C10J
ATOMIC STRUCTURE
(6 lectures)
Introduction
The Atomic Structure course is considered as an important part of the core
course for Introductory Chemistry as concepts which are learnt here will be
employed in the subsequent sections on Physical, Inorganic (e.g. bonding and
antibonding orbitals) and Organic Chemistry (e.g. hybridization). This covers
basic and fundamental principles which are common to all areas of Chemistry.
You are encouraged at this early stage in the degree programme, not to form, or
create, an imaginary boundary which separates the different disciplines of
Chemistry, but to search for and make the links which will make understanding
the different topics more enjoyable.
Atomic structure refers to the configuration of the constituents of the atom.
When we speak of electronic configuration we deal with information that tells
us where in the atom the electron(s) is (are) found. The chemistry of an element
is determined largely by its electronic configuration, so it is very important to
get as much information as possible about the behaviour of electrons in atoms.
It is a good idea to start with the simplest atom, which is Hydrogen, and try to
find out as much as possible about it. Here we refer to the isotope with 1 proton
and 1 electron. Do we know everything about the electron in the hydrogen
atom? Do we really understand what is meant by the phrase “The electron in the
ground state hydrogen atom resides in the 1s orbital?” Do we know where the
electron is in the hydrogen atom? How does it move about in the atom? These
are some of the questions which we will try to address in this section. After we
have gained information about the behaviour of the electron in the ground and
excited states of the hydrogen atom, then we will try to apply what we have
learned to the higher atoms.
Recommended Text: The Elements of Physical Chemistry - P. W. Atkins, 3rd
Ed, Oxford University Press (2001).
Contact for Dr. P. Maragh:
paul.maragh@uwimona.edu.jm
1
This Atomic Structure Unit is delivered in seven sessions which are:
Session 1:
Session 2:
Session 3:
Session 4:
Session 5:
Session 6:
Session 7:
The electronic structure of atoms
Behaviour Of The Electron In The Constant Energy
Hydrogen Atom
The Schrödinger Wave Equation and the Ground & Excited
States of Hydrogen
Quantum Numbers
The Electronic Structure of Higher Atoms
Spherical Polar Coordinates, Radial & Angular
Wavefunctions
Angular Momentum & Space Quantization and Ionization
Enthalpies
In this unit we will be looking at the behaviour of the electron in the
ground state hydrogen atom and will describe the behaviour in terms of its wave
properties. We then look at the behaviour of the electron in excited states of
hydrogen and explain the behaviour and properties of many-electron atoms by
applying the information gained from hydrogen.
Learning objectives
At the end of this unit you will be able to:
• Calculate the electrostatic and gravitational forces between two bodies or
particles
• State the Heisenberg Uncertainty Principle and calculate the uncertainty
in position or velocity of a particle or body
• Define the de Broglie wavelength and calculate same for particles and
bodies
• Explain interference and diffraction in light and electrons
• Explain the terms wavefunction, Eigenfunction and Hamiltonian operator
as they appear in the Schrödinger Wave Equation
• Sketch the radial wavefunctions for the 1s, 2s and 2p orbitals
• Sketch the Radial Distribution Functions for 1s, 2s and 2p orbitals
• Define and depict radial and angular nodes on orbitals
• Define and give examples of principal, orbital angular momentum,
magnetic and spin quantum numbers
• Calculate the energy of the levels and the emission lines in the hydrogen
atom
• Explain the Orbital Approximation and apply it to the Helium atom
2
• State the Pauli Exclusion Principle, and rationalize it in terms of the
relative stability of different electronic configurations (e.g. Lithium).
• State Hund’s rule and explain it in terms of the relative stability of the
different electronic configurations of sub-shells (e.g. Carbon)
• Define Cartesian and Spherical Polar coordinates
• State advantages of expressing wavefunctions in Spherical Polar
coordinates
• Define radial wavefunction and angular wavefunction
• Calculate and plot the hydrogen 1s Radial wavefunction
• Define and calculate orbital angular momentum of an electron in different
orbitals
• Define and explain Space Quantization
• Define Ionization Enthalpy and explain its trend across the Li – Ne
period.
3
SESSION 1: The Electronic Structure of Atoms
Overview: In this session, we will be looking at:
• the types and importance of forces, and their importance, that exist
between the proton and electron in the hydrogen atom.
• the applicability of the Heisenberg Uncertainty Principle to objects
of various sizes.
The Hydrogen Atom
(a) Orbits and Classical views of the atom:
The Hydrogen (H) atom
consists of two particles, a positively charged proton and a negatively charged
electron, moving about one another, under the influence of their mutual
electrostatic attraction. In this classical view of the hydrogen atom, (with the
electron moving around the proton in a fixed orbit), the electron should dissipate
(lose) energy in the form of radiation as it moves and therefore follow a spiral
path and collapse into the positive nucleus. This does not happen and so
classical physics (a branch of Physics) does not explain the behaviour of the
electron in the atom.
-
+
Electrostatic attraction is similar to gravitational attraction in that the force
involved in each case is inversely proportional to the square of the separating
distance. (You are asked to calculate and compare these forces in the tutorials!)
(i)
Electrostatic Force of Attraction:
If two (opposite) charges q1 and q2, in a vacuum, are separated by a
distance r, then the electrostatic force (of attraction) between them
is given by
FE =
q1 q 2
4πε o r 2
…(1)
where εo (= 8.854 x 10-12 C2 J-1 m-1) is the permittivity of a vacuum.
4
(ii)
Gravitational Force of Attraction:
If two bodies of mass m1 and m2, are separated by a distance r, then
the gravitational force of attraction between them is given by
FG =
Gm1 m 2
r2
…(2)
where G (= 6.672 x 10-11 N m2 kg-2) is the gravitational constant.
Example 1. Calculate the electrostatic force of attraction between two point
charges, +20 C and -25 C separated by 0.20 m in a vacuum.
Solution
Using equation (1)
FE =
=
q1q2
4πε o r 2
(+20C )(−60C )
4(3.142)(8.854 x10−12 C 2 J −1m −1 )(0.20m) 2
= 2.69 x1014 N
Example 2. Calculate the gravitational force of attraction between the earth and
the moon, making use of the following data:
Mass of earth, mearth = 5.97 x 1024 kg
Mass of moon, mmoon= 7.36 x 1022 kg
separating distance r = 3.84 x 108 m.
