Download JEE MCQs for JEE Chemistry: Qualitative Salt Analysis
Explore reliable objective questions for Qualitative Salt Analysis tailored for JEE learners. Utilizing these Chemistry multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.
Chapter-wise Objective Questions: Qualitative Salt Analysis
Navigate directly to the 50 objective questions for Qualitative Salt Analysis using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.
Multiple Choice Questions
(a) ethyl bromide
(b) sodium bromide
(c) calcium chloride
(d) sodium chloride
Show Answer & Explanation
Answer: (a) ethyl bromide
Explanation:
1. Ethyl bromide is covalent and does not release free bromide ions.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation
Answer: (b) 2
Explanation:
1. \( \ce{Na2CO3 + 2H2O -> 2NaOH + H2CO3} \) gives two hydroxide ions per formula unit.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) sulphur is present in the mixture as an impurity
(b) the third group radicals are precipitated
(c) the \( \ce{H2S} \) is oxidised by some acid radicals
(d) the fourth group radicals are precipitated as sulphides
Show Answer & Explanation
Answer: (c) the \( \ce{H2S} \) is oxidised by some acid radicals
Explanation:
1. Oxidising acid radicals (such as nitrate, chromate or ferric ions) oxidise \( \ce{H2S} \) to free sulphur, which appears as a yellow turbidity.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) calcium oxalate
(b) calcium hydroxide
(c) calcium oxide
(d) calcium carbonate
Show Answer & Explanation
Answer: (a) calcium oxalate
Explanation:
1. Oxide, carbonate and hydroxide all react with acetic acid.
2. Calcium oxalate is insoluble in weak acids such as acetic acid.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) \( \ce{(C2H5)2B4O7} \)
(b) \( \ce{(C2H5)3BO3} \)
(c) \( \ce{H2B4O7} \)
(d) \( \ce{H3BO3} \)
Show Answer & Explanation
Answer: (b) \( \ce{(C2H5)3BO3} \)
Explanation:
1. Boric acid esterifies with ethanol to give triethyl borate, which burns with a green-edged flame.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) chromic sulphate
(b) chromyl chloride
(c) chromic chloride
(d) chromous chloride
Show Answer & Explanation
Answer: (b) chromyl chloride
Explanation:
1. Chromyl chloride, \( \ce{CrO2Cl2} \), is an orange-red volatile liquid.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) the solution will become dark green
(b) the solution will remain colourless
(c) a yellow solution will be obtained
(d) a yellow precipitate will be obtained
Show Answer & Explanation
Answer: (d) a yellow precipitate will be obtained
Explanation:
1. NaOH converts chromyl chloride to chromate, which with lead acetate gives yellow lead chromate.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) V
(b) Mn
(c) Co
(d) Ni
Show Answer & Explanation
Answer: (d) Ni
Explanation:
1. Nickel(II) forms the red bis(dimethylglyoximato)nickel(II) chelate.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) \( \ce{CuI} \)
(b) \( \ce{NaCl} \)
(c) \( \ce{MgCl2} \)
(d) \( \ce{CuF2} \)
Show Answer & Explanation
Answer: (d) \( \ce{CuF2} \)
Explanation:
1. Cu(II) fluoride is blue due to d-d transitions.
2. CuI is white, and NaCl and \( \ce{MgCl2} \) are colourless.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) \( \ce{CaCO3} \) is soluble in \( \ce{Na2CO3} \)
(b) \( \ce{Na2CO3} \) increases the solubility of fifth group carbonates,
(c) \( \ce{MgCO3} \) will be precipitated out in fifth group
(d) none of these
Show Answer & Explanation
Answer: (c) \( \ce{MgCO3} \) will be precipitated out in fifth group
Explanation:
1. Sodium carbonate provides a high carbonate concentration and would precipitate magnesium carbonate in group V, though magnesium belongs to group VI.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) \( \ce{SO4^{2-}} \)
(b) \( \ce{Cl^-} \)
(c) \( \ce{CO3^{2-}} \)
(d) \( \ce{NO3^-} \)
Show Answer & Explanation
Answer: (d) \( \ce{NO3^-} \)
Explanation:
1. All nitrates are soluble, so nitrate cannot be precipitated.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) lead chloride
(b) mercuric chloride
(c) mercurous chloride
(d) silver chloride
Show Answer & Explanation
Answer: (c) mercurous chloride
Explanation:
1. \( \ce{Hg2Cl2 + 2NH4OH -> Hg + Hg(NH2)Cl + NH4Cl + 2H2O} \): finely divided mercury makes the residue black.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) \( \ce{Ca^{2+}} \)
(b) \( \ce{Mg^{2+}} \)
(c) \( \ce{Pb^{2+}} \)
(d) \( \ce{Ba^{2+}} \)
Show Answer & Explanation
Answer: (b) \( \ce{Mg^{2+}} \)
Explanation:
1. Magnesium salts give no flame colour.
2. Ba and Ca give green and brick red; \( \ce{PbCl2} \) dissolves poorly in cold water.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) \( \ce{CO2} \) and \( \ce{CO2} \)
(b) \( \ce{CO2} \) and CO
(c) CO and \( \ce{CO2} \)
(d) CO and CO
Show Answer & Explanation
Answer: (a) \( \ce{CO2} \) and \( \ce{CO2} \)
Explanation:
1. \( \ce{Ca(OH)2 + CO2 -> CaCO3} \) dissolves as \( \ce{Ca(HCO3)2} \) in excess \( \ce{CO2} \).
