# A metallic element X forms a carbonate with the formula X(CO_3)_2. What would the corresponding formula be of the fluoride compound of element X?

$X {F}_{4}$
The starting species is $X {\left(C {O}_{3}\right)}_{2}$. And thus, separating the ions, we get ${X}^{4 +}$ and $2 \times C {O}_{3}^{2 -}$.
Fluorine is an excellent oxidant and gives ${F}^{-}$ ions upon reduction, so to preserve electrical neutrality we get $X {F}_{4}$.