Chemistry - Carbonyl compounds Question with Solution | TestHub

Mol ratio of Na, Cu and F

$=\frac{1\text{.}08}{0\text{.}539}\text{:}\frac{0\text{.}539}{0\text{.}539}\text{:}\frac{2\text{.}16}{0\text{.}539}=2\text{:}1\text{:}4$

Empirical formula = Na₂CuF₄: Van't Hoff factor (i)

= 3 (given)

Hence, formula of the compound :

Na₂[CuF₄]$\to$2Na⁺+ [CuF₄]^2-