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

List I	List II

Antimicrobials	Names

(A) Narrow Spectrum
Antibiotic	(I) Furacin

(B) Antiseptic	(II) Sulphur dioxide

(C) Disinfectants
(III) Penicillin-G

(D) Broad spectrum antibiotic	(IV) Chloramphenicol

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

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

Solution:
Narrow spectrum antibiotic are effective over less pathogens. Antiseptic are used for surface wounds. Disinfectants are used to sanitize surfaces of objects. Broad spectrum antibiotics work on broad variety of pathogens.
(A) Narrow spectrum antibiotic - penicillin-G
(B) Antiseptic - Furacine
(C) Disinfectants - sulphur dioxide
(D) Broad spectrum antibiotics - chloramphenicol

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Match List I with List II

List I	List II

Elements	Colour imparted to the flame

A	K	I	Brick Red

B	Ca	II	Violet

C	Sr	III	Apple Green

D	Ba	IV	Crimson Red

Choose the correct answer from the options given below:

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

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

Solution:
The colours of the flame are produced because of the movement of the electrons in the metal ions present in their salts. Calcium exhibits a brick-red colour. Strontium exhibits crimson colour. And the barium exhibits an apple green colour.

K is an alkali metal. Potassium is colourless or white but when it is heated, it produces a violet flame due to the excitation of the electrons to the higher energy orbitals. Potassium burns with a light purple-coloured flame.

So, the correct option is C

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Match List-I with List-II.

List-I	List-II

(Molecule)	(Bond order)

(a)	Ne2	(i)	1

(b)	N2	(ii)	2

(c)	F2	(iii)	0

(d)	O2	(iv)	3

Choose the correct answer from the options given below:

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

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

Solution:
Bond order = 12Ã¢â¬â¹ (number of electrons in bonding MO -number of electrons in anti-bonding MO)

(a) Ne2= Total eΘ=20

B.O. =10-102=0

(b) N2= Total e⊝=14

B.O. =10-42=3

(c) F2= Total e⊝=18

B.O. =10-82=1

(d) O2= Total e⊝=16

B.O. =10-62=2

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Match List I with List II.

List I	List II

Reaction	Reagents

(A)	Hoffmann
Degradation	(I)	Conc.KOH,Δ

(B)	Clemmensen
reduction	(II)	CHCl3,NaOH/H3O+

(C)	Cannizzaro reaction	(III)	Br2,NaOH

(D)	Reimer-Tiemann
reaction	(IV)	Zn-Hg/HCl

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

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

Solution:
(A) Hoffmann degradation reaction: Hoffmann developed a method for the preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide.

O ∥R-C-NH2 + Br2 + 4NaOH → R-NH2 + Na2CO3 + 2NaBr + 2H2O

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

(C) Cannizzaro reaction: Aldehydes which do not have an α-hydrogen atom, undergo self-oxidation and reduction (disproportionation) reaction on heating with concentrated alkali.

(D) Reimer-Tiemann reaction: On treating phenol with chloroform in the presence of sodium hydroxide, a –CHO group is introduced at ortho position of benzene ring. This reaction is known as Reimer-Tiemann reaction.

So, the correct matching is given in the following table:

Reactions

Reagent used

(A) Hoffmann degradation	Br2/NaOH

(B) Clemmensen reduction	Zn-Hg/HCl

(C) Cannizzaro reaction	conc.KOH/ KOH/∆

(D) Reimer-Tiemann reaction	CHCl3,NaOH/H3O+

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Match List - I with List - II :

List - I (Metal ion)	List - II (Group in Qualitative analysis)

(a)	Mn2+	(i)	Group - III

(b)	As3+	(ii)	Group - IIA

(c)	Cu2+	(iii)	Group - IV

(d)	Al3+	(iv)	Group - IIB

Choose the most appropriate answer from the options given below :

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

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

Solution:
IIA Group Cations : Hg2+,Pb2+,Bi3+,Cu2+,Cd*2+
IIB Group Cations : As3+,Sb3+,Sn2+,Sn4+
IIIrd Group Cations : Al3+,Cr3+,Fe3+
IV th Group Cations : Zn2+,Mn2+,Ni2+,Co2+

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (31 Aug Shift 2)
Match List I with List II

List I

(molecules/ions)

List II

(No. of lone pairs of e-
on central atom)

(A)	IF7	I.	Three

(B)	ICl4-	II.	One

(C)	XeF6	III.	Two

(D)	XeF2	IV.	Zero

Choose the correct answer from the options given below:

