Exchange Energy

The exactly half-filled and fully filled orbitals have greater stability than other configurations. The reason of their stability are symmetry and exchange energy.
Exchange Energy = Calculated Orbital Energy – Actual Orbital Energy

Symmetry
The half-filled and fully-filled orbitals are more symmetrical than any other configuration. Electrons in the same subshell have equal energy (degenerate) but different spatial distribution. Consequently, their shielding of one- another is relatively small and the electrons are more strongly attracted by the nucleus. Symmetry leads to greater stability.

Exchange Energy
The electrons present in the different orbitals of the same sub-shell can exchange their positions. Each such exchange leads to the decrease in energy known as Exchange Energy. Greater the number of exchanges, greater the exchange energy and hence greater the stability. As the number of exchanges which takes place in the half-filled and fully-filled orbitals is maximum, the exchange energy is maximum and hence maximum stability.

The exchange energy is according to Hund's rule that electrons which enter orbitals of equal energy have parallel spins as far as possible. In other words, the extra stability of half-filled and completely filled subshell is due to-
A. Relatively small shielding
B. Smaller coulombic repulsion energy, and
C. Larger exchange energy.

The Exchange Energy [E_ex] is the energy released when the antiparallel electrons belonging to a degenerate (same energy) subshell are made to have parallel spins.

Explanation
Let's explain by taking examples of both the Cr (atomic number = 24) and Cu (atomic number = 29).
E_ex = K. n(n-1)/2
Where 'n' represents the number of electrons having the same spins and occupying the the degenerate 'd' subshell. K is Exchange constant and has different values for different metals. K depends upon the metal and not the electron.
The word exchange means electrons should belong to a degenerate system- p, d, f sub-shells and they should result in a Parallel Pair and not a pair with opposite spins. 's' Electron can not be exchange between a 3d and 4s electron as they differ in energies.

Now, Cr can have two possible configurations-
A. 3d⁴ 4s² and
B. 3d⁵ 4s¹

Here we have to decide 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. Note that 3d and 4s are not degenerate. Calculate E_ex for both the probable Configurations as-
A. E_ex = K x 4 (3/2) = 6K
B. E_ex = K x 5 (4/2) = 10K
Difference in Exchange energies = 4K
As the amount of energy released in the form of E_ex in 'B' is more than that in 'A', the 3d⁵ 4s¹ configuration should be more stable.

Now, Cu can have two possible configurations-
C. 3d⁹ 4s² and
D. 3d¹⁰ 4s¹

Here also we have to decide as to why the latter should be more stable. Again calculate E_ex in both the probable configurations.
Here the systems have two different types of spins. So calculate E_ex for both types of electrons in 'C', 3d⁹ 4s². Here 3d has 5 electrons with +1/2 (↑) spins while the remaining four electrons are with -1/2 (↓) spins.
So, E_ex from 5 (↑) electrons = K x 5 (4/2) = 10K and
E_ex from 4 (↓) electrons = K x 4 (3/2) = 6K.
So the total E_ex for this configuration= 16K.
In 'D', five 3d electrons are with +1/2 (↑) spins and the remaining five electrons are with -1/2 (↓) spins.
Now, E_ex from 5 (↑) electrons = K x 5 (4/2) = 10K and
E_ex from 5 (↓) electrons = K x 5 (4/2) = 10K.
So the total E_ex for this configuration = 20K and the difference in Exchange energies = 4K. As the amount of energy released in the form of E_ex in 'D' is more than that in 'C', the 3d¹⁰ 4s¹ configuration should be more stable.
In a group from top to bottom, the exchange energy decreases because the valence electrons in the nucleus already have a lot of space to move due to their existence in larger orbitals, thus they don't need the help of exchange energy.