Match List-I with List-II
 

List-I (Ion) List-II
(Group Number in Cation Analysis)
A. Co²⁺ I. Group-I
B. Mg²⁺ II. Group-III
C. Pb²⁺ III. Group-IV
D. Al³⁺ IV. Group-VI

Choose the correct answer from the options given below:

This question was previously asked in
NEET 2025 Official Paper (Held On: 04 May, 2025)
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  1. A-III, B-IV, C-II, D-I
  2. A-III, B-IV, C-I, D-II
  3. A-III, B-II, C-IV, D-I
  4. A-III, B-II, C-I, D-IV

Answer (Detailed Solution Below)

Option 2 : A-III, B-IV, C-I, D-II
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Detailed Solution

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CONCEPT:

Group Analysis of Cations

  • The classification of cations into different groups is based on their solubility and behavior in qualitative analysis.
  • Group I cations form insoluble chlorides when treated with dilute hydrochloric acid.
  • Group II cations form insoluble sulfides when treated with hydrogen sulfide in an acidic medium.
  • Group III cations form insoluble hydroxides when treated with ammonium hydroxide.
  • Group IV cations form insoluble phosphates when treated with ammonium molybdate in an acidic medium.
  • Group V cations do not form precipitates under normal conditions and are analyzed last.

Match the Group with their respective Cations and Group Reagents.

Group Cations Group Reagent
Group zero NH4+ None
Group-I Pb2+ Dilute HCl
Group-II Pb2+, Cu2+, As3+ H2S gas in presence of dil. HCl
Group-III Al3+, Fe3+ NH4OH in presence of NH4Cl
Group-IV Co2+, Ni2+, Mn2+, Zn2+ H2S in presence of NH4OH
Group-V Ba2+, Sr2+, Ca2+ (NH4)2CO3 in presence of NH4OH
Group-VI Mg2+ None

 

EXPLANATION:

  • Co2+ (Cobalt ion): This ion is typically classified as Group-IV in cation analysis because it forms hydroxides when treated with ammonium hydroxide.
  • Mg2+ (Magnesium ion): This ion belongs to Group-VI as it forms sulfides in the presence of hydrogen sulfide in an acidic medium.
  • Pb2+ (Lead ion): This ion is usually classified as Group-I because it forms phosphates when treated with ammonium molybdate in an acidic medium.
  • Al3+ (Aluminum ion): This ion is classified as Group-III because it forms insoluble chlorides in the presence of dilute hydrochloric acid.

Therefore, the correct answer is A-III, B-IV, C-I, D-II

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