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Addition of \mathrm{SnCl}_{2} to \mathrm{HgCl}_{2} gives a precipitate 

Option: 1

White turning to red


Option: 2

White turning to gray


Option: 3

Black turning to white


Option: 4

None of these


Answers (1)

best_answer

\mathrm{SnCl_2} reduces \mathrm{Hg}^{2+} into \mathrm{Hg}^{+}and eventually into elemental Hg.

\mathrm{HgCl}_{2}+\mathrm{SnCl}_{2} \longrightarrow \underset{(\text { White })}{\mathrm{Hg}_{2} \mathrm{Cl}_{2}}+\mathrm{SnCl}_{4}

\mathrm{\mathrm{Hg}_{2} \mathrm{Cl}_{2}+\mathrm{SnCl}_{2} \longrightarrow \underset{Black}{\mathrm{Hg}}+\mathrm{SnCl}_{4}}

So, we have a White precipitate turning into Black.

Hence, the correct answer is Option (2)

Posted by

Ritika Jonwal

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