datavorous/entrance-exam-dataset huggingface dataset:
Match list I with list II and select the correct answer using the codes given below the lists.

List I (Pair of isomers)	List II (Type of isomerism)
(p)	(II)[CrNH3)6[CoCN)6(I)[CoNH3)6[CrCN)6	1.	Ionization
(q)	(IV)[PtBr2NH3)4Cl2(III)[PtCl2NH3)4Br2	2.	Hydrate
(r)	(VI)[CoNCSNH3)5Cl2(V)[CoSCNNH3)5Cl2	3.	Coordination
(s)	(VIII)[CrCl2H2O)4Cl.2H2O(VII)[CrH2O)6Cl3]	4.	Geometrical
5.	Linkage isomerism

Options:
A

p	q	r	s
4	1	5	2

B

p	q	r	s
1	3	2	5

C

p	q	r	s
3	1	5	2

D

p	q	r	s
1	3	5	2

The correct answer is:

p	q	r	s
3	1	5	2

Solution:
Compounds (I) and (II) both contain complex cation and anion, so they show coordination isomerism.
Compounds (III) and (IV) are obtained by the interchange of groups outside the coordination sphere and the ligands within the coordination sphere. Therefore, they given different ions in the solution. Hence, they show ionization isomerism.
Compounds (V) and (VI) are linkage isomers, because the ambident ligands, SCN has two donor atoms, i.e., it has two possibilities of attachment (SCN and NCS) with the central metal. Hence, they show linkage isomerism.
Compounds (VII) and (VIII) are obtained by the replacement of a coordinated group such as Cl- by water of hydration. Hence, they show hydrate isomerism.

Topic: Chemistry, Coordination Compounds, JEE Main
Match list I with list II and select the correct answer using the codes given below the lists.

List I (Reactions)
List II (Products)

(p)
CH2=CHCOOH+H2/Niâ
1.
trans- PhCH=CHOOH

(q)
cis-CH3CH=CHCOOH+Br2/CCl4â
2.
racemic-2,3-dibromobutanoic acid

(r)
C 6 H 5 CH(OH)CH 2 COOHâ+â H + â Î
3.
CH3CH2COOH

(s)
o-HOOCC6H4CH2OH+H+ââÎâ
4.

Options:
A (p)-3, (q)-4, (r)-2, (s)-1
B (p)-2, (q)-3, (r)-1, (s)-4
C (p)-3, (q)-2, (r)-1, (s)-4
D (p)-3, (q)-2, (r)-4, (s)-1

The correct answer is:
(p)-3, (q)-2, (r)-1, (s)-4

Solution:
The reactions are as follows

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Match List I with List II and select the correct answer using the codes given below the lists :

Options:
A I-B, II-D, III-A, IV-E
B I-A, II-D, III-C, IV-B
C I-C, II-B, III-A, IV-E
D I-B, II-E, III-D, IV-C

The correct answer is:
I-B, II-D, III-A, IV-E

Solution:
The correct option is 1 I-B, II-D, III-A, IV-E

Topic: Physics, Ray Optics, JEE Main
Match list I with list II and select this correct answer using the codes given below the lists.

List I (Type of glass)
List II (Property/use)

(p)
Borosilicate glass
1.
Very high transparency

(q)
Calcium-alkali silicate glass
2.
Cheap laboratory glass wares

(r)
Lead glass
3.
Optical glass

(s)
Soda glass
4.
Domestic glass for windows

5.
Low coefficient of expansion

Options:
A (p)-2, (q)-5, (r)-3, (s)-4
B (p)-2, (q)-1, (r)-5, (s)-4
C (p)-5, (q)-2, (r)-1, (s)-3
D (p)-5, (q)-4, (r)-3, (s)-2

