`SnCl_(2)` gives a precipitate with a solution of `HgCl_(2)` In this process `HgCl_(2)` is :

Addition of SnCl_(2) to HgCl_(2) gives precipitate:

Addition of SnCl_(2) to HgCl_(2) gives precipitate:

When excess of SnCl_(2) is added to a solution of HgCl_(2) , a white precipitate turning to gray is obtained. The grey colour is due to

On heating aniline with alcoholic CS_(2) in the presence of HgCl_(2) gives :

Give the shape of HgCl_2

PH_(3)+HgCl_(2) to

Which gives black precipitate on reaction with CS_2 followed by addition of HgCl_2 ?

What is the product obtained in the reaction of HgCl_(2) and Hg(CN)_(2) ?