Transcript Chapter 6

Chapter 6
Chemical & Physical Properties of the
Elements and the Periodic Table
Review Quiz Chapter 6
• Heats of (kJ/mol) conversion.
• ∆H summation formula.
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).
Alkali
Metals
Alkaline Earth
Metals
Transition
Elements
(Metals)
Halogens
Noble
Gases
Trends in the Periodic Table
• The periodic table can be used to predict:
– Covalent radii (atomic size)
– Ionic radii (ionic size)
– First Ionization energy
Covalent radius
• Covalent radius is essentially the size of
an atom.
Covalent Radii (atomic radii)
Atomic Radius
Ionic Radius
• Ionic Radius is the size of an ion.
Isoelectronic Series
•
•
Substances are isoelectronic if they
have the same electron configuration.
Name two isoelectronic species.
Ionization Energy
• Ionization energy is the energy needed to
remove an electron from an atom or ion.
First Ionization Energy
• First Ionization energy is the energy
needed to remove the first electron from
an atom.
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.
Ionization Energies in kJ/mol
1
H
2
3
4
5
6
7
1312
He 2372 5250
Li
520
7297 11810
Be 899
1757 14845 21000
B
800
2426 3659
C
1086 2352 4619
6221
37820 47260
N
1402 2855 4576
7473
9442
25020 32820
53250 64340
Write the equation representing the first
ionization energy of hydrogen.
First Ionization Energy of H
• H + 1312 kJ → H+ + e-
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 Coulomb’s Law.
Coulomb's law helps describe the forces
that bind electrons to an atomic nucleus.
• Based on Coulomb’s
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.
Effective Nuclear Charge
and 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
Effective Nuclear Charge
and Coulomb's law
• By applying Coulomb’s law we can better
understand the force of attraction between
the nucleus and an electron which is
essentially the effective nuclear charge.
Effective Nuclear Charge (Zeff)
•
We can estimate the effective nuclear
charge of an atom by using the following:
1. The nuclear charge (Z)
2. The shielding effect
3. Electron repulsions
The Nuclear Charge (Z)
• Based on the number of protons in the
nucleus.
– Example: Carbon vs. Nitrogen
The Nuclear Charge (Z)
The greater the number of protons in the nucleus
the greater the effective nuclear charge.
Nuclear Charge and Zeff
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 coulomb’s law to explain
relative ionization energies.
Shielding Effect.
•
Shielding can be understood by
examining the electron configuration for
an atom or ion.
Shielding Effect
Energy Levels vs. Sublevels
•
•
Energy levels have the greatest effect on
shielding.
Sublevels increase shielding but to a far
lesser extent.
Ionization Energies in kJ/mol
1
H
2
3
4
5
6
7
1312
He 2372 5250
Li
520
7297 11810
Be 899
1757 14845 21000
B
800
2426 3659
C
1086 2352 4619
6221
37820 47260
N
1402 2855 4576
7473
9442
25020 32820
53250 64340
Zeff can help us
explain the ionization
energies.
Explain the first ionization energies of
Be and B
A
Explain the first ionization energies of
Be and Mg
Effective Nuclear Charge can be
used to help explain atomic radius.
Atomic Radius
Explain the difference in atomic radii
for Li and Be. Which are 1.52 and
1.11 angstroms respectively.
Explain the difference in atomic radii
for Li and Na. Which are 1.52 and
1.86 angstroms respectively.
Effective Nuclear Charge can be
used to help explain atomic radius.
• Based on nuclear charge and shielding.
Nitrogen vs. Oxygen
First Ionization Energy
Electron Repulsions:
Paired 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.
Electron Repulsions:
Paired 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.
Nitrogen vs. Oxygen
First Ionization Energy
Nitrogen vs. Oxygen
First Ionization Energy
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.
• He = 2370 kJ/mol
• Li = 520 kJ/mol
Stability
Schmability
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.
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 < 4p < 4d < 4f.
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.
Reactivity of Metals
• Which alkali metal would you expect ot be
the most reactive?
• Explain the trend in the reactivity of the
alkali metals?
Alkali Metals in Water Accurate
Lab - Spectrophotometry of
Cobalt(II)
Lab - Spectrophotometry of Cobalt(II)
The Beer – Lambert Equation
Beer’s Law
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.
Beer – Lambert Law
A = ƐCl
• where A is the absorbance, Ɛ is the molar
absorption coefficient, C is the molar
concentration (molarity), and 1 is the sample
length.
In this lab you will prepare solutions of CoCl2
and use Beers Law to determine [Co2+]
How can we use the slope of the line to
determine Ɛ, the molar absorption coefficient?
A = ƐCl
• Slope = A/C = Ɛ x l
• If the path length is
1cm then: Slope =
A/C = Ɛ
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.
Why do we use absorbance
instead of transmittance?
Homework
• Write up the Lab Summary (Due tomorrow).
• Complete the pre-lab assignment on a
separate sheet of paper (Due tomorrow).
– You will need a sheet of graph paper for the prelab assignment.
• Finish your homework for Chapter 6 (Due
Wednesday).