When KI (excess) is added to a.CuSO4 b.HgCl2 c.Pb(NO3)2 1)a white ppt. of Cul in I, an orange ppt. H g I 2 , in II and a yellow ppt. P b I 2 , in III 2)a white ppt. of CuI in I, an orange ppt. dissoloving to H g I 2 − 4 in II, and an yellow ppt. of P b I 2 in III 3)a white ppt. of C u I , H g I 2 and P b I 2 in each case 4)none of the above is correct

Options

A. a white ppt. of Cul in I, an orange ppt. `HgI_2`, in II and a yellow ppt. `PbI_2`, in III
B. a white ppt. of CuI in I, an orange ppt. dissoloving to `HgI_4^(2-)` in II, and an yellow ppt. of `PbI_2` in III
C. a white ppt. of `CuI, HgI_2" and " PbI_2` in each case
D. none of the above is correct

Answer

B

Explanation

I `2CuSO_4 + 4KI to underset (" White PPt.")(Cu_2I_2) + I_2 + 2 K_2 SO_4`
`HgCI_2 + 2KIto 2KCl + underset ("orangeppt.")( HgI_2 darr) ,
II HgI_2 + 2KI to underset ("soluble")(K_2 HgI_4)`
III: `Pb(NO_3)_2 + 2KI to underset("Yellow ppt.")(PbI_2 darr) + 2KNO_3`