For the N2O3 Lewis structure we 28 valence
electrons. Nitrogen is the least electronegative
element so we'll put both of the Nitrogens
there in the center and then we'll put the
Oxygens around them. We have three oxygens.
We'll put two electrons between atoms to form
chemical bonds. We've used 8 valence electrons.
And then we'll go around the oxygens on the
outside. So we have eight, ten ... and 26
and we'll put the last pair right here on
the Nitrogen. 28.
So we've used all 28 valence electrons, and
you can see the oxygens, they all have eight
valence electrons. So their outer shells are
full. But the Nitrogens, each Nitrogen only
has six valence electrons. So we need to form
double bonds in order to give those Nitrogens
octets.
Let's move two of these valence electrons,
here from the Oxygen, in with the Nitrogen
to form a double bond. The Oxygen still has
eight valence electrons, but now the Nitrogen
has eight valence electrons as well. We'll
do the same thing on the other side. Again
the Oxygen has an octet and the Nitrogen also has a full outer shell with eight valence
electrons. And we're also still using 28 valence electrons.
If we calculated the formal charges we'd find
that they were zero on all the atoms except
the Nitrogen and the Oxygen. We'd like our
formal charges to be as close to zero as possible
for each of the atoms.
We could draw this structure a number of ways: for instance, we could put the double bond
between the Nitrogens, and that would work
as well. However, this is the Lewis structure
for N2O3 that has the formal charges closest
to zero. And it makes sense: the -1 is on
the Oxygen, it's more electronegative than
Nitrogen. And the net formal charge for the
entire molecule, we add the -1 and the +1
up, is zero. And that makes sense because
N2O3 is not a charged ion. It's just a regular
molecule.
So that's the Lewis structure for N2O3, this
is Dr. B, and thanks for watching.
