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Principles of Qualitative Analysis — JEE Previous Year Questions

Every Principles of Qualitative Analysis question asked in JEE Main and JEE Advanced across the last 186 papers — 62 questions, each with its correct answer. Free to read, no account needed.

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62

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58/186

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31%

All 62 Principles of Qualitative Analysis questions

Most recent papers first.

Q1·ChemistrySingle correctJEE Main 2026
A paper dipped in a dil. H₂SO₄ solution of 'X' upon treatment with SO₂ gas turns into green. The compound 'X' is :
  1. (A)KI-starch
  2. (B)KMnO₄
  3. (C)Pb(CH₃COO)₂
  4. (D)K₂Cr₂O₇

Correct answer: (D)

Step-by-step solution →
Q2·ChemistrySingle correctJEE Main 2026
Among Fe3+\mathrm{Fe^{3+}}Fe3+, Pb2+\mathrm{Pb^{2+}}Pb2+, Cu2+\mathrm{Cu^{2+}}Cu2+ and Mn2+\mathrm{Mn^{2+}}Mn2+, identify the one that gets precipitated out while passing H2S\mathrm{H_{2}S}H2​S in presence of NH4OH\mathrm{NH_{4}OH}NH4​OH 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

Correct answer: (D)

Step-by-step solution →
Q3·ChemistrySingle correctJEE Main 2026
A salt with few drops of conc. HCl gives apple green colour in flame test. The group precipitate of the salt is dissolved in acetic acid and treated with K2CrO4\mathrm{K_{2}CrO_{4}}K2​CrO4​ to give yellow precipitate. When the sodium carbonate extract of the salt solution is heated with conc. HNO3\mathrm{HNO_{3}}HNO3​ and ammonium molybdate, it resulted a canary yellow precipitate. The cation and anion present in the salt are respectively,
  1. (A)Ca2+\mathrm{Ca}^{2+}Ca2+ and SO42−\mathrm{SO_{4}^{2-}}SO42−​
  2. (B)Ba2+\mathrm{Ba}^{2+}Ba2+ and PO43−\mathrm{PO_{4}^{3-}}PO43−​
  3. (C)Mn2+\mathrm{Mn}^{2+}Mn2+ and PO43−\mathrm{PO_{4}^{3-}}PO43−​
  4. (D)Ba2+\mathrm{Ba}^{2+}Ba2+ and SO42−\mathrm{SO_{4}^{2-}}SO42−​

Correct answer: (B)

Step-by-step solution →
Q4·ChemistrySingle correctJEE Main 2026
Given below are two statements: Statement I: Griss–Ilosvay test is used for the detection of nitrite ion, which involves the use of sulphanilic acid and α–naphthylamine reagent. Statement II: In the above test, sulphanilic acid is diazotized by the acidified nitrite ion, which on further coupling with α–naphthylamine forms an azo-dye. In the light of the above statements, choose the correct answer from the options given below
  1. (A)Statement I is false but Statement II is true
  2. (B)Both Statement I and Statement II are true
  3. (C)Statement I is true but Statement II is false
  4. (D)Both Statement I and Statement II are false

Correct answer: (B)

Step-by-step solution →
Q5·ChemistrySingle correctJEE Main 2026
Consider three metal chlorides x, 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. x, y and z are respectively
  1. (A)MgCl2MgCl_2MgCl2​, AgCl and AlCl3AlCl_3AlCl3​
  2. (B)AgCl, Hg2Cl2Hg_2Cl_2Hg2​Cl2​ and PbCl2PbCl_2PbCl2​
  3. (C)AlCl3AlCl_3AlCl3​, PbCl2PbCl_2PbCl2​ and BaCl2BaCl_2BaCl2​
  4. (D)CuCl2CuCl_2CuCl2​, AgCl and PbCl2PbCl_2PbCl2​

Correct answer: (D)

Step-by-step solution →
Q6·ChemistrySingle correctJEE Main 2026
In the Group analysis of cations , Ba2+\text{Ba}^{2+}Ba2+ & Ca2+\text{Ca}^{2+}Ca2+ are precipitated respectively as
  1. (A)sulphide & sulphide
  2. (B)hydroxide & carbonate
  3. (C)carbonate & carbonate
  4. (D)chromate & sulphide

Correct answer: (C)

Step-by-step solution →
Q7·ChemistrySingle correctJEE Advanced 2025
The correct match of the group reagents in List-I for precipitating the metal ion given in List-II from solutions, is
List-IList-II
P.Passing H2SH_2SH2​S in the presence of NH4OHNH_4OHNH4​OH1.Cu2+Cu^{2+}Cu2+
Q.(NH4)2CO3(NH_4)_2CO_3(NH4​)2​CO3​ in the presence of NH4OHNH_4OHNH4​OH2.Al3+Al^{3+}Al3+
R.NH4OHNH_4OHNH4​OH in the presence of NH4ClNH_4ClNH4​Cl3.Mn2+Mn^{2+}Mn2+
S.Passing H2SH_2SH2​S in the presence of dilute HCl4.Ba2+Ba^{2+}Ba2+
5.Mg2+Mg^{2+}Mg2+
  1. (A)P → 3; Q → 4; R → 2: S → 1
  2. (B)P → 4; Q → 2; R → 3: S → 1
  3. (C)P → 3; Q → 4; R → 1: S → 5
  4. (D)P → 5; Q → 3; R → 2: S → 4

Correct answer: (A)

Step-by-step solution →
Q8·ChemistrySingle correctJEE Advanced 2025
During sodium nitroprusside test of sulphide ion in an aqueous solution, one of the ligands coordinated to the metal ion is converted to
  1. (A)NOS−^{-}−
  2. (B)SCN−^{-}−
  3. (C)SNO−^{-}−
  4. (D)NCS−^{-}−

