[vijay.v.nenmeli]

Hello,
I've read that exchange energy is the energy released when one electron jumps into a degenerate orbital 2 from orbital 1 with the electron from the electron from orbital 2 going to orbital 1. But, if energy keeps getting released, should'nt the atom keep losing energy and eventually be left with zero energy? Also, in what form is this energy emitted (I'm guessing light)?
Thanks,

[mjc123]

Energy does not "keep getting released" as the electrons continually physically switch places. Exchange energy (a purely quantum phenomenon with no classical analogue, so it is naturally hard to understand) is the energy difference between the actual electronic state and the product of the two individual electron wavefunctions.
For example, consider an atom with two electrons a and b in orbitals 1 and 2 respectively; call the wavefunction a1b2. If you exchange the electrons between the orbitals you get the state a2b1. (Note that orbitals 1 and 2 do not have to be degenerate, but the states a1b2 and a2b1 are degenerate.) The energy of these states is E₁ + E₂ + J, where E₁ and E₂ are the energies of the single-electron states 1 and 2 and J is the Coulombic repulsion between an electron in state 1 and one in state 2. Now if you solve the Schrodinger equation for the two-electron case, it turns out that the states a1b2 and a2b1 are not solutions, but rather the functions ψ_± = 1/sqrt(2)*(a1b2 ± a2b1), with energies E₁ + E₂ + J ± K, where K is called the exchange energy. (This is the first-order perturbation theory; there are further refinements. Atkins, Molecular Quantum Mechanics describes it quite well.)
That may look a bit complicated, but the key point is that exchange energy is a quantum interaction between the two degenerate states resulting in one state with lower energy and another with higher. It is not electrons actually swapping places and releasing energy to the surroundings.

[Enthalpy]

Not paraphrasing mjc123 is hard...

The simplest representation would have been one electron with the same energy as if it were alone on orbital 1, the other as if it were alone on orbital 2, and their electrostatic repulsion computed between orbitals as if these didn't adapt.

Though, two electrons around one nucleus rearrange themselves. If one electron can be located in some smaller volume around the nucleus, the other probably isn't there, because they repel an other. The result is that the global energy is lower than if the repulsion worked between orbitals that didn't adapt.

It's a fundamental idea of quantum mechanics that only one wavefunction Ψ(r₁, r₂) is written for both electrons in positions r₁ and r₂. It differs from two wavefunctions Ψ(r₁) and Ψ(r₂) whose product would not include this adaptation.

No movement is needed here. This is abstract for me too. I wrote "If one electron can be located..." and not "when" because, if these electrons form a so-called "stationary" solution (they often do), there really is no movement. At any time, Ψ gives a probability to find the electrons in smaller volumes around positions r₁ and r₂, you can also derive a probability to find any electron in a smaller volume around position r, and these probabilities don't depend on time at all. This absence of movement explains why electrons as orbitals don't radiate. It's the big breakthrough of quantum mechanics to explain the atom.

By the way, classical electrons too would rearrange themselves in a similar situation, resulting from the same repulsion energy q²/|r₁ - r₂|/4πε, and you would observe that they find a combination of positions or trajectories, farther apart, with an energy lower than if computed without adaptation. Quantum mechanics only adds the wave behaviour which permits, among others, stationary solutions.

The solution Ψ(r₁, r₂) has no algebraic expression. What mathematical trick is used to approximate it - and has given it a name - is useful to know but shouldn't let misunderstand the unique Ψ for several particles.