Question

An inorganic halide (A) gives the following reactions:
i.) The cation of (A) on reaction with ${{H}_{2}}S $ in HCl medium, gives a black ppt. of (B). (A) neither gives ppt. with HCl nor blue colour with ${{K}_{4}}Fe{{(CN)}_{6}}$.
ii.) (B) on heating with dil. HCl gives black compounds (A) and gas (C) which gives a black ppt. with lead acetate solution.
ii.) The anion of (A) gives chromyl chloride test.
iv.) (B) dissolves in hot dil. $HN{{O}_{3}}$ to give solution (D). (D) gives a ring test.
v.) When $N{{H}_{4}}OH$ solution is added to (D), a white precipitate (E) is formed. (E) dissolves the minimum amount of dil.HCl to give a solution of (A). Aqueous solution of (A) on addition of water gives a whitish turbidity (F).
vi.) Aqueous solution of (A) on warming with alkaline sodium stannite gives a black precipitate of (G) and sodium stannate. The metal (G) dissolves in hydrochloric acid to give solution of A.
Identify (F) and (G) and give balanced chemical equations of reactions.
a.) (F) BiOCl (G) Bi
b.) (F) SiOCl (G) Si
c.) (F) PbOCl (G) Pb
d.) None of these

Answer 1

Hint: $P{{b}^{2+}}$ gives ppt with HCl. Molecules containing chloride ions give chromyl chloride test. Ring test is given by molecules having nitrate ions. Hydrogen Sulphide with lead acetate produces lead Sulphide as black ppt. the cation of SiOCl does not give a black ppt with Hydrogen Sulphide.

Complete Solution :
-as chromyl chloride test is given by anion chloride ion and $B{{i}^{+3}}$ does not give ppt with HCl and blue color with potassium ferrocyanide, so cation is $B{{i}^{3+}}$. (A) is $BiC{{l}_{3}}$.
So cation of (A) is $B{{i}^{3+}}$ which with Hydrogen Sulphide in acidic medium gives ppt of (B) $B{{i}_{2}}{{S}_{3}}$
\[2BiC{{l}_{3}}+3{{H}_{2}}S\xrightarrow{{}}B{{i}_{2}}{{S}_{3}}+6HCl\]

-(B) $B{{i}_{2}}{{S}_{3}}$ on heating with dil. HCl gives black compound (A) $BiC{{l}_{3}}$ and a gas (C) Hydrogen Sulphide which gives a black ppt. with lead acetate solution.
\[B{{i}_{2}}{{S}_{3}}\xrightarrow{HCl}BiC{{l}_{3}}+{{H}_{2}}S\] (C)
Chromyl chloride detects presence of chloride ions.

(B) $B{{i}_{2}}{{S}_{3}}$ dissolves in hot dil. $HN{{O}_{3}}$ to give solution (D) $Bi{{(N{{O}_{3}})}_{3}}$. (D) $Bi{{(N{{O}_{3}})}_{3}}$ gives a ring test due to nitrate ions.
\[B{{i}_{2}}{{S}_{3}}+8HN{{O}_{3}}\xrightarrow{\Delta }2Bi{{(N{{O}_{3}})}_{3}}+2NO+3S+4{{H}_{2}}O\]
When $N{{H}_{4}}OH$ solution is added to (D), a white precipitate (E ) $Bi{{(OH)}_{3}}$ is formed.
$Bi{{(N{{O}_{3}})}_{3}}+3N{{H}_{4}}OH\xrightarrow{{}}Bi{{(OH)}_{3}}+3N{{H}_{4}}N{{O}_{3}}$

(E) $Bi{{(OH)}_{3}}$ dissolves in minimum amount of dil.HCl to give a solution of (A).
\[Bi{{(OH)}_{3}}+3HCl\xrightarrow{{}}BiC{{l}_{3}}+3{{H}_{2}}O\]
Aqueous solution of (A) on addition of water gives a whitish turbidity (F) $BiOCl$.
\[BiC{{l}_{3}}+{{H}_{2}}O\xrightarrow{{}}BiOCl+2HCl\]

Aqueous solution of (A) $BiC{{l}_{3}}$ on warming with alkaline sodium stannite gives a black precipitate of (G) Bi and sodium stannate.
$BiC{{l}_{3}}+2NaOH+N{{a}_{2}}[Sn{{O}_{2}}]\xrightarrow{{}}Bi+NaSn{{O}_{3}}+{{H}_{2}}O+3NaCl$
The metal (G) Bi dissolves in hydrochloric acid to give solution of A.
$2Bi+6HCl\xrightarrow{\Delta }2BiC{{l}_{3}}+3{{H}_{2}}$
So, (F) $BiOCl$ (G) $Bi$
So, the correct answer is “Option B”.

Note: Ring test is also known as common nitrate test, when iron sulphate is added to solution having nitrate ion followed by slow addition of sulphuric acid, brown ring is formed at junction. Chromyl chloride test is used to confirm presence of chloride ions. Solution is heated with acidified potassium chromate which forms chromyl chloride with red fumes.