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

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

Solution:
From the following lewis structures of IF7, ICl4-, XeF6 & XeF2:
, , and
we can deduce the number of lone pairs as well as the hybridisation of the central atom:
IF7:zero lone pair sp3d3ICl4-:two lone pair sp3d2XeF6:one lone pair sp3d2XeF2:three lone pair sp3d

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Match List I with List II:

List I (Monomer)	List II (Polymer)

(A)	Tetrafluoroethene	(I)	Orlon

(B)	Acrylonitrile	(II)	Natural rubber

(C)	Caprolactam	(III)	Teflon

(D)	Isoprene	(IV)	Nylon-6

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

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

Solution:
Orlon is a man-made polymer of acrylonitrile. Thus, its monomer is vinyl cyanide or Prop-2-ene-1-nitrile. Natural Rubber is obtained from latex which is a sap of a tree. Natural rubber is a polymer of isoprene. Teflon is made of a chemical compound named polytetrafluoroethylene (PTFE) which is a synthetic fluoropolymer. Nylon-6 is produced by ring-opening chain growth polymerisation of caprolactam in the presence of water vapour and an acid catalyst at the melt.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2023 (15 Apr Shift 1)
Match List I with List II.

List I (Monomer Unit)	List II (Polymer)

(a)	Caprolactum	(i)	Natural rubber

(b)	2-Chloro-1,3-butadiene	(ii)	Buna-N

(c)	Isoprene	(iii)	Nylon 6

(d)	Acrylonitrile	(iv)	Neoprene

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→ii,b→i,c→iv,d→iii
D a→iv,b→iii,c→ii,d→i

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

Solution:
(a) Caprolactum is the monomeric unit of polymer Nylon-6.

(b) 2-Chlorobuta-1, 3-diene is the monomeric unit of polymer neoprene.
(c) Isoprene is the monomeric unit of polymer natural rubber.
(d) CH2=CH-CN (Acrylonitrile) is one of the monomeric unit of polymer Buna-N.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2021 (24 Feb Shift 1)
Match List I with List II

List I (Name of polymer)	List II (Uses)

A	Glyptal	I	Flexible pipes

B	Neoprene	II	Synthetic wool

C	Acrilan	III	Paints and Lacquers

D	LDP	IV	Gaskets

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-II, D-I
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-II, D-I

Solution:
(A) Glyptal polymer is used as a manufacture of paints and Lacquers

(B) Neoprene is formed by the free radical polymerisation of chloroprene. It is used for manufacturing conveyor belts, gaskets and hoses.

(C) Polyacrylonitrile is used as a substitute for wool in making commercial fibres as orlon or acrilan.

(D) Low density polythene (LDP) is chemically inert and tough but flexible and a poor conductor of electricity. Hence, it is used in the insulation of electricity carrying wires and manufacture of squeeze bottles, toys and flexible pipes.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Match List I with List II

List I
Natural amino acid	List II
One Letter
Code

(A)	Glutamic acid	(I)	Q

(B)	Glutamine	(II)	W

(C)	Tyrosine	(III)	E

(D)	Tryptophan	(IV)	Y

Choose the correct answer from the options given below:

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

Solution:
Amino acids are organic compounds that contain amine and carboxyl functional groups, as well as a side chain specific to each amino acid. There are 20 standard amino acids that are used to create proteins.

One letter codes of the given natural amino acids are
Natural amino acid One Letter Code
(A) Glutamic acid (III) E
(B) Glutamine (I) Q
(C) Tyrosine (IV) Y
(D) Tryptophan (II) W
∴ Correct match is
(A)-III, B-I, (C)-IV, (D)-II

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Match List-I with List-II.

List-I
(Ore)	List-II
(Element Present)

a	Kernite	p	Tin

b	Cassiterite	q	Boron

c	Calamine	r	Fluorine

d	Cryolite	s	Zinc

Choose the most appropriate answer from the options given below:

Options:
A a→r,b→p,c→q,d→s
B a→q,b→s,c→p,d→r
C a→p,b→r,c→s,d→q
D a→q,b→p,c→s,d→r

The correct answer is:
a→q,b→p,c→s,d→r

Solution:
Kernite =Na2B4O7·4H2O

The chemical formula for Kernite is =Na2B4O7·4H2O. It contains boron.

Cassiterite =SnO2

Cassiterite is an ore of Sn.It is also called tinstone and has molecular formula of SnO2Ã¢â¬â¹.

Calamine =ZnCO3

Calamine is an ore of zinc. It is zinc carbonate ZnCO3Ã¢â¬â¹.