Correct answer: (A)

Step-by-step solution →
Q9·ChemistrySingle correctJEE Main 2025
When a salt is treated with sodium hydroxide solution it gives gas X. On passing the gas X through reagent Y a brown coloured precipitate is formed. X and Y respectively, are:
  1. (A)X = NH3_33​ and Y = HgO
  2. (B)X = NH3_33​ and Y = K2_22​HgI4_44​ + KOH
  3. (C)X = NH4_44​Cl and Y = KOH
  4. (D)X = HCl and Y = NH4_44​Cl

Correct answer: (B)

Step-by-step solution →
Q10·ChemistrySingle correctJEE Main 2025
Formation of Na4[Fe(CN)5NOS]Na_4[Fe(CN)_5NOS]Na4​[Fe(CN)5​NOS], a purple coloured complex formed by addition of sodium nitroprusside in sodium carbonate extract of salt, indicates the presence of:
  1. (A)Sodium ion
  2. (B)Sulphate ion
  3. (C)Sulphide ion
  4. (D)Sulphite ion

Correct answer: (C)

Step-by-step solution →
Q11·ChemistrySingle correctJEE Main 2025
Choose the correct tests with their respective observations. (A) CuSO4CuSO_4CuSO4​ (acidified with acetic acid) + K4[Fe(CN)6]→+\,K_4[Fe(CN)_6]\to+K4​[Fe(CN)6​]→ Chocolate brown precipitate. (B) FeCl3+K4[Fe(CN)6]→FeCl_3+K_4[Fe(CN)_6]\toFeCl3​+K4​[Fe(CN)6​]→ Prussian blue precipitate. (C) ZnCl2+K4[Fe(CN)6]ZnCl_2+K_4[Fe(CN)_6]ZnCl2​+K4​[Fe(CN)6​], neutralised with NH4OH→NH_4OH\toNH4​OH→ White or bluish white precipitate. (D) MgCl2+K4[Fe(CN)6]→MgCl_2+K_4[Fe(CN)_6]\toMgCl2​+K4​[Fe(CN)6​]→ Blue precipitate. (E) BaCl2+K4[Fe(CN)6]BaCl_2+K_4[Fe(CN)_6]BaCl2​+K4​[Fe(CN)6​], neutralised with NaOH→NaOH\toNaOH→ White precipitate. Choose the correct answer from the options given below:
  1. (A)A, D and E only
  2. (B)B, D and E only
  3. (C)A, B and C only
  4. (D)C, D and E only

Correct answer: (C)

Step-by-step solution →
Q12·ChemistrySingle correctJEE Main 2025
Find the compound "A" from the following reaction sequences. A→aqua-regiaB→(1) KNO3/NH4OH, (2) AcOHA\xrightarrow{\text{aqua-regia}}B\xrightarrow{\text{(1) }KNO_3/NH_4OH,\ \text{(2) AcOH}}Aaqua-regia​B(1) KNO3​/NH4​OH, (2) AcOH​ yellow ppt (K3[Co(NO2)6]↓K_3[Co(NO_2)_6]\downarrowK3​[Co(NO2​)6​]↓ yellow).
  1. (A)ZnS
  2. (B)CoS
  3. (C)MnS
  4. (D)NiS

Correct answer: (B)

Step-by-step solution →
Q13·ChemistryIntegerJEE Main 2025
Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with K4[Fe(CN)6]K_4[Fe(CN)_6]K4​[Fe(CN)6​] is : Cu2+, Fe3+, Ba2+, Ca2+, NH4+, Mg2+, Zn2+Cu^{2+},\ Fe^{3+},\ Ba^{2+},\ Ca^{2+},\ NH_4^+,\ Mg^{2+},\ Zn^{2+}Cu2+, Fe3+, Ba2+, Ca2+, NH4+​, Mg2+, Zn2+

Correct answer: 3

Step-by-step solution →
Q14·ChemistrySingle correctJEE Main 2025
Identify A, B and C in the given below reaction sequence: A→HNO3Pb(NO3)2→H2SO4B→(1) ammonium acetate, (2) acetic acid, (3) K2CrO4CA\xrightarrow{HNO_3}Pb(NO_3)_2\xrightarrow{H_2SO_4}B\xrightarrow{\text{(1) ammonium acetate, (2) acetic acid, (3) }K_2CrO_4}CAHNO3​​Pb(NO3​)2​H2​SO4​​B(1) ammonium acetate, (2) acetic acid, (3) K2​CrO4​​C (Yellow ppt).
  1. (A)PbCl2, PbSO4, PbCrO4PbCl_2,\ PbSO_4,\ PbCrO_4PbCl2​, PbSO4​, PbCrO4​
  2. (B)PbS, PbSO4, PbCrO4PbS,\ PbSO_4,\ PbCrO_4PbS, PbSO4​, PbCrO4​
  3. (C)PbS, PbSO4, Pb(CH3COO)2PbS,\ PbSO_4,\ Pb(CH_3COO)_2PbS, PbSO4​, Pb(CH3​COO)2​
  4. (D)PbCl2, Pb(SO4)2, PbCrO4PbCl_2,\ Pb(SO_4)_2,\ PbCrO_4PbCl2​, Pb(SO4​)2​, PbCrO4​

Correct answer: (B)

Step-by-step solution →
Q15·ChemistryNumericalJEE Main 2024
Consider the following test for a group-IV cation: M2++H2S→AM^{2+}+H_2S\rightarrow AM2++H2​S→A (Black precipitate) + byproduct; A+A+A+ aqua regia →B+NOCl+S+H2O\rightarrow B+NOCl+S+H_2O→B+NOCl+S+H2​O; B+KNO2+CH3COOH→CB+KNO_2+CH_3COOH\rightarrow CB+KNO2​+CH3​COOH→C + byproduct. The spin-only magnetic moment value of the metal complex C is _______ BM (nearest integer).

