Sankalp NEET Full Test-3 Question-48 Solution

Question: 48. 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:

(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: Option (2)

Explanation:

(based on qualitative inorganic analysis – Group separation):

In classical salt analysis, cations are grouped according to the reagent that precipitates them.

Group-wise classification of given ions

A. Ca²⁺ → Group III

Group III cations are precipitated as hydroxides by NH₄OH in presence of NH₄Cl.

Calcium belongs to Group III (alkaline earth metals) in qualitative analysis.

B. Mg²⁺ → Group IV

Magnesium ions are not precipitated in Group III.

They are precipitated in Group IV as MgNH₄PO₄ using Na₂HPO₄ in presence of NH₄OH and NH₄Cl.

C. Pb²⁺ → Group I

Group I cations form insoluble chlorides with dilute HCl.

Lead(II) chloride (PbCl₂) is sparingly soluble and hence Pb²⁺ belongs to Group I.

D. Al³⁺ → Group III

Aluminium forms Al(OH)₃, precipitated by NH₄OH in presence of NH₄Cl.

Hence it belongs to Group III.

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