Match List I with List II : | LIST I (Precipitating reagent and conditions) | | LIST II (Cation) | | |---|---|---|---| | A. | $\mathrm{NH}_{4} \mathrm{Cl}+\mathrm{NH}_{4} \mathrm{OH}$ | I. | $\mathrm{Mn}^{2+}$ | | B. | $\mathrm{NH}_{4} \mathrm{OH}+\mathrm{Na}_{2} \mathrm{CO}_{3}$ | II. | $\mathrm{Pb}^{2+}$ | | C. | $\mathrm{NH}_{4} \mathrm{OH}+\mathrm{NH}_{4} \mathrm{Cl}+\mathrm{H}_{2} \mathrm{~S}$ gas | III. | $\mathrm{Al}^{3+}$ | | D. | dilute HCl | IV. | $\mathrm{Sr}^{2+}$ | Choose the correct answer from the options given below:

A. precipitates group III as the hydroxide (III). B. precipitates group V as the carbonate (IV). C. precipitates group IV as the sulphide (I). D. dilute HCl precipitates group I as the chloride (II). So A-III, B-IV, C-I, D-II.

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Practical and Analytical Chemistry

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Match List I with List II :

LIST I (Precipitating reagent and conditions)		LIST II (Cation)
A.	$NH_{4} Cl + NH_{4} OH$	I.	$Mn^{2 +}$
B.	$NH_{4} OH + Na_{2} CO_{3}$	II.	$Pb^{2 +}$
C.	$NH_{4} OH + NH_{4} Cl + H_{2} S$ gas	III.	$Al^{3 +}$
D.	dilute HCl	IV.	$Sr^{2 +}$

Choose the correct answer from the options given below:

Answer: A → III, B → IV, C → I, D → II

Solution

N H_{4} Cl + N H_{4} OH

precipitates group III as the hydroxide

\to

A l^{3 +}

(III).
B.

N H_{4} OH + N a_{2} C O_{3}

precipitates group V as the carbonate

\to

S r^{2 +}

(IV).
C.

N H_{4} OH + N H_{4} Cl + H_{2} S

precipitates group IV as the sulphide

\to

M n^{2 +}

(I).
D. dilute HCl precipitates group I as the chloride

\to

P b^{2 +}

(II).
So A-III, B-IV, C-I, D-II.

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Match the columns

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JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Practical and Analytical Chemistry

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