A solution containing $A{s}^{3+},C{d}^{2+},N{i}^{2+}$ and $Z{n}^{2+}$  is made alkaline with dilute $N{H}_{4}OH$ and treated with $H_{2}S$. The precipitate obtained will consist of 

$A{s}^{3+}$and$C{d}^{2+}$ are the radicals of group II, whereas $N{i}^{2+}$and $Z{n}^{2+}$are the radicals of group IV. The solubility product of group IV radicals is higher as compared to group II. $N{H}_{4}OH$ Increase the ionisation of $H_{2}S$ by removing $H^{+}$ of $H_{2}S$ unionisable water.

$H_{2}S\rightleftharpoons 2H^{+}+S^{2-};H^{+}+O{H}^{-}\to H_{2}O$

Thus , excess of sulphide ions are present which leads to the precipitation of all the four ions.

Note: HCI decreases ionisation of  $H_{2}S$ whereas $N{H}_{4}OH$ increases the ionisation of $H_{2}S$.