A) Extraction of $$Al$$ from its ore follows these steps:
Leaching, precipitation, calcination and electrolytic reduction (molten state)
$$Al_2O_3(s)+3H_2O(l)+2NaOH(aq)\xrightarrow {Leaching} 2Na[Al(OH)_4](aq)$$
$$2Na[Al(OH)_4](aq)+CO_2(g)\rightarrow Al_2O_3.xH_2O(s)+2NaHCO_3(aq)$$
$$Al_2O_3.xH_2O(s)\xrightarrow [Calcination]{1470K} Al_2O_3(s)+xH_2O(g)$$
Electrolytic reduction of molten pure $$Al_2O_3$$ is mixed with $$Na_2AlF_6$$ or $$CaF_2$$
This method is known as Hall-Heroult process
$$Cathode: Al^{3+}(melt)+3e^-\rightarrow Al(l)$$
$$Anode: C(s)+O^{2-}(melt)\rightarrow CO(g)+2e^-$$
$$C(s)+2O^{2-}(melt)\rightarrow CO_2(g)+4e^-$$
B) In case of $$Ag$$, extraction follows leaching and displacement method as:
Leaching:
$$Ag_2S+4NaCN\rightleftharpoons 2Na[Ag(CN)_2]+Na_2S$$
$$Na_2S+2O_2\rightarrow Na_2SO_4$$
Displacement by zinc in aqueous solution:
$$2Na[Ag(CN)_2]+Zn\rightarrow Na_2[Zn(CN)_4]+2Ag\downarrow$$
C) Extraction of $$Cu$$ follows these as:
Roasting:
$$2PbS+3O_2\xrightarrow {\Delta} 2PbO+2SO_2$$
Self reduction:
$$2PbO+PbS\rightarrow 2Pb+SO_2$$
D) Extraction of $$Mg$$ follows:
Calcination:
$$MgCl_2.6H_2O\xrightarrow [Dry HCl(g)]{\Delta (calcination)} MgCl_2+5H_2O$$
It is not made anhydrous by simple heating because it gets hydrolysed
$$MgCl_2.6H_2O\xrightarrow {\Delta} MgO.5H_2O+2HCl$$
Electrolytic reduction:
Electrolytic reduction of molten anhydrous carnalite.
$$MgCl_2\rightleftharpoons Mg^{2+}+2Cl^-$$
At cathode: $$Mg^{2+}+2e^-\rightarrow Mg$$ (99% pure);
At anode: $$2Cl^-\rightarrow Cl_2+2e^-$$
Option D represents the incorrect statement.
Ans-option D