Foundation
The atomic molecular theory provides us a
particulate understanding of matter. Each element is characterized
as consisting of identical, indestructible atoms with atomic
weights which have been determined. Compounds consist of identical
molecules, each made up from a specific number of atoms of each of
the component elements. We also know that atoms have a nuclear
structure, meaning that all of the positive charge and virtually
all of the mass of the atom are concentrated in a nucleus which is
a very small fraction of the volume of the atom. Finally, we know
that the electrons in the atom are arranged in "shells" about the nucleus, with each shell farther from the nucleus that
the previous. The electrons in outer shells are more weakly
attached to the atom than the electrons in the inner shells, and
only a limited number of electrons can fit in each shell.
Goals
The shell model of the atom is a good start in
understanding the differences in the chemical properties of the
atoms of different elements. For example, we can understand the
periodicity of chemical and physical properties from our model,
since elements in the same group have the same number of electrons
in the valence shell.
However, there are many details missing from
our description. Other than a very crude calculation of
"distance" of the shells from the nucleus, we have no
description of what the differences are between the electrons in
different shells. What precisely is a "shell?"
Most importantly, the arrangement of elements
into groups and the periodicity of chemical properties both depend
on the concept that a shell is "filled" by a certain
number of electrons. Looking at the number of elements in each
period, the number of electrons which fills a shell depends on
which shell is being filled. In some cases, a shell is filled by
eight electrons, in others, it appears to be 18 electrons. What
determines how many electrons can "fit" in a shell? Why
is there a limit at all?
Finally, a closer look at the ionization
energies
here reveals that
our shell model must be incomplete. Our model implies that the
elements of the second period from Lithium to Neon have their
valence electrons in the second shell. With increasing nuclear
charge, the ionization energy of these atoms should increase from
Lithium to Neon. As a general trend, this is true, but there are
variations. Note that the ionization energy of Oxygen atoms is less
than that of Nitrogen atoms. We need to pursue additional detail in
our model of the structure of the atom.
Observation 1: The Spectrum of Hydrogen
To begin, we need to know a little about
light. All forms of electromagnetic radiation travel as an
oscillating wave, with an electric field component perpendicular to
a magnetic field component. As a wave, the radiation can be
characterized by its "wavelength", symbolized as λλ, which is the
distance between adjacent peaks in the wave.
Different wavelengths correspond to different forms of
electromagnetic radiation. For example, microwave radiation has
wavelength in the range of
10-2
10
-2
to
10-3
10
-3
meters, whereas x-ray
radiation has wavelength in the range
10-9
10
-9
to
10-10
10
-10
meters.
Radiation which is visible to the human eye has wavelength in the
very narrow range from 3.8×10-7
3.8-7 to
7.8×10-7
7.8-7 meters.
Radiation can also be characterized by the
frequency of the electromagnetic wave, which is the number of peaks
in the wave which pass a point in space per second. Frequency is
symbolized by νν. The speed which light travels in a vacuum
in the same for all forms of electromagnetic radiation,
c=2.997×108ms
c
2.9978
m
s
. As such, we can relate the frequency of
light to the wavelength of light by the equation
λ
(
m
)
×
ν
(
s
-1
)
=
c
(
m
s
)
λ
(
m
)
×
ν
(
s
-1
)
=
c
(
m
s
)
(1)
The longer the wavelength λλ, the lower
the frequency νν. This makes sense when we remember that
light travels at a fixed speed. When the wavelength is longer,
fewer peaks will pass a point in space in a second. From this
equation, there is a specific relationship between frequency and
wavelength, and either or both can be used to characterize the
properties of radiation.
With this background in hand, we can use our
understanding of light to pursue more data about the energies of
electrons in atoms. Ionization energies tell us how much energy is
required to remove an electron from an atom, but do not tell what
happens if an electron changes its energy in an atom. To analyze
this, we need a means to measure the energies gained or lost by an
atom. One way to do so is to analyze the "spectrum" of
an atom, which is the set of frequencies of light emitted by the
atom. Since hydrogen is the simplest atom, we analyze the hydrogen
spectrum first. We find that, if we pass a current of electricity
through a sample of hydrogen gas, light is emitted. Careful
analysis shows that, although some of this light is emitted by
H
2
H
2
molecules, some of the light is also emitted by H atoms. Since
light is a form of energy, then these H atoms must release energy
supplied to them by the electrons in the current.
