Chapter 2 Atomic Structure
History of
Atomic Models
• Understanding
atomic structure is a
first step to
understand structure
of the matter
• The so-called electron
density is actually the
probability density of
electron wave!
2.1 The Schrödinger equation and its
solution for one-electron atoms
2.1.1 The Schrödinger equation
The Hamiltonian Operator of one-electron atoms
H atom, He+ and Li2+
Hˆ  Tˆn  Tˆe  Vˆn e
2 2 2 2 ˆ

n 
 e  Vn e
2mn
2me
The reduced mass
M  me
1836me


 0.999me
M  me (1  1836)me
n
e
Consider that the electron approximately surrounds the
atomic nucleus, the Hamilton operator can be simplified as
Hˆ  Tˆe  Vˆn e
2 2 ˆ

 e  Vn e
2me
2 2
2
2
Ĥ  ( 2  2  2 )  V̂n -e
2m e x
y
z
Vˆn e
r
2
Ze

4 0 r
r x y z
2
The Schrödinger equation
Separation of variables ?
2
2
Hˆ   E
e
Spherical polar coordinates
r  x2  y2  z 2
cos  
z
x2  y2  z2
tg  y / x
d = dx dy dz
= r2 sin dr d d
r  distance from origin.
  angle drop from the z-axis.
  angle from the x-axis (on x-y plane)
(x,y,z)  (r,)
x,y,z  r, 
r, Rr
d = dx dy dz
= r2 sin dr d d
Partial derivatives
r x y z
2
cos 
2
x = r sin cos
y = r sin sin
z = r cos
2
z
x y z
2
2
tg  y / x
2
1
2
x
r 1 2
2
2
 ( x  y  z ) 2 x   sin  cos 
r
x 2


1 2
zx
2
2 3 / 2
 sin 
 z ( )( x  y  z ) 2 x   3
x
2
r
 cos  cos 

x
r
Spherical polar coordinates

r 
 
 
( ) ( )
( )
x
x r
x 
x p
 sin  cos 
 cos  cos  
sin  


r
r
 r sin  

 cos  sin  
cos  
 sin  sin  



r
y
r
 r sin  
 sin  

 cos 


z
r
r 
1  2 
1


1
2
  2
(r
) 2
(sin 
) 2
r

r r
r sin  
r sin 2   2
2
Vn  e
Ze 2

(4 0 )r
The Schrödinger equation


 2 8 2 me
1  2 
1
1
Ze 2

(r
) 2
(sin 
) 2 2
(E 
)  0
2
2
2
r

r r
r sin  
r sin  
h
4 0 r
2.1.2 The solution --- separation of variables
Substitute r, Rr into the equation


 2 8 2 me
1  2 
1
1
Ze 2

(r
) 2
(sin 
) 2 2
(E 
)  0
2
2
2
r

r r
r sin  
r sin  
h
4 0 r
and multiply with
r2
R(r )( ) ( )
1  2 R(r )
1

( )
1
 2  ( ) 8 2 me
ze 2
2
(r
)
(sin 
)

(
E

)
r
0
2
2
2
R (r ) r
r
( ) sin  

 ( ) sin  
h
4 0 r
ze 2

( )
 2  ( )
1  2 R(r ) 8 2 me
1
1
2
(r
)
(E 
)r  [
(sin 
)
]
2
2
2
R (r ) r
r
h
( ) sin  

 ( ) sin  
4 0 r
1
1
1  2 R(r ) 8 2 me
ze 2

( )
 2  ( )
2
(r
)
(E 
)r  [
(sin 
)
]
2
2
2
4 0 r
r
( ) sin  

 ( ) sin  
R (r ) r
h
let
ze 2
1  2 R(r ) 8 2 me
2
E
r
(r
)
(
)


2
R(r ) r
h
4 0 r
r
1
1

( )
 2  ( )
(sin 
)
 
2
2
( ) sin  

 ( ) sin  
multiplied
RRadial
eqpart
(
.)
Angular
sin 2 
sin  
1  2  ( )
( )
2
(sin 
)
sin




2
( ) 

 ( ) 
2
sin  
1
( )

 ( )
(sin 
)   sin 2   
( ) 

 ( )  2
sin  
( )
let
(sin 
)   sin 2   m 2
(
( ) 

1  2  ( )
2
m


 ( )  2
eq.)
part
a. () equation
Its complex form:
d 2  ( )
2

m
 ( )  0
2
d
 =Aei|m|, let m= |m|,  =Aeim 
Normalization

2
0
  m d  
A
*
m
2
0
A2 e im eim d  1
1
2
1 im
1
i
m 
e 
cos m 
cos m
2
2
2
 m ( )   m (  2 )
eim  eim (  2 )  eim  eim 2
eim 2  1
cos 2m  i sin 2m  1
The values of m must be
m  0,1,2,     
m : magnetic quantum
number
complex function
 m  Aeim
(m  0,1,2,    )
 m  Aeim  A cos m  iA sin m
  m  Ae im  A cos m  iA sin m
its linear combination
1   m    m  2 A cos m  B cos m
 2   m    m  2iA sin m  B' sin m
normalization condition

