He and hydrogenoid ions
The one nucleus-electron system
1
topic
• Mathematic required.
• Schrödinger for a hydrogenoid
• Orbital s
• Orbital p
2
Two prerequisites
Our world is 3D!
We need to calculate integrals and derivatives in
full space (3D).
A system of one atom has spherical symmetry.
Spherical units are appropriate.
Y= Y(r,q,f) rather than Y= Y(x,y,z)
except that D was defined in cartesian units.
3
Spherical units
x = r sin q. cos f
y = r sin q. sin f
z = r cos q
z
z
P
q
y
y
f
x
P'
x
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derivation
dY/dr for fixed q and f
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derivation
If we know making one derivation, we know how
to make others, to make second derivatives, and
the we know calculating the Laplacian, D
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Integration in space
dV = r2 sinq drdqdf
rsin q d f
dr
rd q
rs in q
rd q
rsin q d f
dr
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Integration limits
r=∞
r=•
ff==
r=0
r=0
q=0
=0 0
ff =
q =0
q=
q= 
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Integration in space
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Integration in space
Integration over q, f and r gives
V = 4/3 r4
Volume of a sphere
The integration over q and f
gives the volume between two spheres of radii r and r+dr:
dV = 4 r2dr
Volume between two
concentric spheres
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Dirac notation
triple integrals
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Spherical symmetry
dV = 4  r2 dR
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Radial density
dP/dR = 4  r2 Y*Y
It is the density of probability of finding a particule (an electron)
at a given distance from a center (nucleus)
It is not the density of probability per volume dP/dV= Y*Y
It is defined relative to a volume that increases with r.
The unit of Y is L-3/2
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Schrödinger for a hydrogenoid (1 nucleus – 1 electron)
The definition of an orbital
atomic orbital: any function Ye(x,y,z)
representing a stationary state of an atomic
electron.
Born-Oppenheimer approximation: decoupling
the motion of N and e
Y(xN,yN,zN,xe,ye,ze)= Y(xN,yN,zN)Y(xe,ye,ze)
mH=1846 me : When e- covers 1m H covers 2.4 cm, C 6.7 mm and Au 1.7 mm
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Schrödinger for a hydrogenoid (1 nucleus – 1 electron)
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Schrödinger for a hydrogenoid (1 nucleus – 1 electron)
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Schrödinger for a hydrogenoid (1 nucleus – 1 electron)
We first look for solutions valid for large r
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Solution for large r
Which of the 2 would you chose ?
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Solution Y = e-ar still valid close to the nucleus
New
To be set to zero
leading to a condition
Already set to zero
on a : a quantification
by taking Ne-ar
due to the potential
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The quantification of a is a
quantification on E
This energy is negative. The electron is stable referred to the free electron
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Energy units
1 Rydberg = 21.8 10-19 Joules =14.14 105 J mole-1
1 Rydberg = 13.606 eV = 0.5 Hartree (atomic
units)
1eV (charge for an electron under potential of 1
Volt)
1eV = 1.602 10-19 Joules = 96.5 KJoules mole-1
(→ n = 8065.5 cm-1)
1eV = = 24.06 Kcal mole-1.
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Atomic units
The energy unit is that of a dipole +/- e of length a0
It is the potential energy for H which is not the total energy
for H (-1/2 a.u.) (E=T+V)
Atomic units : h/2 =1 and 1/4e0 =1
length
a0,
Bohr
radius
charge
e, electron
charge
1.602 10-19C
mass
m, electron
mass
9.11 10-41 Kg
energy
Hartree
– The Schrödinger equation becomes simpler
22
Normalization of Ne-ar
From math textbooks
The density of probability is maxima at the nucleus
and decreases with the distance to the nucleus.
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Radial density of probability
a0/Z is the most probable distance to the nucleus; it was found by Niels Bohr
using a planetary model.
The radial density close to zero refers to a dense volume but very small;
far to zero, it corresponds to a large volume but an empty one
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Orbital 1s
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Average distance to the nucleus
Distance:
Operator r
larger than a0/Z
From math textbooks
In an average value, the weight of heavy values dominates:
(half+double)/2 = 1.25 > 1)
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Distances the nucleus
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Excited states
We have obtained a solution using Y = e-ar; it
corresponds to the ground state.
There are other quantified levels still lower
than E=0 (classical domain where the e is
no more attached to the nucleus)o
We can search for other spherical function
Y= NnPn (r)e-ar where Pn(r) is a polynom of r of
degree n-1
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Orbital ns
Principal quantum number
E2s = Z2 /4 E1s (H)
Ens = Z2 /n2 E1s (H)
Nodes: spheres for solution of equation Pn=O
n-1 solutions
Average distance
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Orbital 2s
E2s = Z2 /4 E1s (H)
A more diffuse orbital:
One nodal surface separating two =regions with opposite phases: the
sphere for r=2a0/Z. Within this sphere the probability of finding the
electron is only 5.4%. The radial density of probability is maximum for
r=0.764a0/Z and r=5.246a0/Z. Between 4.426 et 7.246 the probability
of finding the electron is 64%.
Average distance 5a0/Z
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Orbital 2s
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Radial distribution
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Resolution of Schrödinger equation
in f,
Solving the equation in q and f leads to define two other
quantum numbers.
They also can be defined using momentum instead of
energy.
For q: Angular Momentum (Secondary, Azimuthal)
Quantum Number (l):
l = 0, ..., n-1.
