Give the expression for the solubility product constant $(K_{sp})$ for $Ba{F}_2.$

$[B{a}^{2+}][F^{-}{]}^2$

$\frac{[B{a}^{2+}][F^{-}{]}^2}{[Ba{F}_2]}$

$\frac{[Ba{F}_2]}{[B{a}^{2+}][F^{-}{]}^2}$

$[B{a}^{2+}][F^{-}]$

$[B{a}^{2+}{]}^2[F^{-}]$