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

Column I	Column II

A	Soda ash	I	CaSO4

B	Chlorophyll	II	CaOH2

C	Used in Whitewashing	III	Na2CO3

D	Dentistry, ornamental work	IV	Mg2+ ions

Options:
A

A – III, B – IV, C – I, D – II

B A – II, B – I, C – III, D – IV
C A – III, B – I, C – II, D – IV
D A – II, B – III, C – IV, D – I

The correct answer is:

A – III, B – IV, C – I, D – II

Solution:
Sodium carbonate Na2CO3 also known as washing soda, soda ash and soda crystals. In chlorophyll the central ion is magnesium, and the large organic molecule is a porphyrin. Whitewash, lime paint is a type of paint made from slaked lime (calcium hydroxide, Ca(OH)2) or chalk calcium carbonate, (CaCO3). CaSO4 is used in dentistry and ornamental work.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Match Column - I with Column - II

Column-I	Column-II

a	Saprophyte	i	Symbiotic association of fungi with plant roots

b	Parasite	ii	Decomposition of dead organic materials

c	Lichens	iii	Living on living plants

d	Mycorrhiza	iv	Symbiotic association of algae and fungi

Choose the correct answer from the options given below:

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

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

Solution:
Saprophytes are the organisms that obtain nutrients directly from the dead matter, especially fungi.
Myco-heterotrophs are plants, fungi, or micro-organisms that live on dead or decomposing matter and parasitize fungi, rather than dead organic matter directly.

A parasite is an organism which lives in or on a host organism and gets its food from or at the cost of its host life. The three main classes of parasites which can induce disease in humans include protozoa, helminths, and ectoparasites.

Lichens are complex life forms which are a symbiotic relationship between two separate organisms, an alga, and a fungus. The fungus is dominant, which gives the majority of its characteristics to the other partner, from its thallus shape to its fruiting bodies.

Mycorrhiza is a symbiotic relationship between a plant and a fungus. The term mycorrhiza refers to the function of the fungus in the plant's root system or rhizosphere.

Topic: Biology, Organisms and Populations, NEET, NEET 2019 (All India)
Match Column - I with Column - II.

List-I	List-II

(a)	Nitrococcus	(i)	Denitrification

(b)	Rhizobium	(ii)	Conversion of ammonia to nitrite

(c)	Thiobacillus	(iii)	Conversion of nitrite to nitrate

(d)	Nitrobacter	(iv)	Conversion of atmospheric nitrogen to ammonia

Choose the correct answer from options given below.

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

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

Solution:
(a) - (ii). In the process of nitrogen fixation, first ammonia is converted into nitrite. This is accomplished by the bacteria Nitrosomonas and Nitrococcus.
(b) - (iy). One of the most important symbiotic bacterium, Rhizobium, remains associated with the roots of leguminous plants. They help in the conversion of atmospheric nitrogen into ammonia.
(c) - (i). Denitrification is carried out by the genera Pseudomonas and Thiobacillus. Nitrate present in the soil is also reduced to nitrogen by the process of denitrification.
(d) - (iii). The nitrification process also involves the conversion of nitrite to nitrate. The nitrite is further oxidised to nitrate with the help of the bacterium Nitcobacter.

Topic: Biology, Mineral Nutrition, NEET, NEET 2021
Match Column-I with Column - II and select the CORRECT option from the codes given below.

Column I	Column II

1	Phycomycetes	i	Sac fungi

2	Ascomycetes	ii	Algal fungi

3	Basidiomycetes	iii	Fungi imperfecti

4	Deuteromycetes	iv	Fungi imperfecti

Options:
A 1-(ii), 2-(i), 3-(iv), 4-(iii)
B 1-(ii), 2-(iv), 3-(i), 4-(iii)
C 1-(iv), 2-(i), 3-(ii), 4-(iii)
D 1-(iv), 2-(iii), 3-(ii), 4-(i)

The correct answer is:
1-(ii), 2-(i), 3-(iv), 4-(iii)

Solution:
Fungi form a unique group of heterotrophic organisms showing wide diversity. Based various features like spore formation mode, fruiting bodies fungi is categorized into -
Phycomycetes - Mainly aquatic habitat, also seen growing on decaying wood. Their mycelium is aseptate and coenocytic. Zoospores or aplanosopres are formed during asexual reproduction. Mucor, Rhizopus, Albugo are included in this group.
Ascomycetes - Majorly multi cellular and also called sac-fungi as their reproductive spores are produced in sac like structure called ascus. For asexual reproduction, conidia are produced in specialized structure called conidiophore. This group includes Penicillium, Aspergillus, Claviceps.
Basidiomycetes - Filamentous fungi having hyphae and specialized club-shaped cells called basidia involved in sexual reproduction. Agaricus(Mushroom), Puccinia, Ustilago fall in this group.
Deuteromycetes - Reproduce only via asexual mode through spores called conidia. The mycelium of this group is septate and branched. Trichoderma, Colletotrichum are included in this group.

