Chemical

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Chapter 6

• Chemical Physical Properties of the Elements
  and the Periodic Table

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Review Quiz Chapter 6

• Heats of (kJ/mol) conversion.
• ?H summation formula.

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Valence Electrons

• The valence electrons are the electrons in the
  outer energy level (valence shell).
• All other electrons are termed core electrons
  (electrons not in the outer energy shell).

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Alkali Metals
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Alkaline Earth Metals
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Transition Elements (Metals)
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Halogens
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Noble Gases
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Trends in the Periodic Table

• The periodic table can be used to predict
• Covalent radii (atomic size)
• Ionic radii (ionic size)
• First Ionization energy

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Covalent radius

• Covalent radius is essentially the size of an
  atom.

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Covalent Radii (atomic radii)
Atomic Radius
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Ionic Radius

• Ionic Radius is the size of an ion.

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Isoelectronic Series

• Substances are isoelectronic if they have the
  same electron configuration.
• Name two isoelectronic species.

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Ionization Energy

• Ionization energy is the energy needed to remove
  an electron from an atom or ion.

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First Ionization Energy

• First Ionization energy is the energy needed to
  remove the first electron from an atom.

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Multiple Ionization Energies

• Second Ionization energy is the energy needed to
  remove the second electron from an atom.
• Third Ionization energy is the energy needed to
  remove the third electron from an atom.
• Etc.

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Ionization Energies in kJ/mol
1 2 3 4 5 6 7
H 1312
He 2372 5250
Li 520 7297 11810
Be 899 1757 14845 21000
B 800 2426 3659 25020 32820
C 1086 2352 4619 6221 37820 47260
N 1402 2855 4576 7473 9442 53250 64340
Write the equation representing the first
ionization energy of hydrogen.

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First Ionization Energy of H

• H 1312 kJ ? H e-

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Effective Nuclear Charge (Zeff)

• You will find many of the notes for effective
  nuclear charge on a sheet in your notebook titled
  Effective Nuclear Charge.
• The effective nuclear charge (Zeff) of an atom is
  basically how well it is able to hold on to its
  most loosely held electron.
• Effective nuclear charge is a direct result of
  Coulombs Law.

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Coulomb's law helps describe the forces that bind
electrons to an atomic nucleus.

• Based on Coulombs Law, the force between two
  charged particles is proportional to the
  magnitude of each of the two charges and
  inversely proportional to the square of the
  distance (radius) between them.

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Effective Nuclear Chargeand Coulomb's law

• There are certain properties that depend upon how
  well the nucleus is holding on to an electron(s).
• These properties include
• Ionization energy
• Atomic and ionic radii
• Electronegativity

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Effective Nuclear Chargeand Coulomb's law

• By applying Coulombs law we can better
  understand the force of attraction between the
  nucleus and an electron which is essentially the
  effective nuclear charge.

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Effective Nuclear Charge (Zeff)

• We can estimate the effective nuclear charge of
  an atom by using the following
• The nuclear charge (Z)
• The shielding effect
• Electron repulsions

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The Nuclear Charge (Z)

• Based on the number of protons in the nucleus.
• Example Carbon vs. Nitrogen

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The Nuclear Charge (Z)
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The greater the number of protons in the nucleus
the greater the effective nuclear charge.
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Nuclear Charge and Zeff
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Shielding Effect.

• Core electrons are generally closer to the
  nucleus than valence electrons, and they are
  considered to shield the valence electrons from
  the full electrostatic attraction of the nucleus.
• This shielding effect can be used in conjunction
  with coulombs law to explain relative ionization
  energies.

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Shielding Effect.

• Shielding can be understood by examining the
  electron configuration for an atom or ion.

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Shielding EffectEnergy Levels vs. Sublevels

• Energy levels have the greatest effect on
  shielding.
• Sublevels increase shielding but to a far lesser
  extent.