Most importantly, if we pass the light emitted
by the hydrogen gas sample through a prism, we can separate the
colors as in a rainbow, each with a characteristic frequency. The
resultant image of separated colors is called the
spectrum of hydrogen. We find in this experiment that
there are only four frequencies (four colors) of light in the
emission that are visible. The most intense of the lines in the
spectrum is bright red, but there are blue and violet lines. It
turns out that there are also many other frequencies of light
emitted which are invisible to the human eye.
Careful observation and analysis reveals that
every frequency in the hydrogen atom spectrum can be predicted by a
very simple formula, called the Rydberg equation:
ν=R×1n2-1m2
ν
×
R
1
n
2
1
m
2
(2)
where RR
is the Rydberg constant (
3.29×1015s-1
3.2915
s
-1
). nn
and mm are integers (1,2,3,...).
Each choice of nn
and mm predicts a single observed
frequency in the hydrogen atom
spectrum.
The atoms of all elements emit radiation when
energized in an electric current, and as do all molecules of all
compounds. However, we find that the specific frequencies of light
emitted are characteristic of each atom or molecule. In other
words, the spectrum of each element is unique to each element or
compound. As a result, the spectrum of each substance can be used
to identify that substance. (Note that the Rydberg equation tells
us only the spectrum of hydrogen.)
Our interest is in the fact that the radiation
emitted by an atom tells us about the amounts of energy which can
be released by an atom. For a hydrogen atom, for example, these
changes in energy must correspond to the amounts of energy which
the electrons inside the atom can gain or lose.
At this point, we need to relate the frequency
of radiation emitted by an atom to the amount of energy lost by the
electron in the atom. We thus examine some observations about the
energy of radiation.
Observation 2: The Photoelectric Effect
When a light source is directed at a metal
surface, it is found under many circumstances that electrons are
ejected from the surface. This phenomenon is called the
"photoelectric effect." These electrons can be
collected to produce a usable electric current. (This effect has a
variety of common practical applications, for example, in
"electric eye" devices.) It is reasonable to expect
that a certain amount of energy is required to liberate an electron
from a metal surface, since the electron is attracted to the
positively charged nuclei in the metal. Thus, in order for the
electron to escape, the light must supply sufficient energy to the
electron to overcome this attraction.
The following experimental observations are
found when studying the photoelectric effect. First, in order for
the effect to be observed, the light must be of at least a minimum
frequency which we call the
threshold frequency,
ν
0
ν
0
. This frequency is a
characteristic for a given metal. That is, it is the same value for
each sample of that metal, but it varies from one metal to the
next. For low frequency light, photoelectrons are not observed in
any number, no matter how intense the light source
is. For light with frequency above
ν
0
ν
0
, the number of
photoelectrons emitted by the metal (measured by the photoelectric
current,
ΦΦ) increases directly with the intensity of the
light. These results are shown in
Figure 1.
Second, we can measure the energies of the
electrons emitted by the metal. For a given metal, all
photoelectrons have the same kinetic energy for a fixed frequency
of light above
ν
0
ν
0
. This fixed kinetic energy is independent
of the intensity of the light source. As the frequency of the light
is increased, the kinetic energy of the emitted electrons increases
proportionally. These results are shown in
Figure 2.
Are these results surprising? To the
physicists at the end of the nineteenth century, the answer was
yes, very surprising indeed. They expected that the energy of the
light source should be determined by its intensity. Hence, the
energy required to eject a photoelectron should be supplied by
light of high intensity, no matter how low the frequency of the
radiation. Thus, there should be no threshold frequency, below
which no electrons are emitted. Moreover, the kinetic energy of the
electrons should increase with intensity, not with light frequency. These predictions are
not observed, so the results are counter to physical
intuition.