2
0
  1 d  
B
real function
1 
2 
*
1
2
0
1

1

1

cos m
sin m
( B cos m ) 2 d  B 2  1
sin 2   cos 2   1
sin 2  2 sin  cos 
cos 2  2 cos 2   1
cos(   )  cos  cos   sin  sin 
sin   sin   2 sin
 
cos
 
2
2
 
 
sin   sin   2 cos
sin
2
2
 
 
cos   cos   2 cos
cos
2
2
cos   cos   2 sin
 
2
sin
 
2
Table The solution of () equation
m
complex form
0
1
0 
2
1
1 i
1 
e
2
1
2
2
 1 
1  i
e
2
1 i 2
2 
e
2
1 i 2
 2 
e
2
real form
1
0 
2

cos
1

 sin
1 

cos
2

sin
2


1

1

1

1

cos 
sin 
cos 2
sin 2
b. () equation
sin  
( )
(sin 
)   sin 2   m 2

( ) 
When =l(l+1), l =0,1,2,3,…… () is a well behaved function,
( )  c(1  x 2 )
m
2
d
l m
dx
l m
[( x 2  1) l ]
x  cos
l : angular momentum
quantum number
1 (2l  1) (l  m )
c l
2 l!
2 (l  m )
Necessary condition :
l m
necessary condition: l  |m|
hence, l = 0, 1, 2, 3,…(s,p,d,f,g,h…)
m=0,(-1,0,1),(-2,-1,0,1,2)……
Examples, calculating ()?
l
0
1
m
( )
0
1
2
0
±1
2
0
±1
±2
6
cos 
2
3
sin 
2
10
(3 cos 2   1)
4
15
sin  cos 
2
15
sin 2 
4
c. Solution of R equation
ze 2
1  2 dR (r ) 8 2 me
2
E
r
(r
)
(

)

2
R(r ) dr
h
r
4 0 r
Solution
R(r )  N  e
2Z
  2 r 
r
na0
  / 2 2 l 1
n l
L
()
o
4 0  2
a0 
 0.529 A
2
me e
a0 is called as Bohr radius
2
{(
n

l
)!}
2 l 1
k 1
k

Ln l (  )   ( 1)
( n  l  1  k )!(2l  1  k )! k !
k 0
n l 1
necessary condition: n  l+1
hence, n = 1, 2, 3,……
l = 0, 1, 2, ……
n: Principal
quantum number
Z 2h 2
h2
Z2
Z2
E   2 2 2  ( 2 2 )  2   R 2
8 mn a0
8 ma0 n
n
R is called as Rydberg constant with the value of 13.6 eV
Example:
n=1, l=0
n=2, l=0
Z
R10  2  
 a0 
R20
3/ 2
e
1  Z


a
2 2 0
 /2




3/ 2
( 2   )e
 / 2
Some wavefunctions of hydrogen atom and hydrogen-like ions
2.2 The physical significance of
quantum numbers
2.2.1 The allowed values of quantum numbers
Quantum Numbers
n
1
l
0
0
ml
0
0 -1 0 +1 0 -1 0 +1
2
3
1
0
1
4
2
0
1
4
9
2
3
0 -1 0 +1
-2 -1 0 +1 +2
n2
1
-2 -1 0 +1 +2
-3 -2 -1 0 +1 +2 +3
16
Problem: What values of the azimuthal (l) and magnetic (m) quantum
numbers are allowed for a principal quantum number (n) of 4? How
many orbitals are allowed for n=4?
Plan: We determine the allowable quantum numbers by the rules given
in the text.
Solution: The l values go from 0 to (n-1), and for n=4 they are:
l = 0,1,2,3. The values for m go from -l to zero to +l
For l = 0,
l = 1,
l = 2,
l = 3,
ml = 0
ml = -1, 0, +1
ml = -2, -1, 0, +1, +2
ml = -3, -2, -1, 0, +1, +2, +3
There are 16 mL values, so there are 16 orbitals for n=4!
as a check, the total number of orbitals for a given value of n is n2, so for
n = 4 there are 42 or 16 orbitals!
2.2.2 The principal quantum number, n
• Also called the “energy” quantum number, indicates the
approximate distance from the nucleus .
• Denotes the electron energy shells around the atom,
and it is derived directly from the Schrodinger equation.
• The higher the value of “n” , the greater the energy of
the orbital.
• Positive integer values of n = 1 , 2 , 3 , etc.
Example: Energy states of a H atom.
n = principal
n = 1, 2, 3 …
n = 1 : ground state
n = 2 : first excited state
n = 3 : second excited state
 1
E n   13.6  2 
n 
E1 = -13.6 eV
E2 = -3.40 eV
E3 = -1.51 eV
E4 = -0.85 eV
E5 = -0.54 eV
•
•
•
E = 0 eV
Example: Energy states of a Li2+ ion.
2
Z
En   R 2
n
R = 13.6 eV
3s
E1s= - 9R
2s
E2s,sp= - (9/4)R
E3s= - R
1s
3d
3p
2p
2.2.3 the azimuthal, orbital angular momentum, l
•Denotes the different energy sublevels within the main level “n”
•Also indicates the shape of the orbitals around the nucleus.
orbital angular momentum  degeneracy,
l = 0, 1, 2 …n –l
l
degeneracy = 2 l + 1
0 1 2 3 4, 5, 6 …
type
s p d f g, h, i …
degeneracy 1 3 5 7 9, 11, 13 …
The operator of angular momentum
Mˆ 2  Mˆ x2  Mˆ y2  Mˆ z2


