datavorous/entrance-exam-dataset huggingface dataset:
Match List - I with List - II

List-II

List-II

A	N2g+3H2g→2NH3g	I	Cu

B	COg+3H2g→CH4g+H2Og	II	Cu/ZnO-Cr2O3

C	COg+H2g→HCHOg	III	FexOy+K2O+Al2O3

D	COg+2H2g→CH3OHg	IV	Ni

Choose the correct answer from the options given below

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

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

Solution:
List-II

List-II

A	N2g+3H2g→2NH3g	III	FexOy+K2O+Al2O3

B	COg+3H2g→CH4g+H2Og	IV	Ni

C	COg+H2g→HCHOg	I	Cu

D	COg+2H2g→CH3OHg	II	Cu/ZnO-Cr2O3

N2g+3H2g→2NH3g is Haber process and catalyst used are FexOy+K2O+Al2O3.

H2O + C → H2 + CO

The product of this is called synthesis gas and can be extremely useful in bonding with metallic heterogeneous catalysts to produce different organic compounds.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Match List - I with List - II

List-II

List-II

A	N2g+3H2g→2NH3g	I	Cu

B	COg+3H2g→CH4g+H2Og	II	Cu/ZnO-Cr2O3

C	COg+H2g→HCHOg	III	FexOy+K2O+Al2O3

D	COg+2H2g→CH3OHg	IV	Ni

Choose the correct answer from the options given below

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

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

Solution:
List-II

List-II

A	N2g+3H2g→2NH3g	III	FexOy+K2O+Al2O3

B	COg+3H2g→CH4g+H2Og	IV	Ni

C	COg+H2g→HCHOg	I	Cu

D	COg+2H2g→CH3OHg	II	Cu/ZnO-Cr2O3

N2g+3H2g→2NH3g is Haber process and catalyst used are FexOy+K2O+Al2O3.

H2O + C → H2 + CO

The product of this is called synthesis gas and can be extremely useful in bonding with metallic heterogeneous catalysts to produce different organic compounds.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2022 (25 Jul Shift 1)
Match List - I with List - II :

List-II	List-II

(A)	PtCl42-	(I)	sp3d

(B)	BrF5	(II)	d2sp3

(C)	PCl5	(III)	dsp2

(D)	CoNH363+	(IV)	sp3d2

Choose the most appropriate answer from the options given below

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

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

Solution:
Compound	Compound	Structure

(A)	PtCl42-	(I)	dsp2	Square planer

(B)	BrF5	(II)	sp3d2	Square pyramidal

(C)	PCl5	(II)	sp3d	trigonal pyramidal

(D)	CoNH363+	(IV)	d2sp3	octahedral

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2022 (24 Jun Shift 1)
Match List I with List II :

s reacted with reagents in List I to form products in List II

products in List II.

List I (Reagent) List II (Product)

A.	I.

B.	HBF4,Δ	II.

C.	Cu, HCl	III.

D.	CuCN/KCN	IV.

Choose the correct answer from the options given below:

Options:
A A-IV, B-III, C-II, D-I
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:
Coupling reaction between benzene diazonium chlorides with aniline in acidic medium takes place to give yellow colour. In this reaction the nitrogen N2 in the diazonium ion is lost and forms the bridge between two benzene rings forming an electrophilic substitution reaction.

Fluoro benzene is prepared by treating benzene diazonium chloride with fluoro boric acid. This reaction produces diazonium fluoroborate which on heating produces fluorobenzene. This reaction is called Balz-Schiemann reaction.

Gattermann reaction is used for obtaining chlorobenzene from benzenediazonium chloride by treating it with Cu/HCl respectively.

Similarly, Benzene diazonium chloride solution on reaction with cuprous cyanide and potassium cyanide gives benzonitrile. This reaction is known as Sandmeyer's reaction.

Topic: Chemistry, Amines, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Match List - I with List-II

List I (Molecule)	List II (Shape)

A.	BrF5	i.	T-shape

B.	H2O	ii.	See saw

C.	ClF3	iii.	Bent

D.	SF4	iv.	Square pyramidal

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

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

Solution:
(A) The steric number of the central Bromine atom is =5 bond pairs+one lone pair =6. Hence the shape is square pyramidal structure and the hybridisation is sp3d2.

(B) H2O molecular geometry is bent because it has two bound atoms and two lone pairs on oxygen, the central atom. The two lone pairs on the oxygen atom push the two hydrogen atoms down further away from the oxygen atom and closer to each other, creating a bent shape and 104.5 degrees angle between the two H atoms.

