"Farooq W" <farooq.w@gmail.com> wrote in message
news:1141993301.713445.44770@z34g2000cwc.googlegroups.com...

My question is about the energy order of TERMS.
First let me ask that is my understanding that L, the total orbital
angular momentum, represents energy, greater the value of L greater is
the energy (of the atom in that state) correct or not?

Now we have a p2 configuration which are , the three p orbitals are
*degenerate*. Arranging electrons in them gives three terms: ^3P, ^1D,
^1S

Now according to the Hund's Rule the term having the highest
multiplicity is the ground state i.e. triplet P here. Next comes the
energy order of singlet D and singlet S. Hund's second says that the
term with maximum L is of lower energy when the multiplicity is same.

This is where I am stuck...why maximum L ( representative of energy)
has lower energy i.e. why the the energy order is ^3P < ^1D < ^1S not
^3P < ^1S < ^1D for a given electronic configuration.


Thanks

M. Farooq

Wait a minute, there is no 1d.
It is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p
Now is your question why the 3d after 4s and not before?

Hund's rule: every orbital in a subshell is singly occupied with one
electron before any one orbital is doubly occupied, and all electrons in
singly occupied orbitals have the same spin.

Hund's rule has to do with degenerate orbials in a given subshell (the
subshell is the SPDF). Since electrons have repulsive forces (like charges
repel), it takes a bit more energy to pair them up in the same orbital than
to put them in separate (degenerate) orbitals (the spin has to be
switched--whatever than really means in quantum talk). At least that is my
non-p-chemist take on it.
Also, IIRC, the degenerate orbital were not really degenerate once you
started filling them (something way back in inorganic about that--subtle
splitting do to interaction).


--------
Dr. Dickie
"Let be be finale of seem.
The only emperor is the emperor of ice-cream."
-- Wallace Stevens


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