Question

Which of the following basic radicals will be precipitated by $H_{2}S$ gas in the presence of $N{H}_3$?

• A. $M{n}^{2+}$
• B. $N{i}^{2+}$
• C. $C{d}^{2+}$
• D. $C{a}^{2+}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $M{n}^{2+}$
B $N{i}^{2+}$
C $C{d}^{2+}$

Option (A),(B),(C) are correct.

$M{n}^{2+}$, $N{i}^{2+}$ and $C{d}^{2+}$radicals will be precipitated by $H_{2}S$ gas in the presence of $N{H}_3$. $C{d}^{2+}$ belongs to II group and due to low $k_{sp}$ value precipitated first and then $M{n}^{2+}$ and $N{i}^{2+}$ get precipitated by $H_{2}S$ gas in presence of solution of $N{H}_3$.
$C{d}^{2+}$ is not precipated by $H_{2}S$ gas but by excess of $N{H}_{4}OH$ solution followed by solid ammonium carbonate.