A mixture consists `(A)` (red solid) and `(B)` (colourless solid) which gives lilac colour in flame.
(a)Mixture gives black precipitate `( C)` on passing `H_(2)S (g)`.
(b)`( C)` is soluble in aquaregia and on evaporation of aquaregia and adding `SnCl_(2)` gives greyish black precipitate `(D)`.
The salt solution with `NH_(4)OH` gives a brown precipitate.
(i)The sodium extract of the salt with `C Cl_(4)//FeCl_(3)` gives a violent layer.
(ii)The sodium extract gives yellow precipitate with `AgNO_(3)` solution which is insoluble in dilute ammonia solution.
Identify `(A)` and `(B)`, and the precipitates `(C )` and `(D)`.
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Correct Answer - A,B,C,D are `HgI_(2) , KI, HgS` and `Hg` respectively.
`(A)+(B)to` lilac colour in flame.So one of the cation may be `K^(+)`.
(a)`(A)+(B)overset(H_(2)S(g))to(C)` black precipitate.
(b)`( C)underset("in aque regia")overset("soluble")to "soluble"overset("Evaporation")to"Residue"overset(SnCl_(2))to"Greyish black precipitate"(D)`
`(A)+(B) overset(NH_(4)OH)to` brown precipitate.So second cation may be `Hg^(2+)`
(i)Sodium carbonate extract of salt `overset(C Cl_(4)//FeCl_(3)) to` violent colour.
(ii)Sodium extract of salt `overset(AgNO_(3))to "yellow precipitate" overset(NH_(3))to` soluble.
So the anion may be `I^(-)`
(a)`HgI_(2)+H_(2)Sto HgS (C )darr("black")+2HI`
(b)`3HgS+6HCl+2HNO_(3)to3HgCl_(2)+2NO+4H_(2)O+S`
`HgCl_(2)+SnCl_(2) to Hg darr (D) ("greyish black")+SnCl_(4)`
`2KI(B)+HgI_(2)(A)toK_(2)[HgI_(4)]`(orange)
`2K_(2)[HgI_(4)]+NH_(3)+3KOH to [HgOHg(NH_(2))I]darr`(brown)
Sodium carbonate extract of salt contains `NaI`
`2NaI+2Fe^(3+)overset(C Cl_(4))toI_(2)uarr("violet")+2Na^(+)+2Fe^(2+)`
`I_(2)` dissolves in `C Cl_(4)` giving violet colour solution.
`AgNO_(3)+NaI toAgI darr("yellow")+NaNO_(3)`
`AgI` is insoluble in ammonia solution.
So,`(A),(B),( C)` and `(D)` are `HgI_(2),KI,HgS` and `Hg` respectively.
`Mg^(2+)+2HCO_(3)^(-)`(from sodium salt)`toMg(HCO_(3))_(2)`(solube)`overset(Delta)toMgCO_(3)darr` (white)`+H_(2)O+CO_(2)`
`Hg^(2+)` gives red precipitate of `Hgl_(2)` which dissolves in excess `KI`forming colourless `[Hgl_(4)]^(-)`(orange solution)
`Pb^(2+)` gives yellow precipitate of `Pbl_(2).Cu^(2+)` gives white precipitate of `Cu_(2)I_(2)` with evolution of iodine.
`[Zm(OH)_(4)]^(2-)overset(H_(2)O-"boiled")toZn(OH)_(2)darr("white")+2OH^(-)`
`Zn(OH)_(2)` precipitate is readily soluble in excess of ammonia and in solutions of ammonium salts due to the formation of tetraamminezinc`(II)`.
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