the quantum mechanical treatment of the hydrogen atom 12 the hydrogen atom.pdfthe total energy of...
TRANSCRIPT
-
The Quantum Mechanical Treatment of the Hydrogen Atom
Motivation:
Hydrogen atom orbitals and energy levels form the basis of an
understanding of larger molecules.
covalent bonding molecular orbitals hybridization
In fact, the results of the quantum mechanical treatment of the
hydrogen atom permeate through our modern description of
chemistry.
For example, we often speak of the
p-bond in a C-C double bond as a
result of the interaction between two p
orbitals on each carbon.
The orbitals we so often speak of come from the quantum
mechanical treatment of the hydrogen atom.
-
H atom ground state
electron density
total electron
density of
benzene
3-D plot of the p-bonding orbital in ethylene from a quantum
chemical calculation.
-
The Hydrogen Atom – The Physical Model
The hydrogen atom is a two-particle system
consisting of a proton and an electron.
We will treat the proton and electron as point masses that interact
via Coulomb’s Law given by (in SI units):
2 1( )
4 o
eV r
rp
r is the distance between the electron and nucleus.
e is the charge of the electron:
o is the vacuum permittivity constant and in SI units has a value of:
191 602 10e . x C
12 2 2 1 38.854188x10 o C s kg m
p+
e-
How does this
potential compare
to harmonic
oscillator?
r
-
The quantum mechanical Hamiltonian for this system is:
p+
e-2 2222
2
1
2ˆ
4 ope
ep
mH
rm
e
p
This is a two-particle system, and just like the Harmonic Oscillator and
Rigid Rotor problem, we can separate the Hamiltonian into a center of
mass coordinate and a relative coordinate system:
translational motion of the center of mass
22ˆ
2CM CM
p e
Hm m
relative or internal motion
2 22
int
1ˆ2 4 o
eH
r p
where is the reduced mass
e p
e p
m m
m m
intˆ ˆ ˆ
total CMH H H
-
2 2 22 2 1ˆ
2 42total CM
op e
eH
rm m p
intˆ ˆ
CMH H
This Hamiltonian is separable, with the solution of the center of mass
Hamiltonian being that of the free particle, we have previously
examined.
intĤ E
We want to find the wave functions that govern the internal motion of
the system.
These solutions are the hydrogen atom wave functions that we are
interested in.
-
2 22
int
1ˆ ˆ2 4 o
eH H
r p
The Hamiltonian operator for the internal energy of the hydrogen atom is:
kinetic energy of
internal motionpotential energy,
Coulomb’s law.
In many textbooks, you will often see the hydrogen-atom Hamiltonian
written with me, the mass of the electron, instead of .2 2
2 1ˆ2 4e o
eH
m rp
Who is correct? Are these other textbooks
incorrect?
e p
e p
m m
m m
-
This problem has the same spherical symmetry as the 3-D rigid rotor,
so this suggests that it would be easier to solve our problem in polar
spherical coordinates.
x
y
z
r
r0
p 0
2π0
What do we need to do to convert
our Hamiltonian into polar
spherical coordinates?
2 22 1ˆ
2 4 o
eH
r p
( , , )r
Schrödinger’s equations for the H atom is Best Solvedin Polar Spherical Coordinates
-
Polar Spherical Coordinates
The Laplacian operator in polar spherical coordinates is:
22 2
2 2 2 2 2
1 1 1sin
sin sinr
r rr r r
(there is no need to memorize the above, but you
should know how to work with it and what it means)
The Integration Element
In Cartesian coordinates the integration over all
space involves the following integration element:
d dx dy dz The equivalent integration element in polar
spherical coordinates is given by:
2 sin d r dr d d dr
rd
-
In polar spherical coordinates the Schrodinger equation becomes:
Unlike the rigid rotor, r is not fixed because the electron is free to movecloser or farther away from the nucleus.
