Electrons of the same spin in degenerate orbitals undergo exchange and make the atom more stable. Why do they release energy during exchange? We can calculate the number of possibilities in which the exchange can occur, can all these possibilities occur at once? What makes the electrons stop exchanging again and again?
I'm a high schooler. When I came across the exchange energy concept, many things were not justified in my book. So kindly help me out :)
For a basic explanation about exchange energy I would refer to "what is the exchange energy of electrons?". So I think you should realise that the exchange interaction might be a little misleading term as it implies electrons hop from one orbital to another gaining some energy in the process. But in reality it is a little more difficult. In essense it is the antisymitrization of the total wavefunction of the electrons that result in the electrons being on average closer together when they have opposite spin resulting in more repulsion hence a higher energy or farther from each otherhas when heaving same spin which lead to lower energies.
A more involved explanation would be to consider the example of a two electron wave function $$\Psi(x_1,s_1,x_2,s_2)= \psi(x_1,x_2)\chi(s_1,s_2)$$ where $\psi(x_1,x_2)$ is the spatial wavefunction of the electrons and $\chi(s_1,s_2)$ the spin wavefunction of electrons 1 and 2. As the total wave function needs to be antisymetric under exchange of the electrons (which means that if you swap $x_1\leftrightarrow x_2$ and $s_1\leftrightarrow s_2$ in the total wavefunction $\Psi$ syou would obtain the same wavefunction but with an extra minus sign $-\Psi$). This requirement is a result from the so called "Spin-statistics theorem" which is a quite advanced theory in quantum physics so for this discussion it's better to accept that fact.
Let's take two spatial orbitals $\psi_a$ and $\psi_b$ as basis for the combined electron spatial orbital $\psi$. We construct a spatially symmetric $+$ and anti-symmetric wavefunction $-$ as $$\psi_+ = \frac{1}{\sqrt{2}}(\psi_a(x_1)\psi_b(x_2) + \psi_b(x_1)\psi_a(x_2))$$ and $$\psi_- = \frac{1}{\sqrt{2}}(\psi_a(x_1)\psi_b(x_2) - \psi_b(x_1)\psi_a(x_2))$$ you can check yourself that $\psi_-$ is antisymetric when swapping $x_1\leftrightarrow x_2$. Calculating the expectation value of the square of the separation between the two electrons $\langle (x_1-x_2)^2\rangle\equiv \langle (\Delta x)^2\rangle$ for the symmetric (+) and antisymmetric (-) case result in $$ \langle (\Delta x)^2\rangle_\pm = \langle (\Delta x)^2\rangle_d \mp 2|\langle x\rangle |^2$$ where $ \langle (\Delta x)^2\rangle_d $ is the square of the separation distance if the particles where distinguisable and thus no exchange possible (for example if the electrons had different colours such that you could distinguish one from the other, which they have not in reality). And thus the difference is in the second term (which is called the exchange integral). The main take away from this calculation is the sign of the second term. For symetric spatial wave functions the electrons are somewhat closer together then the distinguisble particle case and for anti-symetric spatial wave functions the electrons are a further appart.
Realizing that the closer the electrons are together the larger the repulsion and thus the higher the energy of the state we see that for a symmetric wave function we expect a higher energy then for an antisymmetric wavefunction. This is the exchange energy.
In the last step we should couple the spatial and the spin together. You might have wondered why we would care about the symmetric spatial wavefunction in the first place as the total wave function of the electrons must be anti-symmetric. However, if we include the spin wavefunction that can also be (anti-)symmetric we can have the following possibilities for a anti-symmetric total wave function: $\Psi_- = \psi_{-}\chi_+$ or $\Psi_- = \psi_{+}\chi_-$ (so not $\Psi_+ = \psi_{+}\chi_+$ or $\Psi_- = \psi_{-}\chi_-$). Hence when electrons have "the same spin" which generally means that they have a symmetric spin wave function, the spatial wave function must be anti-symetric and the electrons lie a bit further appart meaning less repulsion and a lower energy. Similarly when the electrons have "opposite spin" which then would mean they have an anti-symetric spin wave function. They must have a symetric spatial wave function leading and electrons are a little bit closer together meaning more repulsion and a higher energy.
At the level of a standard AP high school chemistry course, the exchange term is a theoretical artifact from describing the many-body electron wave function in terms of products of single-particle orbitals, and has nothing to do with "exchanging electrons." Intuitively, the Pauli exclusion principle prevents electrons from occupying the same orbital: if two electrons have the same spin, the only way they can satisfy the Pauli exclusion principle is if they occupy different spatial positions, and this 'extra' repulsion (that the electrons get 'for free') typically[1] lowers the 'average' potential energy from the electrostatic interaction (at the cost of forcing the electrons to occupy at least one higher energy single-particle orbital beyond the ground state orbital.)
There are too many intermediate/prerequisite ideas to give a more complete answer at this stage without also furnishing a complete course in quantum mechanics, but here's a quick outline of the relevant topics and the approximate order in which you can learn them:
[1] States with very different angular momenta or that otherwise oscillate rapidly are the one possible exception to this: exchange 'interactions' are somewhat unusual.
