Practical Chemistry JEE Main previous year questions with solutions

5 solved JEE Main questions on Practical Chemistry, free to read — no sign-in needed. The full chapter has 42 questions; sign in to attempt the remaining 37 in the exam simulator.

1. Q1JEE Main 2026 (04 Apr, Shift 2)Identification of Basic Radicals
   Among $Fe^{3+}$, $Pb^{2+}$, $Cu^{2+}$ and $Mn^{2+}$, identify the one that gets precipitated out while passing $H_2S$ in presence of $NH_4OH$ as group reagent. The highest possible oxidation state of the corresponding metal is
   1. A.$+3$
   2. B.$+4$
   3. C.$+2$
   4. D.$+7$
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   Answer: (D)

   The group reagent $H_2S$ in the presence of $NH_4OH$ is used for the precipitation of Group IV cations in qualitative analysis. Among the given ions, $Mn^{2+}$ belongs to Group IV and is precipitated as $MnS$. $Pb^{2+}$ and $Cu^{2+}$ belong to Group II and are precipitated by $H_2S$ in an acidic medium. $Fe^{3+}$ belongs to Group III and is precipitated as $Fe(OH)_3$ by $NH_4OH$ in the presence of $NH_4Cl$. The metal corresponding to the precipitated ion is Manganese ($Mn$). The electronic configuration of $Mn$ is $[Ar] 3d^5 4s^2$. By losing all its valence electrons, it can exhibit a maximum oxidation state of $+7$ (for example, in $KMnO_4$). Answer: $+7$
2. Q2JEE Main 2024 (01 Feb, Shift 1)Preliminary Tests
   The number of white coloured salts among the following is ....... $\left(A\right){SrSO}_{4}\left(B\right)Mg\left({NH}_{4}\right){PO}_{4}\left(C\right){BaCrO}_{4}$ $\left(D\right)Mn(OH{)}_{2}\left(E\right){PbSO}_{4}\left(F\right){PbCrO}_{4}$ $\left(G\right)AgBr\left(H\right){PbI}_{2}\left(I\right){CaC}_{2}{O}_{4}$ $\left(J\right)\left[Fe(OH{)}_{2}\left({CH}_{3}COO\right)\right]$
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   Answer: 5

   The colour of the salts are as follows, ${SrSO}_{4}$- white $Mg\left({NH}_{4}\right){PO}_{4}$ - white ${BaCrO}_{4}$ - yellow $Mn(OH{)}_{2}$ - white ${PbSO}_{4}$ - white ${PbCrO}_{4}$ - yellow $AgBr$ - pale yellow ${PbI}_{2}$ - yellow ${CaC}_{2}{O}_{4}$ - white $\left[Fe(OH{)}_{2}\left({CH}_{3}COO\right)\right]$ – Brown Red Hence the number of salts which are white in colour is $5$.
3. Q3JEE Main 2026 (24 Jan, Shift 2)Identification of Basic Radicals
   In the Group analysis of cations, $\mathrm{Ba}^{2+} ~\&~ \mathrm{Ca}^{2+}$ are precipitated respectively as
   1. A.sulphide & sulphide
   2. B.chromate & sulphide
   3. C.carbonate & carbonate
   4. D.hydroxide & carbonate
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   Answer: (C)

   In Group V of cation analysis, both Ba²⁺ and Ca²⁺ are precipitated by adding Na₂CO₃ or (NH₄)₂CO₃ solution. Ba²⁺ precipitates as BaCO₃ (white precipitate) and Ca²⁺ precipitates as CaCO₃ (white precipitate). Both form insoluble carbonates under these conditions, making the answer carbonate & carbonate.
4. Q4JEE Main 2026 (24 Jan, Shift 1)Identification of Basic Radicals
   Consider three metal chlorides $\mathrm{x}, \mathrm{y}$ and z, where x is water soluble at room temperature, y is sparingly soluble in water at room temperature and z is soluble in hot water. $\mathrm{x}, \mathrm{y}$ and z are respectively
   1. A.$\mathrm{CuCl}_{2}, \mathrm{AgCl}$ and $\mathrm{PbCl}_{2}$
   2. B.$\mathrm{AlCl}_{3}, \mathrm{PbCl}_{2}$ and $\mathrm{BaCl}_{2}$
   3. C.$\mathrm{MgCl}_{2}, \mathrm{AgCl}$ and $\mathrm{AlCl}_{3}$
   4. D.$\mathrm{AgCl}, \mathrm{Hg}_{2} \mathrm{Cl}_{2}$ and $\mathrm{PbCl}_{2}$
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   Answer: (A)

   Metal chloride solubility characteristics: x must be water soluble at room temperature, y must be sparingly soluble at room temperature, and z must be soluble in hot water but not in cold water. CuCl₂ is highly soluble in water (x). AgCl is notoriously insoluble in water and sparingly soluble at any temperature, classifying it as sparingly soluble (y). PbCl₂ exhibits unusual solubility behavior—it is sparingly soluble in cold water but its solubility increases with temperature, making it soluble in hot water (z). This combination perfectly matches all three criteria.
5. Q5JEE Main 2026 (21 Jan, Shift 2)Method of preparation for Aldehydes
   By usual analysis, 1.00 g of compound $(\mathrm{X})$ gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound $(X)$ is : (nearest integer) (Given, molar mass in $\mathrm{g} \mathrm{mol}^{-1}: \mathrm{O}=16, \mathrm{Mg}=24, \mathrm{P}=31$)
   1. A.30
   2. B.50
   3. C.40
   4. D.20
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   Answer: (B)

   Given: 1.00 g of compound (X) produces 1.79 g of Mg₂P₂O₇. Molar mass of Mg₂P₂O₇ = 2(24) + 2(31) + 7(16) = 222 g/mol. Moles of Mg₂P₂O₇ = 1.79 / 222 = 0.00806 mol. Since each mole of Mg₂P₂O₇ contains 2 moles of P, moles of P = 2 × 0.00806 = 0.01612 mol. Mass of P = 0.01612 × 31 = 0.4997 g ≈ 0.50 g. Percentage of phosphorus = (0.50 / 1.00) × 100 = 50%.