We can account for these results in a
straightforward but perhaps non-obvious manner. (Einstein provided
the explanation in 1905.) Since the kinetic energy of the emitted
photoelectrons increases proportionally with increases in the
frequency of the light above the threshold frequency, we can conclude from
conservation of total energy that the energy supplied by the light
to the ejected electron must be proportional to its frequency:
E∝ν
∝
E
ν
. This does not immediately account for the
existence of the threshold frequency, though, since it would still
seem to be the case that even low frequency light would possess
high energy if the intensity were sufficient. By this reasoning,
high intensity, low frequency light should therefore produce as
many photoelectrons as are produced by low intensity, high
frequency light. But this is not observed.
This is a very challenging puzzle, and an
analogy helps to reveal the subtle answer. Imagine trying to knock
pieces out of a wall by throwing objects at it. We discover that,
no matter how many ping pong balls we throw, we cannot knock out a
piece of the wall. On the other hand, only a single bowling ball is
required to accomplish the task. The results of this
"experiment" are similar to the observations of the
photoelectric effect: very little high frequency light can
accomplish what an enormous amount of low frequency light cannot.
The key to understanding our imaginary experiment is knowing that,
although there are many more ping-pong balls than bowling balls, it
is only the impact of each individual particle with the wall which
determines what happens.
Reasoning from this analogy, we must conclude
that the energy of the light is supplied in "bundles"
or "packets" of constant energy, which we will call
photons. We have already concluded that the light
supplies energy to the electron which is proportional to the light
frequency. Now we can say that the energy of each photon is
proportional to the frequency of the light. The intensity of the
light is proportional to the number of these packets. This now
accounts for the threshold frequency in a straightforward way. For
a photon to dislodge a photoelectron, it must have sufficient
energy, by itself, to supply to the electron to overcome its
attraction to the metal. Although increasing the intensity of the
light does increase the total energy of the light, it does not
increase the energy of an individual photon. Therefore, if the
frequency of the light is too low, the photon energy is too low to
eject an electron. Referring back to the analogy, we can say that a
single bowling bowl can accomplish what many ping-pong balls
cannot, and a single high frequency photon can accomplish what many
low frequency photons cannot.
The important conclusion for our purposes is
that
light energy is quantized into packets of
energy. The amount of energy in each photon is given by
Einstein’s equation,
E=hν
E
h
ν
(3)
where hh
is a constant called Planck’s
constant.
Quantized Energy Levels in Hydrogen Atoms
We can combine the observation of the hydrogen
atom spectrum with our deduction that light energy is quantized
into packets to reach an important conclusion. Each frequency of
light in the spectrum corresponds to a particular energy of light
and, therefore, to a particular energy
loss by a hydrogen atom, since this light
energy is quantized into packets. Furthermore, since only certain
frequencies are observed, then only certain energy losses are
possible. This is only reasonable if the energy of each hydrogen
atom is restricted to certain specific values. If the hydrogen atom
could possess any energy, then it could lose any amount of energy
and emit a photon of any energy and frequency. But this is not
observed. Therefore, the energy of the electron in a hydrogen atom
must be restricted to certain
energy levels.
The Hydrogen atom spectrum also tells us what these
energy levels are. Recall that the frequencies of radiation emitted
by Hydrogen atoms are given by the
Rydberg equation. Each choice of the positive integers
nn and
mm
predicts a single observed frequency in the hydrogen atom
spectrum.
Each emitted frequency must correspond to an
energy
hν
h
ν
by
Einstein’s equation. This
photon energy must be the
difference between two energy levels for a
hydrogen electron, since that is the amount of energy released by
the electron moving from one level to the other. If the energies of
the two levels are
EmEm and
EnEn, then we can write that
hν=Em-En
h
ν
Em
En
(4)
By comparing this to the Rydberg equation,
each energy level must be given by the formula
E
n
=-hR1n2
E
n
h
R
1
n
2
(5)
We can draw two conclusions. First, the
electron in a hydrogen atom can exist only with certain energies,
corresponding to motion in what we now call a state or an
orbital. Second, the energy of a state can be
characterized by an integer
quantum number, n = 1, 2, 3, ... which determines its
energy.
