chem 59-250 the electronic structure of atoms classical hydrogen-like atoms: + - atomic scale: 10...

Chem 59-250

The Electronic Structure of Atoms

Classical Hydrogen-like atoms:

+

-

Atomic Scale: 10-10 m or 1 Å

Proton mass : Electron mass 1836 : 1

Problems with classical interpretation: - Should not be stable (electron should spiral into nucleus)- Atomic spectra wrong (discrete lines instead of continuum)

Chem 59-250

Quantum Chemical Interpretation:e- are best described by wave functions and quantum numbers.

Quantum numbers for electrons:

n = principal quantum number

l = azimuthal quantum number(orbital angular momentum)

ml = magnetic quantum number

ms = spin magnetic quantum number

“Quantum” means not continuous.

Chem 59-250Energy levels and n

E h Hl h

R

n n1 1

2 2

where: nl < nh

Spectral transitions predicted by:

Chem 59-250

R

e

2

4

2 4

0

2 2

h

Bohr Model of the H atom

= reduced mass (nucleus and electron)Z = nuclear charge (1 for H)e = charge of electron0 = permittivity of a vacuumh = Plank’s constant

EZ

n Rn

Rn

2

2 2

1

R = 13.6 eV

radius(n) = n2a0

a0 = Bohr radius = 0.529 Å

Chem 59-250

n = principal Energy, Size SHELL

n = 1, 2, 3 …

n = 1 : ground state

n = 2 : first excited state

n = 3 : second excited state

E n n

13 6

12.

E1 = -13.6 eV

E2 = -3.40 eV

E3 = -1.51 eV

E4 = -0.85 eV

E5 = -0.54 eV•••E = 0 eV

Chem 59-250

l = azimuthal, orbital angular momentum

degeneracy, shape SUBSHELL

l = 0, 1, 2 …n –1 degeneracy = 2 l + 1

l 0 1 2 3 4, 5, 6 …

type s p d f g, h, i …

degeneracy 1 3 5 7 9, 11, 13 …

s = “sharp”p = “principal”d = “diffuse”

Chem 59-250

ml = magnetic type, “orientation in space” ORBITAL

ml = 0, ±1, ±2 …±l

# of orbitals in a subshell = 2 l + 1

l ml orbital

0 0 s

1 0 pz

±1 px

±1 py

2 0 d2z2-x2 -y2 = dz2

±1 dxz

±1 dyz

±2 dx2 –y2

±2 dxy1s orbital

Chem 59-250

Wave functions and OrbitalsH E

= Rnl(r)Ylml(,)

Rnl(r) – radial function

Ylml(,) – angular function

polar coordinates

= wave function

2 = probability (electron) density

4r22 = radial distribution function

Chem 59-250

Nodes: surfaces where there is 0 probability of finding an electron

Number of nodes = n - 1 Number of radial nodes = n - l –1Number of angular nodes = l 0 for s orbitals

1 for p orbitals2 for d orbitals (except dz2)

A website demonstrating nodes for 2D wavefunctions can be found at:http://www.kettering.edu/~drussell/Demos/MembraneCircle/Circle.html

Chem 59-250

s orbitals

p orbitals (the shading is backwards)

Chem 59-250

d orbitals

Chem 59-250

Summary of quantum numbers needed for H:

n Energy of shell: , size: radius(n) = n2a0

l type (degeneracy) of subshell: s(1), p(3), d(5), f(7) ….

ml type and orientation of orbitals in a subshell

EZ

n Rn

2

2

Energy level diagram:

Chem 59-250

+

--

Many electron atomsE

Zn R

n

2

2

E1 1

R

2 2

2

He, Z = 2

Predict: E1 = -54.4 eV

Actual: E1 = -24.6 eV

Something is wrong with the Bohr Model!

Chem 59-250

ms = spin magnetic electron spin

ms = ±½ (-½ = ) (+½ = )

Pauli exclusion principle:

Each electron must have a unique set of quantum numbers.

Two electrons in the same orbital must have opposite spins.

Electron spin is a purely quantum mechanical concept.

N

S

H

Chem 59-250

N

S

H

N

S

He

Energy level diagram for He. Electron configuration: 1s2

paramagnetic – one (more) unpaired electrons

diamagnetic – all paired electrons

Ene

rgy

1

2

3

0 1 2

n

l

Chem 59-250

Effective Nuclear Charge, Z*

The presence of other electrons around a nucleus “screens” an electron from the full charge of the nucleus.

