let's see if we might be able to do another
Lewis structure that has more zeroes for the
formal charges.
When I see this +2 charge here on the Sulfur,
I know that I can move electrons from the
outer valence electrons of the Oxygen into
the middle.
If I do that twice, if I move these two into
the middle to form double bonds, and get rid
of them, I think that'll get rid of the positive
2 charge.
Let's try that and then recalculate our formal
charges.
So I've moved electrons from the outside of
these two green Oxygens into the middle to
form double bonds.
Let's see how that changes the formal charges.
So for Sulfur, 6 minus zero, there are no
nonbonding; and now we have 2, 4, 6, 8, 10,
12 total bonding electrons.
Six minus 6, that gives us zero.
So the formal charge on the Sulfur is zero.
If we look at the green Oxygens, you can see
that we have 6 minus 4 of the nonbonding,
and then 4 bonding; we divide by 2.
Six minus 4 minus 2 is zero.
Finally, looking at the blue Oxygens, we have
6 minus 6 nonbonding, and then 2 bonding divided
by 2.
Six minus minus 6 minus 1 is minus 1.
So at this point, we see that we have mostly
zeroes.
But if you look, you have a negative 1 and
a negative 1 here.
That works well with this.
Since more of the formal charges are zero,
this is a better structure for SO4 2-.
Since it's an ion, there's one last thing
we need to do.
We need to put brackets around it to show
that it's an ion and the charge of the ion.
We add our 2 minus right there.
And that is the Lewis structure for SO4 2-.
It was a bit of work, but we have the best
structure here.
Our formal charges are in good shape.
We've used all the valence electrons.
So that's it.
This is Dr. B., and thanks for watching.
