Match List - I with List - II, match the gas evolved during each reaction. .tg border-collapse:collapse;border-spacing:0; .tg tdborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal; .tg thborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal; .tg .tg-c3owborder-color:inherit;text-align:center;vertical-align:top .tg .tg-7bttborder-color:inherit;font-weight:bold;text-align:center;vertical-align:top .tg .tg-fymrborder-color:inherit;font-weight:bold;text-align:left;vertical-align:top .tg .tg-0pkyborder-color:inherit;text-align:left;vertical-align:top List - I List - II (A) (NH_4)_2 Cr_2 O_7 \stackrel\Delta\longrightarrow (I) H_2 (B) KMnO_4+HCl arrow (II) N_2 (C) Al+NaOH+H_2 O arrow (III) O_2 (D) NaNO_3 \stackrel\Delta\longrightarrow (IV) Cl_2 Choose the correct answer from the options given below :

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

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

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

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

D and F Block Elements PYQ — JEE Main 2022 · 28 Jul · Shift 1 · Q5 with Solution

Match List - I with List - II, match the gas evolved during each reaction.

	List - I		List - II
(A)	$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \stackrel{\Delta}{\longrightarrow}$	(I)	$\mathrm{H}_{2}$
(B)	$\mathrm{KMnO}_{4}+\mathrm{HCl} \rightarrow$	(II)	$\mathrm{N}_{2}$
(C)	$\mathrm{Al}+\mathrm{NaOH}+\mathrm{H}_{2} \mathrm{O} \rightarrow$	(III)	$\mathrm{O}_{2}$
(D)	$\mathrm{NaNO}_{3} \stackrel{\Delta}{\longrightarrow}$	(IV)	$\mathrm{Cl}_{2}$

Choose the correct answer from the options given below :

A
$(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{III}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{IV})$
B
$(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{IV}),(\mathrm{D})-(\mathrm{II})$
C
$(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{III})$
D
$(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{II})$

View written solutionFree

Correct answer: C

We identify the gas evolved in each reaction.
Reaction (A): $\left(NH_4\right)_2Cr_2O_7 \xrightarrow{\Delta} Cr_2O_3 + N_2 + 4H_2O$ Hence, the gas evolved is: $N_2$ So, $(A) \to (II)$
Reaction (B): Acidified permanganate oxidizes hydrochloric acid to chlorine: $2KMnO_4 + 16HCl \rightarrow 2MnCl_2 + 2KCl + 8H_2O + 5Cl_2$ Hence, the gas evolved is: $Cl_2$ So, $(B) \to (IV)$
Reaction (C): Aluminium reacts with NaOH and water to give sodium aluminate and hydrogen: $2Al + 2NaOH + 2H_2O \rightarrow 2NaAlO_2 + 3H_2$ (or in aqueous form, sodium tetrahydroxoaluminate may be written)

Hence, the gas evolved is: $H_2$ So, $(C) \to (I)$
Reaction (D): On heating sodium nitrate: $2NaNO_3 \xrightarrow{\Delta} 2NaNO_2 + O_2$ Hence, the gas evolved is: $O_2$ So, $(D) \to (III)$
Final matching: $(A)-(II),\ (B)-(IV),\ (C)-(I),\ (D)-(III)$
Comparing with the options, this corresponds to: $\boxed{\text{Option C}}$
Comparison with stored correct answer: Stored correct answer = C, which matches our derived answer.

Match List - I with List - II, match the gas evolved during each reaction.

	List - I		List - II
(A)	$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \stackrel{\Delta}{\longrightarrow}$	(I)	$\mathrm{H}_{2}$
(B)	$\mathrm{KMnO}_{4}+\mathrm{HCl} \rightarrow$	(II)	$\mathrm{N}_{2}$
(C)	$\mathrm{Al}+\mathrm{NaOH}+\mathrm{H}_{2} \mathrm{O} \rightarrow$	(III)	$\mathrm{O}_{2}$
(D)	$\mathrm{NaNO}_{3} \stackrel{\Delta}{\longrightarrow}$	(IV)	$\mathrm{Cl}_{2}$

Choose the correct answer from the options given below :

A
$(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{III}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{IV})$
B
$(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{IV}),(\mathrm{D})-(\mathrm{II})$
C
$(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{III})$
D
$(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{II})$

View written solutionFree

Correct answer: C

We identify the gas evolved in each reaction.
Reaction (A): $\left(NH_4\right)_2Cr_2O_7 \xrightarrow{\Delta} Cr_2O_3 + N_2 + 4H_2O$ Hence, the gas evolved is: $N_2$ So, $(A) \to (II)$
Reaction (B): Acidified permanganate oxidizes hydrochloric acid to chlorine: $2KMnO_4 + 16HCl \rightarrow 2MnCl_2 + 2KCl + 8H_2O + 5Cl_2$ Hence, the gas evolved is: $Cl_2$ So, $(B) \to (IV)$
Reaction (C): Aluminium reacts with NaOH and water to give sodium aluminate and hydrogen: $2Al + 2NaOH + 2H_2O \rightarrow 2NaAlO_2 + 3H_2$ (or in aqueous form, sodium tetrahydroxoaluminate may be written)

Hence, the gas evolved is: $H_2$ So, $(C) \to (I)$
Reaction (D): On heating sodium nitrate: $2NaNO_3 \xrightarrow{\Delta} 2NaNO_2 + O_2$ Hence, the gas evolved is: $O_2$ So, $(D) \to (III)$
Final matching: $(A)-(II),\ (B)-(IV),\ (C)-(I),\ (D)-(III)$
Comparing with the options, this corresponds to: $\boxed{\text{Option C}}$
Comparison with stored correct answer: Stored correct answer = C, which matches our derived answer.