Qualitative Analysis - Result Question 32
####33. Statement I A very dilute acidic solution of $Cd^{2+}$ and $Ni^{2+}$ gives yellow precipitate of $CdS$ on passing $H _2 S$.
Statement II Solubility product of CdS is more than that of NiS.
(1989, 2M)
Passage Based Questions
Passage 1
An aqueous solution of metal ion $M _1$ reacts separately with reagents $Q$ and $R$ in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion $M _2$ always forms tetrahedral complexes with these reagents. Aqueous solution of $M _2$ on reaction with reagent $S$ gives white precipitate which dissolves in excess of $S$. The reactions are summarised in the scheme given below
Tetrahedral $\underset{\text { excess }}{\stackrel{Q}{\longleftarrow}} M _1 \xrightarrow[\text { excess }]{\stackrel{R}{\leftrightarrows}}$ Square planar
Tetrahedral $\underset{\text { excess }}{\stackrel{Q}{Q}} M _2 \underset{\text { excess }}{\stackrel{R}{\leftrightarrows}}$ Tetrahedral
$$ \begin{aligned} & S \text {, stoichiometric amount } \end{aligned} $$
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Answer:
Correct Answer: 33. $Fe^{3+}$
Solution:
- Cation $Cd^{2+}$ belongs to group II while $Ni^{2+}$ belongs to group III of analytical group. Group II radicals are precipitated by passing $H _2 S(g)$ through acidic solution of salt but radicals of group III are precipitated by passing $H _2 S(g)$ in $NH _3 / NH _4 Cl$ buffer solution of salt due to greater solubility products of later salts.
