Question

Colour of ${Cr}^{n+}$ ion solution has a green colour due to $d-d$ transition in its orbits then the minimum value of $n$ is:

Answer 1

$C{r}^{n+}⟶C{r}^{4+}4s^{0}3d^2$ ; $C{r}^{5+}4s^{0}3d^1$

$C{r}^{3+}4s^{0}3d^3$

More is the number of unpaired $e^{-}$ more $d-d$ transition is possible.