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1
He and hydrogenoid ions
The one nucleus-electron system
2
topic
• Mathematic required.
• Schrödinger for a hydrogenoid
• Orbital s
• Orbital p
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Two prerequisitesOur 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.
rrather than x,y,z
except that was defined in cartesian units.
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Spherical unitsx = r sin . cos y = r sin . sin z = r cos
P'
P
x
y
z
z
y
x
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derivationd/dr for fixed and
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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,
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Integration in space
rsin rddr
rsind
drrd rsind
dV = r2 sin drdd
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Integration limits
=0
=
=0
r=
r=0
r = 0
r = ∞
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Integration in space
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Integration in spaceIntegration over , and r gives
V = 4/3 r4
The integration over and gives the volume between two spheres of radii r and r+dr:
dV = 4 r2dr
Volume of a sphere
Volume between two concentric spheres
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Dirac notationtriple integrals
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Spherical symmetry
dV = 4 r2 dR
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Radial density dP/dR = 4 r2 *
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=
It is defined relative to a volume that increases with r.
The unit of 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 e(x,y,z) representing a stationary state of an atomic electron.
Born-Oppenheimer approximation: decoupling the motion of N and e
(xN,yN,zN,xe,ye,ze)= (xN,yN,zN)(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 e-r still valid close to the nucleus
Already set to zero
by taking Ne-r
New
To be set to zero
leading to a condition
on : a quantification
due to the potential
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The quantification of 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
(→ = 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 1/40 =1
– The Schrödinger equation becomes simpler
length charge mass energy
a0, Bohr
radius
e, electron charge
1.602 10-19C
m, electron mass
9.11 10-41 Kg
Hartree
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Normalization of Ne-r
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
In an average value, the weight of heavy values dominates:
(half+double)/2 = 1.25 > 1)
larger than a0/Z
Distance:
Operator r
From math textbooks
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Distances the nucleus
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Excited states
We have obtained a solution using e-r; 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
NnPn (r)e-r where Pn(r) is a polynom of r of degree n-1
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Orbital ns
Ens = Z2 /n2 E1s (H)
E2s = Z2 /4 E1s (H)
Nodes: spheres for solution of equation Pn=O n-1 solutions
Average distance
Principal quantum number
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Orbital 2s
=
E2s = Z2 /4 E1s (H)
Average distance 5a0/Z
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%.
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Orbital 2s
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Radial distribution
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Resolution of Schrödinger equation in ,
Solving the equation in and leads to define two other quantum numbers.
They also can be defined using momentum instead of energy.
For : Angular Momentum (Secondary, Azimuthal) Quantum Number (l): l = 0, ..., n-1.
For : Magnetic Quantum Number (ml):
ml = -l, ..., 0, ..., +l.
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Resolution of Schrödinger equation in ,the magnetic quantum number
P'
P
x
y
z
z
y
x
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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
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
The letter indicates the secondary Quantum number, l
The index indicates the magnetic Quantum number
The number indicates the Principal Quantum number
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Degenerate orbitalsE= Z2/n2 (E1sH)
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
Depends only on the
principal Quantum number
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 ∞)
cos = 0 corresponds to =/2 : the xy plane
or z/r=0 : the xy plane
The 2pz orbital is antisymmetric relative to this plane cos(-)=-cos
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 cos. It does not depend on and has the z axis for symmetry axis.
The angular contribution to the density of probability varies like cos2
Within cones:
Full space 2 cones: a diabolo
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angle Density inside
The half cone(1-cos(l)3)/2
Part of the volume
(1-cos l)/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 cos. It does not depend on and has the z axis for symmetry axis.
The angular contribution to the density of probability varies like cos2
Within cones: angle Density
inside The half cone(cosl)3-1)/2
Part of the volume
(cos l-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
Probability is 87.5% in half of the space
22.5% in the other half
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--
++
O
M'
M
zSpatial representation of the angular part.
Let us draw all the points M with the same contribution of the angular part to the density
The angular part of the probability is OM = cos2
All the M points belong to two spheres that touch at O
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Isodensities, isolevels
Ptotale/cos2l
Ptotale/cos2
Ptotale
CA'
B'B
A
r
R2
BB'A'
A
C Ptotale/cos2l
Ptotale/cos2
Ptotale
CA'
B'B
A
r
R2
BB'A'
A
C
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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
2 is proportional to x2/r2+ y2/r2+ z2/r2=1and 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
= N radial function angular function rl (polynom of degree n-l-r)
n-l-r nodes l nodes
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yzxzxy
z2
x2-y
2
z
y
z
x
y
x
zy
x
Orbitales d
d orbitals
Clover, the forth lobe is the lucky one; clubs have three
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3d orbitals
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Compare the average radius of 1s for the hydrogenoides whose nuclei
are H and Pb.
Pb207
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Make comments on the 1s orbital the atom Pb?
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Name these orbitals
Spherical coordinates x = r sin cos ; y = r sin sin ;z = r cos
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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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