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Zinc II Chloride

An element A ...

Question

An element (A) belongs to group number $12$ and period number $4$ , bluish white in colour reacts with Conc. $H_{2} S O_{4}$ to give its salt (B) with the liberation of $S O_{2}$ gas. Compound (B) reacts with sodium bicarbonate to give (C) which is used as ointment for curing skin diseases. Identify A, B and C and explain the reactions.

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Solution

A bluish white metal (A) present in $4$ th period and $12$ th group is Zinc.
Zinc on treatment with conc. $H_{2} S O_{4}$ gives Zinc sulphate $(Z n S O_{4})$ compound (C) with evolution of $S O_{2}$ .
$Z n + C o n c . 2 H_{2} S O_{4} \to Z n S O_{4} (C) + S O_{2} ↑ + 2 H_{2} O$
Zinc carbonate occurs in nature as calamine.
When $N a H C O_{3}$ is addd to zinc sulphate solution, $Z n C O_{3}$ is obtained.
Zinc carbonate occurs in nature as calamine.
When $N a H C O_{3}$ is added to zinc. sulphite solution, $Z n C O_{3}$ is obtained.
$Z n S O_{4} + 2 N a H C O_{3} \to Z n C O_{3} (C) + N a_{2} S O_{4} + H_{2} O + C O_{2} ↑$
Compound
Molecular formula
Name
A
Zn
Zinc
B
$Z n S O_{4}$
Zinc sulphate
C
$Z n C O_{3}$
Zinc Carbonate

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