Ionization energy period trend

- [Instructor] In this
video, let's look at

the periodic trends for ionization energy.

So, for this period,

as we go across from lithium,

all the way over to neon,

so as we go this way,
across our periodic table,

we can see, in general,
there's an increase

in the ionization energy.

So, lithium is positive
520 kilojoules per mole.

Beryllium's goes up to
900 kilojoules per mole,

and then again, in general,
we see this increase

in ionization energies going over to neon.

So, going across a period,

there's an increase in
the ionization energy.

And that's because,

as we go across our period,

there's an increase in the
effective nuclear charge.

So, increase in Z effective.

And remember, the formula for that is

the effective nuclear charge is equal to

the actual number of protons, which is Z,

and from that we subtract S,

which is the average
number of inner electrons

shielding our outer electrons.

So, let's examine this in more detail,

looking at lithium and beryllium.

Lithium has atomic number three,

so three protons in the nucleus,

so positive three charge,

and lithium's electron
configuration we know

is one s two, two s one.

So, two electrons in our one s orbital,

and one electron in the two s orbital.

Beryllium has one more proton

and one more electron.

So one more proton in the
nucleus, so a plus 4 charge,

and for beryllium, the
electron configuration is

one s two, two s two.

So two electrons in the one s orbital,

and then two electrons
in the two s orbital.

Let's calculate the
effective nuclear charge

for both of these,

and first, we'll start with lithium.

So for lithium,

lithium has a plus three
charge in the nucleus,

so the effective nuclear charge
is equal to positive three,

and from that we subtract
the average number of inner

electrons shielding our outer electrons,

in this case, we have these two inner,

or core electrons, that are
shielding our outer electron,

our valence electron, from this
full positive three charge.

So we know that like charges repel,

so this electron is going to repel

this electron a little bit,

and this electron repels this electron.

And these two inner core
electrons of lithium

have a shielding effect,

they protect the outer electron

from the full positive three charge.

So there's two shielding electrons,

so for a quick effective
nuclear charge calculation

positive three minus two
gives us a value of plus one

for the effective nuclear charge.

So, it's like this outer
electron of lithium

is feeling a nuclear charge of plus one,

which pulls it toward the nucleus, right?

So, there's an attractive
force between the outer

electron and our nucleus.

Now, the actual calculation for this um,

Z is--

S I should say, does not
have to be an integer,

and the actual value for
lithium is approximately

one point three, but our
quick, crude calculation

tells us positive one.

Let's do the same
calculation for beryllium,

so the effective nuclear
charge for beryllium

is equal to the number of protons, right,

which for beryllium is positive four,

and from that, we subtract
the number of inner electrons

that are shielding the outer electrons.

So, it's a similar situation,

we have two inner electrons
that are shielding

this outer electron, they're repelling

this outer electron,

shielding the outer electron
from the full positive

four charge of the nucleus.

SO we say there are two inner electrons,

so the effective nuclear charge is

positive four minus two,

giving us an effective nuclear
charge of positive two.

In reality, the effective
nuclear charge is

approximately one point nine,

and that's because beryllium
has another electron

in its two s orbital over here,

which does effect this
electron a little bit.

It repels it a little bit,

and so it actually deceases
the effective nuclear charge

to about, one point nine.

But again, for a quick
calculation, positive two works.

So, the outer electron for beryllium,

let's just choose this one again,

is feeling an effective nuclear charge

of positive two, which means that,

it's going to be pulled
closer to the nucleus,

there's a greater attractive
force on this outer electron

for beryllium, as compared
to this outer electron

for lithium.

The effective nuclear
charge is only plus one

for this outer electron,

and because of this, the
beryllium atom is smaller, right?

The two s orbital gets
smaller, and the atom itself

is smaller.

Beryllium is smaller than lithium.

So this outer electron here,

let me switch colors again,

this outer electron for
beryllium is closer to

the nucleus than the outer
electron for lithium.

It feels a greater attractive force,

and therefore it takes more energy

to pull this electron away from
the neutral beryllium atom,

and that's the reason for
the higher ionization energy.

So beryllium has an ionization energy

of positive 900 kilojoules per mole,

compared to lithium's of
520 kilojoules per mole.

So it has to do with the
effective nuclear charge.

So far we've compared
lithium and beryllium

and we saw that the ionization energy

went from positive 520 kilojoules per mole

to 900 kilojoules per mole,

and we said that was because of the

increased effective nuclear
charge for beryllium,

but as we go from beryllium to boron,

there's still an increased
effective nuclear charge,

but notice our ionization energy goes

from 900 kilojoules per mole for beryllium

to only 800 kilojoules per mole for boron,

so there's a slight decrease
in the ionization energy.

And let's look at the electron
configuration of boron

to see if we can explain that.

Boron has five electrons,

so the electron
configuration is one s two,

two s two, and two p one.

So that fifth electron
goes into a two p orbital,

and the two p orbital is higher in energy

than a two s orbital, which means

the electron in the two p orbital

is on average, further
away from the nucleus

that the two electron
in the two s orbital.

So if we just sketch
this out really quickly,

let's say that's my two s orbital,

I have two electrons in there,

and this one electron in the two p orbital

is on average further
away from the nucleus.

So, those two electrons
in the two s orbital

actually can repel this
electron in the two p orbital.

So, there's a little bit
extra shielding there

of the two p electron
from the full attraction

of the nucleus, right?

So, even though we have
five protons in the nucleus,

and a positive five charge for boron,

the fact that these two s electrons

add a little bit of extra shielding means

it's easier to pull this electron away.

So, it turns out to be a
little bit easier to pull

this electron in the two p orbital away

due to these two s electrons.

And that's the reason
for this slight decrease

in ionization energy.

As we go from boron to carbon,

we see an increase in ionization energy,

from carbon to nitrogen,

an increase in ionization energy.

Again, we attribute that to increased

effective nuclear charge,

but when we go from nitrogen to oxygen,

we see a slight decrease again.

From about 1400 kilojoules per mole,

down to about 1300 kilojoules
per mole for oxygen.

So, let's see if we can
explain that by writing out

some electron configurations
for nitrogen and oxygen.

Nitrogen has seven
electrons to think about.

So it's electron configuration is

one s two, two s two, and two p three.

So that takes care of all seven electrons.

For oxygen, we have another electron, so

one s two, two s two, two p four

is the electron configuration for oxygen.

Let's just draw using orbital notation

the two s orbital and the two p orbital.

So for nitrogen, here's our two s orbital.

We have two electrons in there,

so let's draw in our two electrons.

And for our two p orbitals,

we have three electrons.

So here are the two p orbitals,

and let's draw in our three electrons

using orbital notation.

Let's do the same thing for oxygen.

So there's the two s orbital
for oxygen, which is full,

so we'll sketch in those two electrons,

and we have four electrons
in the two p orbitals.

So let me draw in the two p orbitals.

There's one electron,

there's two, there's three,

and notice what happens when
we add the fourth electron.

We're adding it to an orbital that already

has an electron in it,

so when I add that fourth
electron to the two p orbital,

it's repelled by the electron
that's already there,

which means it's easier to
remove one of those electrons,

so electrons have like charges,

and like charges repel.

And so that's the reason for this

slight decrease in ionization energy.

So, it turns out to be a little bit easier

to remove an electron from an oxygen atom,

than nitrogen, due to this repulsion

in this two p orbital.

From there on, we see
our general trend again.

The ionization energy for
fluorine is up to 1681,

and then again for
neon, we see an increase

in the ionization energy due to

the increased effective nuclear charge.