Correct answer: 0

Step-by-step solution →
Q16·ChemistrySingle correctJEE Main 2024
In qualitative test for identification of presence of phosphorous, the compound is heated with an oxidising agent, which is further treated with nitric acid and ammonium molybdate respectively. The yellow coloured precipitate obtained is:
  1. (A)Na3PO4⋅12MoO3\text{Na}_3\text{PO}_4\cdot 12\text{MoO}_3Na3​PO4​⋅12MoO3​
  2. (B)(NH4)3PO4⋅12(NH4)2MoO4(\text{NH}_4)_3\text{PO}_4\cdot 12(\text{NH}_4)_2\text{MoO}_4(NH4​)3​PO4​⋅12(NH4​)2​MoO4​
  3. (C)(NH4)3PO4⋅12MoO3(\text{NH}_4)_3\text{PO}_4\cdot 12\text{MoO}_3(NH4​)3​PO4​⋅12MoO3​
  4. (D)MoPO4⋅21NH4NO3\text{MoPO}_4\cdot 21\text{NH}_4\text{NO}_3MoPO4​⋅21NH4​NO3​

Correct answer: (C)

Step-by-step solution →
Q17·ChemistrySingle correctJEE Main 2024
Match List-I (dry tests in qualitative analysis) with List-II (reaction sequence involved, M is metal). Choose the correct answer from the options given below:
List-I (Name of the test)List-II (Reaction sequence)
A.Borax bead testI.MCO3→MO→CO/ΔMCO_3\to MO\xrightarrow{CO/\Delta}MCO3​→MOCO/Δ​ coloured residue with CoO
B.Charcoal cavity testII.MCO3→MCl2→M2+MCO_3\to MCl_2\to M^{2+}MCO3​→MCl2​→M2+ (volatile chloride coloured flame)
C.Cobalt nitrate testIII.MSO4→Na2CO3/ΔM(BO2)2MSO_4\xrightarrow{Na_2CO_3/\Delta}M(BO_2)_2MSO4​Na2​CO3​/Δ​M(BO2​)2​ (coloured borate bead)
D.Flame testIV.MSO4→Na2CO3/ΔMCO3→MO→MMSO_4\xrightarrow{Na_2CO_3/\Delta}MCO_3\to MO\to MMSO4​Na2​CO3​/Δ​MCO3​→MO→M (reduction to metal in cavity)
  1. (A)A-III, B-I, C-IV, D-II
  2. (B)A-III, B-II, C-I, D-IV
  3. (C)A-III, B-I, C-II, D-IV
  4. (D)A-III, B-IV, C-I, D-II

Correct answer: (D)

Step-by-step solution →
Q18·ChemistrySingle correctJEE Main 2024
During the detection of acidic radical present in a salt, a student gets a pale yellow precipitate soluble with difficulty in NH4OHNH_4OHNH4​OH solution when sodium carbonate extract was first acidified with dil. HNO3HNO_3HNO3​ and then AgNO3AgNO_3AgNO3​ solution was added. This indicates presence of:
  1. (A)Br−Br^-Br−
  2. (B)CO32−CO_3^{2-}CO32−​
  3. (C)I−I^-I−
  4. (D)Cl−Cl^-Cl−

Correct answer: (A)

Step-by-step solution →
Q19·ChemistrySingle correctJEE Main 2024
Match List-I (Precipitating reagent and conditions) with List-II (Cation). Choose the correct answer from the options given below:
List-I (Precipitating reagent and conditions)List-II (Cation)
A.NH4Cl+NH4OHNH_4Cl + NH_4OHNH4​Cl+NH4​OHI.Mn2+Mn^{2+}Mn2+
B.NH4OH+Na2CO3NH_4OH + Na_2CO_3NH4​OH+Na2​CO3​II.Pb2+Pb^{2+}Pb2+
C.NH4OH+NH4Cl+H2SNH_4OH + NH_4Cl + H_2SNH4​OH+NH4​Cl+H2​S gasIII.Al3+Al^{3+}Al3+
D.dilute HClIV.Sr2+Sr^{2+}Sr2+
  1. (A)A-IV, B-III, C-II, D-I
  2. (B)A-IV, B-III, C-I, D-II
  3. (C)A-III, B-IV, C-I, D-II
  4. (D)A-III, B-IV, C-II, D-I

Correct answer: (C)

Step-by-step solution →
Q20·ChemistrySingle correctJEE Main 2024
In the precipitation of the iron group (III) in qualitative analysis, ammonium chloride is added before adding ammonium hydroxide to:
  1. (A)prevent interference by phosphate ions
  2. (B)decrease concentration of −OH^-OH−OH ions
  3. (C)increase concentration of Cl−Cl^-Cl− ions
  4. (D)increase concentration of NH4+NH_4^+NH4+​ ions

Correct answer: (B)

Step-by-step solution →
Q21·ChemistryNumericalJEE Main 2024
The number of coloured salts among the following is ___. (A) SrSO4SrSO_4SrSO4​ (B) Mg(NH4)PO4Mg(NH_4)PO_4Mg(NH4​)PO4​ (C) BaCrO4BaCrO_4BaCrO4​ (D) Mn(OH)2Mn(OH)_2Mn(OH)2​ (E) PbSO4PbSO_4PbSO4​ (F) PbCrO4PbCrO_4PbCrO4​ (G) AgBrAgBrAgBr (H) PbI2PbI_2PbI2​ (I) CaC2O4CaC_2O_4CaC2​O4​ (J) [Fe(OH)2(CH3COO)][Fe(OH)_2(CH_3COO)][Fe(OH)2​(CH3​COO)]