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Ionization Energies in kJ/mol
1 2 3 4 5 6 7
H 1312
He 2372 5250
Li 520 7297 11810
Be 899 1757 14845 21000
B 800 2426 3659 25020 32820
C 1086 2352 4619 6221 37820 47260
N 1402 2855 4576 7473 9442 53250 64340

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Zeff can help us explain the ionization energies.
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Explain the first ionization energies of Be and B
A
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Explain the first ionization energies ofBe and
Mg
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Effective Nuclear Charge can be used to help
explain atomic radius.
Atomic Radius
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Explain the difference in atomic radii for Li and
Be. Which are 1.52 and 1.11 angstroms
respectively.
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Explain the difference in atomic radii for Li and
Na. Which are 1.52 and 1.86 angstroms
respectively.
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Effective Nuclear Charge can be used to help
explain atomic radius.

• Based on nuclear charge and shielding.

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Nitrogen vs. OxygenFirst Ionization Energy
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Electron RepulsionsPaired vs. Unpaired Electrons

• Differences in electron electron repulsion
  result from the pairing of electrons within the
  orbitals of a particular subshell.
• This pairing of electrons is responsible for the
  differences in ionization energy for electrons
  within the same subshell.

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Electron RepulsionsPaired vs. Unpaired Electrons

• A paired electron has increased electron
  electron repulsion acting upon it which acts to
  lessen the hold of the nucleus on a paired
  electron lowering the effective nuclear charge.
• Therefore it is easier (takes less energy) to
  remove a paired electron than it does to remove
  an unpaired electron.
• We check the pairing of electrons in the outer
  sublevel by writing an orbital filling diagram.

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Nitrogen vs. OxygenFirst Ionization Energy
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Nitrogen vs. OxygenFirst Ionization Energy
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It is much harder to remove an electron from
helium than it is Li. This is Illustrated by
their respective ionization energies given below.
Explain.
Stability Schmability

• He 2370 kJ/mol
• Li 520 kJ/mol

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Penetration Effect

• Electrons in a higher energy level can often
  penetrate (dive) through lower energy levels
  because of the attraction that the nucleus has on
  them.
• Smaller sublevels can penetrate closer to the
  nucleus than larger sublevels.

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Explain the relative energies of the sublevels
within the fourth energy level.

• The s sublevel penetrates closer to the nucleus
  followed by the p, d and the f has the least
  penetration. The closer to the nucleus the lower
  the energy and therefore the relative energies of
  the sublevels in the fourth energy level is
• 4s lt 4p lt 4d lt 4f.

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Explain why a 4s sublevel has a lower energy than
3d.

• A 4s sublevel penetrates closer towards the
  nucleus than does a 3d so even though the 3d is
  part of the third energy level the 4s on average
  is closer to the nucleus and is therefore lower
  in energy than the 3d.

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Reactivity of Metals

• Which alkali metal would you expect ot be the
  most reactive?
• Explain the trend in the reactivity of the alkali
  metals?

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Alkali Metals in Water Accurate
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Lab - Spectrophotometry of Cobalt(II)
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Lab - Spectrophotometry of Cobalt(II)The Beer
Lambert EquationBeers Law
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Beer Lambert Law

• The amount of light absorbed by a solution can be
  used to measure the concentration of the
  absorbing molecule in that solution by using the
  Beer Lambert Law.

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Beer Lambert Law

• A ?Cl
• where A is the absorbance, ? is the molar
  absorption coefficient, C is the molar
  concentration (molarity), and 1 is the sample
  length.

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In this lab you will prepare solutions of CoCl2
and use Beers Law to determine Co2
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How can we use the slope of the line to determine
?, the molar absorption coefficient?
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A ?Cl
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Transmittance

• A -logT
• The transmittance is the percentage of the light
  in the original light beam that passes through
  the sample and reaches the detector.

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Why do we use absorbance instead of
transmittance?
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Homework

• Write up the Lab Summary.
• Complete the pre-lab assignment on a separate
  sheet of paper.
• You will need a sheet of graph paper for the
  pre-lab assignment.
• Finish your homework for Chapter 6.