The correct answer is:
(p)-5, (q)-4, (r)-3, (s)-2

Solution:
A. Borosilicate glasses contains B and sometimes Al. It can withstand heat changes without cracking. Such glasses are widely used for making good quality laboratory equipments. This glass has low coefficient of expansion.
B. Calcium-alkali-silicate glass is made by heating alkali metal carbonate, CaCO3 and SiO2 . This glass is used as domestic glass for windows and bottles.
C. Most of CaO can be replaced by PbO giving lead glass, which has higher reflective index and is used for making optical parts (lenses) and glass ornaments.
D. Soda glass is obtained by heating Na2CO3 , CaCO3 and SiO2 . It is used for cheap laboratory glass wares.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Match List - I with List - II.
\(\begin{array}{|c|c|c|c|}
\hline & \text { (Test) } & & \text { (Identification) } \\
\hline \text { (A) } & \text { Bayer's test MARKS } & \text { (I) } & \text { Phenol VIARKS } \\
\hline \text { (B) } & \text { Ceric ammonium nitrate test } & \text { (II) } & \text { Aldehyde } \\
\hline \text { (C) } & \text { Phthalein dye test } R \mathrm{~K} & \text { (III) } & \text { Alcoholic-OH group } \\
\hline \text { (D) } & \text { Schiff's test } & \text { (IV) } & \text { Unsaturation } \\
\hline
\end{array}\)
Choose the correct answer from the options given below :

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

The correct answer is:
(A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Solution:
(A) Bayer's test $\rightarrow$ Unsaturation
(B) Ceric ammonium nitrate test $\rightarrow$ Alcoholic-OH group
(C) Phthalein dye test $\rightarrow$ Phenol
(D) Schiff's test $\rightarrow$ Aldehyde

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Match List - I with List - II.
\(\begin{array}{|l|c|c|c|}\hline & \text{ List-I } && \text{ List-II } \\\hline (a) & { \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} } & (i) & 5.92 \mathrm{BM} \\\hline (b) & { \left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} } & (ii) & 0 \mathrm{BM} \\\hline (c) & { \left[\mathrm{Fe}(\mathrm{CN})_{6}\right] 4- } & (iii) & 4.90 \mathrm{BM} \\\hline (d) & { \left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} } & (iv) & 1.73 \mathrm{BM} \\\hline\end{array}\)
Choose the correct answer from the options given below.

Options:
A (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
B (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
C (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
D (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)

The correct answer is:
(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)

Solution:
Fe(CN)63- has Fe3+ central ion d5

Electronic configuration (SFL)

t2geg2,2,10,0

n=1

Magnetic moment:

μ=nn+2 BM

=1(1+2)

=3

=1.73 BM

FeH2O63+: has d5 electrons.

Electronic configuration (WFL)

t2ge1,1,1g1,1

n=5

μ=5(5+2)=35

=5.92 B.M.

Fe(CN)64- has d6 electrons.
Electronic configuration (SFL)

t2g2,2,1eg0,0

n=0

μ=0 B.M.

FeH2O62+ has d6 electron
Electronic configuration: (WFL)

t2g2,2,1eg1,1

n=4

μ=4(4+2)=4.90 B.M.

Topic: Chemistry, Coordination Compounds, JEE Main
Match List - I with List - II.
\(\begin{array}{ll}
\text{List - I} & \text{List - II} \\
\text{Reaction} & \text{Type of redox reaction} \\
\text{(A) } \mathrm{Li} & \text{(I) } 589.2 \\
\text{(B) } \mathrm{Na} & \text{(II) } 455.5 \\
\text{(C) } \mathrm{Rb} & \text{(III) } 670.8 \\
\text{(D) } \mathrm{Cs} & \text{(IV) } 780.0 \\
\end{array}\)
Choose the correct answer from the options given below :

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

The correct answer is:
(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

Solution:
Fact Based

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Match List - I with List - II.
\(\begin{array}{ll}
\text{List - I} & \text{List - II} \\
\text{Reaction} & \text{Type of redox reaction} \\
\text{(A) } \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{(\mathrm{g})} & \text{(I) Decomposition} \\
\text{(B) } 2 \mathrm{~Pb}\left(\mathrm{NO}_3\right)_{2(\mathrm{~s})} \rightarrow 2 \mathrm{PbO}_{(\mathrm{s})}+4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} & \text{(II) Displacement} \\
\text{(C) } 2 \mathrm{Na}_{(\mathrm{s})}+2 \mathrm{H}_2 \mathrm{O}_{\text {(I) }} \rightarrow 2 \mathrm{NaOH}_{(\text {aq. }}+\mathrm{H}_{2(\mathrm{~g})} & \text{(III) Disproportionation} \\
\text{(D) } 2 \mathrm{NO}_{2(\text { g })}+2^{-} \mathrm{OH}( aq. ) \rightarrow \mathrm{NO}_{2(\text { aq. })}^{-}+\mathrm{NO}_{3(\text { aq. })}^{-}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} & \text{(IV) Combination}
\end{array}\)
Choose the correct answer from the options given below :

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

The correct answer is:
(A)-(IV), (B)-(I), (C)-(II), (D)-(III)