ˆ ( , , ) ( , , )H r E r
2
2
2
2
22
2 2 2
1 1sin
sin s2 i
1
n
1
4 o
er
r r rr rE
r
p
Therefore, we cannot simplify the Laplacian further. This is unfortunately
the Schrödinger equation we have to work with!
The Hamiltonian is not separable, but if we multiply both sides of the
SE by 2r2, it becomes pseudo-separable.
-
2
2 2
2 22 2 22 1 2
1 1sin
sin si 4n o
r er r E
r r r
p
The above is pseudo-separable, so we can collect angular components
( and ) on one side and radial components on the other side.
depends only on and
22 2
2 2 22
2 2
2 12
4
1 1sin
sin sino
r er r E
r r r
p
angular dependence
(depends on and )radial dependence
(depends on r)
Collecting sides gives:
This suggests that we can write our total hydrogen atom wave function
as a product of a radial and angular functions.
-
The Hydrogen Atom Wave Function can be Written as a
Product of Radial and Angular Components
Using this form of the wave function and substituting into the last
equation on the previous slide and rearranging gives:
2 2 22
2 2
22
21 2 1 24
1 1sin
sin sino
r er r E R
R r r rA
A
p
2 22 2 21 2 1 2 ( )
( ) 4 o
r er r E R r lhs
R r r r r
p
Because the lhs depends only on the variable r, and the rhs depends only
on the angular variables and , each side must be equal to a constant.
2 2
2 2
1 1sin ( , )
( , ) sin sinrhs A
A
),(),,( AR(r)r
-
The Angular Part of the Hydrogen Atom Wave Function
Let’s first examine the angular part of the hydrogen atom Schrödinger
equation:2 2
2 2
1 1sin ( , )
( , ) sin sinA
A
Multiplying both sides by A gives:
22
2 2
1 1sin ( , ) ( , )
sin sinA A
The above is simply an eigenvalue equation, where the operator is the L2
operator.
2L̂
2ˆ ( , ) ( , )L A A
Haven’t we already solved these already? What is A?
-
Spherical Harmonics Make up the Angular Component of
the Total Hydrogen Atom Wave Function
The spherical harmonics are eigenfunctions of the L2 operator, recall:
22ˆ (( , ) ( , )1)m mY YL ...2,1,0
,...2,1,0m
2 ( , )ˆ ( , )AL A
Comparing the above and our angular Schrödinger equation
tells us that:
( , ) ( , )mA Y 2 ( 1)
So the angular component of the total wave function of the hydrogen
atom are the spherical harmonics:
),()(),,( , mYrRr
-
200Y
210Y
21 1Y
xy
z
220Y
This is why the angular plots of the spherical
harmonics looked like our familiar s, p and d
orbitals.
Some of them don’t, like the donut below (to be
discussed shortly).
We have solved for the angular part of
the total hydrogen atom wave function,
now we need to solve for the radial
component.
-
Solving for the Radial Component of the H-Atom Wave Function
From a few slides back, we had an expression for the radial wave
function R(r),2 2
2 2 21 2 1 2 ( )( ) 4 o
r er r E R r
R r r r r
p
From the solution of the angular part, we know that rearranging
2 ( 1) equation and dividing through by 2r2 gives:
2 2 22
2 2
1 ( 1) 1( ) ( )
4 22 o
er R r ER r
r r rr rp
This is known as the radial Schrödinger equation for the hydrogen
atom. This is the only new equation we have to solve for to get our total
wave function for the hydrogen atom.
Again, we will not explicitly solve this equation. The solutions will simply
be presented and analyzed.
-
Radial Wave Functions of the Hydrogen Atom
)(rRnnormalization
constant
exponential decay
function:polynomial
in ronare
/
The solution of the radial Schrödinger equation gives what the radial
wave function that depends on a new quantum number, n.
)(rRn...3,2,1 n
0, 1, 2... 1n
Notice the new restriction on the quantum number ℓ (why?) and that that
radial wave function does not depend on the quantum number m.
