In the previous lecture we made enormous progress in the development of our Lewis
Structure model of how molecules form and what determines the types of molecules
which will form. Remember, that our Lewis Structure model
is based on the octet rule. An experimentally derived rule that says
that atoms typically bond so as to complete an octet of valence electrons, 8
electrons in their valence shell. And that they do so by sharing electrons
with other atoms in pairs. We call a shared pair of electrons a
covalent bond. We also learned that carbon atoms in
particular are able to form single bonds by sharing a single pair of electrons,
double bonds by forming, or by sharing two pairs of electrons and triple bonds
by sharing three pairs of electrons. We need to extend though, beyond just
doing carbon atoms and hydrogen atoms, which is all we did in the previous
lecture. Let's try fluorine for example.
Fluorine atoms we know, have seven valence electrons, and the valence is
equal to 1. There's the experimentally derived octet
rule, 7 plus 1 being 8. What will our Louis structure look like
for fluorine? Well, let's see.
It's got seven valance electrons, and we need to draw a structure that accurately
represents the fact that it has a valance of 1.
If we were to simply draw the Lewis structure to have 7 electrons sort of
widely distributed around the fluorine, that would suggest that there would be
seven unpaired, unshared electrons. And we might think that the valance of
fluorine was 7, it's not, it's only 1. So this is not an appropriate structure.
Instead, since we know that paired electrons typically are already going to
not form bonds, then and we only want 1 unshared, unpaired electrons.
I'm going to draw a flourine to look like this, because this gives both a single
vacancy a single place where we can form a bond and the other electrons here are
already paired and therefore probably not going to form bonds.
We can use this model to account for example for the stability hydrogen
fluoride molecule because if we then take this fluorine, which we have drawn, as
so. And bring in a hydrogen with its single
electron. Then we wind up with HF with a single
bond in here as we would have expected. It also accounts for the stability of the
F to molecule remember that fluorine is 1 of the diatomic gases that we studied
early on. This suggests that we take the fluorine
drawn as follows and a second fluorine and let them share there unshared,
unpaired, electrons from before. And again form a single bond, and in fact
the experimental data for Fluorine F2 is consistent with this bond strengthened,
bond length being a single bond. So now we have a model for Fluorine.
Let's try Oxygen. Oxygen you recall has 6 valence
electrons, we derived that from our ionization energy data, and it has a
valence of 2 such as in water. There's the octet rule again.
6 plus 2 is equal to 8. How shall we draw Oxygen?
Based upon the lessons learned from flourine we're going to draw oxygen with
2 unshared, unpaired electrons and the remaining 4 electrons existing in pairs.
And that gives us the opportunity for example if we want to draw water to draw
it by taking the two unshared paired electrons, unpaired electrons.
and pairing them up with an electron from each of the 2 hydrogens to form single
bonds to each hydrogen, with 2 long pairs on the oxygens.
Similarly, it helps us understand the strength of the bond in the oxygen
diatomic molecule, oxygen was also one of the diatomic gases we learned about early
on. So we'll take the two oxygens and now if
I were to take the remaining electrons on either side I'd probably will want to
share those so that I can complete an octet of both, for both oxygen atoms and
correspondinly then we'll drawn oxygen with a double bond, and each oxygen
having 2 lone pairs on it. And again, the data for oxygen suggests
that oxygen is doubly bonded. How about nitrogen?
Turns out the bond in nitrogen is even stronger than the bond in O2.
So we will then consider what the nitrogen looks like, again based upon our
lessons from oxygen, and from fluorine. Were not going to draw this with five
electrons scattered around here, rather we will pair 2 of the electrons, and
leave 3 of the electrons unpaired. That allows us to account for
example,fFor the valance of 3 of a nitrogen atom and for example, the
molecule ammonia will take the nitrogen, one pair of electrons, 3 unshared,
unpaired electrons. We then share those electrons with each
of 3 hydrogen atoms resulting in 3 single bonds, in the NH3 molecule.
The nitrogen now has completed its octet. Furthermore, we can look at the M2
diatomic molecule and if we now simply take each nitrogen to have 1 lone pair on
it and share the remaining electrons, we wind up with something that looks like a
triple bond between the nitrogens in much the way we had triple bonds in the carbon
atoms and in fact, the experimental data is consistent with the into bond being an
extremely strong bond consistent with a triple bond.
Well what we can now do is extend molecules in which we combined all the
different types of atoms together and we can begin to develop a systematic means
of drawing these molecular structures. And here's how our systematic method is
actually going to work. Let's consider a particular molecule
CH3NH2, methylamine. This is actually a stable molecule.
It's an important molecule as well, in all kinds of syntheses.
The first thing that we want to do is actually figure out how many available
electrons are there in this molecule to be shared or to be alone in each of the
atoms. To do that what we need to do is count up
the number of valence electrons in the carbon atoms, the number of valence
electrons in the nitrogen atom, plus one each for the hydrogen atoms.
So the number is 4 plus that's for the carbon, plus 5 for the nitrogen, plus 5
times one, where there's one from each of the hydrogens, 4 plus 5 is 9 plus 5 more
gives us a total of 14 available electrons.
This is an extremely important calculation, to do, because at the end of
our determination to molecular structure, the number of electrons that we have
drawn had better actually be 14 or we've made a mistake.
How many electrons do we need for every atom from the main group to have a, an
octet of electrons or for each hydrogen to have 2?
The number that we need is equal to, let's see, carbon needs 8, nitrogen needs
8 each hydrogen needs 2. That's a total of 26.
So we need 26 electrons, to satisfy the valence, of each of the atoms in
Methalamine. But there's no way we've got 26, we've
only got 14. So how can we possibly do this?
The answer is, we've got to share electrons.
How many do we have to share? That turns out to be an easy calculation.
It's the number we need minus the number that we have.
So for example, in this particular case is 26 minus 14, we're going to share a
total of 12 electrons. How are we going to share those 12
electrons? The answer is we're going to calculate
the number of bonds. The number of bonds needs to be half of
the number of shared electrons since we share electrons in pairs so, it needs to
be 6. Now that we know that there are 6 bonds
present, let's see if we can draw them. Well let's see,carbon is clearly going to
be bonded to nitrogen can't can't form this molecule without that.
And then, beyond that, there are going to be 5 hydrogens.
We know carbon has a valence of 4, so it probably wants3 hydrogens around it.
Nitrogen has a valence of 3, so it probably wants 2 hydrogens around it.
At this point it would appear that we've actually completed our Lewis Structure,
so lets check and see. We have2, 4, 6, 8, 10, 12 electrons
remember each of these bonds is a shared pair of electrons, but wait that's only
12, the number that were available were 14, there has to be 2 more electrons.
Furthermore, nitrogen has not satisfied the octet rule.
It has only 6 electrons around it. We can solve this problem by saying the
two extra electrons here that are not yet shown become a lone pair on the nitrogen
atom. This systematic process that we've drawn
on this particular slide is a good way to solve problems for an enormous number of
molecules, in fact, it works rather well in general.
Let's do another example. This particular example is going to
simply be called formaldehyde. It is CH2O.