ˆ
M x  i( y  z )
z
y


ˆ
M y  i( z  x )
x
z


M z  i( x  y )
y
x


)
M x  i(sin 
 cot  cos 





)
M y  i(cos 
 cot  sin 




M z  i



1 
1 
M   [
(sin 
) 2
]
2

sin  
sin  

2
2
2
The operator of angular momentum
2



1
1
2
2
M   [
(sin 
) 2
]
2

sin  
sin  

1

( )
1
 2  ( )
(sin 
)
 
2
2
 ( ) sin  

 ( ) sin  
2
Yl ,m ( ,  )
1 
1  Yl ,m ( ,  )
(sin 
)
   Yl ,m ( ,  )
2
2
sin  

sin 

2
Y
Yl ,m ( ,  )

(

,

)

1

1
l ,m
2
2
Yl ,m ( ,  )

 [
(sin 
)
]


2
2
sin  

sin 


M 2 Yl ,m ( ,  )    2Yl ,m ( ,  )  l ( l  1)  2Yl ,m ( ,  )
where Yl ,m ( ,  )  ( ) ( );   l ( l  1)

M 2   2 
2  l (l  1) 2
  l (l  1)
M  l (l  1)
（l  0,1,2,3   ）
When there exist an angular momentum, there is
also a magnetic moment.
e

M
2me
e
 
l (l  1)  l (l  1)  e
2me
e
e 
 9.274 10  24 J  T 1
2me
Bohr magneton
2.2.4 Magnetic Quantum Number , m
• Denotes the direction or orientation of an orbital
ml = magnetic  type, “orientation in space”  ORBITAL
ml = 0, ±1, ±2 …±l
number of orbitals in a subshell
=2l+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
1s orbital
±2 dxy
2.2.4 Magnetic Quantum Number , m
• Determine the z-component of the orbital angular
momentum of the atom.
• Determine the component z of the magnetic moment in
the direction of the magnetic field.



ˆ
M z  i  ( x  y )  i 
y
x


M z   

M z   i 




R  iR
Aeim
  i 
 iR




 iRAimeim  mRAeim  m

M z   m
  m
M z  m
z  
e
em
Mz 
  m e
2me
2me
Hence an atomic electron that
possesses angular momentum
interacts with an external
magnetic field B.
Space quantization
Space quantization of
orbital angular
momentum. Here the
orbital quantum
number is l=2 and
there are accordingly
l2
2l+1=5 possible values
| L | l (l  1)  6
of the magnetic
quantum number ml,
with each value
corresponding to a
different orientation
relative to the z axis.
Lz  m
2.3 The wave function and electron
cloud
2.3.1 The wave-functions of hydrogen-like ions
3
Z
 1s 
e
3
 a0

Zr
a0
 2s
3
1 Z
Zr

(2  )e
3
4 2 a0
a0

Zr
2 a0
r=2a0
r=4a0
Fig. (left) The -r and 2-r diagram of the 1s state of the
hydrogen atom. (right) The -r of the 2s state .
A Radial
Probability
Distribution
of Apples
2.3.2 The radial distribution function
• The probability of finding electron in the region of space
r + dr, θ + dθ, φ + dφ
d   r 2 sin drd d 
 d  Rnl ( r )  Yl ( ,  ) r sin drd d 
2
2
2
m
2
• What is the probability of electron at r + dr,(in a thin
spherical shell centered at the origin)?
Rnl ( r ) r dr 
2
2
2
0
D  r Rnl ( r )
2
2


0
2
Yl ( ,  ) sin  d d  r 2 Rnl2 ( r )dr
m
Normalized spherical harmonics
Radial distribution function
 = wave function
2 = probability density
r2R2 = radial probability
function function
Calculating the most
probable radius
r 2 Rn2,l ( r )
D ( r )
0
r
Expectation value of r
 r    *r d


0

2
0
0
 

rR | Yl ,m | r dr sin  d d   R r dr
2
n ,l
2
2
0
2 3
n ,l
Example: please derive the radial probability distribution
function of the H 1s orbital and the radium of its maximum.
1
 1s 
e
3
 a0
r
a0
Its radical probability distribution function is
D ( r )1s  
2
0


0
2
4r
| 1s | r dr sin  d d  4 r | 1s |  3 e
a0
2
2
2
To derive the radium of its maximum, we have
dD ( r )1s
8r
0 3 e
dr
a0

2r
a0
r
(1  )  0
a0
r
 r  0,  1   0  r = a 0
a0
2
2 r
a0
• A spherical surface with
the electron density of
zero is called a node, or
nodal surface.
The radial distribution
function has (n-l) maxima
and (n-l-1) nodal surfaces
Fig. The radial distribution diagram (D(r) - r)
2.3.3 The angular function ( Ylm() )
 (r , ,  )  R  RY ( ,  )
Angular part
Ylm ( ,  )   l ( ) m ( )
It indicates the angular distribution function of an atomic
orbital and is the eigenfunction of M2 and Mz operators.
1 
1 