The general form of the radial wave function is:
All hydrogen atom wave functions decay exponentially away from the
nucleus.
-
3/ 2
10
1 /2 o
o
r aR e
a
3/ 2
20
1 1 / 2(1 )
22o
o o
r r aR e
a a
3/ 2 2
30 2
2 1 2 2 / 31
33 3 27
o
o o o
r r r aR e
a a a
5/ 2
21
1 1 / 2
2 6o
o
r aR re
a
7 / 2
232
4 1 / 3
81 30o
o
r aR r e
a
3/ 2 2
31 2
8 1 / 3
27 6 6
o
o o o
r r r aR e
a a a
Normalized Radial Functions of the Hydrogen Atom
)(rRn...3,2,1 n
0, 1, 2... 1n n = 1
n = 2
n = 3
Bohr radius ao = 0h2/pe2
= 0.0529 nm
-
r (ao)
r (ao)
r (ao)
The radial wave functions are plotted as
a function of r in units of
Bohr radii ao = (oh2)/(pe2)
3/ 2
10
1 /2 o
o
r aR e
a
5/ 2
21
1 1 / 2
2 6o
o
r aR re
a
r (ao)
Plots of Radial Wave Functions
-
The Energy of the Hydrogen Atom
2 2 22
2 2
1 ( 1) 1( ) ( ) 0
4 22 o
er R r E R r
r r rr rp
The energy of the hydrogen atom can be derived by solving the radial
Schrödinger equation. It is given by:
4 2
2 2 2 2 2
1 1
832 o oo
e eE
an n
pp - ...3,2,1 n
The total energy of the hydrogen atom depends only on the quantum
number n, and not on the other two quantum numbers, ℓ and m.
NOTE that E is the total energy (kinetic + electrical potential energy)of the hydrogen atom, not just the ‘radial energy’.
-
The energy levels are essentially
in exact agreement with the
discrete line spectra of atomic
hydrogen.1 2
2
2 22 1
1 1
8n ,n
o o
eΔE
a n np
One of the most powerful experimental verifications of the quantum
mechanical description of the hydrogen atom involves the emission
spectra of hydrogen atoms. Selection rule for H-atom electronic
transitions:
∆n unrestricted
Lyman Balmer Paschen
Ultraviolet Visible Infrared
656.5
nm
Experimental Verification of the Hydrogen-Atom Solution
-
The numerical value of the ground state energy
(13.59670 eV) is given by:
The Ground State Energy and the Ionization Energy
2
1 2
11311.9 kJ/mol
8 1o o
eE
ap -
The ionization energy is the energy required to
remove an electron from the ground state of an
atom.
Experimentally, the hydrogen atom ionization energy
is 1311.9 kJ/mol.
This is another powerful way the quantum
mechanical description of the hydrogen atom can be
confirmed experimentally.
n=1
n=2
n=3
n=4n=0
energ
y
1311.9 kJ/mol
-
2
2
1
8 o o
eE
a np - ...3,2,1 n
The hydrogen atom energy levels are depicted to the
right.
n = 1
n = 2
0
energ
y
n = 3
n =4
Notice that the energies are all negative, and as
n increases, the energy tends to zero.
Does this make sense?
-
Lagoon Nebula using the VLT (very large telescope) in Chile
-
Trifid Nebula
Red: H alpha
(n = 3 to n = 2)
Blue: scattered
light
Dark brown:
dust
-
Rosette Nebula using the VLT
-
),),,( ((r)YRr mnmn
n = 1, 2, 3…
ℓ = 0, 1, 2…(n 1)m = 0, ±1, ±2…±ℓ
Total Wave Functions for “Hydrogen-Like” Atoms(One-Electron Systems: H, He+, Li2+, Be3+, …)
The total wave function of the hydrogen atom is the product of the
radial wave function R(r) and the spherical harmonic Ynl(,):
Notice that there are three distinct quantum numbers that characterize
the state of a hydrogen atom.