Solution
From equation (2):
Gm1 m 2
r2
(6.672 x10 −11 N .m 2 .kg − 2 )(5.97 x10 27 kg )(7.36 x10 22 kg )
=
(3.84 x10 8 m) 2
FG =
= 1.99 x10 23 N
In the H-atom, the electrostatic force is ~ 1039 times the gravitational force!
Hence the gravitational force is considered negligible and insignificant.
5
According to Classical Mechanics, used to explain the motion of large bodies,
the proton-electron system should behave similarly to that of the Earth – Moon
system i.e. they should both move about a common centre of gravity in circles
or ellipses.
The mass of the proton is approximately 2000 times (1836 times to be exact) the
mass of the electron. In this case, the centre of gravity of the proton-electron
system is so close to (actually inside) the proton that the proton can be assumed
to be in a fixed position and the electron orbiting it. If the position q, and
velocity v, of the electron are known at some time, then we should be able to
calculate the orbit in which it is moving and predict its future motion.
This is the point where we run into serious problems, as it is impossible to
measure both the position and velocity of the electron, simultaneously, with any
reasonable accuracy. If one is determined accurately then the accuracy with
which the other is measured is sacrificed.
This leads us to the Heisenberg Uncertainty Principle (HUP), which explains
this observation:
The Heisenberg Uncertainty Principle (HUP)
If q and v are the position and velocity of an electron respectively and δq and δv
are their respective uncertainties then, by the HUP,
δq.δv ≥
(h =
where
h
2me
…(3)
h
)
2π
h, Planck’s constant
= 6.626 x 10–34 J s, and
me, mass of the electron = 9.109 x 10-31 kg.
If one looks closely at the inequality (3), it can be seen that the right hand side is
a constant (for a given mass), which is to say that, the product of the
uncertainties cannot be less than the given value of h/4πme. If δq is very small
(i.e. q is known with high accuracy) then δv has to increase in value in order to
maintain the inequality, and vice versa.
6
The inequality can be rearranged and written as
δq.δp ≥ h/4π
…(4)
where p (= mv) is the linear momentum, and δp = m δv.
We will now draw on some experimental evidence, which will help to
shed some light on the matter of determining both the electron’s position and
momentum simultaneously with great accuracy. To locate the electron, one has
to use photons of a certain minimum energy.
Example 3. Determine the uncertainty in the path of flight of a 10.0 g-bullet if
its velocity is 500 ± 1 m s-1.
(N.B. In doing calculations, if all parameters are expressed in SI units then
the answer will come out with SI units).
Solution
From equation (4),
δq.δp ≥ h/4π
but
δp = δ(mv) = mδv (assuming that the uncertainty in the momentum is
as a result of the uncertainty in the velocity, and also
assuming negligible uncertainty in the mass).
and,
and
δq ≥
δq.δv ≥ h/4πm
δq ≥ h/4πmδv.
6.626x10−34 Js
4 x3.142x 0.0100kgx1ms −1
= 5.27 x10− 29 m
This result tells us that there is very small deviation of the flight path
(predicted position) of the bullet, which allows for high accuracy in prediction
of the bullet’s position. Hence, a marksman is able shoot a ‘bullseye’! This
small uncertainty is associated with objects much larger than electrons, atoms
and molecules.
Remember that all electromagnetic radiation consists of photons and the
energy of a photon is given by
E = hν
…(5)
7
where ν is the frequency (in s-1) of the radiation.
Locating the electron with photons can be compared to locating a
mosquito in the night with the light from a flashlight. Both cases are using
photons to locate an object, but in the case of the mosquito the photons have a
comparatively much smaller mass (much smaller momentum). This means that
the momentum from the light photons will not be enough to significantly disturb
the flight path of the mosquito, and one can see EXACTLY where the mosquito
is located at some specific point in time. Let us now turn our attention to the
case of the electron, which is a much, much smaller mass with a much, much
smaller momentum. To locate the electron with high accuracy we need to use
photons of high energy. High energy, here, means photons of higher velocity
and hence greater momentum, which will offer greater disturbances, upon
impact, to the path of the electron. If we need to be more accurate in locating
the electron, we will have to use photons of higher energy, and then there will
be even more disturbances in the momentum of the electron. To cause fewer
disturbances in the momentum then we will have to use photons of lower
energy, and this will cause us to locate (see) the electron less accurately.
Locating the electron with pinpoint accuracy (i.e. δq = 0) would require δv = ∞,
which would mean that we know nothing at all about its velocity.
This difficulty or uncertainty exists for all systems. For the earth – moon
system where the mass of the moon (mmoon) = 7.4 x 10 22 kg, δq.δv is negligible
in the case of the moon, but is significant in the case of the proton-electron
system where the mass of the electron is 9.1 x 10-31 kg. This implies that the
orbit (or path) of the electron cannot be calculated (with a reasonable degree of
accuracy).
ALL IS NOT LOST!
At this point it would seem useless to investigate the matter any further,
but as scientists we seek out another observable which we can determine with
greater accuracy and try and get as much information from (or about) it as
possible. There are other properties of the H atom, which can be measured
accurately, and do not depend on the simultaneous position and velocity of the
electron. Such a useful physical property is energy.
The Law of Conservation of Energy tells us that the total energy of an
isolated H atom must have some fixed value, regardless of the positions and
velocities of the proton and electron.
Total Energy = Kinetic Energytotal
(due to motion)
8
+ Potential Energytotal
(due to attraction)
(6)
The HUP restricts the determination of the total energy of the proton and of the
electron separately but we can determine the sum. So, the logical thing to do
here is to make some observations on the atom and its components while
maintaining the total energy at a constant value.
9
SESSION 2: Behaviour Of The Electron In The Constant Energy
Hydrogen Atom
Overview: In this session we will be:
• looking at the regularities in the baviour of the electron in the
hydrogen atom
• doing a short introduction to waves and their interaction
• looking at matter in terms of both wave and particulate properties
The first part of the discussion will be focusing on the hydrogen atom in
its lowest possible energy state, referred to as the ground state.
A very important regularity appears in measurements of the position of
the electron in the H atom. What is found, is that we can give a definite estimate
of the likelihood that the electron will be found in some specified region of
space.