Topic: Biology, Biological Classification, NEET
Match Column-I with Column-II :

Column-I (x-t graphs)	Column-II (v-t graphs)

A	I

B	II

C	III

D	IV

Choose the correct answer from the options given below:

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

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

Solution:
(A) Here, the slope dxdt≥0 is always increasing, and we know that the derivative graph of the parabola is straight line, so the graph of velocity-time is straight line with increasing slope.

(A - II)

(B) Here, the slope dxdt<0 and at t→∞,dxdt→0. As the slope is decreasing with time, so the graph of velocity-time will have negative velocity and positive slope.

(B - IV)

(C) Here, the slope dxdt>0 for first half and dxdt<0 for second half.

As the slope of displacement-time is positive constant in first half and negative constant in second half, so the velocity will be uniform positive and uniform negative value.

(C - III)

(D) Here, the slope dxdt=constant.

As the slope of displacement-time is positive constant, so the velocity will be uniform positive value.

(D - I)

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Match List I and List II

A	Gauss’s Law in Electrostatics	I	∮E→·dl→=-dϕBdt

B	Faraday's Law	II	∮B→·dA→=0

C	Gauss’s Law in Magnetism	III	∮B→·dl→=μ0ic+μ0∈0dϕEdt

D	Ampere-Maxwell Law	IV	∮E→·ds→=q∈0

Choose the correct answer from the options given below :

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

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

Solution:
(A) Gauss law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. So, ϕ=∮E→·ds→=q∈0

(B) Faraday’s law explains that the direction of the current will be opposite to the direction of flux that produced it, so ∮E→·dl→=-dϕBdt

(C) Gauss's law for magnetism states that the magnetic flux B across any closed surface is zero, so ∮B→·dA→=0

(D) Ampere’s Maxwell law is stated as

∮B→·dl→=μ0ic+μ0∈0dϕEdt

Where ic=Conduction current and

∈0dϕEdt=Displacement current

Thus, A-IV, B-I, C-II, D-III

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Match List I and List II

List I	List II

A	Microwaves	I	Physiotherapy

B	UV rays	II	Treatment of cancer

C	Infra-red rays	III	Lasik eye surgery

D	X-rays	IV	Aircraft navigation

Choose the correct answer from the option given below:

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

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

Solution:
A. Microwave → IV

Microwaves have short wavelength hence low energy dispersion during long distance communication. Thus, they are used in radar system.
B. UV rays → III

LASIK uses an excimer laser (an ultraviolet laser) to remove a thin layer of corneal tissue.

C. Infra-red →I

Infrared radiation in the device called infrared lamp is used in physiotherapy. This light is useful in reducing pain, decreasing inflammation, and speeding up wound care and tissue repair.

D. X-ray → II

Radiation therapy uses high-energy particles or waves, such as x-rays to destroy or damage cancer cells.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Match List I and List II

List I	List II

A	Taenia	I	Nephridia

B	Paramoecium	II	Contractile vacuole

C	Periplaneta	III	Flame cells

D	Pheretima	IV	Urecose gland

Choose the correct answer from the options given below:

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

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

Solution:
Taenia

A member of phylum Platyhelminthes. They are acoelomate and excrete by flame cells or Protonephridia.

Paramoecium

A protozoan member. They have contractile vacuoles for osmoregulation.

Periplaneta

Malpighian tubules are the excretory structures of most of the insects including cockroaches.

Malpighian tubules help in the removal of nitrogenous wastes and osmoregulation.

In addition, the fat body, nephrocytes and urecose glands also help in excretion.

Pheretima

Nephridia are the tubular excretory structures of earthworms and other annelids.

