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.