M   [
(sin 
) 2
]
2
sin  
sin  


2
2
2
2
ˆ
M Ylm ( ,  )  l (l  1) Ylm ( ,  )
Mˆ zYlm ( ,  )  mYlm ( ,  )
2
Angular function ( Ylm() )
s-Orbital (l=0, 2l+1=1)
Y00
Y0, 0
1

4
1
s
4
p-Orbital (l=1, 2l+1=3)
Y10
Y10 
Y1,1 
Y11
3
cos 
4
3
1 i
sin  
e
2
2
3
1  i
Y1, 1 
sin  
e
2
2
pz 
3
cos
4
px 
3
sin  cos 
4
py 
3
sin  sin 
4
1
3
(Y1,1  Y1, 1 ) 
sin  (e i  e i )
2
4 
i
1
3
(Y1,1  Y1, 1 )  i
sin  (ei  e i )
2
4 
p-Orbital
px
py
pz
d-Orbital (l=2, 2l+1=5)
Y20
d z2 
1
5
4 
(3cos 2   1)
Y21
d xz 
1 15
d yz 
1 15
4
4


Y22
sin 2 cos 
d x2  y 2 
sin 2 sin 
d xy 
1 15
4
1 15
4


sin 2  cos 2
sin 2  sin 2
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 dxy
d-Orbital
d z2 
1
5
4 
d xz 
(3cos 2   1)
1 15
4

d x2  y 2 
sin 2 cos 
d yz 
1 15
4

sin 2  cos 2
d xy 
1 15
4

sin 2 sin 
1 15
4

sin 2  sin 2
Nodes
A node is a surface on which an electron is not found.
For a given orbital, the total number of nodes equals n-1.
The number of angular nodes is l.
2.4 The structure of manyelectron atoms
(multi-electron atoms)
2.4.1 Schroedinger equation of many-electron atoms
For hydrogen - like atoms :
Hˆ   E
2
2
h
Ze
Hˆ   2  2 
8 m
4 0 r
for
He
a.u.
e 1
a0  1
a.u.
a.u.
h
1
2
4 0  1

in atom unit (a.u.)
1
Z
Hˆ    2 
2
r
me  1
( Hˆ  Tˆ  Vˆ )
a.u.
a.u.
atom
Z Z
1
1
2
2
ˆ
H   (1   2 )  (  ) 
r1 r2
r12
2
r12  ( x1  x2 ) 2  ( y1  y2 ) 2  ( z1  z 2 ) 2
2
2
2
2
2
2






2  2  2  2 


2
2
2
z 2
x1 y1 z1 x2
y2
Z Z
1
Vˆ  (  ) 
r1 r2
r12
r12
e-1
r1
+
e-2
r2
for many  electron atoms
1
1
Z
2
ˆ
H    i  

2 i
ri i  j rij
i
if
1
rij
Hˆ 
is

i
omitted
1 2 Z
(  i  ) 
ri
2
   1  2  3   
n

Hˆ i
i
 
E  E1  E 2  E 3     
i 1
i
E
i
i
Many electron atoms
He, Z = 2
-
+
 Z2 
En   R  2 
n 
 22 
E1   R  2 
1 
Predict: E1 = -54.4 eV
Actual: E1 = -24.6 eV
Something is wrong with the Bohr Model!
Self-consistent-field (SCF) method
Central field method
1 2
H (1, 2  n)   hi   [ i  V (i )]
2
i
i


 (1,2,3 n)  1 (1) 2 (2)  n (n)

hi  i   i i
E   1   2   n
Slater approximate method for central field
Z i
Z i
ˆ
Vi   