Suppose the electron and the proton are separated by a distance r, and r is
measured a large number of times with the H atom in the same state (same total
energy). The value of r will vary (as suggested by the uncertainty principle), and
the probability of r having some value a is always the same. In other words the
probability of getting some value of r between a and a + δa is always the same.
We can express this by saying that the probability is given by P(a)δr, where P(r)
is a function of r that we can determine from our measurements. This would
suggest that we could specify a given state of the atom by specifying the
corresponding function P, which can be determined by experiment and which
does in fact vary with the energy of the atom.
This is some achievement, but is not enough since it does not take into
account that very small particles exhibit wavelike behaviour. In fact, electrons
undergo diffraction and interference just like light.
This takes us to the concept of Wave-Particle Duality, where waves are seen to
exhibit particle-like properties and particles exhibit wave-like properties, i.e.
matter has a dual nature.
Matter Waves
It was put forward, by Louis de Broglie in 1924, that any particle travelling with
linear momentum p (= mv) should have a wavelength λ given by the relation:
10
λ=
h
.
p
(7)
Since p = mv, then it is expected that for a given value of the velocity v, smaller
particles will have a larger wavelength. The display of wave property by a
particle occurs when the particle is bounded in a region which has dimensions
that are comparable to the natural wavelength of the particle. The diameter of a
ground state hydrogen atom is about 1 Å (= 10-10 m = 100 pm). An electron
traveling with a speed of one-third the speed of light has a natural wavelength of
about 7 pm. This therefore suggests that an electron bounded within the
hydrogen atom (or other atoms) will display wavelike properties. This
observation is crucial in that it suggests that wave properties should become
more important for smaller particles. This therefore means that classical
mechanics is not able to explain the motion of very small particles. To account
for the wave-like nature exhibited by these small particles, Wave Mechanics is
the tool of choice.
Exercise 4. Calculate the de Broglie wavelength of a 50.0-kg man running at a
velocity of 10.0 m s-1.
Solution
λ = h/p = h/mv = 6.626 x 10-34 J s/(50.0 kg x 10.0 m s-1) = 1.33 x 10-36 m.
The value of this wavelength (which aids us in the prediction of wave-like
characteristics) is extremely small, and confirms that the wave properties of the
man can be ignored.
Self Test. Calculate the de Broglie wavelength of the electron mentioned above.
Waves, Waveforms and Wavefunctions
Not every student will have done physics, and so might not have met the
topic on waves. Before we move on, we must have, at least, a basic
understanding of the behaviour and interaction of waves. This basic
understanding of waves should help you to appreciate the wave concept. Waves
can be considered simply:
The waves (or waveforms) depicted by a and b below (left) are generated at the
same time and are doing exactly the same thing at any point in time, and are
said to be in phase. The interaction of waves is called interference. If these two
11
waveforms interact then a new waveform c, is produced with twice the
amplitude of the original waveforms, but with the same wavelength λ. This type
of interaction where a waveform is produced with greater amplitude is called
constructive interference. This compares to a brighter light or louder sound.
a
c
b
λ
λ
The waves (or waveforms) depicted by a and b below are generated at the same
time and are doing exactly the opposite thing at the same time, and are said to
be (completely) out of phase. The interaction of these two waveforms leads to
complete cancellation. This type of interaction is called destructive
interference. This would result in darkness if they were light waves.
a
complete
cancellation
b
12
This forms the basis for wave interaction. It must be noted, however, that
completely constructive or destructive interference does not occur in every case
of interaction of two waves. An intermediate phase difference will lead to
intermediate interaction intensity. This is demonstrated by the Young’s Double
Slit experiment:
1
s
*
o
2
screen
S is a monochromatic (single wavelength) light source. Slits 1 and 2 are
equidistant from the slit O, which means that points on the same wave front will
arrive at slits 1 and 2 simultaneously, and 1 and 2 will now become two new
identical monochromatic sources of light. What is observed on the screen is a
very bright spot (or line) where it intersects the dashed line, with a series of
symmetrical light and dark spots going up and down from the centre, with the
light spots becoming progressively less intense. These bright spots occur when
the waves from the two sources (slits 1 and 2) reach a spot on the screen and are
in phase. Similarly the dark spots occur when they reach a spot and are (exactly)
out of phase. This display of light and dark spots is known as a diffraction
pattern, which is as a consequence of the wave nature of light.
If this experiment is repeated (as done by Davisson & Germer) with a
source of monoenergetic electrons with the same wavelength, a similar
diffraction pattern is observed. Not that you would see bright and dark spots but,
regions of high and low electron density would be seen. This confirms that
electrons do have wave properties.
One cannot tell where a given electron will hit the screen because of the
HUP, but if one uses a many electron beam then the proportion hitting some
particular area will have a definite value given by the corresponding probability
function P. The analogy with light indicates that electronic behaviour cannot
13
only be explained by P but with a wave function ψ (psi – pronounced “sigh”), of
which P is the square.
This does not mean that electrons are waves, but that their behaviour is
governed by mathematical equations similar to those for wave motion. A
mathematical representation describing a given wave motion is called a
wavefunction, ψ. This wavefunction can be positive or negative. The intensity
of the wave motion (e.g. the brightness of light or the loudness of sound) at
some point in space is a function of ψ2 at that point.
For two interacting wave forms with respective wavefunctions ψ1 and ψ2,
the wavefunction describing the new wave form in the region of space where
they interact is given by, ψ12 = ψ1 + ψ2. The resulting intensity is therefore
given by
ψ122 = (ψ1 + ψ2)2 = ψ12 + ψ22 + 2ψ1ψ2.
(8)
This result shows us that the intensity of the new waveform is different from
that of the sum of the intensities of the two separate waveforms, i.e.
(ψ1 + ψ2)2 ≠ ψ12 + ψ22
(9)
Since ψ1 and ψ2 can be positive or negative, so can be their product ψ1ψ2. It can
therefore be seen that when both ψ1 and ψ2 have the same sign then,
(ψ1 + ψ2)2 > ψ12 + ψ22
and when they are of opposite signs then,
(ψ1 + ψ2)2 < ψ12 + ψ22.
It can be reasoned as follows:
if ψ2 is negative, then ψ1 + ψ2 becomes ψ1 + -ψ2 which is equal to ψ1 – ψ2.
Therefore,
(ψ1 – ψ2)2 = ψ12 + ψ22 – 2ψ1ψ2
which is less than (ψ1 + ψ2)2 by 4ψ1ψ2.