These conclusions are reinforced by similar
observations of spectra produced by passing a current through other
elements. Only specific frequencies are observed for each atom,
although only the hydrogen frequencies obey the Rydberg
formula.
We conclude that the energies of electrons in
atoms are "quantized," that is, restricted to certain
values. We now need to relate this quantization of energy to the
existence of shells, as developed in a
previous study.
Observation 3: Photoelectron Spectroscopy of Multi-Electron
Atoms
The ionization energy of an atom tells us the
energy of the electron or electrons which are at highest energy in
the atom and are thus easiest to remove from the atom. To further
analyze the energies of the electrons more tightly bound to the
nucleus, we introduce a new experiment. The photoelectric effect
can be applied to ionize atoms in a gas, in a process often called
photoionization. We shine light on an atom and measure
the minimum frequency of light, corresponding to a minimum energy,
which will ionize an electron from an atom. When the frequency of
light is too low, the photons in that light do not have enough
energy to ionize electrons from an atom. As we increase the
frequency of the light, we find a threshold at which electrons
begin to ionize. Above this threshold, the energy
hν
h
ν
of the
light of frequency νν is greater than the energy required to
ionize the atom, and the excess energy is retained by the ionized
electron as kinetic energy.
In photoelectron spectroscopy, we measure the
kinetic energy of the electrons which are ionized by light. This
provides a means of measuring the ionization energy of the
electrons. By conservation of energy, the energy of the light is
equal to the ionization energy IEIE plus the kinetic energy KEKE of the
ionized electron:
hν=IE+KE
h
ν
IE
KE
(6)
Thus, if we use a known frequency νν and
measure KEKE, we can determine IEIE. The more tightly bound an electron
is to the atom, the higher the ionization energy and the smaller
the kinetic energy of the ionized electron. If an atom has more
than one electron and these electrons have different energies, then
for a given frequency of light, we can expect electrons to be
ejected with different kinetic energies. The higher kinetic
energies correspond to the weakly bound outer electrons, and the
lower kinetic energies correspond to the tightly bound inner
electrons.
The ionization energies for the first twenty
elements are given in
Table 1. We note that there is a single
ionization energy for hydrogen and helium. This is consistent with
the shell model of these atoms since, in both of these atoms, the
electron or electrons are in the innermost shell. The energies of
these electrons correspond to the
n=1n1 energy level of the hydrogen
atom. In lithium and beryllium, there are two ionization energies.
Again, this is consistent with the shell model, since now there are
electrons in both of the first two shells. Note also that the
ionization energy of the inner shell electrons increases as we go
from hydrogen to lithium to beryllium, because of the increase in
nuclear charge. The lower energy electrons correspond to the
n=1n1
energy level of hydrogen and the higher energy electrons correspond
to the
n=2n2 energy level.
| Element |
Ionization Energy (MJ/mol) |
| H |
1.31 |
|
|
|
|
|
| He |
2.37 |
|
|
|
|
|
| Li |
6.26 |
0.52 |
|
|
|
|
| Be |
11.5 |
0.90 |
|
|
|
|
| B |
19.3 |
1.36 |
0.80 |
|
|
|
| C |
28.6 |
1.72 |
1.09 |
|
|
|
| N |
39.6 |
2.45 |
1.40 |
|
|
|
| O |
52.6 |
3.12 |
1.31 |
|
|
|
| F |
67.2 |
3.88 |
1.68 |
|
|
|
| Ne |
84.0 |
4.68 |
2.08 |
|
|
|
| Na |
104 |
6.84 |
3.67 |
0.50 |
|
|
| Mg |
126 |
9.07 |
5.31 |
0.74 |
|
|
| Al |
151 |
12.1 |
7.79 |
1.09 |
0.58 |
|
| Si |
178 |
15.1 |
10.3 |
1.46 |
0.79 |
|
| P |
208 |
18.7 |
13.5 |
1.95 |
1.01 |
|
| S |
239 |
22.7 |
16.5 |
2.05 |
1.00 |
|
| Cl |
273 |
26.8 |
20.2 |
2.44 |
1.25 |
|
| Ar |
309 |
31.5 |
24.1 |
2.82 |
1.52 |
|
| K |
347 |
37.1 |
29.1 |
3.93 |
2.38 |
0.42 |
| Ca |
390 |
42.7 |
34.0 |
4.65 |
2.9 |
0.59 |
Surprisingly, though, boron has three
ionization energies, which does not seem consistent with the shell
model. From the hydrogen atom energy levels, we would have expected
that all n=2n2 electrons would have the same energy. We can note that
the two smaller ionization energies in boron are comparable in
magnitude and smaller by more than a factor of ten than the
ionization energy of the electrons in the inner shell. Thus, the
electrons in the outer n=2n2 shell apparently have comparable
energies, but they are not identical. The separation of the second
shell into two groups of electrons with two comparable but
different energies is apparent for elements boron to neon.