ri ri
ri
Z *  Z 
Rn ,l  (2 )
n
1
2
r
n 1 r
e
Z i

n*
Ei  (
: Screen constant
The presence of other electrons around
a nucleus “screens” an electron from
the full charge of the nucleus.
Z*: Effective nuclear charge
n*: Effective principal quantum number
Z i 2  e
) 
*
4 0 a0
n
2
Z i 2
 R( * )
n
Z i 2
 －13.6eV  ( * )
n
n*=n (when n  3)
n*=3.7 (when n = 4)
n*=4.0 (when n = 5)
How about n > 5 ???
Lithium , Z = 3
Predicted: E1 = -30.6 eV
Actual: E1 = -5.4 eV
 Z *2 
En   R  2 
 n 
 Z *2 
-5.4  -13.6 2 
 2 
-
Z* 
-
+
-
5.4  22
13.6
Z* = 1.26
1.26 = 3 - 
 = 1.74
a. Screen (shielding) constant
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 in the right shells (in higher subshells and shells) of
an electron do not shield it.
3. For ns or np valence electrons:
a) each other electron in the same group contributes 0.35
(0.30 for 1s)
b) each electron in an n-1 group contributes 0.85
c) each electron in an n-2 or lower group contributes 1.00
4. For nd or nf valence electrons:
a) each other electron in the same group contributes 0.35
b) each electron in a lower group (to the left) contributes 1.00
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
Example : O, Z = 8
Electron configuration: 1s2 2s2 2p4
a) (1s2) (2s2 2p4)
b)  = (2 * 0.85) + (5 * 0.35) = 3.45
1s
Z* = Z - 
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.
Example: Ni, Z = 28
Electron configuration: 1s2 2s2 2p6 3s2 3p6 3d8 4s2
(1s2) (2s2 2p6) (3s2 3p6) (3d8) (4s2)
For a 3d electron:
 = (18 * 1.00) + (7 * 0.35) = 20.45
1s,2s,2p,3s,3p
Z* = Z - 
Z* = Z - 
3d
Z* = 28 – 20.45 = 7.55
For a 4s electron:
 = (10 * 1.00) + (16 * 0.85) + (1 * 0.35) = 23.95
1s,2s,2p
Z* = Z - 
3s,3p,3d
4s
Z* = 28 – 23.95 = 4.05
B. Approximation of the atomic orbital energy
Example: Mg, 1s22s22p63s2
Z i 2
En   R  ( * )
n
(12   ) 2
(12  0.3) 2
E1s   R
 136.89 R   R
 136.89 R
2
2
1
1
E2s,2 p
(12  0.85  2  0.35  7) 2
7.85 2
 R
 R
 15.40 R
2
4
2
(12  1.00  2  0.85  8  0.35) 2
E3s   R
 0.9025 R
2
3
2.4.2 The ionization energy and
the affinity energy
I. Ionization energy:
The minimum energy required to remove an electron
from one of its orbitals (in the gas phase).

A( g )  A ( g )  e
The first ionization energy

I 1  E ( A )  E ( A)
The second ionization energy
2

I 2  E( A )  E( A )
Periodic Trends in Ionization Potentials
II. Estimation of ionization energy
Example: CC+, 1s22s22p2  1s22s22p1
I1=E=E(C+)-E(C)
Z i 2
En   R  ( * )
n
E (C  )  2 E1S  3E2' S , 2 P
E (C )  2 E1S  4 E2 S , 2 P
for C 
Z 2*S , 2 P  Z    6  (2  0.35  2  0.85)  3.60
for C
Z 2*S , 2 P  Z    6  (3  0.35  2  0.85)  3.25
I1  3 E
'
2S ,2P
 4 E2 S , 2 P
I actual  11.22eV
3.6 2
3.25 2
 [3  ( )  4  (
) ]  R  11.44eV
2
2
Example: Fe+ for lacking of a K electron, Z = 26
Electron configuration: 1s1 2s2 2p6 3s2 3p6 4s2 3d6
(1s1) (2s2 2p6) (3s2 3p6) (3d6) (4s2)
1s:
26 – 0*0.30 = 26.00
Z* = Z - 
2s,2p: 26 – 7*0.35 – 1*0.85 = 22.70
3s,3p: 26 – 7*0.35 – 8*0.85 – 1*1.0 = 15.75
3d:
26 – 5*0.35 – 17*1.0 = 7.25
4s:
26 – 1*0.35 – 14*0.85 – 9*1.0 = 4.75
J. C. Slater, Phys. Rev. 36(1930)57
Example: Fe, Z = 26
Electron configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d6
(1s2) (2s2 2p6) (3s2 3p6) (3d6) (4s2)
1s:
26 –0.30 = 25.70
Z* = Z - 
2s,2p: 26 – 7*0.35 – 2*0.85 = 21.85
3s,3p: 26 – 7*0.35 – 8*0.85 – 2*1.0 = 14.75
3d:
26 – 5*0.35 – 18*1.0 = 6.25
4s:
26 – 1*0.35 – 14*0.85 – 10*1.0 = 3.75
J. C. Slater, Phys. Rev. 36(1930)57
*
Z
Estimation of ionization energy
2
E



13.6
(
)
n
Example: FeFe+
n*
Fe(Z*)
Fe+(Z*)
(1s)2-1 : 25.70
26.00
(2s,2p)8 : 21.85
22.70
(3s, 3p)8: 14.75
15.75
(3d)6 :
(4s)2 :
6.25
3.75
7.25
4.75
E(Fe)=-2(25.70/1)2-8(21.85/2)2
-8(14.75/3)2-6(6.25/3)2
-2(3.75/3.7)2= -2497.3
E(Fe+)=-2(26.00/1)2-8(22.70/2)2
-8(15.75/3)2-6(7.25/3)2
-2(4.75/3.7)2= -1965.4
J. C. Slater, Phys. Rev. 36(1930)57
I1K = E = E(Fe+)-E(Fe)
= -1965.4+2497.3=531.9 eV (524.0 expt)
III. Electron Affinity
The electron affinity is the energy change that occurs
when an electron is added to a gaseous atom .
B(g) + e-  B- (g) + A
• Electron affinity usually increases as the radii of atoms
decrease.
• Electron affinity decreases from the top to the bottom
of the periodic table.
1. EA of an atom can be empirically predicted using the Slater’s rules.
2. In practice, EA of an atom can be measured by measuring the first
ionization potential of its monoanion in gas phase!
IV. The Electronegativity
• Electronegativity was proposed by Pauling to
evaluate the comparative attraction of the
bonding electrons by the atoms.
It can be concluded that:
1. The electronegativities of metals are small while those
of non-metal are large.
2. Generally, the electronegativity increases from left to
right across the periodic table but it decrease from top
to bottom within a group.
3. Elements with great difference in electronegativities
tend to form ionic bonds.
2.4.3 The building up principle
and the electronic configurations
I. The building up principle (for the ground state)
a. Pauli exclusion principle.
Every orbital may contain only two electrons of
opposite spins.
b. The principle of minimum energy.
Whilst being compatible with the Pauli principle,
electrons occupy the orbital with the lowest energy first.
For multi-electron atoms:
The energy level can be estimated by n+0.7l. (G.X. Xu proposed.)
1s
2s
2p
3s
3p
3d
4s
4p
4d
4f
1.0
2.0
2.7
3.0
3.7
4.4
4.0
4.7
5.4
6.1
5s
5p
5d
5f
6s
6p
6d
6f
5.0
5.7
6.4
7.1
6.0
6.7
7.4
8.1
Therefore, the sequence of the atomic orbitals is: 1s,
2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 5p, 6s, 4f, 5d, 6p
Relative Energies for Shells and Orbitals
•The orbitals have
different energies
and for the d and f
orbitals, the
energies overlap sorbital energies in
the next principal
level.