We, therefore label the total wave function with these three quantum
numbers, n, ℓ and m, usually in this order.
0 0, 1, 100
mn 1 2, 3, 132 mn
-
Considering all possible combinations of ℓ and m
consistent with n, we find that the degree of
degeneracy, g, of hydrogen atoms is:
Degeneracies of the Hydrogen-Like Atom States
H(Z = 1), He+ (Z = 2), Li2+(Z = 3), Be3+(Z = 4), …
ℓ = 0, 1, 2…(n-1)m = 0, ±1, ±2…±ℓ
The energy levels are highly degenerate because the
energy depends only on the quantum number n.
100
200 210 211 21-1
300 310 311 31-1 320 321 32-1 322 32-2
Energ
y
g = n2
2
00
2
222
0
4 1
8
1
8)1,atom H(
na
e
nh
eZEn
2
00
22
222
0
42 1
8
1
8)(
na
eZ
nh
eZZEn
n = 1, 2, 3, …
-
3/ 2
100
1 1 / o
o
r ae
a
p
3/ 2
200
1 1 / 22
4 2o
o o
r r ae
a a
p
5/ 2
21 1
1 1 / 2sin
8o
o
r a ire e
a
p
3/ 2 2
300
1 1 18 / 3(27 2 )
81 3o
o o o
r r r ae
a a a
p
3/ 2 2
31 1
1 1 6 / 3( ) sin
81o
o o o
r r r a ie ea a a
p
7 / 2
232 1
1 1 / 3sin cos
81o
o
r a ir e e
a
p
7 / 2
2 232 2
1 1 / 3 2sin
162o
o
r a ir e e
a
p
n = 2, ℓ = 0
n = 1, ℓ = 0
n = 2, ℓ = 15/ 2
210
1 1 / 2cos
4 2o
o
r are
a
p
n = 3, ℓ = 0
n = 3, ℓ = 1
n = 3, ℓ = 27 / 2
2 2320
1 1 / 3(3cos 1)
81 6o
o
r ar e
a
p
Complete Hydrogen Atom Wave Functions
wave functions Ynlm(r,,)for n = 1, 2, and 3
3/ 2 21/ 2
310
1 2 1 6 / 3( ) cos
81o
o o o
r r r ae
a a a
p
-
Probability distribution functions
Y*nlm(r,,) Ynlm(r,,) r2sindrdd
for n = 1, 2, 3
-
Show that the hydrogen atom wave function Y100 is normalized.
o
2/3
o
100
/11 are
a
p
Show that the hydrogen atom wave function Y100 is orthogonal to
the Y211 wave function.
p
ie
arre
a
sin
2/1
8
1o
2/5
o
211
-
The One-Electron Wave Function
The hydrogen atom wave functions were derived by considering the
two-body (proton + electron) system as an effective “one-body” system.
Due to the large mass difference of the proton and electron, the
hydrogen atom Hamiltonian and wave functions essentially describe the
motion of the electron.Why?
A One-Electron Wave Function is Called an Orbital
The hydrogen atom wave functions are often called orbitals, a term
frequently used in chemistry. Formally an orbital has the following
definition:
An orbital is the wave function describing a
single electron. In other words, an orbital is
a one-electron wave function.
-
What do Hydrogen Atom Wave Functions ‘Look’ Like?
The hydrogen atom wave functions are 3-dimensional functions of space.
In other words, associated with every point in space, is a value of that
wave function. There are many different ways to visualize wave
functions.
3D-Isosurface Plots
When a chemist thinks of a 3-D representation of a
orbital, such as a 210 or 2pz orbital, they usuallyhave a 3-D isosurface plot in mind.
An isosurface plot is 3-dimensional surface of which
the function has the same value. Thus, associated
with each isosurface plot, should be a value of that
surface. r = 0.02 a.u.