We can approximate the energy of the electrons by modifying the Bohr equation to account for the lower “effective” nuclear charge:

EZ

n Rn

*2

2

Z* = Z -

Z* is the effective nuclear chargeZ is the atomic number is the shielding or screening constant

Chem 59-250

Helium , Z = 2

Predicted: E1 = -54.4 eV

Actual: E1 = -24.6 eV - 24.6 -13.61

2

2

Z *

Z *.

.

24 6 1

13 6

2

Z* = 1.34

1.34 = 2 -

= 0.66

+

--

EZ

n Rn

*2

2

Chem 59-250

Lithium , Z = 3

Predicted: E2 = -30.6 eV

Actual: E2 = -5.4 eV -5.4 -13.62

2

2

Z *

Z *.

.

5 4 2

13 6

2

Z* = 1.26

1.26 = 3 -

= 1.74

+ -

-

-

EZ

n Rn

*2

2

Chem 59-250We want to be able to predict and Z*

Slater’s rules for the prediction of for an electron:

1. Group electron configuration as follows:(1s)(2s,2p)(3s,3p)(3d)(4s,4p)(4d)(4f)(5s,5p) etc.

2. Electrons to the right (in higher subshells and shells) of an electron do not shield it.

3. If the electron of interest is an ns or np electron:a) each other electron in the same group contributes 0.35 (0.30 for 1s)b) each electron in an n-1 group contributes 0.85c) each electron in an n-2 or lower group contributes 1.00

4. If the electron of interest is an nd or nf electron:a) each other electron in the same group contributes 0.35b) each electron in a lower group (to the left) contributes 1.00

Chem 59-250

Z* = Z -

Example with a valence electron on oxygen: O, Z = 8 Electron configuration: 1s2 2s2 2p4

a) (1s2) (2s2 2p4)

= (2 * 0.85) + (5 * 0.35) = 3.45 1s 2s,2p

Z* = Z -

Z* = 8 – 3.45 = 4.55

This electron is actually held with about 57% of the force that one would expect for a +8 nucleus.

Chem 59-250

Z* = Z -

Example with two electrons for nickel: Ni, Z = 28 Electron configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d8

(1s2) (2s2 2p6) (3s2 3p6) (3d8) (4s2)

For a 3d electron: = (18 * 1.00) + (7 * 0.35) = 20.45 1s,2s,2p,3s,3p 3d

Z* = Z - Z* = 28 – 20.45 = 7.55

For a 4s electron: = (10 * 1.00) + (16 * 0.85) + (1 * 0.35) = 23.95 1s,2s,2p 3s,3p,3d 4s

Z* = Z - Z* = 28 – 23.95 = 4.05

Chem 59-250

The basis of Slater’s rules for

s and p orbitals have better “penetration” to the nucleus than d (or f) orbitals for any given value of n

i.e. there is a greater probability of s and p electrons being near the nucleus

This means:

1. ns and np orbitals completely shield nd orbitals

2. (n-1) s and p orbitals don’t completely shield n s and p orbitals

Chem 59-250 Periodicity of Effective Nuclear Charge

H1.00

He1.65

Li1.30

Na2.20

K2.20

Rb2.20Cs

2.20Ba

2.85

Sr2.85

Ca2.85

Mg2.85

Be1.95

B2.60

Al3.50Ga

5.00

In5.00Tl

5.00Pb

5.65

Sn5.65

Ge5.65

Si4.15

C3.25

N3.90

P4.80As

6.30

Sb6.30Bi

6.30Po

6.95

Te6.95

Se6.95

S5.45

O4.55

F5.20

Cl6.10Br

7.60

I7.60At

7.60Rn

8.25

Xe8.25

Kr8.25

Ar6.75

Ne5.85

Z* on valence electrons

Chem 59-250 Shielding and Effective Nuclear Charge

The energy of valence electrons in an atom/ion changes with the loss of addition of an electron.

Slater’s rules are only approximate and can give poor predictions. For example:

They ignore the differences in penetration between s and p orbitals. Real s and p orbitals do not have the same energy.

They assume that all electrons in lower shells shield outer electrons equally effectively.

Z* can be used to estimate ionization energy:

H 13.6ie

2

2

Z *

n

Chem 59-250 Shielding and Effective Nuclear Charge

Effective nuclear charge can be used to rationalize properties such as the size of atoms and ions.

Cp*2Be Cp*2B+

Be and B+ are isoelectronic (1s2 2s2) but very different because of effective nuclear charge.