WEBVTT

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

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the periodic trends for ionization energy.

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So, for this period,

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as we go across from lithium,

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all the way over to neon,

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

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

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in the ionization energy.

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So, lithium is positive
520 kilojoules per mole.

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Beryllium's goes up to
900 kilojoules per mole,

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and then again, in general,
we see this increase

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in ionization energies going over to neon.

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So, going across a period,

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there's an increase in
the ionization energy.

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And that's because,

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as we go across our period,

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there's an increase in the
effective nuclear charge.

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So, increase in Z effective.

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And remember, the formula for that is

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the effective nuclear charge is equal to

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the actual number of protons, which is Z,

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and from that we subtract S,

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which is the average
number of inner electrons

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shielding our outer electrons.

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So, let's examine this in more detail,

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looking at lithium and beryllium.

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Lithium has atomic number three,

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so three protons in the nucleus,

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so positive three charge,

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and lithium's electron
configuration we know

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is one s two, two s one.

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So, two electrons in our one s orbital,

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and one electron in the two s orbital.

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Beryllium has one more proton

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and one more electron.

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

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and for beryllium, the
electron configuration is

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one s two, two s two.

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So two electrons in the one s orbital,

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and then two electrons
in the two s orbital.

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Let's calculate the
effective nuclear charge

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for both of these,

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and first, we'll start with lithium.

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So for lithium,

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lithium has a plus three
charge in the nucleus,

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so the effective nuclear charge
is equal to positive three,

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and from that we subtract
the average number of inner

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electrons shielding our outer electrons,

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in this case, we have these two inner,

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or core electrons, that are
shielding our outer electron,

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our valence electron, from this
full positive three charge.

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So we know that like charges repel,

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so this electron is going to repel

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this electron a little bit,

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and this electron repels this electron.

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And these two inner core
electrons of lithium

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have a shielding effect,

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they protect the outer electron

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from the full positive three charge.

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So there's two shielding electrons,

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so for a quick effective
nuclear charge calculation

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positive three minus two
gives us a value of plus one

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for the effective nuclear charge.

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So, it's like this outer
electron of lithium

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is feeling a nuclear charge of plus one,

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which pulls it toward the nucleus, right?

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So, there's an attractive
force between the outer

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electron and our nucleus.

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Now, the actual calculation for this um,

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Z is--

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

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and the actual value for
lithium is approximately

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one point three, but our
quick, crude calculation

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tells us positive one.

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Let's do the same
calculation for beryllium,

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so the effective nuclear
charge for beryllium

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is equal to the number of protons, right,

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which for beryllium is positive four,

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and from that, we subtract
the number of inner electrons

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that are shielding the outer electrons.

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So, it's a similar situation,

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we have two inner electrons
that are shielding

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this outer electron, they're repelling

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this outer electron,

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shielding the outer electron
from the full positive

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four charge of the nucleus.

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SO we say there are two inner electrons,

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so the effective nuclear charge is

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positive four minus two,

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giving us an effective nuclear
charge of positive two.

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In reality, the effective
nuclear charge is

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approximately one point nine,

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and that's because beryllium
has another electron

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in its two s orbital over here,

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which does effect this
electron a little bit.

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It repels it a little bit,

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and so it actually deceases
the effective nuclear charge

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to about, one point nine.

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

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So, the outer electron for beryllium,

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let's just choose this one again,

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is feeling an effective nuclear charge

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of positive two, which means that,

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

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there's a greater attractive
force on this outer electron

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for beryllium, as compared
to this outer electron

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for lithium.

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The effective nuclear
charge is only plus one

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for this outer electron,

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

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The two s orbital gets
smaller, and the atom itself

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is smaller.

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Beryllium is smaller than lithium.

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So this outer electron here,

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let me switch colors again,

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this outer electron for
beryllium is closer to

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the nucleus than the outer
electron for lithium.

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It feels a greater attractive force,

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and therefore it takes more energy

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to pull this electron away from
the neutral beryllium atom,

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

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So beryllium has an ionization energy

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of positive 900 kilojoules per mole,

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compared to lithium's of
520 kilojoules per mole.

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So it has to do with the
effective nuclear charge.

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So far we've compared
lithium and beryllium

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and we saw that the ionization energy

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went from positive 520 kilojoules per mole

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to 900 kilojoules per mole,

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and we said that was because of the

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increased effective nuclear
charge for beryllium,

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but as we go from beryllium to boron,

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there's still an increased
effective nuclear charge,

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but notice our ionization energy goes

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from 900 kilojoules per mole for beryllium

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to only 800 kilojoules per mole for boron,

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

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And let's look at the electron
configuration of boron

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to see if we can explain that.

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Boron has five electrons,

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so the electron
configuration is one s two,

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two s two, and two p one.

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So that fifth electron
goes into a two p orbital,

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and the two p orbital is higher in energy

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than a two s orbital, which means

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the electron in the two p orbital

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is on average, further
away from the nucleus

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that the two electron
in the two s orbital.

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So if we just sketch
this out really quickly,

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

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I have two electrons in there,

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and this one electron in the two p orbital

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is on average further
away from the nucleus.

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So, those two electrons
in the two s orbital

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actually can repel this
electron in the two p orbital.

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So, there's a little bit
extra shielding there

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of the two p electron
from the full attraction

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of the nucleus, right?

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

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and a positive five charge for boron,

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the fact that these two s electrons

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add a little bit of extra shielding means

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it's easier to pull this electron away.

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

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this electron in the two p orbital away

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due to these two s electrons.

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And that's the reason
for this slight decrease

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in ionization energy.

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As we go from boron to carbon,

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we see an increase in ionization energy,

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from carbon to nitrogen,

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an increase in ionization energy.

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Again, we attribute that to increased

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effective nuclear charge,

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but when we go from nitrogen to oxygen,

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we see a slight decrease again.

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From about 1400 kilojoules per mole,

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down to about 1300 kilojoules
per mole for oxygen.

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

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some electron configurations
for nitrogen and oxygen.

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Nitrogen has seven
electrons to think about.

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So it's electron configuration is

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

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So that takes care of all seven electrons.

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For oxygen, we have another electron, so

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one s two, two s two, two p four

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is the electron configuration for oxygen.

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Let's just draw using orbital notation

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the two s orbital and the two p orbital.

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

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We have two electrons in there,

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so let's draw in our two electrons.

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And for our two p orbitals,

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we have three electrons.

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So here are the two p orbitals,

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and let's draw in our three electrons

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using orbital notation.

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Let's do the same thing for oxygen.

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

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so we'll sketch in those two electrons,

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and we have four electrons
in the two p orbitals.

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So let me draw in the two p orbitals.

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There's one electron,

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there's two, there's three,

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and notice what happens when
we add the fourth electron.

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We're adding it to an orbital that already

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has an electron in it,

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

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

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

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so electrons have like charges,

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and like charges repel.

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And so that's the reason for this

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slight decrease in ionization energy.

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So, it turns out to be a little bit easier

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to remove an electron from an oxygen atom,

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than nitrogen, due to this repulsion

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in this two p orbital.

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From there on, we see
our general trend again.

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The ionization energy for
fluorine is up to 1681,

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and then again for
neon, we see an increase

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in the ionization energy due to

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the increased effective nuclear charge.