Question

Which is more stable $F{e^{2 + }}$ or $F{e^{3 + }}$?

Answer 1

We know The Latin name for iron is ferrum; accordingly, the ferric and ferrous demonstrate the presence of iron in the compound. The - ic and - ous endings show the oxidation conditions of the iron particles in the compound. Ferric methods the iron particle has lost three electrons to frame $F{e^{ + 3}}$ , and ferrous methods the iron iota has lost two electrons to shape $F{e^{ + 2}}$. Present day terminology utilizes Roman numerals after a change of metal's name to show what charge it has. For instance, ferric would be $Fe\left( {III} \right)$ demonstrating $F{e^{ + 3}}$ , and ferrous would be $Fe\left( {II} \right)$ showing $F{e^{ + 2}}$ . We should take a gander at how these two iron particles join with oxygen.

Complete answer:
As we know, the ferric ion is steadier than the ferrous ion. This is clarified with the assistance of the electronic setup. In $F{e^{ + 3}}$ particles, there are five 3d half-filled orbitals and are more even than $F{e^{ + 2}}$. Though in $F{e^{ + 2}}$ particles there are four 3d half-filled orbitals and one orbital is filled.

Note:
We need to know that the $F{e^{2 + }}$ to $F{e^{3 + }}$ oxidation potential is $- {\text{ }}0.77V$ consequently $F{e^{2 + }}$ is steadier than $F{e^{3 + }}$ however on the off chance that you take a gander at setup it is actually inverse and you will discover $F{e^{3 + }}$ more steady. Be that as it may, water has a decrease capability of $- {\text{ }}1.2V$. Iron (III) salts don't immediately oxidize water to create oxygen. Then again, oxygen does suddenly oxidize iron (II) to iron (III). Solutions of iron (II) salts are thermodynamically precarious under vigorous (oxygen-rich) conditions. The explanation for this is that the oxidation possibilities are estimated as for hydrogen terminals so don’t treat them appropriately so $F{e^{3 + }}$ is more steady.