(C) The molecular geometry of ClF3 is said to be T-shaped. ClF3 acquires such a shape because two lone pairs take up equatorial positions as they require more space and greater repulsions. They are arranged in a trigonal bipyramidal shape with 3 bonds and 2 lone pairs.

(D) SF4 has an sp3d type of hybridisation. In a molecule of SF4, each S atom is bonded to four fluorine atoms.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (30 Jan Shift 1)
Match List-I and List-II.

List-I	List-II

A.	Elastomeric polymer	I.	Urea formaldehyde resin

B.	Fibre polymer	II.	Polystyrene

C.	Thermosetting polymer	III.	Polyester

D.	Thermoplastic polymer	IV.	Neoprene

Choose the correct answer from the options given below:

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

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

Solution:
(A) Rubber is a natural polymer and possesses elastic properties. It is also termed as elastomeric polymer. Neoprene is a family of synthetic rubbers that are produced by the polymerization of chloroprene. So, the correct matching of elastomeric polymer is neoprene.

(B) Polyesters are the polycondensation products of dicarboxylic acids and diols. Dacron or terylene is the best-known example of polyester. Dacron fibre is crease-resistant and is used in blending with cotton and wool fibres and also as glass reinforcing materials in safety helmets, etc. So, the correct matching of fibre polymer is polyester.

(C) Thermosetting polymer which cannot be reused or remoulded. Urea formaldehyde resin and phenol-formaldehyde resins are the thermosetting polymers.

(D) Thermoplastics can soften upon heating and return to their original form on cooling. They are easily moulded and extruded into films, fibres and packaging. Polystyrene is an example of a thermoplastic polymer.

So the correct matching is A-IV, B-III, C-I, D-II.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2023 (29 Jan Shift 2)
Match List-I and List-II :

List-I	List-II

a	Haematite	i	Al2O3·xH2O

b	Bauxite	ii	Fe2O3

c	Magnetite	iii	CuCO3·Cu(OH)2

d	Malachite	iv	Fe3O4

Choose the correct answer from the options given below :

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

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

Solution:
Ore	Formula

a	Haematite	Fe2O3

b	Bauxite	Al2O3·xH2O

c	Magnetite	Fe3O4

d	Malachite	CuCO3·Cu(OH)2

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Match List-I and List-II

List-I	List-II

A.	Osmosis	I.	Solvent molecules pass through semi permeable membrane towards solvent side.

B.	Reverse osmosis	II.	Movement of charged colloidal particles under the influence of applied electric potential towards oppositely charged electrodes

C.	Electro osmosis	III.	Solvent molecules pass through semi permeable membrane towards solution side

D.	Electrophoresis	IV.	Dispersion medium moves in an electric field.

Choose the correct answer from the options given below:

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

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

Solution:
(A) Osmosis: It is a process by which the molecules of a solvent pass from a solution of low concentration to a solution of high concentration through a semi-permeable membrane.

(B) Reverse osmosis: The direction of osmosis can be reversed if a pressure larger than the osmotic pressure is applied to the solution side. Now the pure solvent flows out of the solution through the semi-permeable membrane. This phenomenon is called reverse osmosis.

(C) Electroosmosis: When electrophoresis, i.e., movement of particles is prevented by some suitable means, it is observed that the dispersion medium begins to move in an electric field. This phenomenon is termed electroosmosis.

(D) Electrophoresis: When an electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied electric potential is called electrophoresis.

So the correct matching is A-III, B-I, C-IV, D-II.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2023 (29 Jan Shift 2)
Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the choices given below the list:

List-I	List-II

A	Franck-Hertz Experiment	(i)	Particle nature of light

B	Photo-electric experiment	(ii)	Discrete energy levels of the atom

C	Davison-Germer Experiment	(iii)	Wave nature of electron

(iv)	Structure of atom

Options:
A A-iv ;B-iii;C-(ii)
B A-iÂ ;B-iv;C-(iii)
C A-iiÂ ;B-iv;C-(iii)
D A-ii ;B-i;C-(iii)

The correct answer is:
A-ii ;B-i;C-(iii)

Solution:
Frank-Hertz's experiment demonstrated the existence of excited states in mercury atoms helping to confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states. The photo-electric experiment demonstrated that photon is the particle of light, which can transfer momentum and energy due to collision. Davidson-Germer's experiment shows the wave nature of electrons.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2015 (04 Apr)
Match List-I with List II of Electromagnetic waves with corresponding wavelength range:

List I	List II

(A)	Microwave	(I)	400 nm to 1 nm

(B)	Ultraviolet	(II)	1 nm to 10–3 nm

(C)	X-Ray	(III)	1 mm to 700 nm

(D)	Infra-red	(IV)	0.1 m to 1 mm

Choose the correct answer from the options given below:

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

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

Solution:
The electromagnetic spectrum is the range of all types of EM radiation. Radiation is energy that travels and spreads out with different energies. In the daily life, two commonly known EM waves are the light wave and the radio wave.