14
You will find this very useful when studying hybridization of orbitals and the
formation of bonds between atoms to form molecules, since both of these
concepts involve the interaction of orbitals which can be on the same atom (as
in hybridization) or on different atoms (as in bond formation).
Exercise 5. Two hydrogen (H) atoms are interacting to form a molecule (H2). If the
wavefunction for each atom is given as ψa, which is allowed to be positive or
negative, derive the intensities associated with the resulting wavefunctions.
Comment on the solutions.
Solution
When the wavefunctions are both positive and both negative the resulting
wavefunctions are:
ψa + ψa = ψa + ψa = 2ψa, and (- ψa) + (-ψa) = -( ψa + ψa) = -2 ψa , respectively
The intensities are given by the square of the wavefunctions, and give the same
solution of 4ψa2 in both cases, i.e. (2ψa)2 = (-2ψa)2 = 4ψa2.
When one is positive and the other is negative, the new wavefunction is:
ψa + (-ψa) = ψa - ψa = 0! The wavefunctions cancel and thus the intensity is also
zero.
The first result signifies constructive interference between the atoms and leads to
increased (shared) electron density between the nuclei and results in bond formation,
(i.e. bonding) and the second result shows destructive interference, or no shared
electron density and results in anti-bonding. (You will meet this topic in the
Inorganic part of the course under the topic of Structure and Bonding, and in the
Organic Chemistry part under Hybridization).
15
SESSION 3: The Schrödinger Wave Equation and the Ground & Excited
States of Hydrogen
Overview: In this session we will be looking at:
• the basic form of the Schrödinger Wave Equation and information
derived from it
• the wavefunctions for the ground and some excited states of
hydrogen
After the introduction to waves and wavefunctions in the previous session, and
now, with a better understanding of the interaction of waves, we can now return
to the ground state hydrogen atom and resume our investigation of the
behaviour of the electron.
What we can conclude here is that since the electron in the hydrogen atom
exhibits wave-like characteristics, then its motion must therefore be described
by a wavefunction, ψ. This wavefunction must contain (or must allow us to
determine) all the information about the electron in the ground state hydrogen
atom. The atom is three-dimensional. The electron is free to move about in this
three-dimensional space. In space, the position of a particle can be located by
the three Cartesian Coordinates x, y and z. The wavefunction ψ can therefore be
described as a function of x, y and z.
The probability of locating the electron in the ground state hydrogen atom is
given by the probability function P. The probability function P is given by the
square of the wavefunction,
P = ψ2.
i.e.
The probability of finding the electron in a region of space with volume dτ (=
dx.dy.dz), is
P = ∫ ψ 2 dτ
or
P=
∫
ψ2 dx.dy.dz
where dτ is a small cube of dimensions dx, dy and dz.
The wavefunction ψ is determined by solving the Schrödinger Wave Equation
(SWE).
16
The Schrödinger Wave Equation (SWE)
The (time-independent) Schrödinger Wave Equation is of the general form
Hψ = Eψ .
In this course you will not be required to solve this equation but at least you
should understand its significance and it mode of operation. It is solved by
applying a special kind of algebra known as operator algebra (i.e. the method
of wave mechanics is operator algebra).
H is the Hamiltonian or energy Operator, which prescribes a series of
mathematical operations associated with kinetic and potential energy to be
performed on the wavefunction ψ.
The wavefunction ψ, is a mathematical expression that describes or defines the
electron in terms of its wave properties. So for the electron in the lowest energy
(ground) state hydrogen atom, the specific wavefunction would describe the
wave properties of the electron under these specific conditions.
The wavefunction appears on both sides of the SWE, and this is for a very
special reason. In order to solve the SWE, the wavefunction must also come out
of (be a part of) the solution. This implies that for a specific state of a system (in
this case the ground state hydrogen atom) the wavefunction is specific and
unique.
The Energy Function E (the total energy), is the net energy of the system
under investigation. Again, for the ground state hydrogen atom, E in this case
would represent the energy of the electron in the field of the proton. For the
wavefunction to be solvable ,E cannot have just any arbitrary value but only one
of a series of discrete, possible, energy values. (i.e. Quantum Theory).
For the hydrogen atom in the ground state H, E and ψ are specific, thus:
H ψ 1 = E1ψ 1
where the subscript ‘1’ denotes the ground state (or first allowed energy state)
of the hydrogen atom.
Classical mechanics states that the energy of any object should be continuous.
17
Solution of the SWE, and therefore Wave Mechanics, predicts that there is a
definite state of minimum (and non-zero) energy of the hydrogen atom for
which ψ is spherically symmetric. Having spherical symmetry, means that, if
determinations of ψ are made at the same distance in all directions around the
H-nucleus, the wavefunction is found to be the same. It is different for a
different distance, but is the same for a given distance.
The following figure depicts the variation of the wavefunction with distance, r
(r = [x2 + y2 + z2] ½) , from the nucleus for the ground state H-atom.
ψ(r)
r
This figure shows that the wavefunction has larger values for small electron –
nucleus separations, and smaller values as the separation increases. The notation
ψ(r), means that ψ is a function of r. In other words, the value of ψ depends only
on the value of r.
We will now recall that the probability of finding the electron at a distance r
from the nucleus is given by the square of ψ at r. A plot of the square of ψ(r)
against r is shown below.
ψ2(r)
r
18
This figure suggests that, in the ground state H-atom, the probability of finding
the electron increases the closer the nucleus is approached. In simple words: on
average, the electron spends most of its time closer to the nucleus.
If several photographs are taken of the H-atom, and if it were possible to see the
electron in these pictures, the electron would be in different places each time.
However, if a long-time exposure photograph is taken, then the resulting picture
would show a blur, with the blur being most dense at the centre of the atom, as
shown below:
The density of the blur in any small volume element, dV, is representative of the
probability of finding the electron in that volume element.
Excited States of Hydrogen Atom
The H-atom can be excited by the absorption of discrete amounts of radiation
energy, to higher energy states, each described by its own characteristic
wavefunction. Because of the obvious analogy between the classical orbits and
the wave mechanical wavefunctions, the latter are referred to as orbitals.
In the H-atom the ground state orbital has the same sign everywhere. This
implies that the sign of the ground state H-atomic orbital must be either all
positive or all negative. Inspection of the SWE reveals that if ψ is a solution,
then -ψ is also a solution. The important thing however is that, whether the
orbital is all positive or all negative, the square gives the same result.
