Download There are a total of n subshells, each specified by an

n  1, 2, 3, 
The allowed energy
levels are quantized
much like or particle in
a box. Since the
energy level decreases
a the square of n, these
levels get closer
together as n gets
larger.
  0,1, 2,  n  1
There are a total of n subshells, each
specified by an angular momentum quantum
number
, and having an angular
momentum of


L    1 
   m  
1
s
2
There are a total of 2  1 orbitals
within each subshell, these can be
thought of as projections of the angular
momentum on the z axis.
The electron has an intrinsic magnetic
moment called “spin”. The orientation of
the angular momentum vector of this
apparent rotation motion can only have a
manitude of ½.

3
S  s( s  1)  

2
z

Moving charges give rise to magnetic fields,
which will then interact. Since the magnetic
moments never align with the “z-axis” the
torque is never zero.
B

y
x

If the orbital angular momentum is zero, you have to think about
how it interacts with the spin, as some magnetic potential energy will
arise from that interaction.
  
J  LS

J  j ( j  1) 
J z  m j
with m j  j , j  1, , j
total angular momentum quantum number :
j    s,   s  1,,   s
The total wavefunction:
  spacespin
must be antisymmetric for electrons!
Chemical properties of an atom are determined by the least tightly
bound electrons.
Factors:
•Occupancy of subshell
•Energy separation between the subshell and the next higher
subshell.
s shell
l=0
Helium and Neon
and Argon are
inert…their outer
subshell is closed.
p shell
l=1
Beryllium and magnesium not inert even
though their outer subshell is
closed…why??
Alright, what do we add next???
Uh-oh…3d doesn’t come next…why???
orbiting electron “sees”
net charge of +e…like
hydrogen
nucleus:
charge +3e
closed s “shell”:
total charge -2e
Note that for higher orbital angular momentum, the energy more nearly equals
to the corresponding levels in hydrogen. The more nuclear charge an electron
“sees”,
2s state more tightly bound since it
has a nonzero probability density
inside the 1s shell and “sees” more
of the nuclear charge.
Note the “pulling down” of the
energy of the low angular momentum
states with respect to hydrogen due
to the penetration of the electric
shielding.
Excited state 4s lies
lower than 3d!!
The energy separating shells becomes smaller with increasing n.
Electrons in lower angular momentum states penetrate shielding more, and thus
are more tightly bound. As the energy levels become closer together, some lower
angular momentum states of higher n may actually have a lower energy.
Order of filling, then, is not what we naively
expect. The degree of charge screening plays
a BIG role.
4s comes BEFORE 3d…
…and apparently 5s comes before 4d, and 6s comes before 5d…
The term with the maximum multiplicity lies lowest in energy.
Also, mutual
repulsion
forces
electrons to
higher energy
states.
Note the filling of
aligned spins before
“doubling up”.
Lanthanide series (or “rare earths” due to low abundance)
Actinide series
Stick them at the bottom to keep things from getting too awkward.
The ionization energies are
periodic…
If you are filling shells in order
of successively higher energies,
why does this happen??
Z2
En  13.6 2 eV
n
The atomic radius is surprisingly
constant. Why does is not scale
with the number of electrons?
E[ K
For a given multiplicity, the term with the largest value of L lies lowest in
energy.
electrons spend less
time near each
other, less coulomb
repulsion.
For atoms with less than half-filled
shells, the level with the lowest
value of J lies lowest in energy.