As such, we conclude from the experimental
data that the second shell of electrons should be described as two
subshells with slightly different energies. For
historical reasons, these subshells are referred to as the as the
"2s" and "2p" subshells, with 2s electrons
slightly lower in energy than 2p electrons. The energies of the 2s
and 2p electrons decrease from boron to neon, consistent with the
increase in the nuclear charge.
Beginning with sodium, we observe four
distinct ionization energies, and beginning with aluminum there are
five. Note for these elements that the fourth and fifth ionization
energies are again roughly a factor of ten smaller than the second
and third ionization energies, which are in turn at least a factor
of ten less than the first ionization energy. Thus, it appears that
there are three shells of electrons for these atoms, consistent
with our previous shell model. As with n=2n2, the n=3n3 shell is again
divided into two subshells, now called the 3s and 3p
subshells.
These data also reveal how many electrons can
reside in each subshell. In each n level, there are two elements
which have only the ionization energy for the s subshell. Hence, s
subshells can hold two electrons. By contrast, there are 6 elements
which have both the s and p subshell ionization energies, so the p
subshell can hold 6 electrons.
The shell and subshell organization of
electron energies can also be observed by measuring the
"electron affinity" of the atoms. Electron affinity is
the energy released when an electron is added to an atom:
A
(
g
)
+
e
-
(
g
)
→
A
-
(
g
)
A
(
g
)
+
e
-
(
g
)
→
A
-
(
g
)
(7)
If there is a strong attraction between the
atom A and the added electron, then a large amount of energy is
released during this reaction, and the electron affinity is a large
positive number. (As a note, this convention is the opposite of the
one usually applied for energy changes in reactions: exothermic
reactions, which give off energy, conventionally have negative
energy changes.)
The electron affinities of the halogens are
large positive values: the electron affinities of F, Cl, and Br are
328.0 kJ/mol, 348.8 kJ/mol, and 324.6 kJ/mol. Thus, the attached
electrons are strongly attracted to the nucleus in each of these
atoms. This is because there is room in the current subshell to add
an additional electron, since each atom has 5 p electrons, and the
core charge felt by the electron in that subshell is large.
By contrast, the electron affinities of the
inert gases are
negative: the addition of an electron to an
inert gas atom actually requires the
input of energy, in effect, to force the
electron into place. This is because the added electron cannot fit
in the current subshell and must be added to a new shell, farther
from the nucleus. As such, the core charge felt by the added
electron is very close to zero.
Similarly, the electron affinities of the
elements Be, Mg, and Ca are all negative. This is again because the
s subshell in these atoms already has two electrons, so the added
electron must go into a higher energy subshell with a much smaller
core charge.
Electron Waves, the Uncertainty Principle, and Electron
Energies
We now have a fairly detailed description of
the energies of the electrons in atoms. What we do not have is a
model which tells us what factors determine the energy of an
electron in a shell or subshell. Nor do we have a model to explain
why these energies are similar but different for electrons in
different subshells.
A complete answer to these questions requires
a development of the quantum theory of electron motion in atoms.
Because the postulates of this quantum theory cannot be readily
developed from experimental observations, we will concern ourselves
with a few important conclusions only.
The first important conclusion is that the
motion of an electron in an atom is described by a wave function.