Correct answer: 5

Step-by-step solution →
Q22·ChemistrySingle correctJEE Main 2024
The compound that is white in color is
  1. (A)ammonium sulphide
  2. (B)lead sulphate
  3. (C)lead iodide
  4. (D)ammonium arsinomolybdate

Correct answer: (B)

Step-by-step solution →
Q23·ChemistrySingle correctJEE Main 2024
Given below are two statements: Statement-I: The gas liberated on warming a salt with dil H2SO4H_2SO_4H2​SO4​, 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 sulphide. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both Statement-I and Statement-II are false
  2. (B)Statement-I is false but Statement-II is true
  3. (C)Statement-I is true but Statement-II is false
  4. (D)Both Statement-I and Statement-II are true.

Correct answer: (D)

Step-by-step solution →
Q24·ChemistrySingle correctJEE Main 2023
In the wet tests for detection of various cations by precipitation, Ba2+Ba^{2+}Ba2+ cations are detected by obtaining precipitate of
  1. (A)Ba(ox)Ba(ox)Ba(ox): Barium oxalate
  2. (B)BaCO3BaCO_3BaCO3​
  3. (C)Ba(OAc)2Ba(OAc)_2Ba(OAc)2​
  4. (D)BaSO4BaSO_4BaSO4​

Correct answer: (B)

Step-by-step solution →
Q25·ChemistrySingle correctJEE Main 2023
When a solution of mixture having two inorganic salts was treated with freshly prepared ferrous sulphate in acidic medium, a dark brown ring was formed whereas on treatment with neutral FeCl3FeCl_3FeCl3​, it gave deep red colour which disappeared on boiling and a brown red ppt was formed. The mixture contains
  1. (A)CH3COO−CH_3COO^-CH3​COO− & NO3−NO_3^-NO3−​
  2. (B)C2O42−C_2O_4^{2-}C2​O42−​ & NO3−NO_3^-NO3−​
  3. (C)SO32−SO_3^{2-}SO32−​ & CH3COO−CH_3COO^-CH3​COO−
  4. (D)SO32−SO_3^{2-}SO32−​ & C2O42−C_2O_4^{2-}C2​O42−​

Correct answer: (A)

Step-by-step solution →
Q26·ChemistrySingle correctJEE Main 2023
Element not present in Nessler's reagent is:
  1. (A)Hg
  2. (B)I
  3. (C)K
  4. (D)N

Correct answer: (D)

Step-by-step solution →
Q27·ChemistrySingle correctJEE Main 2023
Formation of which complex, among the following, is not a confirmatory test of Pb2+Pb^{2+}Pb2+ ions
  1. (A)lead chromate
  2. (B)lead iodide
  3. (C)lead nitrate
  4. (D)lead sulphate

Correct answer: (C)

Step-by-step solution →
Q28·ChemistrySingle correctJEE Main 2023
Which element is not present in Nessler's reagent?
  1. (A)Oxygen
  2. (B)Potassium
  3. (C)Mercury
  4. (D)Iodine

Correct answer: (A)

Step-by-step solution →
Q29·ChemistrySingle correctJEE Main 2023
In the wet tests for identification of various cations by precipitation, which transition element cation doesn't belong to group IV in qualitative inorganic analysis?
  1. (A)Ni2+^{2+}2+
  2. (B)Zn2+^{2+}2+
  3. (C)Co2+^{2+}2+
  4. (D)Fe3+^{3+}3+

Correct answer: (D)

Step-by-step solution →
Q30·ChemistrySingle correctJEE Main 2023
During the qualitative analysis of SO32−SO_3^{2-}SO32−​ using dilute H2SO4H_2SO_4H2​SO4​, SO2SO_2SO2​ gas is evolved which turns K2Cr2O7K_2Cr_2O_7K2​Cr2​O7​ solution (acidified with dilute H2SO4H_2SO_4H2​SO4​):
  1. (A)green
  2. (B)blue
  3. (C)red
  4. (D)black

Correct answer: (A)

Step-by-step solution →
Q31·ChemistrySingle correctJEE Main 2023
The formula for Nessler's reagent is:
  1. (A)HgI2HgI_2HgI2​
  2. (B)K2HgI4K_2HgI_4K2​HgI4​
  3. (C)KHgI3KHgI_3KHgI3​
  4. (D)KHg2I2KHg_2I_2KHg2​I2​

Correct answer: (B)

Step-by-step solution →
Q32·ChemistrySingle correctJEE Main 2023
During the borax bead test with CuSO4_44​, a blue green colour of the bead was observed in oxidising flame due to the formation of:
  1. (A)CuO
  2. (B)Cu(BO2_22​)2_22​
  3. (C)Cu3_33​B2_22​
  4. (D)Cu

Correct answer: (B)

Step-by-step solution →
Q33·ChemistrySingle correctJEE Main 2023
A chloride salt solution acidified with dil. HNO3HNO_3HNO3​ gives a curdy white precipitate, [A], on addition of AgNO3AgNO_3AgNO3​. [A] on treatment with NH4OHNH_4OHNH4​OH gives a clear solution, B. A and B are respectively
  1. (A)AgClAgClAgCl & (NH4)[Ag(OH)2](NH_4)[Ag(OH)_2](NH4​)[Ag(OH)2​]
  2. (B)AgClAgClAgCl & [Ag(NH3)2]Cl[Ag(NH_3)_2]Cl[Ag(NH3​)2​]Cl
  3. (C)H[AgCl3]H[AgCl_3]H[AgCl3​] & (NH4)[Ag(OH)2](NH_4)[Ag(OH)_2](NH4​)[Ag(OH)2​]
  4. (D)H[AgCl3]H[AgCl_3]H[AgCl3​] & [Ag(NH3)2]Cl[Ag(NH_3)_2]Cl[Ag(NH3​)2​]Cl