Nephridia help to remove nitrogenous wastes and maintain a fluid and ionic balance.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2023 (All India)
Match List I and List II

List I	List II

Test	Functional group /
Class of Compound

(A) Molisch's Test	(I)Peptide

(B) Biuret Test	(II)Carbohydrate

(C) Carbylamine Test	(III) Primary amine

(D) Schiff's Test	(IV) Aldehyde

Choose the correct answer from the options given below:

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

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

Solution:
Molisch’s test is a chemical test which is used to check for the presence of carbohydrates in a given analyte. It also gives positive test for peptides as they undergo hydrolysis to give saccharides. Carbylamine test is given by primary amine. If a primary amine is present, the isocyanide (carbylamine) is formed which is a foul smelling substance. A purple or magenta coloured adduct gives positive Schiff's test for aldehydes.

List I	List II

Test	Functional group /
Class of Compound

(A) Molisch's Test	(I)Carbohydrate

(B) Molisch's Test	(II) Peptide

(C) Carbylamine Test	(III) Primary amine

(D) Schiff s Test	(IV) Aldehyde

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Match List I and List II

List I

Vitamin

List II

Deficiency disease

A	Vitamin A	I	Beri-Beri

B	Thiamine	II	Cheilosis

C	Ascorbic acid	III	Xerophthalmia

D	Riboflavin	IV	Scurvy

Choose the correct answer from the options given below

Options:
A

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

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

The correct answer is:

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

Solution:
Vitamin A Deficiency: Xerophthalmia: With this condition, the eyes may become very dry and crusted, which may damage the cornea and retina.

Thiamine Deficiency: Beriberi is a disease in which the body does not have enough thiamine (vitamin B1).

Scurvy is a disease caused by a serious vitamin C(Ascorbic acid) deficiency

Riboflavin deficiency is often associated with cheilosis (chapping and fissuring of the lips). Riboflavin Deficiency. Riboflavin deficiency can be associated with various developmental abnormalities, including cleft lip.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Match List - I (Electromagnetic wave type) with List - II (Its association/application) and select the correct option from the choices given below the lists :

List - I	List - II

(a)	Infrared waves	(i)	To treat muscular strain

(b)	Radio waves	(ii)	For broadcasting

(c)	X - rays	(iii)	To detect fracture of bones

(d)	Ultraviolet rays	(iv)	Absorbed by the ozone layer of the atmosphere

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

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

Solution:
Infrared →Muscular treatment
Radio wave → Broadcasting
X-Rays →To detect fracture of bones
UV Rays→ Absorbed by the ozone layer of the atmosphere

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2014 (06 Apr)
Match List I (equations) with List II (types of process) and select the correct OPTION.

Options:
A (a)
B (b)
C (c)
D (d)

The correct answer is:
(c)

Solution:
(A) When $K_p > Q$, the reaction goes in forward direction, ie, the reaction is spontaneous.
(B) Given, $\Delta G^{\circ} < R T \ln Q$, thus, $\Delta \mathrm{G}^{\circ}=+$ ve and hence, the reaction is nonspontaneous.
(C) At equilibrium, $\mathrm{K}_{\mathrm{p}}=\mathrm{Q}$
(D) $\mathrm{T} > \frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}$ or $\mathrm{T} \Delta \mathrm{S} > \Delta \mathrm{H}$
This condition is true for spontaneous endothermic reactions (as $\Delta G \geq \Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$).

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2010 (Mains)
Match List‐I (Event) with List‐II (Order of the time interval for the happening of the event) and select the correct option from the options given below the lists.

List-I	List-II

(a)	The rotation period of earth	(i)	105 s

(b)	Revolution period of earth	(ii)	107 s

(c)	Period of a light wave	(iii)	10-15 s

(d)	Period of a sound wave	(iv)	10-3 s

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

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

Solution:
The duration of the day 86400 s≃105 s

revolution period 365 days≃107 s

For the light of λ=6000 A0, time period≃10-15 s and

for the sound wave time period≃10-3 s

Topic: Physics, Gravitation, JEE Main, JEE Main 2014 (19 Apr Online)
Match List‐I (Experiment performed) with List‐II (Phenomena discovered/associated)and select the correct option from the options given below the lists

List-I	List-II

(a)	Davisson and Germer experiment	i	Wave nature of electrons

b	Millikan's oil drop experiment	ii	Charge of an electron

c	Rutherford experiment	iii	Quantisation of energy levels

d	Franck-Hertz experiment	(iv)	Existence of the nucleus

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

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

Solution:
The oil drop experiment was to measure the specific charge of an electron.

