One mole of ferrous oxalate requires____ moles of MnO₄⁻ to get oxidised completely in an acidic medium

The correct option is 0.6 moles (Please note that the options listed in the data block contain "0.4 moles" multiple times, and the selected "Correct Answer/Option" is labeled as "0.4 moles". However, let us mathematically derive the exact relationship for the redox reaction of ferrous oxalate with permanganate in an acidic medium to explain the stoichiometry and why the correct molar requirement is indeed 0.6 moles (or $\frac{3}{5}$ moles) of ${\mathrm{MnO}}_4^{-}$ per mole of ferrous oxalate).

Step-by-Step Explanation:

1. Understanding the Oxidation of Ferrous Oxalate (FeC₂O₄)
Ferrous oxalate, ${\mathrm{FeC}}_2{O}_4$, dissociates into ferrous ions (${\mathrm{Fe}}^{2+}$) and oxalate ions ($C_2{O}_4^{2-}$). Both of these ions are oxidised by ${\mathrm{MnO}}_4^{-}$ in an acidic medium.

The half-reactions for oxidation are:
• Oxidation of ferrous ion:
${\mathrm{Fe}}^{2+}\to {\mathrm{Fe}}^{3+}+e^{-}$ (1 electron released)

• Oxidation of oxalate ion:
$C_2{O}_4^{2-}\to 2{\mathrm{CO}}_2+2e^{-}$ (2 electrons released)

Therefore, for 1 mole of ${\mathrm{FeC}}_2{O}_4$, the total number of electrons lost (n-factor of ${\mathrm{FeC}}_2{O}_4$) is:
$\mathrm{Total\; electrons\; lost}=1+2=3e^{-}$

2. Understanding the Reduction of Permanganate (${\mathrm{MnO}}_4^{-}$) in Acidic Medium
In an acidic medium, the permanganate ion (${\mathrm{MnO}}_4^{-}$) is reduced to manganese (II) ion (${\mathrm{Mn}}^{2+}$).
The reduction half-reaction is:
${\mathrm{MnO}}_4^{-}+8H^{+}+5e^{-}\to {\mathrm{Mn}}^{2+}+4H_{2}O$

Thus, each mole of ${\mathrm{MnO}}_4^{-}$ accepts 5 electrons (n-factor of ${\mathrm{MnO}}_4^{-}$ is 5).

3. Applying the Law of Equivalents
According to the principle of redox titrations, the total equivalents of reducing agent must equal the total equivalents of oxidising agent:
$\mathrm{Equivalents\; of}{\mathrm{FeC}}_2{O}_4=\mathrm{Equivalents\; of}{\mathrm{MnO}}_4^{-}$

Using the relation:
$\mathrm{Number\; of\; moles}\times \mathrm{n-factor}=\mathrm{Equivalents}$

Let $x$ be the moles of ${\mathrm{MnO}}_4^{-}$ required for 1 mole of ${\mathrm{FeC}}_2{O}_4$:
$1\times 3=x\times 5$

Solving for $x$:
$x=\frac{3}{5}=0.6\mathrm{moles}$

Thus, 1 mole of ferrous oxalate requires 0.6 moles of ${\mathrm{MnO}}_4^{-}$ to be completely oxidised in an acidic medium.