Practical Chemistry JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Practical Chemistry, free to read — no sign-in needed. The full chapter has 18 questions; sign in to attempt the remaining 13 in the exam simulator.

1. Q1JEE Advanced Adv 2025 (Paper 1)
   The correct match of the group reagents in List-I for precipitating the metal ion given in List-II from solutions, is LIST - I LIST - II (P) Passing $\mathrm{H}_2 \mathrm{~S}$ in the presence of $\mathrm{NH}_4 \mathrm{OH}$ (1) $\mathrm{Cu}^{2+}$ (Q) $\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3$ in the presence of $\mathrm{NH}_4 \mathrm{OH}$ (2) $\mathrm{Al}^{3+}$ (R) $\mathrm{NH}_4 \mathrm{OH}$ in the presence of $\mathrm{NH}_4 \mathrm{Cl}$ (3) $\mathrm{Mn}^{2+}$ (S) Passing $\mathrm{H}_2 \mathrm{~S}$ in the presence of dilute HCl (4) $\mathrm{Ba}^{2+}$ (5) $\mathrm{Mg}^{2+}$
   1. A.$\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 4 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 1$
   2. B.$\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 1$
   3. C.$\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 4 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 5$
   4. D.$\mathrm{P} \rightarrow 5 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 4$
   Show answer & solution

   Answer: (A)

   $\mathrm{Mn}^{+2} \xrightarrow{\mathrm{H}_2 \mathrm{~S}+\mathrm{NH}_4 \mathrm{OH}} \underset{\text { Pink/buff ppt. }}{\mathrm{MnS} \downarrow}$ $\mathrm{Ba}^{+2} \xrightarrow{\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3+\mathrm{NH}_4 \mathrm{OH}} \underset{\text { White ppt. }}{\mathrm{BaCO}_3 \downarrow}$ $\mathrm{Al}^{+3} \xrightarrow{\mathrm{NH}_4 \mathrm{Cl}+\mathrm{NH}_4 \mathrm{OH}} \underset{\text { White ppt. }}{\mathrm{Al}(\mathrm{OH})_3 \downarrow}$ $\mathrm{Cu}^{+2} \xrightarrow{\mathrm{H}_2 \mathrm{~S}+\mathrm{HCl} \text { (dil.) }} \underset{\text { Black ppt. }}{\mathrm{CuS} \downarrow}$
2. Q2JEE Advanced Adv 2021 (Paper 1)
   A mixture of two salts is used to prepare a solution $S,$ which gives the following results: $Whiteprecipitate\left(s\right)only←_{Roomtemperature}^{DiluteNaOH(aq)}S_{\left(aq.solutionofthesalts\right)}\rightarrow _{Roomtemperature}^{DiluteHCl\left(aq\right)}Whiteprecipitate\left(s\right)only$ The correct option $\left(s\right)$ for the salt mixture is(are)
   1. A.$Pb{\left({NO}_{3}\right)}_{2}$ and $Zn{\left({NO}_{3}\right)}_{2}$
   2. B.$Pb{\left({NO}_{3}\right)}_{2}$ and $Bi{\left({NO}_{3}\right)}_{3}$
   3. C.${AgNO}_{3}$ and $Bi{\left({NO}_{3}\right)}_{3}$
   4. D.$Pb{\left({NO}_{3}\right)}_{2}$ and $Hg{\left({NO}_{3}\right)}_{2}$
   Show answer & solution

   Answer: A,B

   $P{b}^{2+}\rightarrow ^{dil.HCl}{PbCl}_{2}$(white precipitate) $P{b}^{2+}\rightarrow ^{dil.NaOH}Pb{\left(OH\right)}_{2}$(white precipitate) ${Bi}^{3+}\rightarrow ^{dilHCl}{BiCl}_{3}$(water-soluble) ${Bi}^{3+}\rightarrow ^{dil.NaOH}Bi{\left(OH\right)}_{3}$(white precipitate) ${Hg}^{2+}\rightarrow ^{dil.HCl}{HgCl}_{2}$(water-soluble) ${Hg}^{2+}\rightarrow ^{dil.NaOH}Hg{\left(OH\right)}_{2}$(yellow precipitate) ${Ag}^{+}\rightarrow ^{dil.HCl}AgCl$(white precipitate) ${Ag}^{+}\rightarrow ^{dil.NaOH}AgOH$(brown precipitate) ${Zn}^{2+}\rightarrow ^{dil.HCl}{ZnCl}_{2}$(water-soluble) ${Zn}^{2+}\rightarrow ^{dil.NaOH}Zn{\left(OH\right)}_{2}$(white precipitate)
3. Q3JEE Advanced Adv 2020 (Paper 1)
   A colorless aqueous solution contains nitrates of two metals, $X$ and $Y$. When it was added to an aqueous solution of $NaCl$, a white precipitate was formed. This precipitate was found to be partly soluble in hot water to give a residue $P$ and a solution $Q$. The residue $P$ was soluble in aq. ${NH}_{3}$ and also in excess sodium thiosulfate. The hot solution $Q$ gave a yellow precipitate with $KI$. The metals $X$ and $Y$, respectively, are
   1. A.$Ag$ and $Pb$
   2. B.$Ag$ and $Cd$
   3. C.$Cd$ and $Pb$
   4. D.$Cd$ and $Zn$
   Show answer & solution

