`HgCl_(2)` and `I_(2)` both when dissolved in water containing `I^(-)` ions the pair of species formed is:

Options

A. `HgI_2 , I^(-)`
B. `HgI_4^(2-)`
C. `Hg_2 I_2 , I^(-)`
D. `HgI_2 , I_3^(-)`

Answer

B

Explanation

In a solution containing `HgCl_2 , I_2 and I^-` , both `HgCl_2` and `I_2` compete for `I^-`.
since formation constant of `[HgI_4]^(2-)` is very large `(1.9xx10^(30))` as compared with `I_3`
`(K_f =700)`
`therefore I^(-)` will preferentially combine with `HgCl_2` .
`HgCl_2 + 2I^(-) to underset("Red ppt")(HgI_2 darr) + 2Cl^(-)`
`HgI_2 + 2I^(-) to underset("soluble")([HgI_4^(2-)]`