s
p
d
f
8
7
6
5
4
3
2
1
1s,2s,2p,3s,3p,4s,
3d,4p,5s,4d,5p,6s,
4f,5d,6p,7s,…
Relative
Energies of
orbitals
The principle of minimum energy
4
3
2
1
I. The building up principle (for the ground state)
a. Pauli exclusion principle.
Every orbital may contain only two electrons of
opposite spins.
b. The principle of minimum energy.
Whilst being compatible with the Pauli principle,
electrons occupy the orbital with the lowest energy first.
c. Hund’s rule.
In degenerate energy states, electrons tend to occupy as
many degenerate orbitals as possible. (The number of
unpaired electrons is a maximum)
II. The electronic configuration (for the ground state)
1s
•
•
•
•
•
•
•
•
•
•
H = 1s1
He = 1s2
Li = 1s2 2s1
Be = 1s2 2s2
B = 1s2 2s2 2p1
C = 1s2 2s2 2p2
N = 1s2 2s2 2p3
O = 1s2 2s2 2p4
F = 1s2 2s2 2p5
Ne = 1s2 2s2 2p6
+
+
2s
2p
The Periodic Table of the Elements
Electronic Structure
H
Li Be
NaMg
K Ca Sc Ti
Rb Sr Y Zr
Cs Ba La Hf
Fr Ra Ac Rf
B C N
Al Si P
V Cr Mn Fe Co Ni Cu Zn Ga Ge As
NbMo Tc Ru Rh Pd Ag Cd In Sn Sb
Ta W Re Os Ir Pt Au Hg Tl Pb Bi
Ha Sg
O
S
Se
Te
Po
F
Cl
Br
I
At
Ce Pr Nd PmSm EuGd Tb DyHo Er Tm Yb Lu
Th Pa U Np PuAmCm Bk Cf Es FmMd NoLr
“ s” Orbitals
“ p” Orbitals
“ d” Orbitals
“ f ” Orbitals
He
Ne
Ar
Kr
Xe
Rn
Electron Configuration using the Periodic Table
2.5 Atomic spectra and spectral
term
2013年诺贝尔化学奖获奖理由：多尺度模型
表彰他们发展了复杂化学体系中的多尺度模型，该研究把计算机模
型应用于化学研究，开拓了崭新研究领域。
化学家们常常会利用塑料短杆和小球来表示分子结构。今天，化学
家们早已开始使用电脑来展示各种模型。在上世纪70年代，Martin
Karplus, Michael Levitt 和Arieh Warshel的工作为这项强大工具的应用
奠定了基础，帮助我们加深对化学过程的理解与预测。时至今日，化学
领域所取得的大部分重要进展都离不开先进计算机模型的帮助。
化学反应极为迅速，在数百万分一秒间，电子已经完成从一个原子
核向另一个原子核的迁移。经典化学已经难以跟上这样的步伐，要想借
助实验方法去描绘化学过程中的每一个小步骤几乎已经是不可能的任务。
借助本次化学奖所奖励的科学家们发展的方法，科学家们得以在计算机
的帮助下揭示一些精妙过程的细节，如废气的催化净化，或是植物绿叶
中发生的光合作用过程。
今年的诺贝尔化学奖成果简单来说便是综合了这两个不同领域方法
的精华，设计出了基于经典物理与量子物理学方法两大领域的方法。举
例而言，为了模拟药物如何在人体内与其靶标蛋白相结合，计算机会利
用量子物理方法计算靶标蛋白质原子与药物发生反应的过程。而这一大
型蛋白质的其余部分则会被借助基于经典物理学的方法进行模拟。
2.5.1 Total Electronic Orbital and Spin Angular Momenta
a. Addition of two angular momenta:
The coupling of two angular momenta characterized by
quantum number j1 and j2 results in a total angular momentum
whose quantum number J has the possible values:
J=j1+j2, j1+j2-1, …, |j1-j2|
2
1
0
-1
-2
j1+j2
j1-j2
Example: Find the possible values of the total-angularmomentum quantum number resulting from the addition
of two angular momenta with the quantum numbers j1 = 2
and j2 = 3.
Solution: j1+j2=2+3=5
|j1-j2|=|2-3|=1
The possible J values are: 5, 4, 3, 2, 1
Example: Find the possible values of the total-angularmomentum quantum number resulting from the addition of
two angular momenta with quantum number j1 = 2 and j2 =
3/2 .
Solution: j1+j2=2+3/2=7/2
|j1-j2|=|2-3/2|=1/2
The possible J values are: 7/2, 5/2, 3/2, 1/2
B. The total electronic orbital angular momentum
The total electronic orbital angular momentum of an n-electron
atom is defined as the vector sum of the angular momentum of
the individual electron:


