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