The other types of EM radiation that make up the electromagnetic spectrum are microwaves, infrared light, ultraviolet light, X-rays and gamma-rays.

The wavelengths of different parts of the electromagnetic spectrum are related by the following relation:

λX<λUV<λIR<λMW

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2023 (15 Apr Shift 1)
Match List-I with .

List-I

Name of reaction

List-II

Reagent used

A	Hell-Volhard Zelinsky reaction	I	NaOH + I2

B	Iodoform reaction	II	(i) CrO2Cl2, CS2 (ii) H2O

C	Etard reaction	III	(i) Br2/red phosphorus(ii) H2O

D	Gatterman-Koch reaction	IV	CO, HCl, anhyd. AlCl3

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→III;B→II;C→I;D→IV
D A→III;B→I;C→IV;D→II

The correct answer is:
A→III;B→I;C→II;D→IV

Solution:
(A) Hell-Volhard-Zelinsky reaction:

Carboxylic acid react with chlorine or bromine in the presence of phosphorus to give α-halogenated carboxylic acids. The halogenation continues till all the hydrogen atom in the α-position are replaced by halogen atoms. This reaction is called HVZ (Hell-Volhard-Zelinsky) reaction.

CH3COOH Acetic acid→-HClCl2, P4 ClCH2COOHChloroacetic acid→-HClCl2, P4Cl2CHCOOHDichloroacetic acid→-HClCl2, P4Cl3CCOOHTrichloroacetic acid

(B) Iodoform reaction:

The Iodoform reaction is a chemical reaction where a iodoformCHI3 is produced by the full iodonation of a methyl ketone (RCOCH3, where R can be either a hydrogen atom, an alkyl or an aryl group) in the presence of a base.

2NaOH+I2→NaOI+NaICH3COCH3+3NaOI→CH3COONa+CHI3+2NaOH

(C) Etard reaction:

In Etard reaction, chromyl chloride oxidises methyl group to a chromium complex, which on hydrolysis gives corresponding Benzaldehyde. The equation for the reaction is:

C6H5-CH3 + CrO2Cl2 →CS2 C6H5-CHOCrOHCl2 →H3O+ C6H5-CHO

(D) Gatterman-Koch reaction

When benzene or its derivative is treated with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride or cuprous chloride, it gives Benzaldehyde or substituted Benzaldehyde. The equation for the reaction is:

C6H5-CH3 →Anh.AlCl3/CuClCO + HCl C6H5-CHO

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2023 (06 Apr Shift 1)
Match List-I with List-II.

1-Bromopropane is reacted with reagents in List-I to give product in List-II

List-I Reagent	List-II Product

A.	KOH (alc)	I.	Nitrile

B.	KCN (alc)	II.	Ester

C.	AgNO2	III.	Alkene

D	H3CCOOAg	IV.	Nitroalkane

Choose the correct answer from the options given below

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

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

Solution:
(A) When 1-bromopropane reacts with alcoholic KOH, the bromine with the alpha H (hydrogen from the carbon alongside the carbon fortified with useful gathering) is eliminated and subsequently forms 1-propene.

(B) When 1-Bromo propane reacts with alc KCN it will form butanenitrile.

(C) When akyl halide reacts with silver nitrite, nitroalkane is formed because the bond between Ag−OAg−O is covalent, the lone pair on nitrogen acts as an attacking site for nucleophilic substitution.

(D) When 1-bromopropane reacts with silver acetate, propyl ethanoate and silver bromide are formed.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Match List-I with List-II:
$\begin{array}{|l|l|l|l|}\hline & \text{List I} & & \text{List II} \\\hline A. & \text{Gene pool} & I. & \text{Stable within a generation} \\\hline B. & \text{Genetic drift} & II. & \text{Change in gene frequency by chance} \\\hline C. & \text{Gene flow} & III. & \text{Transfer of genes into or out of population} \\\hline D & \text{Gene frequency} & IV. & \text{Total number of genes and their alleles} \\\hline \end{array}$
Chose the correct answer from the option given below:

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

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

Solution:
$\begin{array}{|l|l|l|l|}\hline & \text{List I} & & \text{List II} \\ \hline A. & \text{Gene pool} & IV. & \text{Total number of genes and their alleles} \\ \hline B. & \text{Genetic drift} & II. & \text{When change in gene frequency occurs by chance} \\ \hline C. & \text{Gene flow} & III. & \begin{array}{l} \text{When gene migration/transfer of genes into or out} \\ \text{of population occurs multiple times.} \end{array} \\ \hline D & \text{Gene frequency} & I. & \begin{array}{l} \text{According to Hardy Weinberg principle, gene} \\ \text{frequency remains stable within generations.} \end{array} \\ \hline \end{array}$
Hence [A-IV, B-II, C-III, D-I] is the correct match.

