to think about constructing Lewis diagrams,

which you've probably seen before.

They're nice ways of visualizing

how the atoms in a molecule are bonded to each other

and what other lone pairs of valence electrons

various atoms might have.

And so let's just start with an example,

then we'll come up with some rules

for trying to draw these Lewis diagrams.

So the first example that we will look at

is silicon tetrafluoride, and tetrafluoride

is just a fancy way of saying four fluorines,

so tetrafluoride.

Now the first step is to say,

"Well, what are the electrons that are of interest to us?"

And if we're talking about the electrons

that are likely to react, we're talking

about the valence electrons, so V.E. for short,

valence electrons.

So first let's think about how many total valence electrons

are involved in silicon tetrafluoride.

Well, to think about that, we could think

about how many valence electrons does silicon have,

and then how many valence electrons

does each of the fluorines have

if they were just free atoms and neutral,

and then multiply that times four,

'cause you have four fluorines.

So let's get out our periodic table of elements,

and then you can see here that silicon,

its outer shell is the third shell,

and in that third shell it has one, two,

three, four valence electrons.

So silicon here has four valence electrons,

and then to that, we're going to add the valence electrons

from the four fluorines.

A free, neutral fluorine atom,

its outer shell is the second shell,

and in that outer shell, it has one, two,

three, four, five, six, seven electrons.

So each of these fluorines has seven valence electrons,

but there are four of them.

So one silicon tetrafluoride molecule

is gonna have four plus 28 valence electrons.

So this is going to be a total of 32.

Now the next step is to think

about how might these be configured?

And as a general rule of thumb,

we'd wanna put the least electronegative atom

that is not hydrogen at the center.

And we've talked about this before,

but you can even see from the periodic table of elements,

fluorine is actually the most electronegative element,

and so we would at least try to put silicon at the center

and make fluorine a terminal atom,

something on the outside.

So let's try to do that.

So let's put silicon in the center,

and then we have to put the four fluorines some place.

Let's just put one fluorine there, one fluorine there,

one fluorine there, and one fluorine there.

Now the next step is, well let's just say

for simplicity that we just have single bonds

between the silicon and each of the fluorines.

So let's do that.

So one bond, a bond, a bond, a bond.

Now each of these covalent bonds,

each of these lines in our Lewis diagram,

they represent two electrons.

So for example, this one right over here

that I'm doing in yellow, that represents two electrons

that are shared by this fluorine and this silicon.

This represents another two electrons

that is shared between this fluorine and the silicon.

This is another two electrons that's shared

between this fluorine and this silicon.

And this is another two electrons

shared between that fluorine and the silicon.

So, so far, how many electrons have we accounted for?

Well, each of these represent two electrons,

so two, four, six, eight electrons.

So if we subtract eight from this, we are left

with 24 electrons to account for, 24 valence electrons.

So now, our general rule of thumb would be,

try to put those on those terminal atoms

with the goal of getting those terminal atoms

to having eight valence electrons.

In general we try to get the octet rule

for any atom except for hydrogen.

Hydrogen, you just need to get to two in that outer shell.

But fluorine, you want to get it to eight.

It already has two that it can share,

so it needs six more, so let's add that.

Two, four, six.

Let's do that again for this fluorine.

Two, four, six.

Do it again for this fluorine.

Two, four, six.

And then last but not least, for this fluorine.

Two, four, and six.

Now how many more electrons are now accounted for?

Well, six in this fluorine, six in this fluorine,

six in this fluorine, six in this fluorine,

so six times four, we've now accounted

for 24 more electrons.

We've now used up all of the valence electrons.

Now that's good, because we wanted to account

for all of the valence electrons.

We wanna represent them somehow in this Lewis diagram.

The next thing to check for is how satisfied

the various atoms are relative to to the octet rule.

We've already seen that the fluorines

are feeling pretty good.

They each have six electrons that are not in a bond,

and then they're able to share two electrons

that are in a bond, so each of them

can kind of feel like they have eight outer electrons,

eight valence electrons hanging out with them.

And then the silicon is able to share in four bonds.

Each of those bonds have two electrons,

so the silicon is also feeling good about the octet rule.

So I would feel very confident

in this being the Lewis diagram,

sometimes called the Lewis structure,

for silicon tetrafluoride.

So just to hit the point home on what we just did,

I will give you these steps,

but hopefully you find them pretty intuitive.

That's why I didn't wanna show you from the beginning.

But as you see, step one was,

find the total number of valence electrons.

We did that.

That's the four from silicon

and then the 28 from the fluorines.

It says add an electron for every negative charge.

Subtract an electron for every positive charge.

We didn't have to do that in this example

because it's a neutral molecule.

Then it says decide the central atom,

which should be the electronegative except for hydrogen.

That's why we picked silicon,

because fluorine is the most electronegative atom.

And then we drew the bonds.

We saw that the bonds accounted for eight electrons,

and we subtracted those electrons

from the total in step one, and that's just

to keep track of the number of valence electrons

that we are accounting for.

And then we had 24 left over.

And then the next step, it says assign the valence electrons

to the terminal atoms.

That's where we assigned these extra lone pair electrons

to the various fluorines, giving them an extra six each

so that they were all able to fulfill the octet rule.

And then we subtracted that from the total,

really just to account, to make sure

that we're using all of our electrons.

It says it right here: subtract the electrons

from the total in step two.

And then we saw that all of our electrons

were accounted for.

But then in step four, it says if necessary,

assign any leftover electrons to the central atom.

We didn't have to do that in this example.

If the central atom has an octet

or exceeds an octet, you are usually done.

In this case, it had an octet, so we felt done.

And it finally says, if a central atom

does not have an octet, create multiple bonds.

Once again, in this example we were able

to stay pretty simple with just single bonds.

But in future examples, we're going to see

where we might have to do some of these more nuanced steps.

So I will leave you there,

and I'll see you in the next example.