JEE MAIN - Chemistry (2023 - 31st January Morning Shift - No. 1)
When $$\mathrm{Cu}^{2+}$$ ion is treated with $$\mathrm{KI}$$, a white precipitate, $$\mathrm{X}$$ appears in solution. The solution is titrated with sodium thiosulphate, the compound $$\mathrm{Y}$$ is formed. $$\mathrm{X}$$ and $$\mathrm{Y}$$ respectively are :
| $$\mathrm{X=CuI_2}$$ | $$\mathrm{Y=Na_2S_4O_6}$$ |
| $$\mathrm{X=Cu_2I_2}$$ | $$\mathrm{Y=Na_2S_4O_6}$$ |
| $$\mathrm{X=CuI_2}$$ | $$\mathrm{Y=Na_2S_2O_3}$$ |
| $$\mathrm{X=Cu_2I_2}$$ | $$\mathrm{Y=Na_2S_4O_5}$$ |
Explanation
$2 \mathrm{Cu}^{2+}+4 \mathrm{KI} \longrightarrow \underset{\text { White ppt. }}{\mathrm{Cu}_{2} \mathrm{I}_{2}}+\mathrm{I}_{2}$
$\mathrm{I}_{2}+\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \longrightarrow 2 \mathrm{Nal}+\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}$
$\mathrm{X}=\mathrm{Cu}_{2} \mathrm{I}_{2}$
$\mathrm{Y}=\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}$
$\mathrm{I}_{2}+\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \longrightarrow 2 \mathrm{Nal}+\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}$
$\mathrm{X}=\mathrm{Cu}_{2} \mathrm{I}_{2}$
$\mathrm{Y}=\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}$
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