Other orbitals consist of parts with opposite signs. Where the orbital changes
sign, ψ = 0 , and it must of course pass through zero. Regions where ψ = 0 are
called nodes. Therefore, the probability of finding the electron there is zero.
19
The first available excited state of the hydrogen atom, has a wavefunction of
spherical symmetry and changes sign once in going from the nucleus outwards.
This is shown in the following figure:
+
+
ψ
ψ
r
–
r
–
(a)
(b)
ψ2
r
(c)
These figures show that the absolute sign (phase) of the wavefunction is
insignificant. The square of both gives the same result, which represents the
probability function P (= ψ2 = (-ψ)2 ). The electron-density profile and a
diagram for the above wavefunction can be represented as:
radial node (ψ = 0)
–
+
electron density profile
wavefunction, ψ (for (a))
20
respectively, where the the electron density is high at the centre, falls off to zero
at the radial node (ψ = 0), increases to a maximum and then falls off as the
electron-nucleus distance increases.
Another of the excited states of the H-atom has a wavefunction that is depicted
in the following figure:
+
r
ψ
r
angular node (ψ = 0)
–
The figure shows that this specific wavefunction has opposite signs on opposite
sides of the nucleus. The square of this wavefunction however is identical on
opposite sides, representing equal distribution of electron density on both sides
of the nucleus. (See following figure)
angular node
nodal plane
21
SESSION 4: Quantum Numbers
Overview:
•
•
•
•
•
In this session we will:
define quantum numbers
see the origin and significance of quantum numbers
determine the separation of the energy levels in hydrogen
looking at shell and sub-shells in hydrogen
looking at the emission spectra of hydrogen
Each atomic orbital (AO) is specified by a principal quantum number n, by
an orbital angular momentum (or azimuthal) quantum number l, and by a
magnetic quantum number ml. All of these three quantum numbers come out
of the solution of the SWE for the individual hydrogen atomic orbitals, and
together they give the ‘address’ of the electron in the atom (i.e. the orbital in
which the electron is located).
The principal quantum number, n is used to calculate the energy of the electron
in the atom, and is equal to “the total number of nodes in an orbital plus one”
and has values:
n = 1, 2, 3, …
The orbital angular momentum quantum number is equal to the number of
angular nodes present in an orbital and has n possible values:
l = 0, 1, 2, 3, … (n –1)
i.e. for a specific value of n, l can have values from a minimum of zero
up to a maximum of (n – 1).
For a given value of l, there are 2l +1 values of the magnetic quantum number
ml in the range:
ml = 0, ±1, ±2 … ±l.
(or +l, l – 1, l – 2, … –l)
text was removed from here and placed above in blue
Shells
In the hydrogen atom, the energy of an orbital depends only on the value of n.
This means that all orbitals with the same value of n, regardless of the values of
l and ml, are degenerate (have the same energy). This is the reason that all
22
orbitals in hydrogen with the same value of n are said to belong to the same
shell. The values of n and the corresponding shell designations are:
n = 1
K
2
L
3
M
4…
N…
Sub-Shells
Orbitals with the same value of n, but different values of l belong to different
sub-shells which are denoted as follows:
l =
0
s
1
p
2
d
3
f
Each sub-shell contains 2l + 1 individual orbitals corresponding to the 2l + 1
values of ml (all with the same value of l).
Table X. Shells, sub-shells and orbitals for a given value of n
shell
n
l
ml
type
sub-orbitals
# of orbitals/shell
K
1
0
0
s
1
1
L
2
0
0
s
1
1
1, 0, -1
p
3
4
M
3
0
0
s
1
1
2
1, 0, -1 2, 1, 0, -1, -2
p
d
3
5
9
It can be seen from the above table that for n = 1, there is only one value of l
(i.e. l = 0), and hence one value of ml and therefore only one orbital. Similarly
for n = 2, there are 4 orbitals and for n = 3 there are 9. In fact, the number of
orbitals for a shell of quantum number n is equal to n2. There is always one sorbital in a shell and (whenever they are present), three p-orbitals and five dorbitals.
Apart from splitting the nth level into individual sub-levels, ml describes how
each orbital is oriented relative to any arbitrary or preferred direction in space
23
(such as that of a magnetic field). The magnetic quantum number, ml, is used to
explain additional lines that appear in the emission line spectra when the atom is
subjected to an external magnetic field.
Spin Quantum Number, ms
There is, in fact, a fourth quantum number which gives the absolute
configuration that the electron is allowed to have in an orbital. The spin
quantum number ms, is allowed the values of +½ and –½, signifying two
possible spin configurations, α and β. At this tme, these can be viewed simply as
an up or down spin or a clockwise or anticlockwise spin. As you will see later on
in the chapter, for two electrons to occupy the same orbital they must have
opposite spins.
Energy Levels in Hydrogen
In the case of hydrogen, as stated previously, the energy depends only on n.
Therefore 2s and 2p correspond to states of the same energy, as shown in the
following diagram:
2S
E
2Px
2Py
2Pz
1S
For hydrogen, when the electron is in the 2s orbital the state is 2S, when the
electron is in the 2px orbital the state is 2Px, and so on.
Separation of Energy Levels in Hydrogen Atom
When an electric discharge is passed through a sample of hydrogen gas, the
atoms are able to absorb energy. In absorbing this energy, electrons are
promoted to higher levels. When the electrons fall back to states with lower
energy giving out energy in the form of radiation (photons), an emission
spectrum is observed. This spectrum is seen to consist of discrete sharp lines
which mean that the electrons are moving between specific energy levels and in
24
doing so are able to absorb or give out a discrete amount of energy. The energy
levels in hydrogen are given quantitatively by:
En = −
hcRH
,
n2
where n (= 1, 2, 3 …) is the principal quantum number and RH (= 109,678 cm-1)
is the Rydberg constant for hydrogen.
In this definition, the ionized atom (H+ + e–) is taken as the zero of energy and
the energy levels lie at some negative value (more stable value) below this zero.
