Number of electrons having the quantum numbers

$n=4,l=0,s=-\frac{1}{2}$in ${\mathrm{Zn}}^{+2}\text{ ion }$ is / are

${\mathrm{Zn}}^{2+}$ has no use $4s\mathit{ }{e}^{-}$.

$1s^{2}2s^{2}2p^{\mathit{6}}3s^{2}3p^{\mathit{6}}4s^{2}3d^{\mathit{10}}$

Since it is Zn2+, it has lost 2 electrons since the previous shell.

Hence, it will be $4{\mathrm{s}}^0$ instead of $4{\mathrm{s}}^2$

Thus, the number of electrons having the quantum numbers is 0.