Chapter 7, Problem 57A

Interpretation Introduction

Interpretation:

This is to be shown that iron can form both +2 and +3 ion.

Concept introduction:

The atoms lose or gain electrons to form octet configuration. This process is said to be ionization. Compounds which contain ions are held together by ionic bond that is formed by ions.

Answer to Problem 57A

Iron will lose 2 electrons from 4s orbital to form ${\text{Fe}}^{\text{2+}}$ and will lose 2 electrons from 4s and 1 electron from 3d to form ${\text{Fe}}^{\text{3+}}$

Explanation of Solution

Iron belongs to group 8 and has atomic number as 26. It’s atomic number is 26. The electronic configuration is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^6{\text{4s}}^{\text{2}}{\text{3d}}^6$ or $\left[\text{Ar}\right]3{\text{d}}^{\text{6}}{\text{4s}}^{\text{2}}$

There are 4 shells which are half filled.

As per Madelung’s Law, filling of electrons is in a specific order as the atomic number increases. But as Iron has two outermost or valence electrons in 4s orbital. It can easily lose these two electrons to form ${\text{Fe}}^{\text{2+}}$

As it is seen that in the next 3d shell which has 6 electrons, there is one orbital having a paired electron while other 4 are half filled. This one of 3d orbital will loose one electron to have all of its 3d orbitals half filled. Hence, ${\text{Fe}}^{\text{2+}}$ will lose one more electron to form ${\text{Fe}}^{\text{3+}}$

Here it loses 2 electrons from 4s and 1 electron from 3d orbital.

Conclusion

Iron will lose 2 electrons from 4s orbital to form ${\text{Fe}}^{\text{2+}}$ and will lose 2 electrons from 4s and 1 electron from 3d to form ${\text{Fe}}^{\text{3+}}$