Solution:
$\begin{aligned} & \mathrm{A} \rightarrow(\mathrm{IV}) \\ & \mathrm{B} \rightarrow(\mathrm{I}) \\ & \mathrm{C} \rightarrow(\mathrm{II}) \\ & \mathrm{D} \rightarrow(\mathrm{III})\end{aligned}$

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Match List I with List II
\(\begin{array}{|l|l|l|l|}
\hline & \text{ List - I (Test) } & & \text{List - II (Observation)} \\
\hline \text { A. } & \mathrm{Br}_2 \text { water test } & \text { I. } & \text { Yellow orange or orange red precipitate formed } \\
\hline \text { B. } & \text { Ceric ammonium nitrate test } & \text { II. } & \text { Reddish orange colour disappears } \\
\hline \text { C. } & \text { Ferric chloride test } & \text { III. } & \text { Red colour appears } \\
\hline \text { D. } & 2,4 \text { - DNP test } & \text { IV. } & \text { Blue, Green, Violet or Red colour appear } \\
\hline
\end{array}\)
Choose the correct answer from the options given below:

Options:
A A-III, B-IV, C-I, D-II
B A-I, B-II, C-III, D-IV
C A-IV, B-I, C-II, D-III
D A-II, B-III, C-IV, D-I

The correct answer is:
A-II, B-III, C-IV, D-I

Solution:
(A) $\mathrm{Br}_2$ water test is test of unsaturation in which reddish orange colour of bromine water disappears.
(B) Alcohols given Red colour with ceric ammonium nitrate.
(C) Phenol gives Violet colour with natural ferric chloride.
(D) Aldehyde & Ketone give Yellow/Orange/Red Colour compounds with 2, 4-DNP i.e., 2, 4-Dinitrophenyl hydrazine.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Match List I with List II
\(\begin{array}{|l|l|l|l|}
\hline & \text{ List - I } & & \text{List - II} \\
\hline \text { A. } & \mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right] & \text { I. } & \mathrm{sp}^3 \\
\hline \text { B. } & {\left[\mathrm{Ni}(\mathrm{CO})_4\right]} & \text { II. } & \mathrm{sp}^3 \mathrm{~d}^2 \\
\hline \text { C. } & {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3} & \text { III. } & \mathrm{dsp}^2 \\
\hline \text { D. } & \mathrm{Na}_3\left[\mathrm{CoF}_6\right] & \text { IV. } & \mathrm{d}^2 \mathrm{sp}^3 \\
\hline
\end{array}\)
Choose the correct answer from the options given below:

Options:
A A-III, B-I, C-IV, D-II
B A-III, B-I, C-II, D-IV
C A-I, B-III, C-II, D-IV
D A-III, B-II, C-IV, D-I

The correct answer is:
A-III, B-I, C-IV, D-II

Solution:

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Match List - I with List - II.
\(\begin{array}{|l|l|l|l|} \hline &\text{ List-I } && \text{ List-II } \\\hline (a) & 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_{3}(\mathrm{~g}) & (i) & \text{Acid rain} \\\hline (b) & \mathrm{HOCl}(\mathrm{g}) \stackrel{\mathrm{hv}}{\longrightarrow} \mathrm{OH}+\dot{\mathrm{Cl}} & (ii) & \text{Smog} \\\hline (c) & \mathrm{CaCO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{CaSO}_{4}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} & (iii) & \text{Ozone} \\\hline (d) & \mathrm{NO}_{2}(\mathrm{~g}) \stackrel{\mathrm{hv}}{\longrightarrow} \mathrm{NO}(\mathrm{g})+\mathrm{O}(\mathrm{g}) & (iv) & \text{depletion} \\\hline\end{array}\)
Choose the correct answer from the options given below.

Options:
A (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
B (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
C (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
D (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)

The correct answer is:
(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)

Solution:
a→(iv)
b→(iii)
c→(i)

d→(ii)

Topic: Chemistry, Environmental Chemistry, JEE Main
Match List I with List II
\(\begin{array}{|l|l|l|l|}
\hline & \text{List - I} & & \text{List - II} \\
& \text{(Compound/Species)} & & \text{(Shape/Geometry)} \\
\hline \text { A. } & \mathrm{SF}_4 & \text { I. } & \text { Tetrahedral } \\
\hline \text { B. } & \mathrm{BrF}_3 & \text { II. } & \text { Pyramidal } \\
\hline \text { C. } & \mathrm{BrO}_3^{-} & \text {III. } & \text { See saw } \\
\hline \text { D. } & \mathrm{NH}_4^{+}{ }^{+} \mathrm{KS} & \text { IV. } & \text { Bent T-Shape } \\
\hline
\end{array}\)
Choose the correct answer from the options given below:

