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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.