Question

Metallic magnesium is prepared by:
(A) Reduction of $ MgO $ by coke
(B) Electrolysis of aqueous solution of $ Mg{(N{O_3})_2} $
(C) Displacement of $ Mg $ by iron from $ MgS{O_4} $ solution
(D) Electrolysis of molten $ MgC{l_2} $

Answer 1

Hint : $ Mg $ a strong chemical reagent, and it forms stable compounds and reacts with $ {O_2} $ and chlorine in both the liquid and gaseous states. It means that the extraction process of $ Mg $ from its raw materials is energy intensive. Commercially there are two methods available for it- electrolysis of $ MgC{l_2} $ and the other is the Pigeon process. Most used worldwide method is electrolysis which accounted for approximately $ 75\% $ of world $ Mg $ production.

Complete Step By Step Answer:
A) Reduction of $ MgO $ by coke-
 $ MgO $ Cannot be reduced by coke because this reaction requires high temperature so thermodynamically not feasible.
B) Electrolysis of aqueous solution of $ Mg{(N{O_3})_2} $ -
An aqueous solution of $ Mg{(N{O_3})_2} $ is electrolyzed as follows-
Reaction at cathode-
 $ 2{H_2}O(l) \to 2{H_{2(g)}} + {O_{2(g)}} $
Reaction at anode-
 $ 4{H^ + }_{(aq)} + 4O{H^ - }_{(aq)} \to 4{H_2}{O_{(l)}} $
We can see from above reactions that metallic magnesium is not prepared by this reactions.
C) Displacement of $ Mg $ by iron from $ MgS{O_4} $ solution-
We know that more reactive metal displaces less reactive metals from a solution. $ Mg $ Is more reactive than $ Fe $ . The reaction $ Mg + FeS{O_4} \to MgS{O_4} + Fe $
Here we can see that $ Mg $ can replace $ Fe $ . So the above statement is not true.
D) Electrolysis of molten $ MgC{l_2} $ -
Preparation of molten $ MgC{l_2} $
 $ {\text{M}}{{\text{g}}^{{\text{2 + }}}}{\text{ + Ca(OH}}{{\text{)}}_{\text{2}}}{\text{ + Mg(OH}}{{\text{)}}_{\text{2}}}{\text{ + CaC}}{{\text{l}}_{\text{2}}}\underrightarrow {{\text{ dil}}{\text{.HCl}}}{\text{ MgC}}{{\text{l}}_{\text{2}}}{\text{.6}}{{\text{H}}_{\text{2}}}{\text{O}} $
 $ {\text{MgC}}{{\text{l}}_{\text{2}}}{\text{.6}}{{\text{H}}_{\text{2}}}{\text{O + MgC}}{{\text{l}}_{\text{2}}}{\text{.2}}{{\text{H}}_{\text{2}}}{\text{O }}\underrightarrow {{\text{dryHcl}}}{\text{ MgC}}{{\text{l}}_{\text{2}}} $
Now the molten mixture is electrolyzed.
Reaction at cathode-
 $ M{g^{2 + }} + 2{e^ - } \to M{g_{(s)}} $
Reaction at anode-
 $ 2C{l^ - } \to C{l_2} $
Here we can see that metallic magnesium is made by electrolysis of molten $ MgC{l_2} $ .
Hence the correct option is (D).

Note :
 $ MgC{l_2} $ Obtained by any of the above given methods is fused and then mixed with $ NaCl $ and $ CaC{l_2} $ in the temperature range of $ 973 - 1023K $ . Electrolysis of $ MgC{l_2} $ consists of a two-step process- Preparation of molten and then electrolysis of molten.