Principles of Qualitative Analysis Previous Year Questions — JEE Main Chemistry
7 past-year Principles of Qualitative Analysis questions from JEE Main (Chemistry).
Q37 (2025)
When a salt is treated with sodium hydroxide solution it gives gas X. On passing gas X through reagent Y a brown coloured precipitate is formed. X and Y respectively, are:
- $\mathrm{X} = \mathrm{NH}_3\text{ and }\mathrm{Y} = \mathrm{HgO}$
- $\mathrm{X} = \mathrm{NH}_3\text{ and }\mathrm{Y} = \mathrm{K}_2\mathrm{HgI}_4 + \mathrm{KOH}$
- $\mathrm{X} = \mathrm{NH}_4\mathrm{Cl}\text{ and }\mathrm{Y} = \mathrm{KOH}$
- $\mathrm{X} = \mathrm{HCl}\text{ and }\mathrm{Y} = \mathrm{NH}_4\mathrm{Cl}$
### Related Formula
$$\mathrm{NH}_4^+ + \mathrm{OH}^- \rightarrow \mathrm{NH}_3\uparrow + \mathrm{H}_2\mathrm{O}$$
### Core Logic
1. Treating an ammonium salt (containing $\mathrm{NH}_4^+$) with $\mathrm{NaOH}$ releases Ammonia gas ($\mathrm{X} = \mathrm{NH}_3$):
$$\mathrm{NH}_4^+ + \mathrm{OH}^- \rightarrow \mathrm{NH}_3\uparrow + \mathrm{H}_2\mathrm{O}$$
2. Passing ammonia gas through Nessler's reagent ($\mathrm{Y} = \mathrm{K}_2[\mathrm{HgI}_4] + \mathrm{KOH}$) produces a distinct brown precipitate (known as iodide of Millon's base):
$$2[\mathrm{HgI}_4]^{2-} + \mathrm{NH}_3 + 3\mathrm{OH}^- \rightarrow \mathrm{HgO} \cdot \mathrm{Hg(NH_2)I} \downarrow (\text{brown}) + 7\mathrm{I}^- + 2\mathrm{H}_2\mathrm{O}$$
Hence, $\mathrm{X}$ is $\mathrm{NH}_3$ and $\mathrm{Y}$ is $\mathrm{K}_2\mathrm{HgI}_4 + \mathrm{KOH}$.
### Pattern Recognition
Classic confirmatory test for ammonium ion: heating with alkaline solutions yields $\mathrm{NH}_3$ gas, which always produces a brown precipitate with alkaline Nessler's reagent.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Qualitative Salt Analysis
Class 12 Chemistry: p-Block Elements
Q1480 (2024)
On passing a gas, 'X', through Nessler's reagent, a brown precipitate is obtained. The gas 'X' is
- $\mathrm{H}_2\mathrm{S}$
- $\mathrm{CO}_2$
- $\mathrm{NH}_3$
- $\mathrm{Cl}_2$
### Related Formula
$$2\mathrm{K}_2[\mathrm{HgI}_4] + \mathrm{NH}_3 + 3\mathrm{KOH} \rightarrow \mathrm{HgO}\cdot\mathrm{Hg}(\mathrm{NH}_2)\mathrm{I} \downarrow + 7\mathrm{KI} + 2\mathrm{H}_2\mathrm{O}$$
### Core Logic
Nessler's reagent ($\text{K}_2[\text{HgI}_4]$ in alkaline solution) reacts explicitly with ammonia gas ($\text{NH}_3$) or ammonium ions to create an insoluble iodide of Millon's base ($\text{HgO}\cdot\text{Hg}(\text{NH}_2)\text{I}$). This characteristic complex showcases a signature brown structural coloration.
### Step 1: Finalization
Thus, the unknown tracking gas 'X' is uniquely identified as ammonia ($\text{NH}_3$).
### Pattern Recognition
Ammonia detection relies directly upon alkaline potassium tetraiodomercurate(II) resulting in brown Millon's base precipitations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Principles of Qualitative Analysis
Q67 (2024)
A reagent which gives brilliant red precipitate with Nickel ions in basic medium is
- sodium nitroprusside
- neutral $\mathrm{FeCl}_3$
- meta-dinitrobenzene
- dimethyl glyoxime
### Related Formula
$$\mathrm{Ni}^{2+} + 2\mathrm{dmg}^- \rightarrow [\mathrm{Ni}(\mathrm{dmg})_2] \downarrow \quad \text{(Rosy Red Complex)}$$
### Core Logic
In basic medium (typically using ammonium hydroxide), nickel($II$) ions react cleanly with dimethylglyoxime ($\text{dmgH}$) to generate a stable, neutral coordination complex containing structural hydrogen bonding. This complex isolates out of solution as a highly signature brilliant rosy red precipitate.
### Step 1: Finalization
Therefore, dimethyl glyoxime is universally utilized for the quantitative and qualitative determination of $\text{Ni}^{2+}$ ions.
### Pattern Recognition
Nickel plus dmg in alkaline solution yields a signature rosy red chelate stabilized via symmetrical internal hydrogen bonds.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Principles of Qualitative Analysis
Q61 (2024)
Given below are two statements:
Statement-I: The gas liberated on warming a salt with dil $H_2SO_4$, turns a piece of paper dipped in lead acetate into black, it is a confirmatory test for sulphide ion.
Statement-II: In statement-I the colour of paper turns black because of formation of lead sulphite.
In the light of the above statements, choose the most appropriate answer from the options given below:
- $\text{Both Statement-I and Statement-II are false}$
- $\text{Statement-I is false but Statement-II is true}$
- $\text{Statement-I is true but Statement-II is false}$
- $\text{Both Statement-I and Statement-II are true}$
### Related Formula
$$Na_2S + H_2SO_4 \rightarrow Na_2SO_4 + H_2S$$
$$(CH_3COO)_2Pb + H_2S \rightarrow PbS + 2CH_3COOH$$
### Core Logic
When a salt containing sulphide is warmed with dilute $H_2SO_4$, $H_2S$ gas is evolved.
$H_2S$ gas turns lead acetate paper black due to the formation of lead sulphide ($PbS$), not lead sulphite.
Thus, Statement I is correct, but Statement II is incorrect.
### Pattern Recognition
Lead acetate paper test is standard for $H_2S$ gas, forming black $PbS$. Always watch out for the trap suffix: sulphite ($SO_3^{2-}$) vs sulphide ($S^{2-}$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Qualitative Analysis
Class 11 Chemistry: p-Block Elements
Q64 (2024)
The compound that is white in color is
- $\text{ammonium sulphide}$
- $\text{lead sulphate}$
- $\text{lead iodide}$
- $\text{ammonium arsinomolybdate}$
### Core Logic
Analyzing the colors of the given salts:
Lead sulphate ($PbSO_4$) is white.
Ammonium sulphide ($(NH_4)_2S$) is soluble/colorless in solution.
Lead iodide ($PbI_2$) is bright yellow.
Ammonium arsinomolybdate is yellow.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Salt Analysis
