Question

Solution 'A' gives a white sulphide precipitate with group IV reagent. Its precipitate dissolves in dil.HCl to give solution 'B'. A bluish white precipitate appears when potassium ferrocyanide $\left(K_{4}Fe\right(CN{)}_6)$ solution is added to the solution 'B' after neutralisation by $N{H}_{4}OH$ solution. The cation present in solution 'A' is

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

Answer 1

The correct option is B $Z{n}^{2+}$

Group IV cations forms sulphide precipitate with $H_{2}S$ which is the group IV reagent.

Confirmation of $M{n}^{2+}$:
Zinc sulphide dissolves in hydrochloric acid to form zinc chloride.
(a). When potassium ferrocyanide $K_{4}Fe(CN{)}_6$ solution is added to the solution after neutralisation by $N{H}_{4}OH$ solution, a white or a bluish white precipitate
of zinc ferrocyanide appears.
$2ZnC{l}_2+K_4\left[Fe\right(CN{)}_6]\to \underset{Zincferrocyanide}{Z{n}_2\left[Fe\right(CN{)}_6]}+4KCl$

(b). On addition of sodium hydroxide solution to zinc chloride, it gives a white precipitate of zinc hydroxide, which is soluble in excess of NaOH solution on heating.
This confirms the presence of Zn2+ ions.
$ZnC{l}_2+2NaOH\to Zn(OH{)}_2+2NaCl$
$Zn(OH{)}_2+2NaOH\to \underset{Sodiumzincate}{N{a}_{2}Zn{O}_2}+2H_{2}O$