Options:
A A-II, B-III, C-I, D-IV
B A-II, B-IV, C-III, D-I
C A-III, B-IV, C-II, D-I
D A-III, B-II, C-IV, D-I

The correct answer is:
A-III, B-IV, C-II, D-I

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (06 Apr Shift 1)
Match List I with List II
\(\begin{array}{|l|l|l|l|}
\hline & \text{List - I} & & \text{List - II} \\
& \text{(Precipitating reagent and conditions)} & & \text{(Cation)} \\
\hline \text { A. } & \mathrm{NH}_4 \mathrm{Cl}+\mathrm{NH}_4 \mathrm{OH} / \mathrm{ARKS} & \text { I. } & \mathrm{Mn}^{2+} \\
\hline \text { B. } & \mathrm{NH}_4 \mathrm{OH}+\mathrm{Na}_2 \mathrm{CO}_3 & \text { II. } & \mathrm{Pb}^{2+} \\
\hline \text { C. } & \mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{Cl}+\mathrm{H}_2 \mathrm{~S} \text { gas } & \text { III. } & \mathrm{Al}^{3+} \\
\hline \text { D. } & \text { dilute } \mathrm{HCl} & \text { IV. } & \mathrm{Sr}^{2+} \\
\hline
\end{array}\)
Choose the correct answer from the options given below:

Options:
A A-III, B-IV, C-I, D-II
B A-III, B-IV, C-II, D-I
C A-IV, B-III, C-II, D-I
D A-IV, B-III, C-I, D-II

The correct answer is:
A-III, B-IV, C-I, D-II

Solution:
Theory based question

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2024 (06 Apr Shift 1)
Match List I with List II
\(\begin{array}{|l|l|l|l|}
\hline & \text{List-I} & & \text{List-II} \\
\hline & \text{(Molecule)} & & \text{(Shape)} \\
\hline \text { A. } & \mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{xH}_2 \mathrm{O} & \text { I. } & \text { Violet } \\
\hline \text { B. } & {\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}^{4-}\right.} & \text { II. } & \text { Blood Red } A \\
\hline \text { C. } & {[\mathrm{Fe}(\mathrm{SCN})]^{2+}} & \text { III. } & \text { Prussian Blue } \\
\hline \text { D. } & \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3 & \text { IV. } & \text { Yellow } \\
\hline
\end{array}\)
Choose the correct answer from the options given below:

Options:
A A-III, B-I, C-II, D-IV
B A-I, B-II, C-III, D-IV
C A-IV, B-I, C-II, D-III
D A-II, B-III, C-IV, D-I

The correct answer is:
A-III, B-I, C-II, D-IV

Solution:
$\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{xH}_2 \mathrm{O} \rightarrow$ Prussian Blue
$\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]^{4-} \rightarrow$ Violet
$[\mathrm{Fe}(\mathrm{SCN})]^{2+} \rightarrow$ Blood Red
$\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3 \rightarrow$ Yellow

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Match List I with List II
\(\begin{array}{|l|l|l|l|}
\hline & \text{List-I} & & \text{List-II} \\
\hline & \text{(Molecule)} & & \text{(Shape)} \\
\hline \text { A. } & \mathrm{NH}_3 & \text { I. } & \text { Square pyramid } \\
\hline \text { B. } & \mathrm{BrF}_5 & \text { II. } & \text { Tetrahedral } \\
\hline \text { C. } & \mathrm{PCl}_5 & \text { III. } & \text { Trigonal pyramidal } \\
\hline \text { D. } & \mathrm{CH}_4 & \text { IV. } & \text { Trigonal bipyramidal } \\
\hline
\end{array}\)
Choose the correct answer from the options given below:

Options:
A A-II, B-IV, C-I, D-III
B A-III, B-I, C-IV, D-II
C A-IV, B-III, C-I, D-II
D A-III, B-IV, C-I, D-II

The correct answer is:
A-III, B-I, C-IV, D-II

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Match List I with List II
\(\begin{array}{|l|l|l|l|}
\hline & \text{List-I} & & \text{List-II} \\
\hline & \text{(Name of the test)} & & \text{(Reaction sequence involved) [M is metal]} \\
\hline \text { A. } & \text { Borax bead test } & \text { I. } & \mathrm{MCO}_3 \rightarrow \mathrm{MO} \xrightarrow{\mathrm{Co}\left(\mathrm{NO}_3\right)_2} \mathrm{CoO} \cdot \mathrm{MO} \\
\hline \text { B. } & \begin{array}{l}
\text { Charcoal } \\
\text { cavity test }
\end{array} & \text { II. } & \mathrm{MCO}_3 \rightarrow \mathrm{MCl}_2 \rightarrow \mathrm{M}^{2+} \\
\hline \text { C } & \begin{array}{l}
\text { Cobalt nitrate } \\
\text { test }
\end{array} & \text { III } & \mathrm{MSO}_4 \xrightarrow[\Delta]{\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7} \mathrm{M}\left(\mathrm{BO}_2\right)_2 \rightarrow \mathrm{MBO}_2 \rightarrow \mathrm{M} \\
\hline \text { D. } & \text { Flame test } & \text { IV } & \mathrm{MSO}_4 \xrightarrow[\Delta]{\mathrm{Na}_2 \mathrm{CO}_3} \mathrm{MCO}_3 \rightarrow \mathrm{MO} \rightarrow \mathrm{M} \\
\hline
\end{array}\)
Choose the correct answer from the options given below:

Options:
A A-III, B-II, C-IV, D-I
B A-III, B-IV, C-I, D-II
C A-III, B-I, C-II, D-IV
D A-III, B-I, C-IV, D-II

The correct answer is:
A-III, B-IV, C-I, D-II

Solution:
Cobalt nitrate test
$\mathrm{MCO}_3 \rightarrow \mathrm{MO} \xrightarrow[+\Delta]{\mathrm{Co}\left(\mathrm{NO}_3\right)_2}$ CoO. $\mathrm{MO}$
Flame test
$\mathrm{MCO}_3 \rightarrow \mathrm{MCl}_2 \rightarrow \mathrm{M}^{2+}$
Borax Bead test
$\mathrm{MSO}_4 \underset{\Delta}{\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7} \mathrm{M}\left(\mathrm{BO}_2\right)_2 \rightarrow \mathrm{MBO}_2 \rightarrow \mathrm{M}$
Charcoal cavity test
$\mathrm{MSO}_4 \xrightarrow[\Delta]{\mathrm{Na}_2 \mathrm{CO}_3} \mathrm{MCO}_3 \rightarrow \mathrm{MO} \rightarrow \mathrm{M}$

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Match List I with List II
\(\begin{array}{llll}
\text{List - I} & \text{List - II} \\
\text{(Pair of compounds)} & \text{(Isomerism)} \\
\text{(A) n-propanol and Isopropanol} & \text{(I) Metamerism} \\
\text{(B) Methoxypropane and ethoxyethane} & \text{(II) Chain Isomerism} \\
\text{(C) Propanone and propanal} & \text{(III) Position Isomerism} \\
\text{(D) Neopentane and Isopentane} & \text{(IV) Functional Isomerism} \\
\end{array}\)
Choose the correct answer from the options given below :

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

The correct answer is:
(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

Solution:
$\qquad$ $\Rightarrow$ Position isomers
$\qquad$ $\Rightarrow$ Metamers
$\qquad$ $\Rightarrow$ Functional isomers
$\qquad$ $\Rightarrow$ Chain isomers

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (05 Apr Shift 2)
Match List I with List II :
\begin{array}{|l|l|l|l|}\hline & List - I & & List - II \\\hline A. & Squamous Epithelium & I. & Goblet cells of alimentary canal \\\hline B. & Ciliated Epithelium & II. & Inner lining of pancreatic ducts \\\hline C. & Glandular Epithelium & III. & Walls of blood vessels \\\hline D. & Compound Epithelium & IV. & Inner surface of Fallopian tubes \\\hline\end{array}
Choose the correct answer from the options given below :

Options:
A A-II, B-III, C-I, D-IV
B A-II, B-IV, C-III, D-I
C A-III, B-I, C-II, D-IV
D A-III, B-IV, C-I, D-II

The correct answer is:
A-III, B-IV, C-I, D-II

Solution:
The correct answer is option (4) as
- The simple squamous epithelium is involved in functions like forming a diffusion boundary and is found in the walls of blood vessels.
- The main function of ciliated epithelium is to move particles in a specific direction over the epithelium and is present in the inner surface of fallopian tubes.
- Goblet cells of the alimentary canal are examples of unicellular glands which consist of isolated glandular cells.
- Compound epithelium is made of more than one layer of cells and is found in the inner lining of pancreatic ducts.