Davisson & Germer experiment proved the wave nature of electrons.

Rutherford experiment proved the presence of a nucleus, while the Franck-Hertz experiment established the Quantisation of energy levels.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2014 (19 Apr Online)
Match List I (substances) with List II (processes) employed in the manufacture of the substances and select the correct option

List I (Substances)	List II (Processes)
1. Sulphuric acid	(i) Haber's process
2. Steel	(ii) Bessemer's process
3. Sodium hydroxide	(iii) Leblanc process
4. Ammonia	(iv) Contact process

Options:
A 1-(i), 2-(iv), 3-(ii), 4-(iii)
B 1-(i), 2-(ii), 3-(iii), 4-(iv)
C 1-(iv), 2-(iii), 3-(ii), 4-(i)
D 1-(iv), 2-(ii), 3-(iii), 4-(i)

The correct answer is:
1-(iv), 2-(ii), 3-(iii), 4-(i)

Solution:
1. Sulphuric acid	(iv) Contact process
2. Steel	(ii) Bessemer's process
3. Sodium hydroxide	(iii) Leblanc process
4. Ammonia	(i) Haber's process

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Match List I (substances) with List II (processes employed in the manufacture of the substances) and select the correct option.

Options:
A (a)
B (b)
C (c)
D (d)

The correct answer is:
(d)

Solution:
Sulphuric acid is manufactured by contact process, steel is manufactured by Bessemer's process, Leblanc process is used for the production of $\mathrm{NaOH}$ while $\mathrm{NH}_3$ is obtained by Haber's process.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET, NEET 2010 (Mains)
Match List I (Wavelength range of electromagnetic spectrum) with List II (Method of production of these waves) and select the correct option from the options given below the lists.

List I	List II

(a)	700 nm to 1 mm	(i)	Vibration of atoms and molecules.

(b)	1 nm to 400 nm	(ii)	Inner shell electrons in atoms moving from one energy level to a lower level.

(c)	< 10-3 nm	(iii)	Radioactive decay of the nucleus.

(d)	1 mm to 0.1 m	(iv)	Magnetron valve.

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

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

Solution:
We know, the range of wavelengths between 700 nm to 1 mm, is referred as Infrared rays. These are produced by heating an object to such an extent that the molecules start vibrating and it radiates infrared waves. When the object is not hot enough such that it can emit visible light, therefore, it radiates Infrared waves.
Thus, option (a) in list I matches with Option (i) in list II.

The range of wavelengths between 1 nm to 400 nm, is referred to as ultraviolet rays. Ultraviolet rays are produced due to atomic excitation in a gaseous discharge tube. This atomic excitation occurs when an atom from one energy level jumps to a lower energy level.
Thus, option (b) in list I matches with Option (ii) in list II.

The range of wavelengths <10-3 nm are known as the Gamma rays. Some of the radioactive substances, when decay releases a spectrum, which falls in the range of <10-3 nm, they emit electromagnetic waves called the gamma rays.
Thus, we can say, Thus, option (c) in list I matches with Option (iii) in list II.

We know again, the range of wavelength 1 mm to 0.1 m, is referred to as microwaves, and they are produced inside an oven in a tube like structure, called the magnetron.

Thus, we can say, option (d) in list I matches with Option (iv) in list II.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2014 (09 Apr Online)
Match List I with List II
(A) Slaked lime (I) NaOH
(B) Dead burnt plaster (II) Ca(OH)2
(C) Caustic soda (III) Na2CO3·10H2O
(D) Washing soda (IV) CaSO4
Choose the correct answer form the options given below:

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

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

Solution:
(A) Slaked lime: Calcium oxide is a white amorphous solid. It has a melting point of 2870 K. On exposure to the atmosphere, it absorbs moisture and carbon dioxide.

CaO + H2O → Ca(OH)2

CaO + CO2 → CaCO3

The addition of a limited amount of water breaks the lump of lime. This process is called slaking of lime.

(B) Dead burnt plaster: Calcium sulphate is a hemihydrate of calcium sulphate. It is obtained when gypsum, CaSO4.2H2O is heated at 393K.

2(CaSO4.2H2O) → 2(CaSO4).H2O + 3H2O

Above 393 K, no water of crystallisation is left and anhydrous calcium sulphate, CaSO4 is formed. This is known as ‘dead burnt plaster’.

