Trends in the Periodic Table
• The trends (atomic radius, ion radius, ionization energy and
electronegativity) can be explained using the concepts of energy
levels and Coulomb’s Law.
Coulomb's Law
• The electrical force between 2
charged objects depends on the
amount of charge on each object and
the distance between them.
In 1783, Charles Coulomb used the
pictured apparatus to determine how
charge and distance affected the
measured force….
Thank you, Mr. Plattner, for your physics notes!
•His conclusion? The
directly
electrical force is _________
related to each of the
charge, q, of the objects.
F ~ q1q2
“is proportional to”
Coulomb also discovered that if he increased the distance
between the charged spheres by a factor of two, the force
four
decreased by a factor of ______!
•His conclusion? The electrical
inversely related to the
force is _________
_______
square of the distance between
the objects.
F
1
~ 2
d
“is proportional to”
Coulomb summarized these
results in equation form in
what is now known as
“Coulomb’s Law”:
kq1q2
Fe  2
d
Where….
d = distance between centers of objects
q = charge on each object
k = the proportionality constant
k  9.0 x10 N  m
9
2
C2
The magnitude of k indicates that electrical forces are very strong.
Coulomb’s Law
Summary:
• Fe between valence electrons and the nucleus increases as
the (+) charge in the nucleus increases… (charge ↑, Fe ↑)
• Fe decreases on the electron (−) and the nucleus (+) as the
distance between them increases… (d ↑, Fe ↓)
Like charges repel; Opposite charges attract
• This equation is NOT on the AP Equation Sheet,
so you will have to memorize these relationships!
Trend 1: Relative Atomic Sizes (Radius)
Atomic Radius
•
•
Atoms increase in size moving down group of elements due to
the fact that electrons are being added to higher energy levels
which are naturally farther from the nucleus.
Moving across a period from left to right, the size decreases.
Why?
•
More electrons are being added to the same energy level which
would not significantly increase the atom’s radius because at the
same time, there is an increase in the number of protons in the
nucleus of the atom.
•
This increases the effective nuclear charge of the atom, and the
nucleus pulls more strongly on the entire electron cloud…more
“Coulombic Force”…so the atom becomes smaller in size.
(Zeff charge ↑, Fe↑)
Effective Nuclear Charge
• The effective nuclear charge (often symbolize as Zeff or Z*) is
the net positive charge experienced by an electron in a multielectron atom.
• The term "effective" is used because the shielding effect of
negatively charged electrons prevents higher orbital electrons
from experiencing the full nuclear charge by the repelling effect
of inner-layer electrons.
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.
Effective Nuclear Charge (Example)
• Consider a sodium cation, a fluorine anion, and a neutral
neon atom. They are “isoelectronic”… each has 10
electrons.
– The number of non-valence electrons is 2…10 total electrons − 8 valence,
but the effective nuclear charge varies because each has a different atomic
number.
Zeff = (nuclear charge – core electrons)
• So, the sodium cation has the largest effective nuclear
charge, and thus the smallest atomic radius since its
electrons would be experiencing the greatest Coulombic
force pulling the electron cloud inward. (Zeff charge ↑, Fe↑)
Trend 2: Atomic Radius vs. Ion Radius
• Removing e−’s : Cations are smaller than the original atom.
• Adding e−’s : Anions are larger than the original atom.
• Why?
Atomic Radius vs. Ion Radius
• For the positive cations, there’s less electronelectron repulsion with the fewer electrons in the
cloud, so the cation radius will be smaller.
– The protons in the nucleus hold the remaining
electrons more strongly as the radius shrinks.
• For the negatively charged anions, there’s more
electron-electron repulsion with additional
electrons in the cloud, so the anion radius will be
larger.
– The nuclear attractive force decreases so the electrons
are less tightly held by the nucleus.
Trend 3:
First Ionization Energies
(Units: kJ/mol)
https://www.youtube.com/watch?v=azQSMhWco8Y (…how IE is actually measured)
Trend 3: Ionization Energy (IE)
• The ionization energy of an atom is the energy required to remove
an electron from the atom in the gaseous phase.
Example: Li (g) + 520 kJ/mol  Li+(g) + 1 e−
• Ionization energy decreases moving down a group.
• The increased distance between electrons and the nucleus and
increased shielding by a full principal energy level means that it requires
less energy to remove an electron.
−The effect of increased nuclear charge is balanced by the effect of
increased shielding, and the number of energy levels becomes the
predominant factor. (d ↑, Fe ↓)
•Ionization energy increases moving across a period.
• The reason for this, as is the case with periodic trends in atomic radii,
is that as the nucleus becomes more positive, the effective nuclear
charge increases its pull on the electrons and it becomes more difficult
to remove an electron. (Zeff charge ↑, Fe ↑)
• Although removing the first electron from an atom requires energy, the
removal of each subsequent electron requires even more energy.
• This means that the second IE is greater than the first, the third IE is
greater than the second, and so on…Why?
• The reason it becomes more difficult to remove additional electrons is
that they’re closer to the nucleus and thus held more strongly by the
positive charge of the protons. (d ↓, Fe↑)
• When you see a huge increase in the subsequent IE, it indicates that
you have decreased the energy level by 1, and the e− being removed is
significantly closer to the nucleus. (d ↓, Fe↑) (It is also a noble gas
configuration which is more stable than usual.)
Important Exceptions to the Ionization Energy Trends
For a test, I would only ask that you could explain why the exception is happening, not
predict that an exception would occur.
• When electron pairing first occurs within an orbital, electron-electron
repulsions increase, so that removing an electron takes less energy
(it’s easier); thus the IE drops at this time.
Example: Nitrogen (1400) vs. Oxygen (1310)
– Less energy is required to remove an electron from oxygen’s valence
in spite of an increasing Zeff because oxygen’s p4 electron is the first
to pair within the orbital.
– The repulsion created lowers the amount of energy required to remove
either electron.
– Differences in electron-electron repulsion are responsible for the differences
in energy between electrons in different orbitals in the same shell…2p3 vs. 2p4
• There is also a drop in ionization energy from s2 to p1—also in spite
of an increasing Zeff.
Example: Magnesium (736) vs. Aluminum (577)
– The p electrons are less tightly held because they do not penetrate the
electron cloud toward the nucleus as well as an s electron does so they
experience less Fe.
Photoelectron Spectroscopy: PES
Measuring and Interpreting Ionization Energy Data
• What is PES?
https://www.youtube.com/watch?v=NRIqXeY1R_I
Trend 4: Electronegativity
Trend 4: Electronegativity
• Electronegativity is a measure of the attraction an atom
has for electrons when it is involved in a chemical bond.
•It is a relative scale from 0 to 4.
Noble Gases =0
Fluorine = 4.0
• Elements that have high ionization energy will also have
high electronegativity since their nuclei strongly attract
electrons.
•The reasoning behind the electronegativity trend is the
same as the reasoning behind the IE trend…larger atoms
are less attracted not only to their own electrons but also
to other electrons in a chemical bond. (d ↑, Fe ↓)