Topic: Biology, Structural Organisation in Animals, NEET, NEET 2024 (Re-NEET)
Match List I with List II : (Both having metallurgical terms)

List-I	List-II

(a)	Concentration of Ag
ore	(i)	Reverberatory furnace

(b)	Blast furnace	(ii)	Pig iron

(c)	Blister copper	(iii)	Leaching with dilute NaCN solution

(d)	Froth floatation
method	(iv)	Sulfide ores

Choose the correct answer from the options given
below:

Options:
A (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
B (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
C (a)-(iv), (b) -(i), (c)-(iii), (d)-(ii)
D (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

The correct answer is:
(a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)

Solution:
(a) Concentration of Ag is performed by leaching with dilute NaCN solution
(b) Pig iron is formed in blast furnace
(c) Blister Cu is produced in Bessemer converter
(d) Froth floatation method is used for sulphide
ores.

Note : During extraction of Cu reverberatory furnace is involved.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (25 Jul Shift 2)
Match List – I with List –II.

Choose the correct answer from options given below:

Options:
A A-III, B-IV, C-I, D-II
B A-IV, B-II, C-I, D-III
C A-IV, B-II, C-III, D-I
D A-III, B-IV, C-II, D-I

The correct answer is:
A-IV, B-II, C-I, D-III

Solution:
DIBAL-H i.e., Di isobutyl aluminium hydride causes partial reduction of cyanide group along with the ester group and further hydrolysis forms corresponding aldehyde as the product. The reaction is as follows,

CH3CH25COOC2H5→ DIBAL- H,H2OCH3CH25CHO

The Clemmensen reduction is a reaction that is used to reduce aldehydes or ketones to alkanes using hydrochloric acid and zinc amalgam.

C6H5COC6H5→Zn(Hg)& conc. HClC6H5CH2C6H5

A Grignard reagent is an organomagnesium compound which can be described by the chemical formula ‘R-Mg-X’ . Here aldehyde will react with Grignard's reagent to give alcohol.

C6H5CHO→H2OCH3MgBrC6H5CH(OH)CH3

CH3COCH2COOC2H5→NaBH4,H+CH3CH(OH)CH2COOC2H5

In this reaction, NaBH4,H+ is used as a reducing agent, to get CH3CH(OH)CH2COOC2H5 as a product.

Hence option B is the answer.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2024 (01 Feb Shift 1)
Match List I with List II
$\text { Choose the correct answer from the options given below: }$

Options:
A A-I. B-IV, C-II, D-III
B A-IV. B-I. C-II. D-III
C A-IV. B-I, C-III, D-II
D A-I. B-IV, C-III, D-II

The correct answer is:
A-I. B-IV, C-II, D-III

Solution:
A - V lags by $90^{\circ}$ from I hence option (I) is correct.
B - V lead by $90^{\circ}$ from I hence option (IV) is correct
$\mathrm{C}-\mathrm{In} \mathrm{LCR}$ resonance $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$. Hence circuit is purely resistive so option (II) is correct
D - In LCR series V is at some angle from I hence
(III) is correct.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2024 (04 Apr Shift 2)
Match List - I with List - II.
Choose the correct answer from the options given below :

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

The correct answer is:
(A)-(II), (B)-(IV), (C)-(I), (D)-(III)

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Match List I with List II
Choose the correct answer from the options given below :-

Options:
A A-II, B-III, C-I, D-IV
B A-I, B-III, C-II, D-IV
C A-II, B-IV, C-III, D-I
D A-I, B-II, C-III, D-IV

The correct answer is:
A-II, B-IV, C-III, D-I

Solution:
The Lyman series lies in the ultraviolet, in chemistry, the Lyman series is a hydrogen spectral series of transitions and resulting ultraviolet emission lines of the hydrogen atom as an electron goes from n≥2 to n=1 (where n is the principal quantum number), the lowest energy level of the electron.

All of the wavelengths in the Balmer series are in the visible region of the electromagnetic spectrum (400 nm to 740 nm). The presence of Hydrogen is detected in astronomy by using the H-Alpha line of the Balmer series; it is also a component of the solar spectrum.

All the wavelength of the Paschen series falls in the Infrared region of the electromagnetic spectrum.

Pfund series lie in the infrared.

A – II, B – IV, C – III, D – I

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2024 (29 Jan Shift 2)