Co-Ordination Compounds Question 301
Question: A compound contains 1.08 mole of Na, 0.539 mole of Cu and 2.16 mole of F. Its aqueous solution shows osmotic pressure which is three times that of urea having same molar concentration. The formula of the compound is:
Options:
A) $ Na_4[ CuF_6 ] $
B) $ Na[ CuF_4 ] $
C) $ Na_2[ CuF_4 ] $
D) $ Na_2[CuF_3] $
Show Answer
Answer:
Correct Answer: C
Solution:
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 $ =Na_2CuF_4 $ : Van’t Hoff factor (i)= 3 (given)
Hence, formula of the compound: $ Na_2[CuF_4]\to 2N{a^{+}}+{{[CuF_4]}^{2-}} $