Shown to the right is the isosurface plot of the
density of the 210 orbital:
2210
-
How ‘big’ the orbitals are when we visualize them depends on what
isosurface value we choose.
Due to the exponential decay
of all H-atom wave functions,
the larger the isosurface value,
the ‘smaller’ the orbital is
depicted in an isosurface plot.
The size of an Orbital in an Isosurface plot is Arbitrary
0 1.5 2.0 2.5
4.0
2.0
1.0 3.00.5
r/a0
|100|2
large isosurface
value
small isosurface
value
When wave functions are plotted, the color
represents the sign of the wave function.
We will discuss complex-valued wave functions
later.
23 zdΨ
-
2200
2300
2320
2210 2
21 1
2310
231 1
232 1
232 2
Isosurface plots of the densities
of the n = 1, 2 and 3 orbitals
2100
z
-
Spectroscopic Designation of States
What about our familiar s, p and d orbitals? There is a relation between
the quantum numbers ℓ and the s, p, d labels we give the atomicorbitals.
s p d f g h
ℓ = 0 1 2 3 4 5
Ignoring the quantum number m for the time being, we have:
1s = 10 n = 1, ℓ = 0
2s = 20 n = 2, ℓ = 0
2p = 21 n = 2, ℓ = 1
3s = 30 n = 3, ℓ = 0
3p = 31 n = 3, ℓ = 1
3d = 32 n = 3, ℓ = 2
Notice that we are not specifying the x, y, z , x2-y2 etc. states yet.
-
Real and Directed Orbitals
In chemistry, we often speak about the s, px, py, pz and the five
d orbitals, dxy, dyz, dxz, dz2 and dx2-y2.
How are these orbitals, that we often use to explain chemical bonding,
related to the atomic hydrogen orbitals? To illustrate this point, consider
the n = 2 states, where we would expect to have:
2s 2 zp 2 yp
2 xp
200 210 211 21 1
Are they
Equivalent?
-
Consider the 200 wave function:3/ 2
200
1 1 / 22
4 2o
o o
r r ae
a a
p
There is no angular dependence, meaning that
this is a perfectly spherical function.
x
y
z
r
r0
p 02π0
Indeed, the n = 2, ℓ = 0, m = 0 wave function is
the familiar 2s orbital.
200 2s
What about the 210 wave function which is
given as:5/ 2
210
1 1 / 2cos
4 2o
o
r are
a
p
-
An isosurface plot of the 210 wave function shown tothe right shows that this is indeed our familiar 2pzorbital. (Here, positive is “red”, negative is “blue”.)
z
5/ 2
2
1 1 / 2cos
4 2zo
po
r are
a
p
cosz r
What about the 21+1 wave function? Is this one of the 2px or 2pyorbitals?
This is a complex function with both real and imaginary components,
making it a bit difficult to plot.
5/ 2
21 1
1 1 / 2sin
8o
o
r a ire e
a
p
We know however that the square of the wave function is always
positive, so let’s try plotting this:
221 1
-
This means that:5
2 2 2211
1 1 /sin
64o
o
r ar e
a
p
We will get the same result for the 21-1 orbital!
52 2 2
21 1
1 1 /sin
64o
o
r ar e
a
p
When we plot this we get a donut isosurface plot!
221 1
So what are all of the 2px, 2py, and other
orbitals we are familiar with?
Are they donuts?
Are they complex?
-
Real and Directed Orbitals are Linear Combinations of
the Complex Set We have Derived
The orbitals we use in chemistry are purely real functions that are
constructed as a linear combination of the wave functions we have
derived.
12121122
1-p ΨΨΨ x
2 211 21 11
2yp -Ψ Ψ Ψ
i
Show this for the 2px orbital.
Show that the 2px orbital as defined above is normalized.
-
The Energies of the Real and Directed Orbitals
Do the real and directed orbitals still satisfy the Schrödinger equation
for the hydrogen atom and what are their energies?
