EFFECTIVE NUCLEAR CHARGE
VALENCE ELECTRONS
>> ATOMIC RADIUS
>> IONIC RADIUS
>> IONIZATION ENERGY
>>ELECTRONEGATIVITY
TERMS WHEN DESCRIBING TRENDS, ETC.
__ electron repulsion
__ electrostatic attraction
__ valence electrons
__ less/more tightly
__ far away from nucleus/further away
__ electrons and protons
__ core electrons
__ electron shielding
__ effective nuclear charge
__ energy level
What are the trends in the 1st ionization energy in the
periodic table? Explain the causes of these trends in terms
of atomic structure.
1st ionization increases from left to right in a period. This is because Zeff
increases from left to right. Since Zeff is increasing, the energy required to
remove an electron also must be higher. 1st ionization energy decreases in a
family as the period increases. With each step down in a family the valence
electron is in a higher energy level. At a higher energy level the electron is
farther from the nucleus. There is less energy required to remove the
electron as one goes down the family.
What are the trends in atomic radius in the periodic table?
Explain the causes of these trends in terms of atomic
structure.
Atomic radius decreases from left to right in a period. This is because Zeff
increases from left to right. Since Zeff is increasing and the distance of the
electron from the nucleus (energy level) is approximately the same, the force
between the electron and nucleus is greater. The greater force pulls the
electron cloud closer to the nucleus. Thus the atomic radius is small. Atomic
radius increases in a group, or family, as the period increases. With each step
down in a family the valence electron is in a higher energy level. At a higher
energy level the electron is farther from the nucleus.
Which elements are exceptions to the trend in ionization
energy across the 2nd period? Explain these exceptions in
terms of atomic structure.
Beryllium is higher than expected. This is because the electron would have to
be removed from the filled 2s orbital. A filled 2s orbital is more stable than a
2p occupied by one electron. The 2p electron is also farther from the nucleus
than the 2s electron. Therefore, it is easier to remove the 1 2p from B than a
paired 2s from Be. Oxygen has a lower IE than expected. The addition of the
next electron in oxygen causes pairing of one of the 2p’s. The nitrogen
arrangement is more stable and has less electron-electron repulsion than the
arrangement in oxygen. Therefore, more energy is required to remove the
nitrogen electron than that in oxygen.
The periodic nature of ionization energy is emphasized in this
diagram. With each new period the ionization energy starts
with a low value. Within each period you will notice that the
pattern is really kind of a zigzag pattern progressing up as you
go across the periodic table. The zigs and zags on that graph
correspond to the sublevels in the energy levels. So far in this
lesson we have presumed that all the electrons in the second
energy level are pretty much the same. Two factors make
that not completely true. One factor is that because s and p
orbitals have different shapes, the electrons in p orbitals have
more energy and are further from the nucleus. The other
factor is that when electrons are paired up in an orbital, they
repel one another somewhat. Those two factors account for
the zigzag nature of the increase in ionization energy.
Nevertheless, as a general trend, from left to right across the
periodic table, ionization energy does increase. Also as you go
down the periodic table, the ionization energy does decrease
for the reasons given.
Which of the following would be expected to have the
lowest first ionization energy?
Ne, O, C, F, Li ; Li
Which of the following elements has the higher ionization
energy?
a. Ti or V; V
b. P or Cl; Cl
c. Mg or Sr; Mg
d. Fe or Os; Fe
e. B or Br; B
Arrange the following atoms in order of increasing
ionization energy: In, F, Si, Sr, Ne.
[Smallest IE] Sr < In < Si < F < Ne [Largest IE]
Which element will have a higher second ionization energy,
magnesium (Mg) or potassium (K)? Why?
a. As or Se; As
Potassium will have a higher 2nd ionization energy, because to remove an
electron from K+ we have to remove a core electron (from the 3p subshell),
whereas for magnesium there are two valence electrons in the 3s subshell.
b. Ru or Rh ; Ru
Choose the best answer from the following: F, O , S
a.
b.
Has the largest ionization energy? F
Largest atomic radius? S
Choose the best answer from the following: Be 2+, He, Li+
a.
b.
c.
Has the smallest atomic radius? Be 2+
Has lowest ionization energy? He
Are isoelectronic? He, Li+, Be 2+
For each of the following sets of atoms, decide which is
larger, which is smaller, and why.
a. Li, C, F
All are in the same period and thus have the same number of energy levels.
Therefore, the important factor is the nuclear charge. Li is the largest
because it has the smallest nuclear charge and pulls the electrons toward the
nucleus less than the others. F is the smallest because it has the largest
nuclear charge and pulls the electrons toward the nucleus more than the
others.
b. Li, Na, K
All are in the same goup and thus have the same effective nuclear charge.
Therefore, the important factor is the number of energy levels. Li is the
smallest because it uses the smallest number of electron energy levels. K is
the largest because it uses the largest number of electron energy levels.
c. Sr or Ba ; Ba
d. F or I; I
e. Ca or Mg ; Ca
f.
Na or S; Na
g. Li or Li+ ; Li
h. S or S2-; S2i.
S2- or K+ ; S2-
Arrange the following ions and atoms in order of increasing
radii.
a.
Mg, S2- , S, Mg2+ , K+, Se2- ; Mg2+, K+, S, Mg, S2-, Se2-
What is the trend in sizes of the ions N3-, O2-, F- and why
does the trend exist?
N3- > O2->F-; the effective nuclear charge is smallest at N3- and largest at F-
Which of the following pairs is described as isoelectronic?
a.
Ca and Be
b.
P- and S2-
c.
H and He
d.
Cu+ and Zn2+
e.
Na and Na+
c. Ge, P, O
All are in different groups and periods, therefore both factors must be taken
into account. Fortunately both factors reinforce one another. Ge is the
largest because it uses the largest number of energy levels and has the
smallest effective nuclear charge. O is the smallest because it uses the
smallest number of energy levels and has the largest effective nuclear
charge.
d. C, N, Si
Not all are in the same group and period, so, again, both factors must be
taken into account. C and N tie for using the smallest number of energy
levels, but N has a higher effective nuclear charge. Therefore, N is the
smallest. C and Si tie for having the lowest effective nuclear charge, but Si
uses more energy levels. Therefore, Si is the largest.
Rank the following species in terms of decreasing atomic
radius: Cl, Tl, As, Sn, Pb
*Tl > Pb > Sn > As > Cl
Arrange the following atoms/ions in order of increasing
atomic radius: Ce4+, Xe, I-, Cs+, Te2-.
Since the ions are isoelectronic the relative sizes of the ions/atoms are
dependent only upon changes in the effective nuclear charge.[Smallest
radius] Ce4+ < Cs+ < Xe < I- < Te2- [Largest radius]
In each of the following pairs of elements, which has the
larger radius?
Cu+2 and Zn2+
Arrange the ions Be2+, F-, Mg2+, S2-, Cl- in order of
increasing size
Be2+, Mg2+, F-, Cl-, S2-
What is the correct order of increasing electronegativity for
Zn, O, F, and Ca?
Ca, Zn, O, F,
Where in the periodic table are the least electronegative
elements found?
Lower left