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Question

On complete reaction of FeCl3 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)


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Solution

Step 1: Forming the complete reaction

Ferric chloride FeCl3 reacts with Oxalic acid (C2H2O4) in an aqueous solution of Potassium hydroxide (KOH).

The product formed is Potassium trioxalatoferrate(III) K3[Fe(C2O4)3], Potassium chloride, and water

The complete reaction is:

FeCl3(aq)Ferricchloride+3H2C2O2(aq)Oxalicacid+6KOH(aq)PotassiumhydroxideK3[Fe(C2O4)3](aq)Potassiumtrioxalatoferrate(III)+3KCl(aq)Potassiumchloride+6H2O(l)Water

Step 2: Calculating the secondary valency of Iron

In the compound, Potassium trioxalatoferrate(III) K3[Fe(C2O4)3], the cation and anions are:

K3[Fe(C2O4)3]Potassiumtrioxalatoferrate(III)[Fe(C2O4)3]3-tris(oxalato)ferrate(III)ion+K3+Potassiumion

Here, the oxalate (C2O4)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 (C2O4)2-is a bidentate ligand, so the secondary valency of the Iron(Fe) is 3×2=6


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