Question

Light green (Compound 'A')$\underset{}{\overset{\Delta }{\to }}$ White Residue (B) $\underset{temp.}{\overset{High}{\to }}C+D+E$

(i) 'D' and 'E' are two acidic gas.

(ii) 'D' is passed through $Hg{Cl}_2$ solution to give yellow ppt.

(iii) 'E' is passed through water first and then $H_{2}S$ is passed, white turbidity is obtained.

(iv) A is water-soluble and addition of $Hg{Cl}_2$ in it, white ppt is obtained but white ppt does not turn into grey on addition of excess solution of 'A'.

'D' and 'E' are respectively.

• A. ${SO}_2$ and ${SO}_3$
• B. ${SO}_3$ and ${SO}_2$
• C. ${SO}_2$ and ${CO}_2$
• D. ${CO}_2$ and $CO$

Answer 1

The correct option is B ${SO}_3$ and ${SO}_2$

Light green compound A is hydrated ferrous sulphate $\left(FeS{O}_4.7H_{2}O\right)$.
When it is heated to 300${}^{o}C$, it loses water of hydration to form white coloured anhydrous ferrous sulphate. This on further heating to high temperature forms a mixture of ferric oxide, sulphur dioxide and sulphur trioxide.
$\underset{LightgreenA}{FeS{O}_4.7H_{2}O}\stackrel{{300}^{o}C}{\to }\underset{anhydrous(white,B)}{2FeS{O}_4}\stackrel{hightemperature}{\to }\underset{C}{F{e}_2{O}_3}+\underset{D}{S{O}_3\uparrow }+\underset{E}{S{O}_2\uparrow }$.
Thus D and E are sulphur trioxide and sulphur dioxide respectively. They are acidic gases.
$\underset{D}{S{O}_3}+HgC{l}_2+H_{2}O\to \underset{yellowprecipitate,basicmercury\left(II\right)sulphate}{HgS{O}_4}+2HCl$
$\underset{E}{S{O}_2}+2H_{2}S\to 2H_{2}O+\underset{whiteturbidity}{3S\downarrow }$.