L   li ; M L   mi
i
i
The total-electronic-orbital-angular-momentum quantum
number L of an atom is indicated by a code letter:
L
0
1
2
3
4
5
6
7
Letter S
P
D
F
G
H
I
K
For a fixed L value, the quantum number ML (MLħ---the z
component of the total electronic orbital angular momentum)
takes on 2L+1 values ranging from –L to L.
Example: Find the possible values of the quantum number L for
states of carbon atom that arise from the electron configuration
1s22s22p3d.
Solution:
s
l=0
p
l=1
d
l=2
Addition of two angular momenta rule
The total-orbital-angular-momentum quantum number ranges
from 1+2 = 3 to |1-2| = 1
L = 3, 2, 1
C. The total electronic spin angular momentum
The total electronic spin angular momentum S of an n-electron
atom is defined as the vector sum of the spins of the individual
electron:


S   ms (i )
i
For a fixed S value, the quantum number MS takes on 2S+1
values ranging from –S to S.
Example: Find the possible values of the quantum number S for
states of carbon atom that arise from the electron configuration
1s22s22p3d.
Solution:
1s electrons: Ms = + ½ - ½ =0
2s electrons: Ms = + ½ - ½ =0
2p electrons: ms = ½
3d electrons: ms = ½
Addition of two angular momenta rule
S = 1, 0
D. The total angular momentum
  
J  LS
J = (L + S), (L+ S) –1,… L- S 
Spin – orbit coupling
2.5.2 Atomic term and term symbol
A set of equal-energy atomic states that arise from the same
electron configuration and that have same value of L and the
same S value constitutes an atomic term.
Term symbol:
2S+1L
Each term consists of (2L+1)(2S+1) atomic states. (spinorbit interaction neglected)
3D
2P:
:
L=2, S=1
L=1, S=1/2
2.5.3 Derivation of Atomic term
a. Configurations of Completely filled subshells
MS=imsi =0
--
S=0
ML=imi =0
--
L=0
Only one term:
1S
b. Nonequivalent electrons.
(2p)1(3p)1
l1=1, l2=1
We need not worry about any restrictions
imposed by the exclusion principle
L= 2, 1, 0
ms1 = ½ ms2 = ½ S=1, 0
3D, 1D, 3P, 1P, 3S,
1S
terms
(np)1 (nd)1
l1=1, l2=2
L= 3, 2, 1
ms1 = ½ ms2 = ½
S=1, 0
3F, 1F, 3D, 1D, 3P, 1P
c. Equivalent electrons.
1s22s22p2

np2 (two electrons in the same subshell)
Equivalent electrons have the same
value of n and the same value of l. Two
electrons should avoid to have the same
four quantum numbers.
The number of microstates:
C  15
2
6
No
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
1





m
0
-1






















ML 
 m (i )
l
i
2
1
1
1
1
0
0
0
0
0
-1
-1
-1
-1
-2
ML
M S   m s (i )
i
0
1
0
0
-1
0
0
0
1
-1
0
0
1
-1
0
MS
1D, 3P, 1S
5
9
1 -> 15
• The terms arising from a subshell containing N electrons are
the same as the terms for a subshell that is N electrons short
of being full.
Term : p0
===
p6
p1
===
p5
p2
===
p4
(nd)2
S=1
Take off-diagonal numbers:
3, 2, 1, 0, -1, -2, -3 3F
3P
1, 0, -1
l1=2
m1=2
1
0
-1
-2
4
3
2
1
0
2
3
2
1
0
-1
1
2
1
0
-1
-2
0
S=0
Take diagonal and off-diagonal
numbers:
1
0
-1
-2
-3
0
-1
-2
-3
-4
-2
||
m2
4, 3, 2, 1, 0, -1, -2, -3, -4
2, 1, 0, -1, -2
0
l2=2
-1
1G
1D
1S
d. Energy level of microstates: (terms).
Hund’s Rule:
1. For terms arising from the same electron configuration the
term with the largest value of S lies lowest.
2. For the same S, the term with the largest L lies lowest.
np3 4S , 2D, 2P
p2 :
3P, 1D, 1S
e. Spin-Orbit interaction.
The spin-orbit interaction splits an atomic term into levels.
The splitting of these levels gives the observed fine structure
in atomic spetra.
2S+1L

