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Ionization energy period trend

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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.
Title:
Ionization energy period trend
Description:

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Video Language:
English
Duration:
10:03

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