The precipitate of  ${\mathrm{CaF}}_2\left({ \mathrm{K}}_{\mathrm{sp}}=1.7\times {10}^{-10}\right)$ is obtained when equal volumes of the following are mixed ?

 For precipitation, $Q_{\mathrm{sp}}$ (product of molar concentration) $>{\mathrm{K}}_{\mathrm{sp}}$ (solubility product).

In water, ${\mathrm{CaF}}_2$ is in equilibrium with its ions as follows:

${\mathrm{CaF}}_2( \mathrm{s})\rightleftharpoons {\mathrm{Ca}}^{2+}\left(\mathrm{aq}\right)+2{ \mathrm{F}}^{-}\left(\mathrm{aq}\right)$

Solubility product of ${\mathrm{CaF}}_2$

${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{2+}\right]\left[{\mathrm{F}}^{-2}\right]=1.7\times {10}^{-10}$

In A), $Q_{sp}=\left[\frac{{10}^{-4}}{2}\right]{\left[\frac{{10}^{-4}}{2}\right]}^2=1.25\times {10}^{-13}$

In B),$Q_{\mathrm{sp}}=\left[\frac{{10}^{-2}}{2}\right]{\left[\frac{{10}^{-3}}{2}\right]}^2=1.25\times {10}^{-9}$

In C), $Q_{\mathrm{sp}}=\left[\frac{{10}^{-5}}{2}\right]{\left[\frac{{10}^{-3}}{2}\right]}^2=1.25\times {10}^{-12}$

In D), $Q_{\mathrm{sp}}=\left[\frac{{10}^{-3}}{2}\right]{\left[\frac{{10}^{-5}}{2}\right]}^2=1.25\times {10}^{-14}$

Only for option B, the ionic product is greater than the solubility product.

Therefore, the correct option is (B).