   Answer: (A)

   $X={AgNO}_{3}$, $Y=Pb{\left({NO}_{3}\right)}_{2}$ ${\text{AgNO}}_{3}{}_{\left(X\right)}+\text{NaCl}\rightarrow \text{AgCl}_{(\text{P})}↓+{\text{NaNO}}_{3}$ $AgCl (\text{s})↓_{(\text{P})}+2{\text{NH}}_{3}\rightarrow {\left[\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}\right]}^{+}_{Soluble}{\text{Cl}}^{-}$ $AgCl_{(P)}+2{Na}_{2}{S}_{2}{O}_{3}\rightarrow {Na}_{3}\left[Ag{\left({S}_{2}{O}_{3}\right)}_{2}\right]_{Solubel}+NaCl$ $Pb{\left({NO}_{3}\right)}_{2}{}_{\left(Y\right)}+2NaCl\rightarrow {PbCl}_{2}{}_{\begin{matrix}White ppt \left(Q\right) \\ (Soluble in hot \\ water)\end{matrix}}\text{↓}+2{NaNO}_{3}$ ${\text{PbCl}}_{2}{}_{\begin{matrix}\left(aq.\right) \\ \left(Q\right)\end{matrix}}+2\text{KI}\rightarrow {\text{PbI}}_{2}{}_{Yellow ppt}+2\text{KCl}$
4. Q4JEE Advanced Adv 2019 (Paper 1)
   The green colour produced in the borax bead test of a chromium(III) salt is due to -
   1. A.$Cr{\left(B{O}_{2}\right)}_{3}$
   2. B.$CrB$
   3. C.$C{r}_{2}{\left({B}_{4}{O}_{7}\right)}_{3}$
   4. D.$C{r}_{2}{O}_{3}$
   Show answer & solution

   Answer: (A)

   Borax bead test ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}{\text{.10H}}_{\text{2}}\text{O}\rightarrow ^{\Delta }{\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}} \rightarrow ^{\Delta }{\text{NaBO}}_{\text{2}} +{\text{B}}_{\text{2}}{\text{O}}_{\text{3}}$ $C{r}_{2}{O}_{3}+{B}_{2}{O}_{3}{\rightarrow ^{\Delta }}_{Greencoloured}^{Cr(B{O}_{2}{)}_{3}}$
5. Q5JEE Advanced Adv 2018 (Paper 2)
   The correct option(s) to distinguish nitrate salts of $M{n}^{2+}andC{u}^{2+}$ taken separately is(are)
   1. A.$M{n}^{2+}$ shows the characteristic green colour in the flame test
   2. B.Only $C{u}^{2+}$ shows the formation of precipitate by passing ${H}_{2}S$ in acidic medium
   3. C.Only $M{n}^{2+}$ shows the formation of precipitate by passing ${H}_{2}S$ in faintly basic medium
   4. D.$C{u}^{2+}/Cu$ has higher reduction potential than $M{n}^{2+}/Mn$ (measured under similar conditions)
   Show answer & solution

   Answer: B,D

   $(A)$ $C{u}^{+2}andM{n}^{+2}$ both gives green colour in flame test and cannot distinguished $(B)$ $C{u}^{+2}$ belongs to group-II of cationic radical will gives ppt. of CuS in acidic medium $(C)$ $C{u}^{+2}andM{n}^{+2}$ both form ppt. in basic medium $(D)$ $C{u}^{+2}/Cu=+0.34v\left(SRP\right)$ $M{n}^{+2}/Mn=-0.1.18v\left(SRP\right)$