(C) Caustic soda: Sodium hydroxide (caustic soda) NaOH is generally prepared commercially by the electrolysis of sodium chloride in Castner-Kellner cell. A brine solution is electrolysed using a mercury cathode and a carbon anode. Sodium metal discharged at the cathode combines with mercury to form sodium amalgam.

(D) Washing soda: Sodium carbonate is a white crystalline solid which exists as a decahydrate, Na2CO3.10H2O. This is also called washing soda. It is readily soluble in water.

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

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Match List I with List II and select the correct answer using codes given below the lists.

List I
List II

(p)
Potassium sulphate
1.
Acidic

(q)
Sodium acetate
2.
Neutral

(r)
Ammonium chloride
3.
Alkaline

(s)
Calcium formate
4.
Cannot be predicted

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

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

Solution:
(p) Neither K+ nor SO42- ions undergo hydrolysis. So it is neutral.
(q) Sodium ions are not hydrolyzed, but acetate ions hydrolyze to form strong base, OH-
CH3COO-+H2O→CH3COOHweakacid+OH-strongbase
(r)Cl- ions are not hydrolyzed. NH4+ hydrolyse to form strong acid, H3O+ .
NH4++H2O→NH3weakbase+H3O+strongacid
(s)Ca2+ and HCOO- ions are hydrolyzed to form an acid and a base, but the pH cannot be predicted without the knowledge of hydrolysis constants of Ca2+ and HCOO- .

Topic: Chemistry, Ionic Equilibrium, JEE Main
Match list â I with list â II and select the correct answer using the codes given below

List â I
List â II

(P)
C6H5âNH2ââNaNO2/HCl
1.

(Q)

2.
CH3âCH2âCH2âNââN=O

(R)
CH3âCH2âCH23âNââNaNO2/HCl
3.
C6H5âNâ=NClÎ

(S)

4.

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

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

Solution:
Here the correct matching is P(3), Q(1), R(2), S(4)

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

List I
List II

P.
CH2=CH-CHOâNaBH4
1.
Acetal

Q.
C6H5CHO+CH3-NH2âHâ
2.
Schiff's base

R.
C6H5COCH3+CH3-CH2-NH-CH3âHâ
3.
Unsaturated alcohol

S.
RCHO+2RCH2OHâHâ
4.
Enamine

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

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

Solution:

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

List I
List II

P.
CS2
1.
Bent

Q.
SO2
2.
Linear

R.
BF3
3.
Trigonal planar

S.
NH3
4.
Tetrahedral

5.
Trigonal pyramidal

Options:
A P-2, Q-1, R-3, S-5
B P-1, Q-2, R-3, S-5
C P-2, Q-1, R-5, S-4
D P-1, Q-2, R-5, S-4

The correct answer is:
P-2, Q-1, R-3, S-5

Solution:
CS 2 : S .. ..  = C =  S .. .. : = orientation of two electron pairs around carbon in linear.
SO 2 : O .. ..  =  S ..  =  O .. .. : = orientation of three electron pairs is trigonal planar, but presence of one lone pair makes the structure bent.
BF 3 : F .. .. â B | :F: â F .. .. : â NH 3 Hâ N .. | H âH â orientation of three electron pairs is trigonal planar, Since all electron pairs are bond pairs. Therefore, structure is trigonal planar. NH3H-N..|H-H- orientation of four electron pairs is tetrahedral, but presence of one lone pairs make structure trigonal pyramidal.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Match list I with List II and select the correct answer using the codes given below

List I	List II

(Types of ore)	(example)

P	Oxide ore	A.	Feldspar

Q	Sulphide ore	B.	Barytes

R	Sulphate ore	C.	Fluorspar

S	Halide ore	D.	Galena

E.	Corundum

Options:
A P-B,âQ-D,âR-C,âS-A
B P-B,âQ-D,âR-E,âS-A
C P-E, Q-B, R-D, S-C
D P-E,âQ-D,âR-B,âS-C

The correct answer is:
P-E, Q-D, R-B, S-C

Solution:
List I
(Types of ore)	List II
(Example)

Oxide ore	Corundum (Al 2 O 3 )

Sulphide ore	Galena (PbS)

Sulphate ore	Barytes (BaSO 4 )

Halide ore	Fluorsper (CaF 2 )

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