Q36E - Chemistry (OpenStax)

Question

Determine the number of unpaired electrons expected for \({\left( {{\rm{Fe}}{{\left( {{\rm{N}}{{\rm{O}}_2}} \right)}_6}} \right)^{3 - }}\)and for \({\left( {Fe{F_6}} \right)^{3 - }}\)in terms of crystal field theory.

Step-by-Step Solution

Answer

First we have\({\left( {{\rm{Fe}}{{\left( {{\rm{N}}{{\rm{O}}_2}} \right)}_6}} \right)^{3 - }}\), where the central atom is the transition metal iron which is attached with \(6\) nitrate ions.

Step 1 0f 4: oxidation state of \(Fe\)

• Let us determine the oxidation state of \(Fe\)as we use \(x\) for the oxidation state of \(Fe\)We know that general oxidation state of nitrate ion is \( - 1,\)
• Hence,  we can determine the oxidation state of \(Fe\)like this:

         \(\begin{align}{\underline{\phantom{xx}}}x + (6 \cdot  - 1) &=   - 3\\x - 6 =  - 3\\x &=  - 3 + 6\\x &=   + 3\end{align}\)

• It means the oxidation state of \(Fe\)is \( + 3.\)
• If we look at the periodic table we can see that the atomic number of Fe is 26 and its electronic configuration is\((Ar)3{d^6}4{s^2}\)
• In oxidation state of \(F{e^{3 + }}\)the electronic configuration is: \((Ar)3{d^5}4{s^0}.\)

Step 2 of 4: Explanation of electron pairs

• We know that ligand contributes a pair of electrons to the metal and since \({\rm{N}}{{\rm{O}}_2}\)is a strong field ligand hence, it causes larger splitting. 
• The magnitude of pairing energy \(\left( P \right)\)is less than crystal field splitting energy in octahedral field: \(P < {\Delta _o}\)which means the electron pairs donated by ligand go to the innermost orbitals.
• From the crystal field diagram we can see that \({\left( {{\rm{Fe}}{{\left( {{\rm{N}}{{\rm{O}}_2}} \right)}_6}} \right)^{3 - }}\)has unpaired electron.

Step 3 of 4: oxidation state

• In\({\left( {{\rm{Fe}}{{\rm{F}}_6}} \right)^{3 - }}\), the central atom is the transition metal iron which is attached with \(6\) fluoride ions.
• We can determine the oxidation state of \(Fe\)as by using\(x\)for the oxidation state of \(Fe\).
• As we know that general oxidation state of fluoride ion is \( - 1\),let us determine the oxidation state of \(Fe\)like this:

     \(\begin{align}{\underline{\phantom{xx}}}x + (6 \cdot  - 1) &=   - 3\\x - 6 &=  - 3\\x& =  - 3 + 6\\x &=   + 3\end{align}\)

• It means that the oxidation state of \(Fe\)is \( + 3.\)
• As per the periodic table,  the atomic number of \(Fe\) is \(26\) and its electronic configuration is \((Ar)3{d^6}4{s^2}.\)
• In oxidation state of \(F{e^{3 + }}\),the electronic configuration is: \((Ar)3{d^5}4{s^0}.\)

Step 4 of 4:Splitting of ligand

• The ligand contributes a pair of electrons to the metal and since \({F^ - }\)is a weak field ligand hence, it causes smaller splitting. 
• The magnitude of pairing energy \((P)\)is bigger than crystal field splitting energy in octahedral field: \({\bf{P}} > {\Delta _a},\)which means that the electron pairs donated by ligand go to the outermost orbitals.
• We could conclude from the crystal field diagram that \({\left( {{\rm{Fe}}{{\rm{F}}_6}} \right)^{3 - }}\)has 5 unpaired electrons.

Result

\({\left( {{\rm{Fe}}{{\left( {{\rm{N}}{{\rm{O}}_2}} \right)}_6}} \right)^{3 - }}\)has\(1\)unpaired electron and \({\left( {{\rm{Fe}}{{\rm{F}}_6}} \right)^{3 - }}\)has \(5\)unpaired electrons.