Topic: Biology, Evolution, NEET, NEET 2024 (Re-NEET)
Match List-I with List-II :
$\begin{array}{|c|c|c|c|}
\hline & \text { List-I } & & \text { List-II } \\
\hline & \text { EM-Wave } & & \text { Wavelength Range } \\
\hline \text { (A) } & \text { Infra-red } & \text { (I) } & < 10^{-3} \mathrm{~nm} \\
\hline \text { (B) } & \text { Ultraviolet } & \text { (II) } & 400 \mathrm{~nm} \text { to } 1 \mathrm{~nm} \\
\hline \text { (C) } & \text { X-rays } S & \text { (III) } & 1 \mathrm{~mm} \text { to } 700 \mathrm{~nm} \\
\hline \text { (D) } & \text { Gamma rays } & \text { (IV) } & 1 \mathrm{~nm} \text { to } 10^{-3} \mathrm{~nm} \\
\hline
\end{array}$

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

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

Solution:
Infrared is the least energetic thus having biggest wavelength $(\lambda) \&$ gamma rays are most energetic thus having smallest wavelength $(\lambda)$.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2024 (05 Apr Shift 2)
Match List-I with List-II.
$\begin{array}{|l|l|l|l|}\hline & \text{List-I} & & \text{List-II} \\\hline A. & \text{Epinephrine} & I. & \text{Hyperglycemia} \\\hline B. & \text{Thyroxine} & II. & \text{Smooth muscle contraction} \\\hline C. & \text{Oxytocin} & III. & \text{Basal metabolic rate} \\\hline D. & \text{Glucagon} & IV. & \text{Emergency hormone} \\\hline \end{array}$
Choose the correct answer from the options given below :

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

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

Solution:
The correct answer is option (3) because
- Epinephrine is an emergency hormone, also known as adrenaline and is secreted from adrenal medulla.
- Thyroxine is secreted from thyroid gland and plays an important role in regulation of basal metabolic rate.
- Oxytocin is a peptide hormone synthesised by hypothalamus and released by posterior pituitary. It causes smooth muscle contraction.
- Glucagon is secreted from $\alpha$-cells of pancreas, acts mainly on hepatocytes and stimulates glycogenolysis resulting in an increased blood sugar level or hyperglycemia.

Topic: Biology, Chemical Coordination and Integration, NEET, NEET 2024 (Re-NEET)
Match List-I with List-II:
$\begin{array}{|l|l|r|l|}\hline & \text{List-I} & & \text{List-II} \\ \hline A. & \text{Residual Volume} & I. & \begin{array}{l} \text{Maximum volume of air that can be breathed in after forced} \\ \text{expiration} \end{array} \\ \hline B. & \text{Vital Capacity} & II. & \text{Volume of air inspired or expired during normal respiration} \\ \hline C. & \text{Expiratory Capacity} & III. & \text{Volume of air remaining in lungs after forcible expiration} \\ \hline D. & \text{Tidal Volume} & IV. & \text{Total volume of air expired after normal inspiration} \\ \hline \end{array}$
Choose the correct answer from the options given below:

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

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

Solution:
Option (3) is the answer of this question because correct match is A-III, B-I, C-IV, D-II.
Option (1), (2) and (4) are incorrect because they are not correctly matched.
Residual volume - Volume of air remaining in lungs after forcible expiration.
Vital capacity - Maximum volume of air that can be breathed in after forced expiration.
Expiratory Capacity - Total volume of air expired after normal inspiration.
Tidal volume - Volume of air inspired or expired during normal respiration.

Topic: Biology, Breathing and Exchange of Gases, NEET, NEET 2024 (Re-NEET)
Match list-I with list-II :

(a)	Be	(i)	Treatment of cancer

(b)	Mg	(ii)	Extraction of metals

(c)	Ca	(iii)	Incendiary bombs and signals

(d)	Ra	(iv)	Windows of X-ray tubes

(v)	Bearings for motor engines.

