Question

On complete reaction of ${\mathrm{FeCl}}_3$ with oxalic acid in an aqueous solution containing KOH, resulted in the formation of product A. The secondary valency of Fe in the product A is …………..(Round off to the Nearest Integer)

Answer 1

Step 1: Forming the complete reaction

Ferric chloride ${\mathrm{FeCl}}_3$ reacts with Oxalic acid $\left({\mathrm{C}}_2{\mathrm{H}}_2{\mathrm{O}}_4\right)$ in an aqueous solution of Potassium hydroxide $\left(\mathrm{KOH}\right)$.

The product formed is Potassium trioxalatoferrate(III) ${\mathrm{K}}_3\left[\mathrm{Fe}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right]$, Potassium chloride, and water

The complete reaction is:

$\underset{\mathrm{Ferric}\mathrm{chloride}}{{\mathrm{FeCl}}_3\left(\mathrm{aq}\right)}+\underset{\mathrm{Oxalic}\mathrm{acid}}{3{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_2\left(\mathrm{aq}\right)}+\underset{\mathrm{Potassium}\mathrm{hydroxide}}{6\mathrm{KOH}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Potassium}\mathrm{trioxalatoferrate}\left(\mathrm{III}\right)}{{\mathrm{K}}_3\left[\mathrm{Fe}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right]\left(\mathrm{aq}\right)}+\underset{\mathrm{Potassium}\mathrm{chloride}}{3\mathrm{KCl}\left(\mathrm{aq}\right)}+\underset{\mathrm{Water}}{6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

Step 2: Calculating the secondary valency of Iron

In the compound, Potassium trioxalatoferrate(III) ${\mathrm{K}}_3\left[\mathrm{Fe}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right]$, the cation and anions are:

$\underset{\mathrm{Potassium}\mathrm{trioxalatoferrate}\left(\mathrm{III}\right)}{{\mathrm{K}}_3\left[\mathrm{Fe}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right]}\to \underset{\mathrm{tris}\left(\mathrm{oxalato}\right)\mathrm{ferrate}\left(\mathrm{III}\right)\mathrm{ion}}{{\left[\mathrm{Fe}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right]}^{3-}}+\underset{\mathrm{Potassium}\mathrm{ion}}{{\mathrm{K}}^{3+}}$

Here, the oxalate ${\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}^{2-}$is a bidentate ligand and 3 Oxalate are attached to iron.

Now we know that, secondary valency is nothing but the number of ligands that are attached to the central metal atom.

Now since the oxalate ${\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}^{2-}$is a bidentate ligand, so the secondary valency of the Iron(Fe) is $3\times 2=6$