Correct answer: (B)

Step-by-step solution →
Q34·ChemistrySingle correctJEE Main 2023
Match List-I (Cations) with List-II (Group reagents). Choose the correct match from the options given below:
List-I (Cations)List-II (Group reagents)
A.Pb2+,Cu2+Pb^{2+}, Cu^{2+}Pb2+,Cu2+i.H2SH_2SH2​S gas in presence of dilute HCl
B.Al3+,Fe3+Al^{3+}, Fe^{3+}Al3+,Fe3+ii.(NH4)2CO3(NH_4)_2CO_3(NH4​)2​CO3​ in presence of NH4OHNH_4OHNH4​OH
C.Co2+,Ni2+Co^{2+}, Ni^{2+}Co2+,Ni2+iii.NH4OHNH_4OHNH4​OH in presence of NH4ClNH_4ClNH4​Cl
D.Ba2+,Ca2+Ba^{2+}, Ca^{2+}Ba2+,Ca2+iv.H2SH_2SH2​S in presence of NH4OHNH_4OHNH4​OH
  1. (A)A-iii, B-i, C-iv, D-ii
  2. (B)A-i, B-iii, C-ii, D-iv
  3. (C)A-iv, B-i, C-iii, D-i
  4. (D)A-i, B-iii, C-iv, D-ii

Correct answer: (D)

Step-by-step solution →
Q35·ChemistrySingle correctJEE Main 2022
Fe3+^{3+}3+ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:
  1. (A)[Fe(H2_{2}2​O)6_{6}6​]2_{2}2​ [Fe(CN)6_{6}6​]
  2. (B)Fe2_{2}2​[Fe(CN)6_{6}6​]2_{2}2​
  3. (C)Fe3_{3}3​[Fe(OH)2_{2}2​(CN)4_{4}4​]2_{2}2​
  4. (D)Fe4_{4}4​[Fe(CN)6_{6}6​]3_{3}3​

Correct answer: (D)

Step-by-step solution →
Q36·ChemistrySingle correctJEE Main 2022
A white precipitate was formed when BaCl2_22​ was added to water extract of an inorganic salt. Further, a gas 'X' with characteristic odour was released when the formed white precipitate was dissolved in dilute HCl. The anion present in the inorganic salt is :
  1. (A)I−^-−
  2. (B)SO32−_3^{2-}32−​
  3. (C)S2−^{2-}2−
  4. (D)NO2−_2^-2−​

Correct answer: (B)

Step-by-step solution →
Q37·ChemistrySingle correctJEE Main 2022
Match List I with List II List I (Anion) A. CO32−_3^{2-}32−​ B. S2−^{2-}2− C. SO32−_3^{2-}32−​ D. NO2−_2^-2−​ List II (Gas evolved on reaction with dil. H2_22​SO4_44​) I. Colourless gas which turns lead acetate paper black II. Colourless gas which turns acidified potassium dichromate solution green. III. Brown fumes which turns acidified KI solution containing starch blue. IV. Colourless gas evolved with brisk effervescence, which turns lime water milky. Choose the correct answer from the options given below:
  1. (A)A-III, B-I, C-II, D-IV
  2. (B)A-II, B-I, C-IV, D-III
  3. (C)A-IV, B-I, C-III, D-II
  4. (D)A-IV, B-I, C-II, D-III

Correct answer: (D)

Step-by-step solution →
Q38·ChemistrySingle correctJEE Main 2022
Which statement is not true with respect to nitrate ion test ?
  1. (A)A dark brown ring is formed at the junction of two solutions.
  2. (B)Ring is formed due to nitroferrous sulphate complex.
  3. (C)The brown complex is [Fe(H2O)5(NO)]SO4[Fe(H_{2}O)_{5} (NO)]SO_{4}[Fe(H2​O)5​(NO)]SO4​.
  4. (D)Heating the nitrate salt with conc. H2SO4H_{2}SO_{4}H2​SO4​, light brown fumes are evolved.

Correct answer: (B)

Step-by-step solution →
Q39·ChemistrySingle correctJEE Main 2022
During the qualitative analysis of salt with cation y2+^{2+}2+ , addition of a reagent (X) to alkaline solution of the salt gives a bright red precipitate. The reagent (X) and the cation (y2+^{2+}2+) present respectively are:
  1. (A)Dimethylglyoxime and Ni2+^{2+}2+
  2. (B)Dimethylglyoxime and Co2+^{2+}2+
  3. (C)Nessler's reagent and Hg2+^{2+}2+
  4. (D)Nessler's reagent and Ni2+^{2+}2+

Correct answer: (A)

Step-by-step solution →
Q40·ChemistrySingle correctJEE Main 2022
In the flame test of a mixture of salts, a green flame with blue centre was observed. Which one of the following cations may be present?
  1. (A)Cu2+^{2+}2+
  2. (B)Sr2+^{2+}2+
  3. (C)Ba2+^{2+}2+
  4. (D)Ca2+^{2+}2+

