PERIODICITY DATA ANALYSIS: TRENDS AND DISCONTINUITIES
Purpose:
 Determine Periodic Trends
 Explain Discontinuities using knowledge of atomic structure and interactions
Definitions:
Period: a horizontal row of the table
What does the period number tell you?
# of principal energy levels
Group / Family: a vertical column of the table
What does the group number tell you?
# of valence electrons
Periodic law: properties repeat (happen periodically) if you arrange elements in order of
increasing atomic number (Z)
Before you begin, define the following terms:
Atomic radius –
½ the distance between nuclei in a diatomic (covalent) molecule
Ionization Energy –
The energy required to completely remove an electron from a gaseous atom or ion in the
ground state.
Electron Affinity –
The energy RELEASED when a gaseous atom or ion in the ground state accepts an extra
electron.
Electronegativity –
A measure of how strongly an atom attracts electrons in a covalent bond.
Effective nuclear charge –
The actual (net) attractive force experienced by an electron in a multi-electron atom. Zeff is the
difference between the number of protons in the nucleus (Z) and the average number of shielding
electrons (S) between the nucleus and the electron in question. (Zeff = Z – S)
Isoelectronic –
Atoms or ions have the same electron configuration
Atomic Radius vs. Z
Ionization Energy vs. Z (eV*)
*The electron volt (symbol eV; also written electronvolt)
is a unit of energy approximately equal to 1.602×10−19 J.
Graph Analysis Questions
Using the graphs provided, your Periodic Table, and your knowledge of atomic structure, answer the
following analysis questions on a separate sheet of paper and attach.
1. Use the details of modern atomic theory to explain each of the following experimental observations.
(a) Within a family such as the alkali metals, the ionic radius increases as the atomic number increases.
The radii of the alkali metal ions increase with increasing atomic number because the number of principal
energy levels increases.
(b) The radius of the chlorine atom is smaller than the radius of the chloride ion, Cl-.
(Radii: Cl atom = 0.99Å; Cl- ion = 1.81Å)
The chloride ion is larger than the chlorine atom because
 electron-electron repulsion increases OR
 shielding increases OR
 the electron-proton ratio increases OR
 the effective nuclear charge decreases
(c) The first ionization energy of aluminum is lower than the first ionization energy of magnesium.
(First ionization energies: 12Mg = 7.6 eV; 13Al = 6.0 eV)
The ionization energy of Mg is greater than that for Al because the 3p orbital is at a higher energy than the 3s orbital.
(d) For magnesium, the difference between the second and third ionization energies is much larger than
the difference between the first and second ionization energies. Ionization energies for Mg:
1st = 7.6 eV
2nd = 14 eV
3rd = 80 eV
The much greater difference between the 2nd and 3rd ionization energies in Mg (relative to the difference between the 1st
and 2nd) is due to the 3rd electron being a non-valence (inner/core) electron.
2. Using the Ionization Energy (kJ mol-1) graph as a reference, briefly (in one to three sentences) explain each
of the following in terms of atomic structure.
(a) In general, there is an increase in the first ionization energy from Li to Ne.
Across the period from Li to Ne the number of protons is increasing in the nucleus while the number of principal energy
levels remains constant. This causes an increase in the effective nuclear charge, and consequently a consequently
stronger attraction for the valence electrons.
(b) The first ionization energy of B is lower than that of Be. (same as 1c)
The electron removed in the case of Be is a 2s electron whereas in the case of B it is a 2p electron. 2p
electrons are higher in energy than 2s electrons and therefore are easier to remove.
(c) The first ionization energy of O is lower than that of N.
The electron ionized in O is paired with another electron in the same orbital, whereas in N the electron comes
from an orbital containing only one electron. The ionization energy of the O electron is less because of the
repulsion between two electrons in the same orbital.
(d) Predict how the first ionization energy of Na compares to those of Li and of Ne. Explain.
The ionization energy of Na will be less than those of both Li and Ne because the electron removed comes from a higher
energy orbital which is farther from the nucleus, and therefore less tightly held
3. Account for each of the following in terms of principles of atom structure, including the number, properties,
and arrangements of subatomic particles.
(a) The second ionization energy of sodium is about three times greater than the second ionization
energy of magnesium.
Na has only one valence electron, whereas Mg has two in 3s. IE2 for Na is very large because the 2nd electron in Na is
taken from the second energy level, which is lower in energy than the second 3s electron taken from Mg.
(b) The difference between the atomic radii of Na and K is relatively large compared to the difference
between the atomic radii of Rb and Cs.
