- [Instructor] In the last
video, we talked about
the definition for electron affinity,
and we also talked about group trends.
In this video we're going
to get into period trends,
but before we do that,
let's go ahead and review
the energy changes associated
with electron affinity.
So remember, electron
affinity is referring to
the energy that is given
off when an electron
is added to a neutral
atom in the gaseous state.
So here we have a neutral
atom in the gaseous state.
We're adding an electron to it,
and that would form an
anion, and most of the time
that's going to give off energy.
And so when we're representing
the energy change,
energy being given off, we
would use a negative sign.
So that negative sign just
means that energy is given off.
Once again these values
are in kilojoules per mole.
So if you go down here to fluorine,
you see a number here, right?
And you see a negative sign
in front of that number,
and the negative sign just
means that that is the energy
that is released when you add an electron
to a neutral atom of fluorine.
When you have a very stable atom
that doesn't want an electron, right?
So this example, you
would have to add energy
in order to force it
to accept that electron
to become an anion like that,
and so since you're adding
energy this time, right?
You would represent that energy
change with a positive sign,
but since it's very difficult
to measure how much energy it
takes to do this, most of the time,
you'll see textbooks just say,
"The energy is some
value greater than zero.
"We don't know exactly what it is."
And actually, even more frequently,
you'll see just zero
written in parentheses.
And so when you see this, that
means that it takes energy
to add an electron, and so those
atoms don't want electrons,
and so we can see that down here.
Neon does not have an
affinity for an electron,
and we'll talk about why in a few minutes.
All right, so now that we
have reviewed what these
numbers mean a little bit,
let's go ahead and think
about the trend for a period.
And we'll start with the
value here for boron.
So boron gives off 27.7
kilojoules per mole of energy
when you add an electron
to the neutral atom.
When you go over here to carbon,
you can see that that number
has increased a little bit.
So carbon gives off even more energy.
And when we go over here to oxygen,
it gives off even more
energy, and finally of course,
fluorine gives off the most out of those.
And so there is a trend, right?
As you move to the right, the
general trend for electron
affinity, there's more
energy being given off,
and so therefore a greater
affinity for that electron.
So as you move across a period,
there's an increase in
the electron affinity.
Remember from the previous video
that the idea of electron affinity
goes to the attraction between
the electron you were adding,
or the negatively charged
electron you were adding
and the positively charged
nucleus of the atom.
All right, the more of an attraction
the nucleus has for the electron, right?
The more affinity the
atom has for the electron,
and therefore the more energy
that will be given off.
And so that's just the
idea of electron affinity
that you have to think
about when you're trying
to explain these trends.
So before we explain this trend,
let's go ahead and fill
in the valence electrons
for the elements in the second
period on our periodic table.
So we'll start with lithium over here,
which we know has one valence electron.
And this is going to represent
the 2S orbital right here,
so you have to already understand
electron configurations
to understand what I'm
going to talk about here.
So lithium with one valence
electron's going to go into
the 2S orbital here like that.
Beryllium has two
valence electrons, right?
So the 2S orbital is going to be
completely filled for beryllium.
We move over here to boron, all right.
Three valence electrons, one, two.
The next electron has to go
into a 2P orbital, right?
So I have these three
lines here representing
the three P orbitals in
the second energy level,
and so boron gets one more electron.
It goes into one of those P orbitals here.
So these represent my P orbitals
in the second energy level,
and there are three of them.
Carbon has four valence electrons,
so one, two, three, and of course
using Hund's Rule, we can't
pair up electrons at this point,
so we have to put our next
electron into our next P orbital.
Nitrogen with five valence electrons.
One, two, three, four, and five like that.
Once again following Hund's Rule.
Oxygen with six.
One, two, three, four, five, and six.
Fluorine with seven.
One, two, three, four,
five, six, and seven,
so for the last two we've had
to start pairing things up.
And finally we get to neon
with eight valence electrons.
One, two, three, four, five,
six, seven, and eight.
So to explain this general trend
for electron affinity,
let's look in more detail
at oxygen and fluorine.
So when you're talking
about electron affinity,
once again, you're talking
about adding an electron
to a neutral atom here,
so if we were to add
an electron to oxygen, it would have to go
into one of those two P orbitals.
So this electron in
magenta that I'm adding,
all right, I'm putting it
into a 2P orbital for oxygen.
For fluorine, same thing, right?
If I'm adding an electron to fluorine,
it also has to go into a 2P orbital.
So why does fluorine give off
more energy when you add an
electron than oxygen does?
Well to think about that, let's go through
the factors we discussed in
the previous video, all right?
So one of those factors was distance.
What is the distance of
that electron in magenta
that you were adding
from the nucleus, right?
Or so the closer that
electron you're adding is,
the more of an effect
the nucleus will have,
the more of an attraction
it'll be between those two.
In this case, distance isn't
really much of a factor
because in both cases, all
right, the electron in magenta
that we are adding is
going into a P orbital,
and so therefore, the electrons
in magenta are approximately
the same distance from the nucleus,
and so we don't have to worry
about distance that much here.
Another factor we discussed
was electron shielding.
All right? So electron shielding.
And we talked about that in reference
to inner shell electrons, so
for both oxygen and fluorine,
since they're in the second period, right?
I didn't draw in the 1S electrons,
but let me go ahead and put those in here.
So there'll be two electrons, right,
in the 1S orbital for oxygen.
