When `PH_(3)` absorbed in `HgCl_(2)` solution the corresponding phosphide is obtained:

PH_(3)+HgCl_(2) to

Assertion : When SnCl_(2) solution is added to HgCl_(2) solution, a milky white precipitate is obtained and on adding excess SnCl_(2) , a black precipitate is formed. Reason : The disproportionation if Hg(II) is easier than its reduction only.

Assertion : When SnCl_(2) solution is added to HgCl_(2) solution, a milky white precipitate is obtained and on adding excess SnCl_(2) , a black precipitate is formed. Reason : The disproportionation if Hg(II) is easier than its reduction only.

When SnCl_(2) is treated with HgCl_(2) , the products formed are

When SnCl_(2) reacts with HgCl_(2) , the product formed are :

PH_(3) can be obtained by:

PH_(3) can be obtained by:

When excess of SnCl_(2) is added to a soin of HgCl_(2) a white ppt turning grey is obtained the grey colour is due to the formation of

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