JEE Chemistry Salt Analysis MCQs Set 04

Practice MCQs for JEE Chemistry Salt Analysis

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Multiple Choice Questions

Question 1: An ammoniacal metal salt solution gives a brilliant red precipitate on addition of dimethylglyoxime. The metal ion is
(a) Cu2+
(b) Co2+
(c) Fe2+
(d) Ni2+
Show Answer & Explanation

Answer: (d) Ni2+

Explanation:
1. Ni2+ in ammoniacal solution forms a bright red, insoluble, chelated complex with dimethylglyoxime, nickel dimethylglyoximate.
2. This is the standard test for nickel.

Teacher's Note:
1. DMG is a specific reagent for Ni2+.
2. The complex has hydrogen bonds between the two ligands.

Question 2: Match the cations with the group reagents.
CationsGroup reaction
(P) Pb2+, Cu2+(i) H2S gas in presence of dilute HCl
(Q) Al3+, Fe3+(ii) (NH4)2CO3 in presence of NH4OH
(R) Co2+, Ni2+(iii) NH4OH in presence of NH4Cl
(S) Ba2+, Ca2+(iv) H2S in presence of NH4OH

(a) P - i, Q - iii, R - ii, S - iv
(b) P - iv, Q - ii, R - iii, S - i
(c) P - iii, Q - i, R - iv, S - ii
(d) P - i, Q - iii, R - iv, S - ii
Show Answer & Explanation

Answer: (d) P - i, Q - iii, R - iv, S - ii

Explanation:
1. Group II (Pb2+, Cu2+) is precipitated by H2S in dilute HCl.
2. Group III (Al3+, Fe3+) is precipitated as hydroxides by NH4OH with NH4Cl.
3. Group IV (Co2+, Ni2+) is precipitated by H2S in ammoniacal medium.
4. Group V (Ba2+, Ca2+) is precipitated by (NH4)2CO3 with NH4OH.

Teacher's Note:
1. Learn the group reagents in order of groups I to VI.
2. NH4Cl suppresses OH- so only group III hydroxides precipitate.

Question 3: A chloride salt solution acidified with dilute HNO3 gives a curdy white precipitate A on addition of AgNO3. A on treatment with NH4OH gives a clear solution B. A and B are respectively
(a) HAgCl2 and [Ag(NH3)2]Cl
(b) HAgCl2 and NH4[Ag(OH)2]
(c) AgCl and [Ag(NH3)2]Cl
(d) AgCl and NH4[Ag(OH)2]
Show Answer & Explanation

Answer: (c) AgCl and [Ag(NH3)2]Cl

Explanation:
1. Cl- with Ag+ gives the curdy white precipitate of AgCl (A).
2. AgCl dissolves in NH4OH due to the complex [Ag(NH3)2]Cl, which gives a clear solution (B).

Teacher's Note:
1. AgCl dissolves in NH3 but AgBr only partly and AgI not at all.
2. The diammine silver(I) ion is linear.

Question 4: During the borax bead test with CuSO4, a blue-green colour of the bead was observed in the oxidising flame due to the formation of
(a) Cu3B2
(b) Cu
(c) Cu(BO2)2
(d) CuO
Show Answer & Explanation

Answer: (c) Cu(BO2)2

Explanation:
1. Borax on heating gives Na2B4O7 gives 2NaBO2 + B2O3.
2. CuSO4 on heating gives CuO, which combines with B2O3 to give Cu(BO2)2, copper metaborate.
3. This is blue-green in the oxidising flame, whereas in the reducing flame the bead becomes colourless or red due to Cu or Cu2O.

Teacher's Note:
1. The coloured bead is a metal metaborate.
2. Cu(BO2)2 is blue in the oxidising flame.

Question 5: In the wet tests for identification of various cations by precipitation, which transition element cation does not belong to group IV in qualitative inorganic analysis?
(a) Fe3+
(b) Zn2+
(c) Co2+
(d) Ni2+
Show Answer & Explanation

Answer: (a) Fe3+

Explanation:
1. Group IV contains Co2+, Ni2+, Zn2+ and Mn2+, precipitated as sulphides by H2S in ammoniacal medium.
2. Fe3+ is precipitated as Fe(OH)3 with NH4OH in the presence of NH4Cl, so it belongs to group III.

Teacher's Note:
1. Group III: Fe3+, Al3+ and Cr3+.
2. Group IV: Zn2+, Mn2+, Co2+ and Ni2+.

Question 6: The formula of Nessler's reagent is
(a) KHg2I2
(b) KHgI3
(c) K2HgI4
(d) HgI2
Show Answer & Explanation

Answer: (c) K2HgI4

Explanation:
1. Nessler's reagent is an alkaline solution of potassium tetraiodomercurate(II), K2[HgI4].
2. It is made by adding KI to HgCl2 until the red HgI2 dissolves.

Teacher's Note:
1. It gives a brown precipitate or colour with ammonia.
2. Test for NH4+ ions.

Question 7: When Cu2+ ion is treated with KI, a white precipitate X appears in solution. The solution is titrated with sodium thiosulphate and the compound Y is formed. X and Y respectively are
(a) X = Cu2I2, Y = Na2S4O5
(b) X = Cu2I2, Y = Na2S4O6
(c) X = CuI2, Y = Na2S4O3
(d) X = CuI2, Y = Na2S4O6
Show Answer & Explanation

Answer: (b) X = Cu2I2, Y = Na2S4O6

Explanation:
1. 2Cu2+ + 4I- gives Cu2I2 (white) + I2.
2. The liberated iodine is titrated with thiosulphate: I2 + 2S2O3 2- gives 2I- + S4O6 2-.
3. So X = Cu2I2 and Y = Na2S4O6 (sodium tetrathionate).

Teacher's Note:
1. CuI2 does not exist because Cu2+ oxidises iodide.
2. This reaction is used in the iodometric estimation of copper.

Question 8: Which element is not present in Nessler's reagent?
(a) Mercury
(b) Potassium
(c) Iodine
(d) Oxygen
Show Answer & Explanation

Answer: (d) Oxygen

Explanation:
1. Nessler's reagent is K2[HgI4] in KOH solution.
2. It contains K, Hg and I but no oxygen.

Teacher's Note:
1. The formula K2HgI4 has no oxygen.
2. The reagent is made alkaline with KOH for the test.

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JEE Chemistry Salt Analysis Objective Test Questions

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