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Match the List -I (containing anions ) with List -II (containing reagent used in testing ) using the codes as given below in the column. {:("List"-I,,"List"-II),(("anions"),,("reagents")),((a)S^(2-),,(p)"Barium chloride solution in presence of HCl"),((b)NO_(3)^(-),,(q)"Sodium nitroprusside"),((c)I^(-),,(r)"chlorine water and chloroform"),((d)SO_(4)^(2-),,(s)"iron"(II)"sulphate solution and conc."H_(2)SOS_(4)):} Code : |
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Answer» `{:(,"(a)","(B)","(c)","(d)"),(,"(q)","(s)","(r)","(p)"):}` `S^(2+)+BaCl_(2)rarr"No precipitate".` `(b)""2NO_(3)^(-)+4H_(2)SO_(4)+6Fe^(2+)rarr6Fe^(3+)+2NOuarr+4SO_(4)^(2-)+4H_(2)O` `[Fe(H_(2)O)_(6)]^(2+)+NOrarrunderset(("Brown complex"))([Fe(H_(2)O)_(5)NO]^(2+)+H_(2)O)` `(c )""2I^(-)+CL_(2)+I_(2)+2Cl^(-)` `""I_(2)+C CI_(4)rarrI_(2)` dissolves forming a violet solution. `(d)""Ba^(2+)+SO_(4)^(2-)rarrBaSO_(4)darr("white")` `HCl`REMOVES the impurities of `S^(2-),I^(-)` and `NO_(3)^(-)` and white precipitate of `BaSO_(4)` is THUS obtained. |
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