This can be directly appreciated by considering the simplest case of two hydrogen atoms, first held at large distance and then brought closer together for the electronic wavefunctions centred on the two atoms to overlap. The consequent splitting of the energies of the two-particle singlet and triplet states is due to exchange (reason why it is called 'exchange splitting'). Through this, the singlet state stabilises as the unique ground state of the diatomic system. For the relevant details, you may consult, for instance, the book Solid State Physics, by Ashcroft and Mermin, pp. 674-679.

But, how can i use this idea to explain the stability of d5 configurations like in Cr=24, its goes from d4 to d5 configuration and the reason is exchange energy. How ? I am not getting any idea to solve  except the fact that half-filled configurations are more stable.

Dear Gourav, your new specific question is far removed from your main original question on this page (your opening "But" suggests that I may have neglected some issue in my previous response). Be it as it may, it turns out that band-structure effects, leading to a particular nested Fermi surface, are of considerable significance to the ground state of the Chromium crystal, which is an incommensurate spin-density-wave state. Patrik Fazekas in Sec. 7.7.5 of his book Lecture Notes on Electron Correlation and Magnetism (World Scientific, Singapore, 1999) considers in some detail the physics of this system.

Thanks a lot sir, I think, first i need to look at the suggestion.

You are welcome Gourav!

The Exchange Energy[Eex] is the energy released when the anti parallel electrons belonging to a degenerate( same energy) subshell are made to have parallel spins. I try to explain this phenomenon by taking examples of both the Cr(24) and Cu(29).

Eex=K.n(n-1)/2

Where n stands for the the number of electrons having the same spins and occupying the the degerate “d” subshell. K is called Exchange constant and has different values for different metals and are reported in the literature.

[I]Now Cr can have two possible configurations-3d^4 4s^2[ 1st case] and 3d^5 4s^1[2nd case] and we are to decide as to why the latter should be more stable. We have four electrons with same spins in the first case and five electrons with the same spin in the DEGENERATE 3d subshell{ Pls. note that 3d and 4s are not degenerate]

Simply calculate Eex for both the probable configurations as:

[ 1st case], Eex=K.4.3/2=6K

[2nd case], Eex=K.5.4/2=10K

Difference in Exchane energies =4K

As the amount of energy released in the form of Eex in the 2nd case is more than that in 1st case, the3d^5 4s^1 configuration should be more stable thus more probable.

[II]Again, Cu can have two possible configurations-3d^9 4s^2[ 1st case] and 3d^10 4s^1[2nd case] and we are to decide as to why the latter should be more stable.

Again calculate Eex in both the probable configurations:

BUT IT IS A BIT DIFFERENT HERE.

Here the systems have two different types of spins

So calculate Eex for both types of electrons in the 1st case

[ 1st case] 3d^9 4s^2

[i] Here 3d has 5 electrons with+1/2[ up] spins while the remaining four electrons are with -1/2[ down] spins.

(a) Eex from 5 (up) electrons= K.5.4/2= 10K/

(b) Eex from 4 (down) electrons= K.4.3/2= 6K/

Total Eex for this configuration= 16K

[ 2nd case] 3d^10 4s^1

[i] Here five 3d electrons are with+1/2[ up] spins and so are the remaining five electrons but with -1/2[ down] spins.

(a)Eex from 5 (up) electrons= K.5.4/2= 10K/

(b) Eex from 5 (down) electrons= K.5.4/2= 10K/

Total Eex for this configuration= 20K/

So difference in Exchange energies=4K/

As K and K/ have different values, so the Eex of the half filled configuration[3d^5 4s^1] of Cr and that of the completely filled subshell configuration[3d^10 4s^1] of Cu should differ and so should be their overall stabilities.

Thanks Manohar for your answer, but conceptual part is still a miss there. The way you have given the example, any system will try to get the same spin because it will always try to  have all the parallel spin... but this is not true. I know the K value should play some role, but from the example it seems that K will remain same for any particular system and then single electron in s will be more stable as it will result in more exchange energy.

I am clear with all the facts you have just pointed out.

Thanks

"Repulsion" among electrons can be roughly speaking split in two main parts: Coulombic repulsion and a penalty due to the fact that electrons are fermions: two electrons with the **same** spin cannot be en the same place. Exchange is nothing but that last piece of electron-electron interaction energy. Exchange favours  electrons that otherwise had to live in a state with the same density of  probability to have parallel spin. Lets say you have e electrons and 5 d orbitals. You have three options: 1) populate one d orbital with two electrons of opposite spin, 2) occupy two orbitals with antiparallel spins and 3) occupy two orbitals with parallel spins. Case 1 is the less favorable because if both electrons "live' in the same d orbital the average distance between them is the smallest of the three cases and  Coulombic repulsion will be the biggest. The difference between cases 2 and 3 is in case three the electrons must to avoid each other more than in case 2 because they have the same spin. Then, the Pauli exclusion principle  makes the average distance between electrons larger in case 3 than in case 2. Therefore, electron-electron repulsion is smaller in case 3. This in and attempt to grasp the concepts, but you must understand that the actual picture is a Fermonic many body problem and that going into the details of quantum mechanics is a must if one wants to understand this fully.

--Carlos