Near the bottom of a group, the increase in the nuclear charge is cancelled out by the increase in shielding electrons,
so the effective nuclear charge remains fairly constant. So as you go from Rb to Cs, the number of protons greatly
increases, but so do the number of shielding electrons. This causes the Zeff of Rb and Cs to be nearly identical, which
gives them very similar radii. Since Na is so small to begin with, going down one period to K, those extra electrons do a
much more effective job of shielding the 19 protons in the K nucleus from the outermost electrons in K, thus giving a
larger difference in the radii of Na and K.
(c) A sample of nickel chloride is attracted to a magnetic field, whereas a sample of solid zinc chloride is
not.
Ni2+ has 8 electrons in the 3d sublevel. The distribution of these electrons in the five 3d orbitals according to Hund's rule
leaves two unpaired electrons, which give nickel chloride its paramagnetic properties. The Zn2+ ion on the other hand has
a full 3d sublevel with 10 electrons. This means that all of these electrons are paired with another electron of opposite
spin in the five 3d orbitals.
4. Use principles of atomic structure to answer each of the following.
(a) The radius of the Ca atom is 0.197 nanometer; the radius of the Ca2+ ion is 0.099 nanometer.
Account for this difference.
The radius of the Ca2+ ion is smaller than the Ca atom because the Ca2+ ion has lost its two valence electrons from the 4s
subshell while the number of protons in the nucleus remains constant. This results in a greater effective nuclear charge
for the Ca2+ ion. In addition, the removal of the two valence electrons reduces electron-electron repulsion in the Ca2+ ion,
so the electron cloud contracts.
(b) The first and second ionization energies of K are 419 kJ/mol and 3052 kJ/mol, and those of Ca are
590 kJ/mol and 1145 kJ/mol, respectively. Account for the observed differences in their
(i) first ionization energies
Ca has a greater first ionization energy than K because Ca has a greater effective nuclear charge (Ca has more protons
in its nucleus than K while having the same number of principal energy levels of electrons to hold on to).
(ii) second ionization energies
(c) The first ionization energy of Mg is 738 kJ mol-1 and that of Al is 578 kJ mol-1. Account for this
difference.
K has a greater second ionization than Ca because the second electron removed from a K atom is in the 3p subshell,
which is closer to the nucleus and experiences a greater attractive force than the second electron removed from a Ca
atom, which is in the 4s subshell.
5. Explain each of the following observations using principles of atomic structure and/or bonding.
(a) Potassium has a lower first ionization energy than lithium.
K has a lower first ionization energy than Li because the valence electron in K experiences a lower effective nuclear
charge due to an increase in the number of shielding electrons.
(b) The ionic radius of N3– is larger than that of O2–.
The N and O2– ions are isoelectronic with one another, which means that they have the same number of
electrons and the same electron configuration. The N3- ion is larger than the O2- ion because there is one less
proton in the Nitrogen nucleus. Because of this, the valence electrons in the N3- ion experience a lower Zeff,
which means they are not held as tightly and the radius of the N3- ion is larger.
3–
(c) A calcium atom is larger than a zinc atom.
A calcium atom is larger than a zinc atom because there are 20 protons in the Ca nucleus, whereas there are 30
protons in the Zn nucleus. Because these two elements are in the same period, they have the same number of
principal energy levels. Since Zn has more protons in its nucleus, its electrons experience a greater Zeff and are
held more tightly, giving Zn a smaller radius than Ca.
(d) Boron has a lower first ionization energy than beryllium. (same as 2b & 1c)
The electron removed in the case of Be is a 2s electron whereas in the case of B it is a 2p electron. 2p
electrons are higher in energy than 2s electrons and therefore are easier to remove.
(e) The first ionization energy of S is less than that of P. (same as 2c)
The electron ionized in S is paired with another electron in the same orbital, whereas in P the electron comes
from an orbital containing only one electron. The ionization energy of the S electron is less because of the
repulsion between two electrons in the same orbital.
7.
The electron affinities of five elements are given below:
Boron:
Carbon:
Nitrogen:
Oxygen:
Fluorine:
26.7 kJ/mol
153.9 kJ/mol
7 kJ/mol
141 kJ/mol
328 kJ/mol
For the elements above, explain the observed trend as atomic number increases. Account for the break in
this trend at nitrogen.
Electron affinity (EA) is defined as the amount of energy released when an atom in the gaseous state accepts an
electron. The general trend when going across a period is one of increasing EA. This trend is observed because
as one goes across a period, there is an increase in the number of protons in the nucleus while the number of
shielding electrons remains constant across a period, thereby causing the added electron to experience a greater
attraction for the nucleus (Zeff) and release more energy when the electron is added on.
A break in this trend occurs at nitrogen. Nitrogen has 3 electrons in the 2p subshell (exactly half-filled, one
electron in each 2p orbital). The addition of an electron to nitrogen is not energetically favorable because the
added electron would experience a repulsive force from the electron already occupying the orbital to which the
extra electron would be added.