Therefore, the first three energy levels in hydrogen would be:
E1 = −
hcR
hcR H
hcR
hcR H
hcRH
= − hcRH , E 2 = − 2 H = −
, and E 3 = − 2 H = −
,
2
4
9
1
2
3
indicating how far below the zero each lies. As the value of n increases the
separation between successive energy levels decreases:
0
n=∞
– hcR/9
n=3
– hcR/4
n=2
– hcR
n=1
The Separation of Energy Levels in the Hydrogen Atom
When the H-atom absorbs or emits energy, the electron has to move between
these specific energy levels and therefore, absorbs or emits a specific amount of
electromagnetic energy. As a consequence, a line spectrum is observed. The
observed spectral emissions (lines) can be explained as follows:
25
Transition from n = 3 to n = 2 (in emission). The difference in energy, ∆E, is
∆E = E3 − E2 =  −
hcRH   hcRH 
1 1
1 1
 − −
 = hcRH  −  = hcRH  2 − 2 
9  
4 

 4 9
2 3 
Emission lines (4 to be exact) in the visible spectrum of hydrogen, occur when
the electron falls from a higher level down to n = 2. The expression
corresponding to these lines is
∆E = hcRH 
1
1
− 2  , where n = 3, 4, 5, 6.
2
n 
2
This visible series of emission lines is known as the Balmer Series. Two other
series: Lyman (ultra-violet) and Paschen (infra-red) occur when emissions
terminate in the n = 1 and the n = 3 levels respectively.
6
5
4
3
2
Paschen (IR)
Balmer
(visible)
1
n
Lyman (UV)
Figure showing the origin of the spectral series in hydrogen atom.
26
It can be seen from the above figure that if the series are placed in order of
increasing energy the result is: Paschen (IR) < Balmer (vis) < Lyman (UV).
Exercise 6. Calculate the energy (in Joules) of the photon emitted when an electron
falls from the n = 4 level to the n = 2 level in the hydrogen atom. Convert the energy
of the photon from Joules to wavenumbers (cm-1).
Solution
1 
 1
Energy = hcRH  2 − 2 
4 
2
= (6.626 x10 −34 J s )(2.998 x1010 cm s −1 )(109678 cm −1 )( 1 4 − 116)
= 4.085 x10 −19 J
The wavenumber of the photon is determined from the expression
1 
1
ν = RH  2 − 2 , where n and m are the quantum numbers of the levels.
m 
n
= (109678 cm −1 )( 1 4 − 116) = 20565 cm −1
The wavenumber of a radiation is the number of wavelengths that can fit in a length of
1 cm. In this example, 20565 wavelengths joined end-to-end would measure 1 cm.
Since we now have a fair understanding of what is happening in the hydrogen
atom, we can now turn to the next higher atom, Helium. This should be more
demanding since we now have to account for the repulsion experienced between
the two electrons and the separate attractions between both electrons and the
positive nucleus. In this section we will try to apply, what we have just learnt
about hydrogen, to see if it can be used to explain the behaviour of the electrons
in the Helium atom
27
SESSION 5: The Electronic Structure of Higher Atoms
Overview: In this session we will:
• apply the information gained in studying hydrogen to derive the
electronic configuration of higher atoms
• define and apply the Orbital Approximation to helium
• define and apply the Pauli Exclusion Principle
• define and apply Hund’s Rule
Helium: The Orbital Approximation
The Helium atom consists of a nucleus (with 2 protons) and 2 electrons moving
about it.
2+
e1
e2
The wavefunction ψHe, which describes this system, must allow for
simultaneous measurements of the positions of both electrons. ψHe must tell the
probability of finding e1 in volume element dτ1 and e2 in volume element dτ2,
where
and
dτ1 (= dx1.dy1.dz1)
dτ2 (= dx2.dy2.dz2)
have the dimensions given in parentheses.
The probability P, of finding e1 in volume element dτ1 and e2 in volume element
dτ2 is,
P = ψHe2. dτ1 .dτ2 ,
where ψHe is now a function of six coordinates.
28
If electrons did not repel each other then each would behave as if the other was
absent (i.e. Hydrogen-like) and could be represented by separate hydrogen-like
wavefunctions, ψ1 and ψ2, but this is NOT THE CASE.
The orbital approximation can be stated thus:
An electron in a many electron atom will feel the effect of, the averaged
charge of the nucleus and all other electrons present in the atom.
In this approximation it is assumed that each electron resides in its own orbital
and the total wavefunction,
ψ = ψ(1).ψ(2)…
where ψ(1) is the wavefunction for electron 1 and ψ(2) is the wavefunction for
electron 2 and, so on, where each individual orbital is hydrogen-like but where
the nuclear charges are modified (to obtain an effective nuclear charge) by the
presence of all other electrons in the atom.
Let us now take a step-by-step application of the orbital approximation to the
ground state Helium atom:
(i)
(ii)
(iii)
(iv)
(v)
There are two electrons, e1 and e2, and a nucleus with a 2+ charge.
Imagine that e1 is stationary at some location in the atom.
Let e2 move freely about and generate an electron charge cloud.
This cloud of charge will be symmetrically distributed about the nucleus.
Take the average of this charge cloud (in (iv) above) and the nuclear
charge – this will give an effective nuclear charge centred on the nucleus.
(vi) e1 will now feel the effect of this effective nuclear charge centred on the
nucleus.
(vii) The result of this approximation is a hydrogen-like wave function for e1.
If this is repeated, where e2 is now stationary and e1 is allowed to generate a
charge cloud, then a similar result is obtained where a hydrogen-like
wavefunction is the solution for e2.
These two resulting ground-state, hydrogen-like wavefunctions must be of the
1S type (i.e. ψ1S) since the ground state of hydrogen is 1S. This solution is
claiming that both electrons in the ground state helium atom reside in the 1S
orbital, and therefore the configuration of the ground state helium atom is
He = (1s)2
(i.e. ψHe = ψ1S ψ1S)
29
Quantum mechanical calculations may be performed to determine the energy of
the ground-state helium atom having the (1s)2 configuration. This calculated
value is in very good agreement with experimental values and confirms the
ground state configuration of the He atom.
This orbital approximation, as the name suggests, is only approximate, but it is a
useful model for discussing the properties of atoms, and is the starting point for
more sophisticated descriptions of atomic structure.
You must recognize that this is a very simplified approach and if it is used
without modification on the Lithium atom then the predicted theoretical result
would be a ground-state configuration of (1s)3, a configuration that gives a
predicted energy which is much less than the experimental value. This points to
the fact that the third electron cannot therefore reside in the 1s orbital, but has to
find other suitable ‘accommodation’ in another orbital of higher energy.