2S+1L
J
J=L+S, L+S-1, …, |L-S|
np3 2P, 2D, 4S
4S
 4S3/2, 2D  2D5/2, 2D3/2 ,
f. The ground state of terms
Hund’s Rule:
3. For the same L and S values, when the number of
electrons is half-filled or less, the term with the smallest J
lies lowest; whereas when the number of electrons is more
than half-filled, the term with the largest J lies lowest.
nd3
L=3, S=3/2  4F3/2
2
1
0
np4
-1
-2
L=1, S=1
1
0
-1
 3P2
The ground-state term
(np)2
3P
0
Refer to Page 55
Hund’s Rules
less than half-filled:
Large S;Large L;Small J
3P
2
(np)4
(nd)4
more than half-filled:
Large S;Large L;Large J
(nd)6
5D
0
5D
4
s1s1s1：
s1s1：3S，1S
s1：L=0；S=1/2
{3S：L=0；S=1；s1：L=0；S=1/2}→
→{L=0；S=3/2，1/2}→{4S→4S3/2，2S →2S1/2}
{1S：L=0；S=0；s1：L=0；S=1/2}→
→{L=0；S=1/2}→{2S →2S1/2}
p1p1p1：
p1p1：3D，1D，3P，1P，3S，1S
p1：L=1；S=1/2
{3D：L=2，S=1；p1：L=1，S=1/2}→{L=3，2，1；S=3/2,1/2}→
→{4F，2F；4D，2D；4P，2P}
{1D：L=2，S=0；p1：L=1，S=1/2}→{L=3，2，1；S=1/2}→
→{2F；2D； 2P}
{3P：L=1，S=1；p1：L=1，S=1/2}→{L=2，1，0；S=3/2,1/2}→
→{4D，2D；4P，2P；4S，2S；}
{1P：L=1，S=0；p1：L=1，S=1/2}→{L=2，1，0；S=1/2}→
→{2D；2P；2S }
{3S：L=0，S=1；p1：L=1，S=1/2}→{L=1；S=3/2，1/2}→
→{ 4D；2P}
{1S：L=0，S=0；p1：L=1，S=1/2}→{L=1；S=1/2}→{2P}
p1p1p1： {4F(1), 4D(2),4P(3), 4S(1), 2F(2) ,2D(4),2P(6), 2S(2) }
p1p1d1：
p1p1：3D，1D，3P，1P，3S，1S
d1：L=2；S=1/2
{3D：L=2，S=1；d1：L=2，S=1/2}→{L=4, 3，2，1, 0 ；
S=3/2，1/2}→{4G ,4F, 4D,4P, 4S; 2G ,2F ,2D,2P,2S }
{1D：L=2，S=0；d1：L=2，S=1/2}→{L=4, 3，2，1, 0 ;S=1/2}→
→{2G ,2F ,2D,2P,2S }
{3P：L=1，S=1；d1：L=2，S=1/2}→{L=3, 2，1；S=3/2，1/2}→
→{4F, 4D,4P ; 2F ,2D, 2P }
{1P：L=1，S=0；d1：L=2，S=1/2}→{L=3, 2, 1；S=1/2}→
→{2F , 2D, 2P}
{3S：L=0，S=1；d1：L=2，S=1/2}→{L=2；S=3/2，1/2}→
→{ 4D；2D}
{1S：L=0，S=0；d1：L=2，S=1/2}→{L=2；S=1/2}→{2D}
p1p1d1： {4G ,4F(2), 4D(3),4P(2), 4S;2G(2) ,2F(4) ,2D(6),2P(4),2S(2) }
p1d1d1：
p1d1：3F,3D,3P, 1F,1D,1P
d1：L=2；S=1/2
{3F：L=3，S=1；d1：L=2，S=1/2}→{L=5, 4, 3，2，1, 0 ；
S=3/2，1/2}→{4H,4G ,4F, 4D,4P ; 2H ,2G ,2F ,2D,2P}
{3D：L=2，S=1；d1：L=2，S=1/2}→{L=4, 3，2，1, 0 ； S=3/2，
1/2}→{4G ,4F, 4D,4P, 4S; 2G ,2F ,2D,2P,2S }
{3P：L=1，S=1；d1：L=2，S=1/2}→{L=3, 2，1；S=3/2，1/2}→
→{4F, 4D,4P ; 2F ,2D, 2P }
{1F：L=3，S=0；d1：L=2，S=1/2}→{L=5,4,3,2,1; S=1/2} →
→{2H ,2G ,2F ,2D,2P}
{1D：L=2，S=0；d1：L=2，S=1/2}→{L=4,3,2,1, 0 ;S=1/2} →
→{2G ,2F ,2D,2P,2S }
{1P：L=1，S=0；d1：L=2，S=1/2}→{L=3, 2, 1；S=1/2}→
→{2F , 2D, 2P}
p1d1d1:{4H(1) ,4G(2) ,4F(3), 4D(3),4P(3), 4S(1);
2H(2) ,2G(4),2F(6),2D(6),2P(6),2S(2) }
1
1S
m
0 －1
1D
1S
0
mJ=0
2
mJ=2
1
0
－1
－2
3P
2
mJ=2
1
0
－1
－2
3P
1
3P
0
1D
(np)2
3P
a
b
c
d
mJ=1
0
－1
mJ=0
e