JEE Main On 8 April 2017 Question 7
Question: A solution containing a group-IV cation gives a precipitate on passing, $ H _2S. $ . A solution of this precipitate in dil. HCl produces a white precipitate with NaOH solution and bluish-white precipitate with basic potassium ferrocyanide. The cation is : [JEE Online 08-04-2017]
Options:
A) $ M{n^{2+}} $
B) $ Z{n^{2+}} $
C) $ N{i^{2+}} $
D) $ C{o^{2+}} $
Show Answer
Answer:
Correct Answer: B
Solution:
- Precipitation with $\mathrm{H}_2 \mathrm{~S}$ :
- When $\mathrm{Zn}^{2+}$ is present in a solution and $\mathrm{H}_2 \mathrm{~S}$ is passed through, a zinc sulfide precipitate is formed:
$$ \mathrm{Zn}^{2+}+\mathrm{H}_2 \mathrm{~S} \rightarrow \mathrm{ZnS} \downarrow+2 \mathrm{H}^{+} $$
- Reaction with Dilute HCl and NaOH :
- The precipitate ZnS dissolves in dilute HCl , forming zinc chloride:
$$ \mathrm{ZnS}+2 \mathrm{HCl} \rightarrow \mathrm{ZnCl}_2+\mathrm{H}_2 \mathrm{~S} \uparrow $$
- When NaOH is added, zinc hydroxide precipitates:
$$ \mathrm{ZnCl}_2+2 \mathrm{NaOH} \rightarrow \mathrm{Zn}(\mathrm{OH})_2 \downarrow+2 \mathrm{NaCl} $$
- Reaction with Basic Potassium Ferrocyanide:
- Zinc reacts with potassium ferrocyanide in basic conditions to form a bluishwhite zinc ferrocyanide precipitate:
$$ 3 \mathrm{Zn}^{2+}+2 \mathrm{~K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightarrow \mathrm{Zn}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2 \downarrow+8 \mathrm{~K}^{+} $$
These reactions confirm the presence of $\mathrm{Zn}^{2+}$. Hence, the correct answer is B$) \mathrm{Zn}^{2+}$.
BETA