Interpretation of the wave motion of electrons is a very
complicated proposition, and we will only deal at present with a
single important consequence, namely the
uncertainty principle. A characteristic of wave motion
is that, unlike a particle, the wave does not have a definite
position at a single point in space. By contrast, the location of a
particle is precise. Therefore, since an electron travels as a
wave, we must conclude that we cannot determine the precise
location of the electron in an atom. This is, for our purposes, the
uncertainty principle of quantum mechanics. We
can make measurements of the location of the
electron, but we find that each measurement results in a different
value. We are then forced to accept that we cannot determine the
precise location. We are allowed, however, to determine a
probability distribution for where the electron is
observed.
This probability distribution is determined by
quantum mechanics. The motion of the electron in a hydrogen atom is
described by a function, often called the
wave function or the
electron orbital and typically designated by the symbol Ψ. Ψ is a function of the position of the electron rr, and quantum mechanics tells us that
|Ψ|2
Ψ
2
is the
probability of observing the electron at the
location rr.
Each electron orbital has an associated
constant value of the electronic energy,
E
n
E
n
, in agreement with our
earlier conclusions. In fact, quantum mechanics exactly predicts
the energy shells and the hydrogen atom spectrum we observe. The
energy of an electron in an orbital is determined primarily by two
characteristics of the orbital. The first, rather intuitive,
property determines the average potential energy of the electron:
an orbital which has substantial probability in regions of low
potential energy will have a low total energy. By Coulomb’s
law, the potential energy arising from nucleus-electron attraction
is lower when the electron is nearer the nucleus. In atoms with
more than one electron, electron-electron repulsion also
contributes to the potential energy, as Coulomb’s law
predicts an increase in potential energy arising from the repulsion
of like charges.
A second orbital characteristic determines the
contribution of kinetic energy, via a more subtle effect arising
out of quantum mechanics. As a consequence of the uncertainty
principle, quantum mechanics predicts that, the more confined an
electron is to a smaller region of space, the higher must be its
average kinetic energy. Since we cannot measure the position of
electron precisely, we define the uncertainty in the measurement as
Δx
Δ
x
. Quantum mechanics also tells us that we cannot measure
the momentum of an electron precisely either, so there is an
uncertainty
Δp
Δ
p
in the momentum. In mathematical detail, the
uncertainty principle states that these uncertainties are related
by an inequality:
ΔxΔp≥h4π
Δ
x
Δ
p
h
4
(8)
where
hh is Planck’s constant,
6.62×10-34Js
6.62-34
J
s
(previously seen in
Einstein’s
equation for the energy of a photon). This inequality reveals that,
when an electron moves in a small area with a correspondingly small
uncertainty
Δx
Δ
x
, the uncertainty in the momentum
Δp
Δ
p
must be large. For
Δp
Δ
p
to be large, the momentum must also be
large, and so must be the kinetic energy.
Therefore, the more compact an orbital is, the
higher will be the average kinetic energy of an electron in that
orbital. This extra kinetic energy, which can be regarded as the
confinement energy, is comparable in magnitude to the
average potential energy of electron-nuclear attraction. Therefore,
in general, an electron orbital provides a compromise, somewhat
localizing the electron in regions of low potential energy but
somewhat delocalizing it to lower its confinement energy.
Electron Orbitals and Subshell Energies
We need to account for the differences in
energies of the electrons in different subshells, since we know
that, in a Hydrogen atom, the orbital energy depends only on the n
quantum number. We recall that, in the Hydrogen atom, there is a
single electron. The energy of that electron is
thus entirely due to its kinetic energy and its attraction to the
nucleus. The situation is different in all atoms containing more
than one electron, because the energy of the electrons is affected
by their mutual repulsion. This repulsion is very difficult to
quantify, but our model must take it into account.
