Valence shell electronic configuration of $C{r}^{2+}$ is:

The correct option is A $4s^{0}3d^4$
Explanation :
Cr has the electronic configuration of: $4s^{1}3d^5$ from which we remove two electrons. So the electronic configuration is: $4s^{0}3d^4$

$C{r}^{2+}$ (22 electrons) : $\left[Ar\right]3d^4$