For f: Magnetic Quantum Number (ml):
ml = -l, ..., 0, ..., +l.
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Resolution of Schrödinger equation
in f,the magnetic quantum number
z
z
P
q
y
f
x
P'
x
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y
Quantum numbers
•
Principal Quantum Number (n): n = 1, 2,3,4, …, ∞
Specifies the energy of an electron and the size of the orbital (the
distance from the nucleus of the peak in a radial probability
distribution plot). All orbitals that have the same value of n are said
to be in the same shell (level).
•
Angular Momentum (Secondary, Azimunthal) Quantum Number
(l): l = 0, ..., n-1.
•
Magnetic Quantum Number (m): m = -l, ..., 0, ..., +l.
•
Spin Quantum Number (ms): ms = +½ or -½.
Specifies the orientation of the spin axis of an electron. An
electron can spin in only one of two directions (sometimes called up
and down).
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Name of orbitals
The letter indicates the secondary
Quantum number, l
1s
2s 2p (2p+1, 2p0, 2p-1)
The number indicates the Principal Quantum number
3s 3p (3p+1, 3p0, 3p-1)
3d (3p+2, 3p+1, 3p0, 3p-1 3p-2 )
4s 4p (4p+1, 4p0, 4p-1)
The index indicates the magnetic
4d (4d+2, 4d+1, 4d0, 4d-1 4d-2 ) Quantum number
4f (4f+3, 4f+2, 4f+1, 4f0, 4f-1 4f-2 , 4f-3
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Degenerate orbitals
Depends only on the
2
2
E= Z /n (E1sH) principal Quantum number
1s
2s 2p (2p+1, 2p0, 2p-1)
3s 3p (3p+1, 3p0, 3p-1)
3d (3p+2, 3p+1, 3p0, 3p-1 3p-2 )
4s 4p (4p+1, 4p0, 4p-1)
4d (4d+2, 4d+1, 4d0, 4d-1 4d-2 )
4f (4f+3, 4f+2, 4f+1, 4f0, 4f-1 4f-2 , 4f-3
1 function
4 functions
9 functions
16 functions
Combination of degenerate functions: still OK for hydrogenoids.
New expressions (same number); real expressions; hybridization.
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Functions 2p
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symmetry of 2pZ
Nodes:
No node for the radial part (except 0 and ∞)
cosq = 0 corresponds to q=/2 : the xy plane
or z/r=0 : the xy plane
The 2pz orbital is antisymmetric relative to this plane cos(-q)=-cosq
The z axis is a C∞ axis
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Directionality of 2pZ
This is the product of a radial function (with no node) by an angular function
cosq. It does not depend on f and has the z axis for symmetry axis.
The angular contribution to the density of probability varies like cosq2
Within cones:
Full space 2 cones: a diabolo
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angle
Density inside
The half cone
(1-cos(ql)3)/2
Part of the
volume
(1-cos ql)/2
15°
0.049
0.017
30°
0.175
0.067
45°
0.323
0.146
60°
0.437
0.25
75°
0.491
0.37
90°
0.5
0.5
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Directionality of 2pZ
This is the product of a radial function (with no node) by an angular function
cosq. It does not depend on f and has the z axis for symmetry axis.
The angular contribution to the density of probability varies like cosq2
Within cones:
Density inside
The half cone
(cosql)3-1)/2
Part of the
volume
(cos ql-1)/2
15°
0.049
0.017
30°
0.175
0.067
45°
0.323
0.146
60°
0.437
0.25
75°
0.491
0.37
90°
0.5
0.5
angle
Probability
is 87.5% in
half of the
space
22.5% in
the other
half
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Spatial representation of the angular part.
z
Let us draw all the points M with
the same contribution of the
angular part to the density
+
-
q
co
q
s
M
The angular part of the probability
is OM = cosq2
All the M points belong to two
spheres that touch at O
M'
si n q
+
O
-
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Isodensities, isolevels
R
2
R
2
C
A'
A
A'
2
/c os
C P tPotale
ql
totale /cos
2
ql
P
B'B'
BB
A
totale
P
totale
2
P t otale
/c os
/cos
q
P t otale
r
q
r
B' B'
B B
A A
2
A'
A'
CC
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2p orbital
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3p orbital
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The 2px and 2py orbitals are
equivalent
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One electron equally distributed
on the three 2p levels
Y2 is proportional to x2/r2+ y2/r2+ z2/r2=1
and thus does not depend on r: spherical symmetry
An orbital p has a direction, like a
vector.
A linear combination of 3 p orbitals,
is another p orbital with a
different axis:
The choice of the x,y, z orbital is
arbitrary
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orbitals
Y= N
radial function
rl (polynom of degree n-l-r)
n-l-r nodes
angular function
l nodes
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d orbitals
y
z
x
2
x -y
2
z
y
Clover, the forth lobe is the
lucky one; clubs have three
2
z
z
x
y
x
50
xy
xz
yz
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3d orbitals
52
Compare the average radius of 1s
for the hydrogenoides whose nuclei
are H and Pb.
207
Pb
82
Make comments on the 1s orbital
the atom Pb?
53
Name these orbitals
Spherical coordinates x = r sin q cos f ; y = r sin q sin f ; z = r cos q
54
Paramagnetism
Is an atom with an odd number of electron
necessarily diamagnetic?
Is an atom with an even number of electron
necessarily paramagnetic?
What is the (l, ml) values for Lithium?
Is Li dia or para? Why?
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Summary
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