Consider the 2py orbital which is a linear combination of the 211
and 21-1 orbitals.
Indeed, any linear combination of degenerate wave functions is also a
solution to the Schrödinger equation with the same energy.
2 211 21 11
2yp -Ψ Ψ Ψ
i
200 210 211Ψ AΨ BΨ CΨ
Show this for the following linear combination:
-
2px orbital directed along the x axis:
p
p
p
cossin1
4
1
)sincossin(cossin1
8
1
esinesin1
28
1
)(2
1
0
0
0
2/
2/5
0
2/
2/5
0
2/
2/5
0
121211p2
ar
ar
iiar
x
rea
iirea
rea
-
2py orbital directed along the y axis:
p
p
p
sinsin1
4
1
)sincossin(cossin1
28
1
esinesin1
28
1
)(2
1
0
0
0
2/
2/5
0
2/
2/5
0
2/
2/5
0
121211p2
ar
ar
iiar
y
rea
iireai
reai
i
-
Is the 2px orbital normalized?
1
)11(2
1
d2
1
d2
1
d2
1
d22
d
*
121
*
121211
*
211
*
121
*
121211
*
121121
*
211211
*
211
121211
*
121211
121211
*
121211
p2
*
p2
xx
-
Any linear combination of degenerate wave functions
has the same energy.
2
2112102002
211221022002
211210200
211210200
211210200
)(
ˆˆˆ
ˆˆˆ
)(ˆˆ
E
CBAE
CEBEAE
HCHBHA
CHBHAH
CBAHH
-
The Real and Directed d-Orbitals
The d-orbitals that we often use in transition metal chemistry are also
normalized linear combinations of the complex functions we have
derived.
What about the n = 3, ℓ = 2, m = 0 hydrogen atom orbital which is a
purely real function?7 / 2
2 2 2320
1 1 / 3(3 cos )
81 6o
o
r ae r r
a
p
z = r cos 2 23z r
So our familiar dz2 orbital is shorthand for a 3z2 r2 orbital.
The other d orbitals we are familiar with are linear combinations of the
remaining four m = ±1 and m = ±2 functions.
-
Linear combinations used to construct nd orbitals:
p
p
p
p
p
2sinsin16
15)()(
2
1d
2cossin16
15)()(
2
1d
sincossin4
15)()(
2
1d
coscossin4
15)()(
2
1d
)1cos3(16
5)(d
2
22222
2
22222)(
21221
21221
2
220
22
2
rRi
rR
rRi
rR
rR
nnnnxy
nnnyxn
nnnnyz
nnnnxz
nnnz
-
The familiar d-orbitals are the ‘real and directed’ d-orbitals.
Here, “positive” is blue and “negative” is yellow.
dx2-y2 dz2
dxy dxz dyz
-
The real and directed 4f orbitals
-
Angular Momentum of the Hydrogen Atom
The hydrogen atom energy levels are
n2-fold degenerate.
g = n2En
erg
y
1, 0, 0n m
2, 0,1, 1,0,1n m
What is different about the states
if the energies are the same?
Each of the states has a different angular momentum. In other words,
these states each have different values for L2 and Lz.
Recall that the total hydrogen atom wave function is a product of the
radial wave function and the spherical harmonics.
),),,( ((r)YRr mnmn
-
The spherical harmonics are eigenfunctions of the L2 and Lz operators:22ˆ (( , ) ( , )1)m mY YL
, ) ( , )ˆ (m mzY YL m
ˆzL i
22 2
2 2
1 1ˆ sinsin sin
L
Because the L2 and Lz operators act only on and ,
We know that the hydrogen atom wave functions are eigenfunctions of
the L2 and Lz operators. Also, we know that each of the hydrogen atom
states has a definite square of the angular momentum and z-component of the angular momentum.
