A) \[N{{a}_{4}}\left[ Cu{{F}_{6}} \right]\]
B) \[Na\left[ Cu{{F}_{4}} \right]\]
C) \[N{{a}_{2}}\left[ Cu{{F}_{4}} \right]\]
D) \[N{{a}_{2}}[Cu{{F}_{3}}]\]
Correct Answer: C
Solution :
[c] Mole ratio of Na, Cu and \[F=\frac{1.08}{0.539}:\frac{0.539}{0.539}:\frac{2.16}{0.539}=2:1:4\] Empirical formula \[=N{{a}_{2}}Cu{{F}_{4}}\] : Van't Hoff factor (i)= 3 (given) Hence, formula of the compound: \[N{{a}_{2}}[Cu{{F}_{4}}]\to 2N{{a}^{+}}+{{[Cu{{F}_{4}}]}^{2-}}\]You need to login to perform this action.
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