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correctJEE Advanced 2021
The reaction of K3_{3}3​[Fe(CN)6_{6}6​] with freshly prepared FeSO4_{4}4​ solution produces a dark blue precipitate called Turnbull's blue. Reaction of K4_{4}4​[Fe(CN)6_{6}6​] with the FeSO4_{4}4​ solution in complete absence of air produces a white precipitate X, which turns blue in air. Mixing the FeSO4_{4}4​ solution with NaNO3_{3}3​, followed by a slow addition of concentrated H2_{2}2​SO4_{4}4​ through the side of the test tube produces a brown ring. Among the following, the brown ring is due to the formation of
  1. (A)[Fe(NO)2_{2}2​(SO4_{4}4​)2_{2}2​]2−^{2-}2−
  2. (B)[Fe(NO)2_{2}2​(H2_{2}2​O)4_{4}4​]3+^{3+}3+
  3. (C)[Fe(NO)4_{4}4​(SO4_{4}4​)2_{2}2​]
  4. (D)[Fe(NO)(H2_{2}2​O)5_{5}5​]2+^{2+}2+

Correct answer: (D)

Step-by-step solution →
Q42·ChemistryMultiple correctJEE Advanced 2021
A mixture of two salts is used to prepare a solution S\mathbf{S}S, which gives the following results : White precipitate(s) only ←Room temperatureDilute NaOH(aq.)\xleftarrow[\text{Room temperature}]{\text{Dilute NaOH(aq.)}}Dilute NaOH(aq.)Room temperature​ S\mathbf{S}S (aq. solution of the salts) →Room temperatureDilute HCl(aq.)\xrightarrow[\text{Room temperature}]{\text{Dilute HCl(aq.)}}Dilute HCl(aq.)Room temperature​ White precipitate(s) only The correct option(s) for the salt mixture is(are)
  1. (A)Pb(NO3)2\mathrm{Pb(NO_3)_2}Pb(NO3​)2​ and Zn(NO3)2\mathrm{Zn(NO_3)_2}Zn(NO3​)2​
  2. (B)Pb(NO3)2\mathrm{Pb(NO_3)_2}Pb(NO3​)2​ and Bi(NO3)3\mathrm{Bi(NO_3)_3}Bi(NO3​)3​
  3. (C)AgNO3\mathrm{AgNO_3}AgNO3​ and Bi(NO3)3\mathrm{Bi(NO_3)_3}Bi(NO3​)3​
  4. (D)Pb(NO3)2\mathrm{Pb(NO_3)_2}Pb(NO3​)2​ and Hg(NO3)2\mathrm{Hg(NO_3)_2}Hg(NO3​)2​

Correct answer: (A), (B)

Step-by-step solution →
Q43·ChemistrySingle correctJEE Advanced 2021
The reaction of K3_{3}3​[Fe(CN)6_{6}6​] with freshly prepared FeSO4_{4}4​ solution produces a dark blue precipitate called Turnbull's blue. Reaction of K4_{4}4​[Fe(CN)6_{6}6​] with the FeSO4_{4}4​ solution in complete absence of air produces a white precipitate X, which turns blue in air. Mixing the FeSO4_{4}4​ solution with NaNO3_{3}3​, followed by a slow addition of concentrated H2_{2}2​SO4_{4}4​ through the side of the test tube produces a brown ring. Precipitate X\mathbf{X}X is
  1. (A)Fe4_{4}4​[Fe(CN)6_{6}6​]3_{3}3​
  2. (B)Fe[Fe(CN)6_{6}6​]
  3. (C)K2_{2}2​Fe[Fe(CN)6_{6}6​]
  4. (D)KFe[Fe(CN)6_{6}6​]

Correct answer: (C)

Step-by-step solution →
Q44·ChemistrySingle correctJEE Main 2021
The potassium ferrocyanide solution gives a Prussian blue colour, when added to :
  1. (A)CoCl3_{3}3​
  2. (B)FeCl2_{2}2​
  3. (C)CoCl2_{2}2​
  4. (D)FeCl3_{3}3​

Correct answer: (D)

Step-by-step solution →
Q45·ChemistrySingle correctJEE Main 2021
Match List-I with List-II : List-I (Metal Ion) (a) Mn2+\mathrm{Mn^{2+}}Mn2+ (b) As3+\mathrm{As^{3+}}As3+ (c) Cu2+\mathrm{Cu^{2+}}Cu2+ (d) Al3+\mathrm{Al^{3+}}Al3+ List-II (Group in Qualitative analysis) (i) Group - III (ii) Group - IIA (iii) Group - IV (iv) Group - IIB Choose the most appropriate answer from the options given below :
  1. (A)(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
  2. (B)(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
  3. (C)(a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)
  4. (D)(a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)

Correct answer: (B)

Step-by-step solution →
Q46·ChemistryNumericalJEE Main 2021
Consider the sulphides HgS, PbS, CuS, Sb2S3\mathrm{Sb_2S_3}Sb2​S3​, As2S3\mathrm{As_2S_3}As2​S3​ and CdS. Number of these sulphides soluble in 50% HNO3\mathrm{HNO_3}HNO3​ is ____.

Correct answer: 4

Step-by-step solution →
Q47·ChemistrySingle correctJEE Main 2021
To an aqueous solution containing ions such as Al3+^{3+}3+,Zn2+^{2+}2+,Ca2+^{2+}2+,Fe3+^{3+}3+,Ni2+^{2+}2+,Ba2+^{2+}2+ and Cu2+^{2+}2+ was added conc. HCl, followed by H2_22​S . The total number of cations precipitated during this reaction is / are:
  1. (A)4
  2. (B)1
  3. (C)3
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q48·ChemistrySingle correctJEE Main 2021
An inorganic compound ‘X’ on treatment with concentrated H2SO4H_{2}SO_{4}H2​SO4​ produces brown fumes and gives dark brown ring with FeSO4FeSO_{4}FeSO4​ in presence of concentrated H2SO4H_{2}SO_{4}H2​SO4​. Also compound ‘X’ gives precipitate ‘Y’, when its solution in dilute HCl is treated with H2SH_{2}SH2​S gas. The precipitate ‘Y’ on treatment with concentrated HNO3HNO_{3}HNO3​ followed by excess of NH4OHNH_{4}OHNH4​OH further gives deep blue coloured solution, Compound ‘X’ is:
  1. (A)Cu(NO3)2Cu(NO_{3})_{2}Cu(NO3​)2​
  2. (B)Pb(NO3)2Pb(NO_{3})_{2}Pb(NO3​)2​
  3. (C)Co(NO3)2Co(NO_{3})_{2}Co(NO3​)2​
  4. (D)Pb(NO2)2Pb(NO_{2})_{2}Pb(NO2​)2​

