A) NaF
B) \[N{{a}_{3}}Al{{F}_{6}}\]
C) \[Al{{F}_{3}}\]
D) none of these
Correct Answer: B
Solution :
Chief ore of aluminium is bauxite \[A{{l}_{2}}{{O}_{3}}.2{{H}_{2}}O.\]Firstly bauxite is concentrated by chemical method. (Baeyer?s method) using \[\text{NaOH}\] in which pure alumina \[(A{{l}_{2}}{{O}_{3}})\] is obtained. Bauxite has impurity of sand \[(Si{{O}_{2}})\] only \[\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\] reacts with \[\text{NaOH}\] \[\underset{(Bauxite)}{\mathop{A{{l}_{2}}{{O}_{3}}}}\,+NaOH\xrightarrow{{}}\underset{(soluble\,in\,water)}{\mathop{NaAl{{O}_{2}}+{{H}_{2}}O}}\,\] Impurities remain in the solution. It is then filtered and the filterate is strongly heated \[\underset{filterate}{\mathop{NaAl{{O}_{2}}}}\,+{{H}_{2}}O\xrightarrow{{}}\underset{(white\,ppt.)}{\mathop{Al{{(OH)}_{3}}}}\,+NaOH\] This solution is again filtered to obtain white precipitate of \[Al{{(OH)}_{3}}\] which is strongly heated to obtain pure \[A{{l}_{2}}{{O}_{3}}.\] \[Al{{(OH)}_{3}}\xrightarrow{\Delta }A{{l}_{2}}{{O}_{3}}+{{H}_{2}}O\] Aluminium being highly electropositive is reduced by electrolytic reduction (Hall's process). During Hall's process along with molten \[\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\] small amount of cryolite \[\text{(N}{{\text{a}}_{3}}\text{Al}{{\text{F}}_{6}}\text{)}\] and fluorspar \[(Ca{{F}_{2}})\] are also added as they decrease the melting point of \[\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\] and increase its electrical conductivity.
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