The question now arises, “Where can the third electron in Li find suitable
‘accommodation’?” This takes us to the Pauli Exclusion Principle.
The Pauli Exclusion Principle
The capacity of an orbital to hold electrons is limited. One can consider an
orbital ψ, as being subdivided into two spin-orbitals (ψα and ψβ, one for each
spin).
ψ = ψα . ψβ,
The Pauli Principle: only one electron can occupy a given spin orbital.
The third electron in Lithium must then seek an orbital of higher energy, which
must be the 2s or the 2p. In hydrogen, the 2s and the 2p are degenerate since it
is a one-electron system (without electronic interaction). This is not so for
multi-electron atoms.
This outer electron in helium does not see the full charge, Z, of the nucleus, due
to screening from the other electrons. It can be considered to experience an
effective nuclear charge Z*, also located at the nucleus. The question is “Will
the electron go to the 2s or to the 2p?” If the electron density distribution
diagrams are reviewed, it will be clear that an electron in the 2s orbital spends
more time near to the nucleus (which is more energetically favourable and
therefore more stable) than an electron in the 2p orbital. The more energetically
favourable ground-state configuration of Lithium is
30
Li = (1s)2(2s)1
[i.e. ψLi = (ψ1S)2.ψ2S]
(where (ψ1S)2 signifies a doubly occupied orbital, and not the square of the
wavefunction).
Atoms Be – Ne:
Hund’s Rule
By extrapolation, the configuration of the Beryllium atom can be reasoned to
be,
Be = (1s)2(2s)2
[i.e. ψBe = (ψ1S)2.(ψ2S)2]
and that of Boron to be,
B = (1s)2(2s)2(2p)1
[i.e. ψB = (ψ1S)2.(ψ2S)2.(ψ2P)]
where the electron in the 2p can reside in any one of the three available and
degenerate 2p orbitals.
When we come to Carbon, there are several available options for the second
electron to be placed in the 2p orbital set. These options are as follows:
(i)
↑
↓
The electrons, with opposite spins, are placed in two different sub-orbitals.
(ii)
↑↓
Both electrons, with opposite spins, are placed in the same sub-orbital.
(iii)
↑
↑
The electrons, with the same spin, are placed in two different sub-orbitals.
(iv)
↓
↓
The electrons, with the same spin, but with the opposite orientation as in (iii),
are placed in two different sub-orbitals.
31
Hund’s Rule will assist us in deciding which of these configuration(s) is(are)
the most stable.
Rule: Other things being equal, the lowest (most stable) energy state is that in
which the maximum number of electrons have parallel (same) spins.
Hund’s Rule points to options (iii) and (iv) as being the most stable of the four.
It is to be noted that, pairs of electrons with parallel (same) spins have their
motions correlated so as to keep them as far apart as possible. This means that
the electrons would have less interaction with each other, when compared to the
case where the electrons are of opposite spins. It should be noted that electrons
being of the same electric charge (i.e. both negatively charged) do not want to
be in the same region of space and would want to interact as little as possible.
This means that for the carbon atom, options (i) and (ii) would be of higher
energy, and therefore less stable, than options (iii) and (iv). This is because in (i)
and (ii) the actual repulsion of the two p-electrons is greater.
The p-orbital configurations for the other atoms in the period are therefore, by
extrapolation:
Nitrogen
↑
↑
↑
↑↓ ↑
↑
Oxygen
Fluorine
↑↓ ↑↓ ↑
Neon
↑↓ ↑↓ ↑↓
32
SESSION 6: Spherical Polar Coordinates, Radial & Angular
Wavefunctions
Overview: In this session we will look at
• Cartesian and Spherical Polar Coordinate systems
• the separation of the total wavefunction into radial and angular
parts
• information on electron density distribution from hydrogenic radial
wavefunctions
• information on orbital shape from hydrogenic angular
wavefunctions
Spherical Polar Coordinates
The usual means of expressing the location of a point in three-dimensional
space is by using the Cartesian (or x-y-z) Coordinate system. This consists of
three mutually perpendicular axes, with the origin (where the three axes
intersect) having the coordinates (0,0,0).
z
x
(0,0,0)
y
In expressing wavefunctions, it is more convenient to use a different type of
coordinate system. This new system, Spherical Polar Coordinates, also has three
coordinates, namely r, θ and φ, one displacement and two angles, as shown in
the following diagram:
33
z
θ
φ
r
y
x
The position of a mass m moving at a distance (radius) r from the nucleus, is
located by the three polar coordinates r, θ and φ.
To cover all space:
r is the distance from the nucleus, and has the range 0 → ∞,
θ is the angle that the radius vector makes with the vertical (or z-axis) and has
the range 0 → π, and
φ is the horizontal angle that the radius vector makes with the xz-plane, and has
the range 0 → 2π.
When a wavefunction ψ, is specified using the spherical polar coordinates, it
can be separated into two parts: a radial part R, which is a function of r and an
angular part Y, that is a function of θ and φ. Thus:
ψ(n,l,ml) = R(r).Y(θ,φ)
sentence taken out The square of the radial wavefunction, R(r)2, gives
information about the variation of the electron density with distance, r, from the
nucleus. The angular wavefunction, Y(θ,φ), gives information about the threedimensional region of space in which the electron can be found (i.e. the shape of
the orbital).
34
Acceptable angular wavefunctions are specified by the two quantum numbers l
and ml, which come out of solutions of the SWE. Normalized angular
wavefunctions are denoted as Yl , m (θ,φ) and are called the spherical harmonics.
l
In the following table there are some examples of Hydrogenic Radial
wavefunctions and Spherical Harmonics.
Let us take a closer look at the R1S function. This is a term involving e-r, which
means that the value of the function decreases as the value of the parameter r
increases. This explains the shape (exponential decrease) of the graph of R1S vs
r. The spherical harmonic Y1S does not depend on the value of the angles θ and
φ, since they do not appear in the expression. This means that Y1S is independent
of θ and φ and therefore is of spherical symmetry. Hence the 1s orbital is
spherical. This is the kind of information that can be obtained by just taking a
quick look at R(r) and Y(θ,φ). The other spherical harmonics look more complex
and one would have to do a three-dimensional plot while varying r, θ and φ to
get the picture. (Please see your text books for actual diagrams).
