A) \[A{{l}_{2}}{{O}_{3}}+HF+NaAl{{F}_{4}}\]
B) \[A{{l}_{2}}{{O}_{3}}+Ca{{F}_{2}}+NaAl{{F}_{4}}\]
C) \[A{{l}_{2}}{{O}_{3}}+N{{a}_{3}}Al{{F}_{6}}+Ca{{F}_{2}}\]
D) \[A{{l}_{2}}{{O}_{3}}+KF+N{{a}_{3}}Al{{F}_{6}}\]
Correct Answer: C
Solution :
[c] Fused alumina \[(A{{l}_{2}}{{O}_{3}})\] is a bad conductor of electricity. Therefore, cryolite \[(N{{a}_{3}}Al{{F}_{6}})\] and fluorspar \[(Ca{{F}_{2}})\] are added to purified alumina which not only make alumina a good conductor of electricity but also reduce the melting point of the mixture to around 1140K.
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