# Exchange energy of d6 configuration

In NCERT Chemistry book, it is given as:

Exchange energy is responsible for the stabilization of energy state. Exchange energy is approximately proportional to the total number of possible pairs of parallel spins in the degenerate orbitals. When several electrons occupy a set of degenerate orbitals, the lowest energy state corresponds to the maximum possible extent of single occupation of orbital and parallel spins (Hund's rule). The loss of exchange energy increases the stability. As the stability increases, the ionization becomes more difficult. There is no loss of exchange energy at $$\ce{(d^6)}$$ configuration. $$\ce{Mn+}$$ has $$\ce{(3d^5) (4s^1)}$$ configuration and configuration of $$\ce{Cr+}$$ is $$\ce{(d^5)}$$, therefore, ionization enthalpy of $$\ce{Mn+}$$ is lower than $$\ce{Cr+}$$. In the same way, $$\ce{Fe^2+}$$ has $$\ce{(d^6)}$$ configuration and $$\ce{Mn^2+}$$ has $$\ce{(d^5)}$$ configuration. Hence, ionization enthalpy of $$\ce{Fe^2+}$$ is lower than the $$\ce{Mn^2+}$$.

But, there are 5 electrons of same spin in $$\ce{Fe^2+}$$. So, the number of exchanges would be $$\mathrm{^5C_2}$$ = 10. Then why is there no loss of exchange energy in $$\ce{Fe^2+}$$ ?

• Possibly the loss of degeneracy in presence of ligands is being discussed. The 5 d-orbitals of Fe which were degenerate will turn into 3 lower energied orbitals and 2 higher energied orbitals in presence of a ligand. Jun 23 at 10:46
• "4 unpaired electrons of same spin" - They don't have to be unpaired. There are 5 electrons of the same spin. Jun 25 at 10:43
• But still why is it zero?
– user112344
Jun 25 at 11:14
• Because you're looking at an ionisation process, and thus the key question is the difference in exchange energy between $\ce{Fe^2+}$ and $\ce{Fe^3+}$. Jun 25 at 11:14
• I used the same logic for ionization energy of Fe(II) and Mn(II), but I am unable to understand why is the exchange energy of Fe(II) is zero.
– user112344
Jun 25 at 11:17

As @orthocresol points out, the key is that you need to compare the exchange energy before vs after the ionization process. Anything that is unchanged by ionization cannot affect ionization energy. This is a general idea for a lot of aspects of chemistry: close the thermodynamical circles carefully, since so much information can be gained from them.

So, let us compare the Fe$$^{2+}$$ to Fe$$^{3+}$$ ionization. Both ions have 5 electrons with the same spin,[1] leading to the same exchange energy. Additionally, Fe$$^{2+}$$ has an electron with the opposite spin, which does not contribute to exchange energy. All in all, no difference in exchange energy, so no effect of the exchange energy in the ionization energy. To reiterate: it is not that the exchange energy is zero, it is the before vs after difference (so, the effect) which is zero.

Now let us see what happens with the Mn$$^{2+}$$ to Mn$$^{3+}$$ ionization. Here we are going from 5 electrons with the same spin to 4 electrons with the same spin,[1] so here we are losing exchange energy between the states before and after ionization, and therefore there this is something extra to pay in the ionization energy.

[1] For simplicity of the argument, we will be assuming "high-spin" configuration, as is implicit in the question.

probably the exchanges happening in the same orbital contribute to the exchange energy. if we say that an electron is moved from 3d to 4s orbital then it is said to be excited(Exchanged is not suitable to be used in this case).

so when you ask me to talk about the Fe2+ ion here is its electronic configuration **1s2 2s2 2p6 3s2 3p6 3d6** so in this case as for stability, the electron tends to excite to the 4s orbital and hence we call this the excitation of electron and not exchange. I hope this would clear your doubt.

cheers!! edit1: you may feel doubted why the electron excites from 3d to 4s so as it gets 3d5 configuration which makes it more stable and more exchanges contribute to more stability.