Hydrogenic Radial Wavefunctions
Orbital
1S
2S
2P
n
1
2
2
(N.B. ρ =
l
Rn,l
0
Z
2
 ao
0
1
2(2) 1 / 2
Z

 ao
2
 (2 − 12 ρ )e − ρ / 4

1
1
4(6) 1 / 2
Z

 ao
 2 −ρ / 4
 ρe

2Zr
)
na o
35



3/ 2
e
1
− ρ
2
3
3
Spherical Harmonics
l
ml
Yl , ml
0
 1 


 4π 
1/ 2
0
1/ 2
0
 3 


 4π 
±1
3
m  
 8π 
1
1
cos θ
1/ 2
(sin θ )e ± iφ
Exercise 7. Use the information in the table above to generate the total
wavefunction of the hydrogen 1S state.
Solution
To generate the total wavefunction for the 1S state of hydrogen (ψ1S) where n = 1
and Z = 1, the radial part is combined with the angular part:
Z
ψ1S = R1S.Y1S = 2
 ao



3/ 2
 1 

= 
3 
 πa o 
e
1
− ρ
2
1/ 2
.e
−
 1 
.

 4π 
1/ 2
 Z  3 / 2  1 1 / 2  − 1 ρ
= 2  
 e 2
4
π
a
 
 o  
r
ao
36
SESSION 7: Angular Momentum, Space Quantization and Ionization
Enthalpies
Overview: In this session we will look at
• the quantization of orbital angular momentum and its implications
• the importance of the magnetic quantum number in spatial
orientation of orbitals within a sub-shell
• the effects of shielding and electron repulsion on the variation of
Ionization Enthalpy across a period
Previously we had looked at the different quantum numbers that are used to
classify the electrons in an atom. l is known as the orbital angular momentum
quantum number. In solving the SWE, l and ml come out of the angular
solutions of ψ and specify the angular momentum (A.M.) of the electron about
the nucleus. The magnitude of the angular momentum is given by:
A.M. = {l(l + 1)}1/2. h
It should be noted that the magnitude of the angular momentum depends only
on the value of l. The magnitude of the A.M. of an electron in the s, p and d
orbitals would therefore be 0, 2 and 6 respectively.
The A.M. is a vector quantity, meaning that it has both magnitude and direction.
The A.M. vector is not allowed to have just any arbitrary orientation. Quantum
Mechanics states that:
A rotating body may not take up any arbitrary orientation with respect to
some specified axis, which is defined by the direction of an externally
applied magnetic or electric field. This is called space quantization.
This means that the electron has to behave in a certain way so that the angular
momentum vector, that it generates, does not take up just any arbitrary
orientation. For a specified z-axis, the z-component of the A.M. vector, for an
electron in an orbital with magnetic quantum number ml, has a value of ml h .
For example, if an electron is in a p-orbital, (l = 1), the angular momentum
vector is allowed to have only one of three possible orientations corresponding
to z-components of -1 h , 0 h and +1 h (since the allowed value of ml = -1, 0,
+1). This implication is that the orientation of a rotating body is quantized. See
diagram:
37
Z
+1
ml
0
–1
The arrows in the diagram indicate the three possible orientations of the angular
momentum vector with respect to the z-axis. The vector is also allowed to rotate
freely about the z-axis, hence the path described by the two arrows (ml = +1, -1)
moving around the z-axis would be of a right-side-up and of an inverted cone,
respectively, and the path of the third arrow (ml = 0) would prescribe a
horizontal circle (see diagram):
ml = +1
ml = 0
ml = – 1
38
Ionization Enthalpies
The First Ionization enthalpy (∆Hion) of an atom (element) is defined as the
energy that must be supplied to the atom in order to excite the most weakly held
electron out of reach of the nucleus.
A
+
∆Hion
A+
→
e–
+
The trend for the values of ionization energy across a period (from left to right)
is for there to be a general increase. However, there are some anomalies along
the period which can be explained by the shielding (screening) effect of
electrons in inner orbitals and the repulsive energy of electrons in the same
orbital. Observe the figure below:
First Ionization Energy
2500
He
Ne
2000
F
1500
N
H
1000
Be
O
C
B
Li
500
0
0
2
4
6
8
10
12
A t o mi c N umb er
Note that for the period Li – Ne, there is almost a straight line increase for the
value of ionization energy. There is, however, a dip at Boron and at Oxygen.
At Lithium, the value is low because the 1S electron density (due to the two 1S
electrons) is shielding the outer 2S electron from the full effects of the Li
nucleus. This 2s electron feels a reduced attraction from the nucleus because of
the shielding and so the electron in the 2S orbital can be removed without much
difficulty.
At Beryllium, the nuclear charge has increased by +1 but the electron is being
placed into the same 2S orbital. Electrons in the same orbital do not effectively
39
shield each other from the nucleus. This additional 2S electron will therefore
feel an increased attraction from the nucleus and so will be more difficult to
remove, hence a higher value of the ionization energy.
At Boron, a slight decrease is observed. This decrease is due to effective
shielding of the outer 2P electron from the nucleus by the inner core of 2S
electrons. An almost linear increase is observed through to Carbon and Nitrogen
as electrons are added to separate p-orbitals as the nuclear charge is increased.
The electrons in the p-orbitals do not screen each other and therefore an increase
is seen here.
At Oxygen there is a decrease. The fourth p-electron has to be placed into an
orbital that is already occupied by a single electron. This electron will
experience electron-electron repulsion in the orbital, and this repulsion will
make it easier to remove one of these electrons, hence the decrease. An increase
is observed to Fluorine and Neon as electron are added to the other p-orbitals.
Summary
You have covered new information about the hydrogen atom (and some of the
higher atoms), and in particular, about the behaviour of the electron in the
different orbitals on the atom. You now have a more modern view of the atom,
and the different theories that are used to assist in explaining its observed
properties. The material covered here is very important and will form the
foundation for building the other Units. This should also serve as a guide but
should not deter you from reading around the topics which are covered within
the Unit, as reading will strengthen your knowledge base.
It is desirable that you make full use of the text books that are available, and
revise the required material before the tutorial sessions, so that you will be
better able to interact with the tutor. If you do this you will find that the tutorials
are more beneficial.
40