Correct answer: (A)

Step-by-step solution →
Q49·ChemistrySingle correctJEE Main 2021
Reagent, 1-naphthylamine and sulphanilic acid in acetic acid is used for the detection of
  1. (A)N2_{2}2​O
  2. (B)NO3−_{3}^{-}3−​
  3. (C)NO
  4. (D)NO2−_{2}^{-}2−​

Correct answer: (D)

Step-by-step solution →
Q50·ChemistrySingle correctJEE Main 2021
On treating a compound with warm dil. H2_22​SO4_44​, gas X is evolved which turns K2_22​Cr2_22​O7_77​ paper acidified with dil. H2_22​SO4_44​ to a green compound Y. X and Y respectively are :
  1. (A)X = SO2_22​, Y = Cr2_22​(SO4_44​)3_33​
  2. (B)X = SO2_22​, Y = Cr2_22​O3_33​
  3. (C)X = SO3_33​, Y = Cr2_22​O3_33​
  4. (D)X = SO3_33​, Y = Cr2_22​(SO4_44​)3_33​

Correct answer: (A)

Step-by-step solution →
Q51·ChemistrySingle correctJEE Main 2021
Given below are two statements : Statement I : Colourless cupric metaborate is reduced to cuprous metaborate in a luminous flame. Statement II : Cuprous metaborate is obtained by heating boric anhydride and copper sulphate in a non-luminous flame. In the light of the above statements, choose the most appropriate answer from the options given below.
  1. (A)Statement I is false but statement II is true.
  2. (B)Statement I is true but Statement II is false.
  3. (C)Both Statement I and Statement II are true.
  4. (D)Both Statement I and Statement II are false.

Correct answer: (D)

Step-by-step solution →
Q52·ChemistrySingle correctJEE Advanced 2020
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. NH3NH_3NH3​ 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

Correct answer: (A)

Step-by-step solution →
Q53·ChemistrySingle correctJEE Advanced 2019
The green colour produced in the borax bead test of a chromium(III) salt is due to
  1. (A)CrB
  2. (B)Cr2_{2}2​O3_{3}3​
  3. (C)Cr2_{2}2​(B4_{4}4​O7_{7}7​)3_{3}3​
  4. (D)Cr(BO2_{2}2​)3_{3}3​

Correct answer: (D)

Step-by-step solution →
Q54·ChemistryMultiple correctJEE Advanced 2018
The correct option(s) to distinguish nitrate salts of Mn2+Mn^{2+}Mn2+ and Cu2+Cu^{2+}Cu2+ taken separately is (are)
  1. (A)Mn2+Mn^{2+}Mn2+ shows the characteristic green colour in the flame test
  2. (B)Only Cu2+Cu^{2+}Cu2+ shows the formation of precipitate by passing H2SH_{2}SH2​S in acidic medium
  3. (C)Only Mn2+Mn^{2+}Mn2+ shows the formation of precipitate by passing H2SH_{2}SH2​S in faintly basic medium
  4. (D)Cu2+/CuCu^{2+}/CuCu2+/Cu has higher reduction potential than Mn2+/MnMn^{2+}/MnMn2+/Mn (measured under similar conditions)

Correct answer: (B), (D)

Step-by-step solution →
Q55·ChemistrySingle correctJEE Advanced 2016
In the following reaction sequence in aqueous solution, the species X, Y and Z, respectively, are S2O32−→Ag+Xclear solution→Ag+Ywhite precipitate→with timeZblack precipitateS_{2}O_{3}^{2-} \xrightarrow{Ag^{+}} \underset{\text{clear solution}}{X} \xrightarrow{Ag^{+}} \underset{\text{white precipitate}}{Y} \xrightarrow{\text{with time}} \underset{\text{black precipitate}}{Z}S2​O32−​Ag+​clear solutionX​Ag+​white precipitateY​with time​black precipitateZ​
  1. (A)[Ag(S2O3)2]3−[Ag(S_{2}O_{3})_{2}]^{3-}[Ag(S2​O3​)2​]3−, Ag2S2O3Ag_{2}S_{2}O_{3}Ag2​S2​O3​, Ag2SAg_{2}SAg2​S
  2. (B)[Ag(S2O3)3]5−[Ag(S_{2}O_{3})_{3}]^{5-}[Ag(S2​O3​)3​]5−, Ag2SO3Ag_{2}SO_{3}Ag2​SO3​, Ag2SAg_{2}SAg2​S
  3. (C)[Ag(SO3)2]3−[Ag(SO_{3})_{2}]^{3-}[Ag(SO3​)2​]3−, Ag2S2O3Ag_{2}S_{2}O_{3}Ag2​S2​O3​, AgAgAg
  4. (D)[Ag(SO3)3]3−[Ag(SO_{3})_{3}]^{3-}[Ag(SO3​)3​]3−, Ag2SO4Ag_{2}SO_{4}Ag2​SO4​, AgAgAg

Correct answer: (A)

