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1 He and hydrogenoid ions The one nucleus-electron system

1 He and hydrogenoid ions The one nucleus-electron system

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Page 1: 1 He and hydrogenoid ions The one nucleus-electron system

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He and hydrogenoid ions

The one nucleus-electron system

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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

82

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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