# Please help me with ?

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If I had N fermions, how could I determine the number of ways I can place these into g distinguishable containers? For example, how could I determine how many ways I could place 54 electrons into 54porbitals?These indistinguishable particles can exchange. The ith particle can exchange with any of the other particles (assume nearly-degenerate orbitals), and this follows for each particle. Each particle has two orientations (up/down).

Up to two particles of opposite spin are allowed in each container, but zero particles in any given container is also allowed. That is, 0−2

particles are allowed per container, in integer increments.

Another way to say this is, how many singlet configuration state functions would I need to describe all possible configurations of 54 electrons in 54 nearly-degenerate p orbitals?

If I had N fermions, how could I determine the number of ways I can place these into g distinguishable containers? For example, how could I determine how many ways I could place 54 electrons into 54porbitals?These indistinguishable particles can exchange. The ith particle can exchange with any of the other particles (assume nearly-degenerate orbitals), and this follows for each particle. Each particle has two orientations (up/down).

Up to two particles of opposite spin are allowed in each container, but zero particles in any given container is also allowed. That is, 0−2

particles are allowed per container, in integer increments.

Another way to say this is, how many singlet configuration state functions would I need to describe all possible configurations of 54 electrons in 54 nearly-degenerate p orbitals?

##### 1 Answer

#### Answer:

Do you want the theoretical permutations, or a realistic orbital-filling pattern?

#### Explanation:

Because the electrons are indistinguishable from each other there is no way to “see” a permutation based on the number of electrons. Putting an electron in one orbit or another ONLY establishes that an electron is in that orbital, not WHICH electron of 54 is there.

Again, physically, the electron orbitals are filled in sequence, so until you get to the valence orbitals, the “location” of the lower level orbitals is irrelevant – ALL lower orbitals are completely filled – and you can’t tell one electron apart from another, other than their orbital position.

So, with the stipulated 54 electrons, that would imply 54 protons as a base element – Xenon. That is a noble gas, with all of its electron orbitals filled completely. The only variations for 54 electrons would be ions of Iodine, Cesium or Barium, and then on through the 6th Period transition elements. In all cases, as stated above, the configuration of the electrons would not change, being essentially equivalent to Xenon, with a charge on the ion due to the difference between the nuclear protons and number of electrons.

If you want to play with the theoretical permutations based on the number of unique (if indistinguishable) electrons and their possible orbital configurations (the four primary quantum numbers), then that is a pretty straight-forward statistical computation. It may be better posted in the Statistics Subject for a better answer than in the Chemistry one.