Match List I with List II

List-I (Conversion)		List-II (Number of Faraday required)
(A)	1 mol of  H2O to O2	(I)	3F
(B)	1 mol of $Mn{O}_4^{-\; <!--\; -\; -->}$ to Mn2+	(II)	$$2F
(C)	1.5 mol of Ca from molten CaCl2	(III)	1F
(D)	1 mol of FeO to Fe2O3	(IV)	5F

Choose the correct answer from the options given below:

To find the correct match between List-I and List-II, we need to calculate the number of Faradays of electricity required for each conversion. The quantity of electricity required for the oxidation or reduction of 1 mole of a substance is given by the relation:
Charge ($Q$) = $n\times F$
where $n$ is the number of moles of electrons transferred per mole of the reactant, and $F$ is Faraday's constant (1 Faraday = charge of 1 mole of electrons).

(A) Conversion of 1 mol of H₂O to O₂:
The oxidation reaction of water is:
$H_{2}O\to \frac{1}{2}{O}_2+2H^{+}+2e^{-}$
Here, the oxidation state of oxygen changes from -2 in H₂O to 0 in O₂.
For 1 mole of H₂O, 2 moles of electrons are released.
Therefore, the number of Faradays required = 2F.
This matches with (II).

(B) Conversion of 1 mol of MnO₄^- to Mn²⁺:
The reduction reaction is:
$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$
In MnO₄^-, the oxidation state of Mn is +7, and it is reduced to +2 in Mn²⁺.
The change in oxidation state per Mn atom = 7 - 2 = 5.
For 1 mole of MnO₄^-, 5 moles of electrons are required.
Therefore, the number of Faradays required = 5F.
This matches with (IV).

(C) Conversion of 1.5 mol of Ca from molten CaCl₂:
The reduction reaction of calcium ion is:
$C{a}^{2+}+2e^{-}\to Ca$
To deposit 1 mole of Ca, 2 moles of electrons are required (2F).
To deposit 1.5 moles of Ca:
Number of Faradays required = $1.5\times 2F=3F$.
This matches with (I).

(D) Conversion of 1 mol of FeO to Fe₂O₃:
The oxidation reaction is:
$2FeO\to F{e}_2{O}_3+2H^{+}$ (or written per mole of FeO: $F{e}^{2+}\to F{e}^{3+}+e^{-}$)
In FeO, the oxidation state of Fe is +2. In Fe₂O₃, the oxidation state of Fe is +3.
The change in oxidation state per Fe atom = 3 - 2 = 1.
For 1 mole of FeO, 1 mole of electrons is released.
Therefore, the number of Faradays required = 1F.
This matches with (III).

Combining all the matches:
A-II, B-IV, C-I, D-III