The same thing for fluorine,
and these electrons, these
inner shell electrons, right?
Would shield these outer shell electrons
from the effect of the nucleus,
but in both cases, we have
the same number of inner shell
electrons for oxygen and fluorine,
and once again, in both
cases, the electron in magenta
is going into 2P orbitals,
similar distance,
so the electron shielding
is pretty much the same
for both oxygen and fluorine
for the magenta electrons.
And so we really have to
move on to the third factor,
which of course is nuclear charge.
All right, so what kind of charge
does the nucleus have?
Well we need to look up the atomic numbers
for these elements, and
we know that oxygen has
an atomic number of eight, all right?
So meaning it has eight
protons in the nucleus,
so it has a charge of
plus eight in the nucleus.
Fluorine has an atomic number of nine,
so there are nine protons in the nucleus.
And now we of course we see a difference.
Fluorine has higher positive
charge in the nucleus, right?
The higher the match
with the positive charge,
the more this positive charge can attract
this electron in magenta.
All right, so the higher the
value for your positive charge,
the more of an attractive force
there is for that electron,
and therefore the fluorine
atom has a higher affinity
for that electron, and therefore
it releases more energy.
And so nuclear charge
helps to explain this trend
that we see, so as you go across a period,
you're always adding protons,
and because you're always
adding protons, you get
an increased attraction
for the nucleus, for the
electron that you were adding.
Therefore you have an
increased electron affinity.
So let's look at some of
the exceptions that we see
because we actually have
a lot of exceptions here
for our period.
And one of them of course
would be neon, right?
So this breaks our trend, right?
We're not releasing more energy.
As a matter of fact,
you have to add energy
for neon to accept an electron.
And the reason for that
is because you already have a completely
full second energy level, right?
You have these eight
electrons here filling all
of your orbitals, so if you
tried to add another electron
to neon, it would have to go
to the third energy level.
That of course is not favorable.
Neon is perfectly happy
and perfectly stable
just the way it is, all right?
So that's why this is an exception.
It's already stable in its
electron configurations,
so adding an electron,
does not make it happy.
All right, let's look at
some of the other exceptions over here.
So let's look at baryllium.
All right, so that's also an exception.
So the value of zero here,
that implies you have
to add energy in order for
it to except an electron.
And it's kind of a
similar idea to neon here,
except that the second
energy level isn't full,
but the 2S orbital is full, right?
So if you wanted to add
another electron to beryllium,
it would have to go into
the higher energy to,
into a higher energy to P orbital,
and that's not as favorable.
So beryllium is kind of already
has a stable electron configuration too.
And so it doesn't want an electron,
so you'd have to force it
in order to accept one.
And then of course for
nitrogen, right, same thing.
With an electron affinity
value of zero here,
you have to force it to
accept an electron as well,
and that's because in this particular
electron configuration, right?
You have one unpaired electron
in each of your two P orbitals,
which is an unusually stable
electron configuration.
And so if you tried to add an electron.
Right, you tried to force
an electron in here,
it's already happy the way it is,
and so you would have to make it do that,
and so again an unusually
stable electron configuration
is the reason for nitrogen not following
our general trend here
for electron affinity.
And so that's just some of the logic
behind those exceptions, but
we have one more exception
to cover here, and that is the difference
in electron affinity between
lithium and boron here.
So lithium actually gives off more energy
than boron does, so lithium has a higher
affinity for an electron.
And to think about why, we need to realize
where that electron is going.
For lithium, if you were
to add an electron to it,
that electron's going into a 2S orbital.
For boron, if you were to
add an electron to boron,
it would go into a 2P orbital,
and so when we think about
these factors, right?
Let's first think about distance.
All right, so the
electron in the 2S orbital
right here for lithium that we're adding
is closer to the nucleus.
And therefore the nucleus has
more of an attractive force
for that electron and the
atom has more affinity for it,
and therefore more energy is given off
when you add that electron
to the 2S orbital.
The 2P orbital, right, so
adding an electron right here
to boron in the 2P orbital,
that's further away
from the nucleus, so there's
not as strong of an attraction.
And so therefore, there's a
lower amount of energy given off
when you add an electron
to that 2P orbital.
So distance kind of explains it.
If we think about electron shielding.
So let's do that next.
So electron shielding,
you can think about both of them as being
in the same periods, so
you're inner shell electrons
don't have much do to
with it, but once again,
the distance in electron shielding play
a role because if you think about the.
Let me go back over here
to this electron and boron.
Right you're adding this
electron to a 2P orbital.
And again, 2P orbitals are
further away from the nucleus,
and so there's a little
bit more electron density
between that electron in the 2P orbital,
or I can think about these electrons,
and there's more electron
density between that electron
in the 2P orbital and the nucleus,
and so therefore, it's more shielded,
and so therefore it's not feeling as much
of an attractive force from the nucleus.
And so there's not as much
of an attractive force,
and so therefore, you have
a smaller amount of energy
released when you add that electron.
That's different from this
electron here, the 2S, right?
The 2S orbital is not as shielded,
and so it feels more the
effect of the nucleus,
and so there's more of
an attractive force,
giving off more energy.
So that takes care of electron shielding.
Now nuclear charge might
make you think of something
different because of course,
boron has more protons
than lithium does, and so that,
that factor is overcome
by the distance factor
and also electron shielding,
so those two turn out
to be the reasons for
this exception that we see
in the trend for electron affinity.
