Chapter 9, Problem 9D.13E

Interpretation Introduction

Interpretation:

An explanation has to be given why when paramagnetic ${\left[\text{Fe}{\left(\text{CN}\right)}_{\text{6}}\right]}^{\text{3-}}$.ion is reduced to ${\left[\text{Fe}{\left(\text{CN}\right)}_{\text{6}}\right]}^{\text{4-}}$ it becomes diamagnetic but when paramagnetic ${\left[\text{Fe}{\left(\text{Cl}\right)}_{\text{4}}\right]}^{\text{-}}$ ion is reduced to ${\left[\text{Fe}{\left(\text{Cl}\right)}_{\text{4}}\right]}^{\text{2-}}$ it remains paramagnetic.

Answer to Problem 9D.13E

If paramagnetic ${\left[\text{Fe}{\left(\text{CN}\right)}_{\text{6}}\right]}^{\text{3-}}$.ion is reduced to ${\left[\text{Fe}{\left(\text{CN}\right)}_{\text{6}}\right]}^{\text{4-}}$ it becomes diamagnetic but when paramagnetic ${\left[\text{Fe}{\left(\text{Cl}\right)}_{\text{4}}\right]}^{\text{-}}$ ion is reduced to ${\left[\text{Fe}{\left(\text{Cl}\right)}_{\text{4}}\right]}^{\text{2-}}$ it remains paramagnetic because ${\left[\text{Fe}{\left(\text{CN}\right)}_{\text{6}}\right]}^{\text{3-}}$ & ${\left[\text{Fe}{\left(\text{CN}\right)}_{\text{6}}\right]}^{\text{4-}}$ are octahedral complexes while ${\left[\text{Fe}{\left(\text{Cl}\right)}_{\text{4}}\right]}^{\text{-}}$ & ${\left[\text{Fe}{\left(\text{Cl}\right)}_{\text{4}}\right]}^{\text{2-}}$ are tetrahedral complexes.

Explanation of Solution

The oxidation number of iron in ${\left[\text{Fe}{\left(\text{CN}\right)}_{\text{6}}\right]}^{\text{3-}}$ complex is +3 and the electronic configuration of ${\text{Fe}}^{\text{3+}}$ is $\left[\text{Ar}\right]{\text{3d}}^5.$

The orbital energy level diagram for ${\text{d}}^5$ configuration is,

Since, cyanide acts as a strong ligand in ${\left[\text{Fe}{\left(\text{CN}\right)}_{\text{6}}\right]}^{\text{3-}}$ complex all the d electrons are paired up in the lower energy levels and there is one unpaired electron.  Thus, it is paramagnetic.

The oxidation number of iron in ${\left[\text{Fe}{\left(\text{CN}\right)}_{\text{6}}\right]}^{\text{4-}}$ complex is +2 and the electronic configuration of ${\text{Fe}}^{\text{2+}}$ is $\left[\text{Ar}\right]{\text{3d}}^6.$

The orbital energy level diagram for ${\text{d}}^6$ configuration is,

Since, cyanide acts as a strong ligand in ${\left[\text{Fe}{\left(\text{CN}\right)}_{\text{6}}\right]}^{\text{4-}}$ complex all the d electrons are paired up in the lower energy levels and there is no unpaired electron.  Thus, it is diamagnetic.

The oxidation number of iron in ${\left[\text{Fe}{\left(\text{Cl}\right)}_{\text{4}}\right]}^{\text{-}}$ complex is +3 and the electronic configuration of ${\text{Fe}}^{\text{3+}}$ is $\left[\text{Ar}\right]{\text{3d}}^5.$

${\left[\text{Fe}{\left(\text{Cl}\right)}_{\text{4}}\right]}^{\text{-}}$ is a tetrahedral complex.  In a tetrahedral complex there are four ligands attached to the central metal.  The d-orbitals split into two different energy levels.  The two three consist of ${\text{d}}_{\text{xy}}\text{,}{\text{d}}_{\text{xz}}\text{,}{\text{d}}_{\text{yz}}$ and the bottom two consist of ${\text{d}}_{{\text{x}}^{\text{2}}{\text{-y}}^{\text{2}}}\text{\&}{\text{d}}_{{\text{z}}^{\text{2}}}$ the reason for this is due to the poor orbital overlap between the metal and the ligand orbitals

The orbital energy level diagram for ${\text{d}}^5$ configuration is,

Since, chloride ion is a weak ligand the electrons are filled according to the Hund’s rule in the d-orbitals and the number of unpaired electron in iron complex is one.  Thus, it is paramagnetic complex.

The oxidation number of iron in ${\left[\text{Fe}{\left(\text{Cl}\right)}_{\text{4}}\right]}^{\text{2-}}$ complex is +2 and the electronic configuration of ${\text{Fe}}^{\text{2+}}$ is $\left[\text{Ar}\right]{\text{3d}}^6.$

The orbital energy level diagram for ${\text{d}}^6$ configuration is,

Since, chloride ion is a weak ligand the electrons are filled according to the Hund’s rule in the d-orbitals and the number of unpaired electron in iron complex is two.  Thus, it is paramagnetic complex.