Show this
-
The Quantum Numbers ℓ and m Define the Angular Momentum of the Hydrogen Atom State
The quantum number ℓ defines the square of the magnitude of theangular momentum vector of the state:
2 2 ( 1)L
The quantum number m defines the z-component of the angularmomentum vector of the state:
zL m
Because the angular momentum of the hydrogen atom is due to the
angular motion of the electron, it is often referred to as the orbital
angular momentum.
Sketch the angular momentum vectors of the
hydrogen atom in a 1s and 2pz orbitals.
-
The H-Atom Quantum Numbers have Common Names
The principle quantum number, specifies the energy of the
hydrogen atom.n
The angular momentum quantum number, (sometimes called
the azimuthal quantum number), specifies the magnitude of the
orbital angular momentum of the state.
ℓ
2 2 ( 1)L 0,1,2... 1n
2
2
1
8 o o
eE
a np - ...3,2,1 n
The magnetic momentum quantum number, specifies the
z-component of the angular momentumm
zL m 0, 1, 2...m
-
The Spatial and Nodal Structure of the Hydrogen Atom Orbitals
When using atomic orbitals to discuss molecular bonding, the spatial
and nodal structure of the wave functions is important
Motivation
For example, the overlap between orbitals on different atoms, either
constructive, destructive or zero overlap, is highly dependent upon the
nodal structure of the orbitals involved.
We will discuss this in terms of the purely real and directed orbitals.
Recall both sets of orbitals are equivalent, and can be obtained from
one another by rotation.
Nodes of the Wave Function
Recall that nodes are where the wave function changes sign and has
zero value.
+
-+ -
-
Radial and Angular Nodes
The number of nodes is determined by the quantum numbers n and ℓ.The total number of nodes is given by:
total number of nodes = n 1
So the ground state hydrogen atom wave function 100 has no nodes.
Check this!
The nodes can be categorized as radial and angular nodes arising from
each component of the total hydrogen atom wave function, R and Y.
number of radial nodes = n – ℓ 1
number of angular nodes = ℓ
-
Angular Nodes
The real and directed orbitals have nodal planes (or equivalent) where
the wave function changes sign. The number of angular nodes is given
by the quantum number ℓ.
ℓ = 0, the ‘s’ orbitals
The s orbitals have no angular dependence, and are spherical wave
functions and therefore have no angular nodes.
ℓ = 1, the ‘p’ orbitals
The p orbitals have one angular plane, each.
+
-x
y
z
+- x
y
z
x
y
z
+ -
-
dxz dxy dyz
x
y
z
+
-
-
+
x
y
z
-++
-
x
y
z
++-
-
+ x
y
z
+
-
-
dx2-y2 dz2
x
y
z
+
+-
ℓ = 2, the ‘d’ orbitals
The d orbitals have two angular planes each.
Where are the two angular nodes of the dz2 orbital?
2
7 / 2
2 2320 3
1 1 / 3(3cos 1)
81 6o
dzo
r ar e
a
p
-
Radial Nodes
Plots of the radial wave function show that the number of radial nodes is
given by:number of radial nodes = n ℓ 1
Below are plotted the 1s, 2s, 3s and 4s radial wave functions.
r(ao)r(ao) r(ao) r(ao)
The radial nodes give isosurface plots of the electron density that are
layered like an onion.
22s
23s
Explain
-
The nodes of the radial wave functions give the higher angular
momentum wave functions a similar structure.
2
2 zp
2
3 zp
Sketched below are 2-D slices of the isosurface plots of the wave
function.
Notice the alternation of the sign of the layers and the larger spatial
extent as n varies from 2 to 4.
2 zp 3 zp 4 zp
+
-
+
-
+
-
-
++-
-
The same layered structure exists for the 4d and 5d orbitals!
The reason we don’t sketch or represent the inner orbitals when
discussing bonding is because it is only the outer ‘layer’ that participates
in the interaction.
4 xzd
+
+ --
-
-
+
--
++
+
-
-
24 zd