Step-by-step solution →
Q56·ChemistryMultiple correctJEE Advanced 2016
The reagent(s) that can selectively precipitate S2−S^{2-}S2− from a mixture of S2−S^{2-}S2− and SO42−SO_{4}^{2-}SO42−​ in aqueous solution is(are)
  1. (A)CuCl2CuCl_{2}CuCl2​
  2. (B)BaCl2BaCl_{2}BaCl2​
  3. (C)Pb(OOCCH3)2Pb(OOCCH_{3})_{2}Pb(OOCCH3​)2​
  4. (D)Na2[Fe(CN)5NO]Na_{2}[Fe(CN)_{5}NO]Na2​[Fe(CN)5​NO]

Correct answer: (A)

Step-by-step solution →
Q57·ChemistryMultiple correctJEE Advanced 2015
The pair(s) of ions where BOTH the ions are precipitated upon passing H2S\mathrm{H_{2}S}H2​S gas in presence of dilute HCl, is(are)
  1. (A)Ba2+\mathrm{Ba^{2+}}Ba2+, Zn2+\mathrm{Zn^{2+}}Zn2+
  2. (B)Bi3+\mathrm{Bi^{3+}}Bi3+, Fe3+\mathrm{Fe^{3+}}Fe3+
  3. (C)Cu2+\mathrm{Cu^{2+}}Cu2+, Pb2+\mathrm{Pb^{2+}}Pb2+
  4. (D)Hg2+\mathrm{Hg^{2+}}Hg2+, Bi3+\mathrm{Bi^{3+}}Bi3+

Correct answer: (C), (D)

Step-by-step solution →
Q58·ChemistryIntegerJEE Advanced 2014
Among PbSPbSPbS, CuSCuSCuS, HgSHgSHgS, MnSMnSMnS, Ag2SAg_2SAg2​S, NiSNiSNiS, CoSCoSCoS, Bi2S3Bi_2S_3Bi2​S3​ and SnS2SnS_2SnS2​, the total number of BLACK coloured sulphides is

Correct answer: 6

Step-by-step solution →
Q59·ChemistrySingle correctJEE Advanced 2014
An aqueous solution of metal ion M1 reacts separately with reagents Q and R in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M2 always forms tetrahedral complexes with these reagents. Aqueous solution of M2 on reaction with reagent S gives white precipitate which dissolves in excess of S. The reactions are summarized in the scheme given below. Reagent S is
  1. (A)K4[Fe(CN)6]\mathrm{K_4[Fe(CN)_6]}K4​[Fe(CN)6​]
  2. (B)Na2HPO4\mathrm{Na_2HPO_4}Na2​HPO4​
  3. (C)K2CrO4\mathrm{K_2CrO_4}K2​CrO4​
  4. (D)KOH

Correct answer: (D)

Step-by-step solution →
Q60·ChemistrySingle correctJEE Advanced 2013
An aqueous solution of a mixture of two inorganic salts, when treated with dilute HCl, gave a precipitate (P) and a filtrate (Q). The precipitate P was found to dissolve in hot water. The filtrate (Q) remained unchanged, when treated with H2S\mathrm{H_2S}H2​S in a dilute mineral acid medium. However, it gave a precipitate (R) with H2S\mathrm{H_2S}H2​S in an ammoniacal medium. The precipitate R gave a coloured solution (S), when treated with H2O2\mathrm{H_2O_2}H2​O2​ in an aqueous NaOH medium. The coloured solution S contains
  1. (A)Fe2(SO4)3\mathrm{Fe_2(SO_4)_3}Fe2​(SO4​)3​
  2. (B)CuSO4\mathrm{CuSO_4}CuSO4​
  3. (C)ZnSO4\mathrm{ZnSO_4}ZnSO4​
  4. (D)Na2CrO4\mathrm{Na_2CrO_4}Na2​CrO4​

Correct answer: (D)

Step-by-step solution →
Q61·ChemistrySingle correctJEE Advanced 2013
Upon treatment with ammoniacal H2SH_2SH2​S, the metal ion that precipitates as a sulfide is
  1. (A)Fe(III)
  2. (B)Al(III)
  3. (C)Mg(II)
  4. (D)Zn(II)

Correct answer: (D)

Step-by-step solution →
Q62·ChemistrySingle correctJEE Advanced 2013
An aqueous solution of a mixture of two inorganic salts, when treated with dilute HCl, gave a precipitate (P) and a filtrate (Q). The precipitate P was found to dissolve in hot water. The filtrate (Q) remained unchanged, when treated with H2S\mathrm{H_2S}H2​S in a dilute mineral acid medium. However, it gave a precipitate (R) with H2S\mathrm{H_2S}H2​S in an ammoniacal medium. The precipitate R gave a coloured solution (S), when treated with H2O2\mathrm{H_2O_2}H2​O2​ in an aqueous NaOH medium. The precipitate P contains
  1. (A)Pb2+\mathrm{Pb^{2+}}Pb2+
  2. (B)Hg22+\mathrm{Hg_2^{2+}}Hg22+​
  3. (C)Ag+\mathrm{Ag^{+}}Ag+
  4. (D)Hg2+\mathrm{Hg^{2+}}Hg2+

Correct answer: (A)

Step-by-step solution →

Principles of Qualitative Analysis — frequently asked

How many questions from Principles of Qualitative Analysis appear in JEE?

Principles of Qualitative Analysis has appeared in 58 of the last 186 JEE Main and JEE Advanced papers — about 31% of them — contributing 62 questions in total across those papers.

Is Principles of Qualitative Analysis an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 31% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Principles of Qualitative Analysis questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Chemistry chapters

  • Coordination Compounds 318
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  • Redox Reactions and Electrochemistry 265
  • Chemical Bonding and Molecular Structure 207
  • d- and f-Block Elements 202
  • Aldehydes and Ketones 199
  • Equilibrium 199
  • Solutions 198

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