2. Heating the bicarbonate gives back \( \ce{CaCO3} \) and releases \( \ce{CO2} \).
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) reducing agent
(b) oxidising agent
(c) iodide ion
(d) iodine
Show Answer & Explanation
Answer: (b) oxidising agent
Explanation:
1. An oxidising agent liberates iodine from iodide, which turns starch blue.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) \( \ce{NH4SCN} \)
(b) \( \ce{AgNO3} \)
(c) \( \ce{BaCl2} \)
(d) none of these
Show Answer & Explanation
Answer: (a) \( \ce{NH4SCN} \)
Explanation:
1. Ferric ions give a blood-red colour with thiocyanate; ferrous ions give none.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) \( \ce{PbS} \)
(b) \( \ce{HgS} \)
(c) \( \ce{CdS} \)
(d) \( \ce{CuS} \)
Show Answer & Explanation
Answer: (c) \( \ce{CdS} \)
Explanation:
1. CdS needs a low hydrogen ion concentration to precipitate completely.
2. HgS, PbS and CuS precipitate even from fairly strong acid.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) \( \ce{PbS} \)
(b) \( \ce{CuS} \)
(c) \( \ce{CdS} \)
(d) \( \ce{SnS} \)
Show Answer & Explanation
Answer: (d) \( \ce{SnS} \)
Explanation:
1. Yellow ammonium sulphide oxidises SnS and dissolves it as thiostannate; CuS, CdS and PbS do not dissolve.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) \( \ce{Bi^{3+}} \)
(b) \( \ce{Cu^+} \)
(c) \( \ce{Hg^{2+}} \)
(d) \( \ce{Pb^{2+}} \)
Show Answer & Explanation
Answer: (a) \( \ce{Bi^{3+}} \)
Explanation:
1. Black \( \ce{BiI3} \) dissolves in excess iodide to give the orange tetraiodobismuthate(III) ion \( \ce{[BiI4]^-} \).
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) \( \ce{Ag^+, NO3^-} \)
(b) \( \ce{Al^{3+}, OH^-} \)
(c) \( \ce{Na^+, S^{2-}} \)
(d) \( \ce{K^+, SO4^{2-}} \)
Show Answer & Explanation
Answer: (b) \( \ce{Al^{3+}, OH^-} \)
Explanation:
1. \( \ce{Al^{3+} + 3OH^- -> Al(OH)3} \) is insoluble.
2. The other combinations give soluble salts.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) I group basic radicals precipitate as chlorides
(b) IV group basic radicals precipitate as sulphides
(c) V group basic radicals precipitate as carbonates
(d) all of these statements are correct
Show Answer & Explanation
Answer: (d) all of these statements are correct
Explanation:
1. Group I ions are precipitated as chlorides, group IV as sulphides in ammoniacal medium, and group V as carbonates.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) Ca, Ba, Sr
(b) Mg
(c) both
(d) none
Show Answer & Explanation
Answer: (a) Ca, Ba, Sr
Explanation:
1. Group V cations Ca, Ba and Sr form insoluble carbonates; Mg belongs to group VI.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) \( \ce{CaCO3} \)
(b) \( \ce{PbCl2} \)
(c) \( \ce{PbSO4} \)
(d) \( \ce{AgCl} \)
Show Answer & Explanation
Answer: (d) \( \ce{AgCl} \)
Explanation:
1. AgCl dissolves in ammonia as \( \ce{[Ag(NH3)2]Cl} \).
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) \( \ce{Fe4[Fe(CN)6]3} \)
(b) \( \ce{Fe(OH)3} \)
(c) \( \ce{KMnO4} \)
(d) \( \ce{K4Fe(CN)6} \)
Show Answer & Explanation
Answer: (a) \( \ce{Fe4[Fe(CN)6]3} \)
Explanation:
1. The blue precipitate is ferric ferrocyanide.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) \( \ce{Cu^{2+}} \)
(b) \( \ce{Zn^{2+}} \)
(c) \( \ce{Pb^{2+}} \)
(d) \( \ce{Fe^{3+}} \)
Show Answer & Explanation
Answer: (c) \( \ce{Pb^{2+}} \)
Explanation:
1. Pb(II) forms PbCl2 (group I) and PbS (group II); the others do not give a chloride precipitate.
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
(a) \( \ce{Pb^{2+}} \)
(b) \( \ce{Ag^+} \)
(c) \( \ce{Cu^+} \)
(d) \( \ce{Sn^{2+}} \)
Show Answer & Explanation
Answer: (d) \( \ce{Sn^{2+}} \)
Explanation:
1. Pb(II), Cu(I) and Ag(I) form insoluble chlorides and sulphides.
2. Sn(II) chloride is soluble, so tin is precipitated only by \( \ce{H2S} \).
Teacher's Note:
1. Link each observation to its reagent: colour, gas evolved and precipitate are the usual clues in salt analysis.
2. Memorise the confirmatory test for each ion; they are asked repeatedly.
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Multiple Choice Questions (MCQs) for JEE Chemistry Qualitative Salt Analysis
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