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41.
How do we differentiate between $$F{e^{3 + }}\,{\text{and}}\,C{r^{3 + }}$$ in group $$III?$$
A
by taking excess of $$N{H_4}OH$$ solution
B
by increasing $$NH_4^ + $$ ion concentration
C
by decreasing $$O{H^ - }$$ ion concentration
D
both (B) and (C)
Answer :
by increasing $$NH_4^ + $$ ion concentration
When we add $$N{H_4}Cl,$$ it suppresses the ionisation of $$N{H_4}OH$$ and prevents the precipitation of higher group hydroxide in gp$$\left( {III} \right).$$
NOTE : Further ferric chloride and chromium chloride form different colour precipitates with $$N{H_4}OH.$$
$$FeC{l_3} + 3N{H_4}OH \to \mathop {Fe{{\left( {OH} \right)}_3} \downarrow }\limits_{{\text{reddish}}\,{\text{brown}}} + 3N{H_4}Cl$$
$$CrC{l_3} + 3N{H_4}OH \to \mathop {Cr{{\left( {OH} \right)}_3}}\limits_{{\text{Bluish}}\,{\text{green}}.} + 3N{H_4}Cl$$
42.
A metal nitrate reacts with $$KI$$ to give a black precipitate which on addition of excess of $$KI$$ is converted into orange colour solution. The cation of the metal nitrate is
A
$$H{g^{2 + }}$$
B
$$B{i^{3 + }}$$
C
$$P{b^{2 + }}$$
D
$$C{u^ + }$$
Answer :
$$B{i^{3 + }}$$
$$Bi{\left( {N{O_3}} \right)_3}\left( {aq} \right) + 3KI\left( {aq} \right) \to $$ $$\mathop {Bi{I_3}\left( s \right)}\limits_{{\text{Black}}} + 3KN{O_3}\left( {aq} \right)$$
$$\eqalign{
& Bi{I_3}\left( s \right) + KI\left( {aq} \right) \to \mathop {K\left[ {Bi{I_4}} \right]}\limits_{{\text{Orange}}} \cr
& {\text{The metal ion is }}B{i^{3 + }}. \cr} $$
43.
The green colour produced in the borax bead test of a chromium (III) salt is due to
44.
An aqueous solution of colourless metal sulphate $$M$$ gives a white precipitate with $$N{H_4}OH.$$ This was soluble in excess of $$N{H_4}OH.$$ On passing $${H_2}S$$ through this solution a white $$ppt.$$ is formed. The metal $$M$$ in the salt is
45.
Which one among the following pairs of ions cannot be separated by $${H_2}S$$ in dilute hydrochloric acid?
A
$$B{i^{3 + }},S{n^{4 + }}$$
B
$$A{l^{3 + }},H{g^{2 + }}$$
C
$$Z{n^{2 + }},C{u^{2 + }}$$
D
$$N{i^{2 + }},C{u^{2 + }}$$
Answer :
$$B{i^{3 + }},S{n^{4 + }}$$
NOTE : The ions of group $$II$$ of salt analysis are
precipitated by $$HCl\,{\text{and}}\,{H_2}S$$ whereas members of group $$IV$$ are precipitated by $${H_2}S$$ in alkaline medium.
$$\because B{i^{3 + }}\,{\text{and}}\,S{n^{4 + }}$$ both belong to group $$II$$
∴ They will be precipitated by $$HCl$$ in presence of $${H_2}S.$$
Both $$B{i^{3 + }}\,{\text{and}}\,S{n^{4 + }}$$ belong to group $$II$$ of qualitative inorganic analysis and will get precipitated by $${H_2}S.$$
46.
Which of the following gives a precipitate with $$Pb{\left( {N{O_3}} \right)_2}$$ but not with $$Ba{\left( {N{O_3}} \right)_2}?$$
47.
$$\left[ X \right] + {H_2}S{O_4} \to \left[ Y \right]$$ a colourless gas with irritating smell, $$\left[ Y \right] + {K_2}C{r_2}{O_7} + {H_2}S{O_4} \to $$ green solution. $$[X]$$ and $$[Y]$$ is :
A
$$SO_3^{2 - },S{O_2}$$
B
$$C{l^ - },HCl$$
C
$${S^{2 - }},{H_2}S$$
D
$$CO_3^{2 - },C{O_2}$$
Answer :
$$SO_3^{2 - },S{O_2}$$
$$S{O_2}$$ and $${H_2}S$$ both being reducing agents, can turn acidified dichromate solution green. $$S{O_2}$$ can be obtained by the action of acid upon sulphite while $${H_2}S$$ is evolved by the action of acid upon sulphide. However, $$S{O_2}$$ has a burning sulphur smell which is irritating. $${H_2}S$$ has rotten egg like smell.
48.
A solution which is $${10^{ - 3}}M$$ each in $$M{n^{2 + }},F{e^{2 + }},Z{n^{2 + }}$$ and $$H{g^{2 + }}$$ is treated with $${10^{ - 16}}M$$ sulphide ion. If $${K_{sp}}$$ of $$MnS,FeS,ZnS$$ and $$HgS$$ are $${10^{ - 15}},{10^{ - 23}},{10^{ - 20}}$$ and $${10^{ - 54}}$$ respectively, which one will precipitate first?
A
$$FeS$$
B
$$MgS$$
C
$$HgS$$
D
$$ZnS$$
Answer :
$$HgS$$
TIPS/Formulae : For precipitation,
Ionic product > solubility product
$$HgS$$ having the lowest $${K_{sp}}$$ among the given compounds will precipitate first.