A simple way to deal with the effect of
electron-electron repulsion is to examine the shell structure of
the atom. The two n=1n1 electrons in beryllium are in a shell with a
comparatively short average distance from the nucleus. Therefore,
the two n=2n2 electrons are in a shell which is, on average,
"outside" of the n=1n1 shell. The n=1n1 electrons are thus
the "core" and the n=2n2 electrons are in the valence
shell. This structure allows us to see in a simple way the effect
of electron-electron repulsion on the energies of the n=2n2
electrons. Each n=2n2 electron is attracted by the +4 charge on the
tiny beryllium nucleus, but is repelled by the two -1 charges from
the inner shell formed by the two n=1n1 electrons. Net, then, an n=2n2
electron effectively "sees" roughly a +2 nuclear
charge. We refer to this +2 as the "core charge" since
it is the net charge on the core resulting from the balance of
attraction to the nucleus and repulsion from the core electrons.
The nucleus is partially "shielded" from the valence
electrons by the core electrons.
This shielding effect does not seem to account
for the difference in ionization energies between 2s and 2p or for
the lower ionization energy of boron compared to beryllium, since,
in each atom, the valence electrons are in the
n=2n2 shell. However,
the shielding effect is not perfect. Recall that we only know the
probabilities for observing the positions of
the electrons. Therefore, we cannot definitely state that the
n=2n2
electrons are outside of the
n=1n1 core. In fact, there is some
probability that an
n=2n2 electron might be found inside the
n=1n1
core, an effect called "core penetration." When an
n=2n2
electron does penetrate the core, it is no longer shielded from the
nucleus. In this case, the
n=2n2 electron is very strongly attracted
to the nucleus and its energy is thus lowered. What is the extent
of this penetration? We must consult quantum theory. The answer is
in
Figure 3, which shows the probability of finding an electron a
distance
rr away from the nucleus for each of the 1s, 2s, and 2p
orbitals. We can see that there is a greater probability (though
small) for the 2s electron to penetrate the core than for the 2p
electron to do so.
As a result of the core penetration, an
electron in a 2s orbital feels a greater "effective nuclear
charge" than just the core charge, which was approximated by
assuming perfect shielding. Thus the effective nuclear charge for a
2s electron is greater than the effective nuclear charge for a 2p
electron. Therefore, the energy of an electron in the 2s orbital in
beryllium is lower than it would be in the 2p orbital.
A detailed analysis from quantum mechanics
gives the following ordering of orbitals in order of increasing
energy:
1s<2s<2p<3s<3p<4s<3d<4p<5s<...
1s
2s
2p
3s
3p
4s
3d
4p
5s
...
(9)
This ordering can be rationalized on the basis
of effective nuclear charge, shielding, and core
penetration.
Review and Discussion Questions
Problem 1The photoelectric effect demonstrates that radiation energy
is quantized into "packets" or photons. Explain how and why this
observation is of significance in understanding the structure of
atoms.
Problem 2Explain how we can know that higher frequency light contains
higher energy photons.
Problem 3Electron affinity is the energy released when an electron is
attached to an atom. If an atom has a positive electron affinity,
the added electron is attracted to the nucleus to form a stable
negative ion. Why doesn't a Beryllium atom have a positive electron
affinity? Explain how this demonstrates that the energy of a 2s
orbital is less than the energy of a 2p orbital.
Problem 4Why does an inert gas atom have a high ionization energy but
a low electron affinity? Why do these properties combine to make
the atoms of inert gases unreactive?
Problem 5Consider electrons from two different subshells in the same
atom. In photoelectron spectroscopy, the lower energy electron has
a higher ionization energy but is observed to have lower kinetic
energy after ionization. Reconcile the lower energy with the higher
ionization energy with the lower kinetic energy.
Problem 6Chlorine atoms have 5 distinct ionization energies. Explain
why. Predict the number of ionization energies for Bromine atoms,
and explain your answer. (Hint: examine the structure of the
periodic table.)
Problem 7
Why does a Bromine atom have a much smaller radius than a
Potassium atom, even though a Br atom has 16 more electrons than
does a K atom?
Problem 8Explain why electrons confined to smaller orbitals are
expected to have higher kinetic energies.
Problem 9
Define "shielding" in the context of
electron-electron repulsion. What is the significance of shielding
in determining the energy of an electron? How is the affected by
core penetration?
"General Chemistry course in Braille."