Choose the most appropriate answer, the option given below :

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

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

Solution:
(a) Be→ it is used in the Windows of X -ray
tubes
(b) Mg→ it is used in the Incendiary bombs
and signals
(c) Ca→ it is used in the Extraction of metals
(d) Ra→ it is used in the Treatment of cancer

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Match List-I with List-II and select the correct answer:

List-I (Ion)	List-II (Shapes)

(a)	ICl2-	(1)	Linear

(b)	BrF2+	(2)	Pyramidal

(c)	ClF4-	(3)	Tetrahedral

(d)	AlCl4-	(4)	Square planar

(5)	Angular

Options:
A a-1; b-2; c-4; d-5
B a-4; b-5; c-2; d-3
C a-1; b-5; c-4; d-3
D a-5; b-1; c-3; d-4

The correct answer is:
a-1; b-5; c-4; d-3

Solution:
(a) ICl2-, BP = 2. LP = 3sp3d; Linear
(b) BrF2+BP = 2, LP = 2sp3; Angular
(c) ClF4-, BP = 4, LP = 2sp3d2; Squareplanar
(d) AICI4-, BP = 4, LP = 0sp3 Tetrahedral

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Match List-I with List-II and select the correct answer.

List-I	List-II

P.	Î± -D and Î² -D glucose	1.	Enantiomers

Q.	D-glucose and D-galactose	2.	Anomers

R.	Erythose and threose	3.	Epimers

S.	D-glyceraldehyde and L-glyceraldehyde	4.	Diastereomers

Options:
A P-3, Q-2, R-1, S-4
B P-2, Q-3, R-4, S-1
C P-3, Q-2, R-4, S-1
D P-2, Q-3, R-1, S-4

The correct answer is:
P-2, Q-3, R-4, S-1

Solution:
List-I	List-II

P.	Î± -D and Î² -D glucose	2.	Anomers

Q.	D-glucose and D-galactose	3.	Epimers

R.	Erythose and threose	4.	Diastereomers

S.	D-glyceraldehyde and L-glyceraldehyde	1.	Enantiomers

Topic: Chemistry, General Organic Chemistry, JEE Main
Match List-I with List-II :
\(\begin{array}{l|l}
\text{List-I} & \text{List-II} \\
\text{(Compounds)} & \text{(Molecular formula)} \\
\hline \text { (a) Borax } & \text { (i) } \mathrm{NaBO}_2 \\
\text { (b) Kernite } & \text { (ii) } \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 4 \mathrm{H}_2 \mathrm{O} \\
\text { (c) Orthoboric acid } & \text { (iii) } \mathrm{H}_3 \mathrm{BO}_3 \\
\text { (d) Borax bead } & \text { (iv) } \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O}
\end{array}\)
Choose the correct answer from the options given below :

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

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

Solution:
(a) Borax $-\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O}$ or $\mathrm{Na}_2\left[\mathrm{~B}_4 \mathrm{O}_5(\mathrm{OH})_4\right] \cdot 8 \mathrm{H}_2 \mathrm{O}$
(b) Kernite $-\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 4 \mathrm{H}_2 \mathrm{O}$
(c) Orthoboric acid $-\mathrm{H}_3 \mathrm{BO}_3$
(d) Borax bead $-\mathrm{NaBO}_2$
(obtained by heating of borax)
$\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O} \stackrel{\Delta}{\stackrel{-10 \mathrm{H}_2 \mathrm{O}}{\longrightarrow}} \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaBO}_2+\mathrm{B}_2 \mathrm{O}_3$
Sodium metaborate

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Match list-I with list-II :
\begin{array}{|l|l|l|l|}
\hline & List-I & & List-II \\
\hline (a) & Mercury & (i) & Vapour phase refining \\
\hline (b) & Copper & (ii) & Distillation refining \\
\hline (c) & Silicon & (iii) & Electrolytic refining \\
\hline (d) & Nickel & (iv) & Zone refining \\
\hline
\end{array}
Choose the most appropriate answer from the option given below:

Options:
A \( a-i, b-i v, c-i i, d-i i i \)
B \( a-i i, b-i i i, c-i, d-i v \)
C \( a-i i, b-i i i, c-i v, d-i \)
D \( a-i i, b-i v, c-i i i, d-i \)

The correct answer is:
\( a-i i, b-i i i, c-i v, d-i \)

Solution:
(a) Mercury → Distillation refining
(b) Copper → Electrolytic refining
(c) Silicon → Zone refining
(d) Nickel → Vapour phase refining

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main