Chemistry Q72 – JEE-MAIN 2025

Home Chemistry Coordination CompoundsQuestion #344

The number of optical isomers exhibited by the iron complex (A) obtained from the following reaction is _______.

$ F e C l_{3} + K O H + H_{2} C_{2} O_{4} \to A $

Enter Calculated Numerical Value:

Clue 1 of 3: Complex Formation

Identify the central metal ion and the ligands involved in the reaction to determine the structure of complex A.

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Step 1: Identify the complex (A)✦ Active

The reactants are $F e C l_{3}$ , $K O H$ , and $H_{2} C_{2} O_{4}$ . $F e C l_{3}$ provides $F e^{3 +}$ ions. $H_{2} C_{2} O_{4}$ (oxalic acid) is a dicarboxylic acid that acts as a bidentate ligand, oxalate ( $C_{2} O_{4}^{2 -}$ ). In the presence of $K O H$ , oxalic acid will be deprotonated to form oxalate ions. Iron(III) typically forms octahedral complexes with a coordination number of 6. Since oxalate is a bidentate ligand, three oxalate ligands will coordinate to the $F e^{3 +}$ ion.

[F e (C_{2} O_{4})_{3}]^{3 -}

Thus, complex A is the trioxalatoferrate(III) ion.

Step 2: Determine the type of isomerism○ Expand

The complex $[F e (C_{2} O_{4})_{3}]^{3 -}$ is an octahedral complex of the type $[M (A A)_{3}]$ , where $M = F e^{3 +}$ and $A A = C_{2} O_{4}^{2 -}$ (a symmetrical bidentate ligand). Complexes of this type are known to be chiral, meaning they are non-superimposable on their mirror images. They do not possess any plane of symmetry or center of inversion.

💡 Teacher's Secret Hint

Recall the conditions for optical isomerism in coordination compounds.

Step 3: Count the number of optical isomers○ Expand

Since the complex $[F e (C_{2} O_{4})_{3}]^{3 -}$ is chiral, it exists as a pair of enantiomers (dextro and levo forms). Therefore, the number of optical isomers exhibited by complex A is 2.

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Chemistry Question 72 – JEE-MAIN 2025

Clue 1 of 3: Complex Formation

🧠 Full Solution Path