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Understanding the Periodic Table
vrije Universiteit amsterdam
PERIODIC TABLE OF THE ELEMENTS
n=5n=4n=3
n=2
llllmnl
n
−−−=−≤
=
,....2,1,1
.......3,2,1
n=1
n εn (eV) l=0 l=1 l=2 l=3 l=… . degeneracy
n n2
.. … .. … ..
4 ε4= -0.85 4s 4p 4d 4f 1+3+5+7 = 16
3 ε3= -1.51 3s 3p 3d 1+3+5 = 9
2 ε2= -3.40 2s 2p 1+3 = 4
1 ε1= -13.6 1s 1
Energy levels (n,l,m) in the hydrogen atom
llllmnl
n
−−−=−≤
=
,....2,1,1
.......3,2,1
n εn (eV) l=0 l=1 l=2
n
.. …..
4 ε4= -0.85 4s 4p 4d
3 ε3= -1.51 3s 3p
3d
2 ε2= -3.40 2s 2p
1 ε1= -13.6 1s
llllmnl
n
−−−=−≤
=
,....2,1,1
.......3,2,1
Why 1s22s2p4
?
Pauli exclusion principle
No more than 2 electrons per
state
Energy levels in an H-like atom
Quantum physicsλπ nr =2
22
24 nmZe
r on
hπε=
( )2
2
4 rZee
rvm
oπε=
rZe
rZemvE
oo πεπε 8421 22
2 −=−=
( ) 2220
42 142 n
emZEnhπε
−=
Atomic radii
n εn (eV) l=0 l=1 l=2
n
.. …..
4 ε4= -0.85 4s 4p 4d
3 ε3= -1.51 3s 3p
3d
2 ε2= -3.40 2s 2p
1 ε1= -13.6 1s
H1s
n εn (eV) l=0 l=1 l=2
n
.. …..
4 ε4= -0.85 4s 4p 4d
3 ε3= -1.51 3s 3p
3d
2 ε2= -3.40 2s 2p
1 ε1= -13.6 1s
He1s2
PERIODIC TABLE OF THE ELEMENTS
n εn (eV) l=0 l=1 l=2
n
.. …..
4 ε4= -0.85 4s 4p 4d
3 ε3= -1.51 3s 3p
3d
2 ε2= -3.40 2s 2p
1 ε1= -13.6 1s
Li1s22s1
n εn (eV) l=0 l=1 l=2
n
.. …..
4 ε4= -0.85 4s 4p 4d
3 ε3= -1.51 3s 3p
3d
2 ε2= -3.40 2s 2p
1 ε1= -13.6 1s
Be1s22s2
n εn (eV) l=0 l=1 l=2
n
.. …..
4 ε4= -0.85 4s 4p 4d
3 ε3= -1.51 3s 3p
3d
2 ε2= -3.40 2s 2p
1 ε1= -13.6 1s
B1s22s22p1
n εn (eV) l=0 l=1 l=2
n
.. …..
4 ε4= -0.85 4s 4p 4d
3 ε3= -1.51 3s 3p
3d
2 ε2= -3.40 2s 2p
1 ε1= -13.6 1s
CIs this the
best possibility ?
n εn (eV) l=0 l=1 l=2
n
.. …..
4 ε4= -0.85 4s 4p 4d
3 ε3= -1.51 3s 3p
3d
2 ε2= -3.40 2s 2p
1 ε1= -13.6 1s
C1s22s22p2
n εn (eV) l=0 l=1 l=2
n
.. …..
4 ε4= -0.85 4s 4p 4d
3 ε3= -1.51 3s 3p
3d
2 ε2= -3.40 2s 2p
1 ε1= -13.6 1s
N1s22s22p3
n εn (eV) l=0 l=1 l=2
n
.. …..
4 ε4= -0.85 4s 4p 4d
3 ε3= -1.51 3s 3p
3d
2 ε2= -3.40 2s 2p
1 ε1= -13.6 1s
O1s22s22p4
Now you know why it is 1s22s2p4
n εn (eV) l=0 l=1 l=2
n
.. …..
4 ε4= -0.85 4s 4p 4d
3 ε3= -1.51 3s 3p
3d
2 ε2= -3.40 2s 2p
1 ε1= -13.6 1s
F1s22s22p5
n εn (eV) l=0 l=1 l=2
n
.. …..
4 ε4= -0.85 4s 4p 4d
3 ε3= -1.51 3s 3p
3d
2 ε2= -3.40 2s 2p
1 ε1= -13.6 1s
Ne1s22s22p6
PERIODIC TABLE OF THE ELEMENTS
Why [Ar]4s23d1
?
n εn (eV) l=0 l=1 l=2
n
.. …..
4 ε4= -0.85 4s 4p 4d
3 ε3= -1.51 3s 3p
3d
2 ε2= -3.40 2s 2p
1 ε1= -13.6 1s
K[Ar]3d1
This is not what
happens
n εn (eV) l=0 l=1 l=2
n
.. …..
4 ε4= -0.85 4s 4p 4d
3 ε3= -1.51 3s 3p
3d
2 ε2= -3.40 2s 2p
1 ε1= -13.6 1s
K[Ar]4s1
In reality the 3d-states are higher than the 4s
n εn (eV) l=0 l=1 l=2
n
.. …..
4 ε4= -0.85 4s 4p 4d
3 ε3= -1.51 3s 3p
3d
2 ε2= -3.40 2s 2p
1 ε1= -13.6 1s
Ca[Ar]4s2
n εn (eV) l=0 l=1 l=2
n
.. …..
4 ε4= -0.85 4s 4p 4d
3 ε3= -1.51 3s 3p
3d
2 ε2= -3.40 2s 2p
1 ε1= -13.6 1s
Sc[Ar]4s23d1