Cryolite =Na3AlF6, which consists of Aluminium, Sodium & Fluorine.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (26 Feb Shift 1)
Match List I with List II

List I

Oxide

List II

Type of bond

A	N2O4	I	1 N = O bond

B	NO2	II	1 N – O –N bond

C	N2O5	III	1 N – N bond

D	N2O	IV	1 N = N / N ≡ N bond

Choose the correct answer from the options given below :

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

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

Solution:
The coplanarity of N2O4 results from a delicate balance of the repulsive forces, which favour the skew structure, and of the effects of bonding which favour the planar structure.

NO2 is a bent molecule. The N2O5 molecule is a trigonal planar molecule. N2O has a linear, unsymmetrical structure that may be thought of as a hybrid of two resonance forms.

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 :

List - I
(Parameter)	List - II (Unit)

(a)	Cell constant	(i)	S cm2 mol-1

(b)	Molar conductivity	(ii)	Dimensionless

(c)	Conductivity	(iii)	m-1

(d)	Degree of dissociation of electrolyte	(iv)	Ω-1 m-1

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

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

Solution:
Cell constant =ℓA=m-1

Conductivity k=1ρ=ℓRA=Ω-1 m-1

Molar Conductivity =λm=k×1000 Molarity =S cm2 mol-1

Degree of dissociation = Number of mole dissociated out of one mole, hence, it is dimensionless.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2021 (31 Aug Shift 2)
Match List-I with List-II

List-I (process)	List-II (catalyst)

(a)	Deacon's process	(i)	ZSM-5

(b)	Contact process	(ii)	CuCl2

(c)	Cracking of hydrocarbons	(iii)	Ni

(d)	Hydrogenation of vegetable oils	(iv)	V2O5

Choose the most appropriate answer from the options given below-

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

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

Solution:
In manufacture of H2SO4 (contact process), V2O5 is used as a catalyst. Ni catalysts enables the hydrogenation of fats. CuCl2 is used as catalyst in Deacon's process. ZSM-5 used as catalyst in cracking of hydrocarbons.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Match List - I with List - II.

List –I
(Processes/Reactions)	List-II
(Catalyst)

A	2SO2g+O2g→2SO3g	I	Fes

B	4NH3g+5O2g→4NOg+6H2Og	II	Pts-Rhs

C	N2g+3H2g→2NH3g	III	V2O5

D	Vegetable oil l+H2→ Vegetable ghee(s)	IV	Nis

Choose the correct answer from the options given below

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

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

Solution:
2SO2g+O2g⟶V2O52SO3g: contact process

4NH3g+5O2g⟶Pts-Rhs4NOg+6H2Og: Ostwald's process

N2g+3H2g⟶Fes2NH3g; Haber's process

Vegetable oil l+H2g⟶Nis vegetable ghee
: Hydrogenation

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2022 (26 Jul Shift 1)
. Match List I with List II :

List I (Reagents used)	List II (Compound with Functional group detected)

A.	Alkaline solution of copper sulphate and sodium citrate	I.

B.	Neutral FeCl3 solution	II.

C.	Alkaline chloroform solution	III.

D.	Potassium iodide and sodium hypochloride	IV

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

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

Solution:
(A) Aldehydes give red precipitate with Fehling's solution.

(B) Phenols give violet colour with neutral ferric chloride solution

(C) Primary amines give isocyanides with chloroform in the presence of alkali. This reaction known as carbylamine reaction.

(D) 1-phenylethanol give haloform reaction with sodium hypochlorite. The products are benzoate and chloroform.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Match List I with List II

LIST I Type	LIST II Name

A	Antifertility drug	I	Norethindrone

B	Tranquilizer	II	Meprobomate

C	Antihistamine	III	Seldane

D	Antibiotic	IV	Ampicillin

Choose the correct answer from the options given below :

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

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

Solution:
Norethindrone is an example of synthetic progesterone derivative most widely used as antifertility drug. Meprobamate, is relatively mild tranquilizer suitable for relieving tension. Synthetic drugs, brompheniramine (Dimetapp) and terfenadine (Seldane), act as antihistamines. Ampicillin and Amoxycillin are synthetic modifications of penicillins. These have broad spectrum antibiotics.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2023 (24 Jan Shift 2)
Match List I with List II

List I
Type of Hydride	List II
Example

A	Electron deficient hydride	I	MgH2

B	Electron rich hydride	II	HF

C	Electron precise hydride	III	B2H6

D	Saline hydride	IV	CH4

Choose the correct answer from the options given below :

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

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

Solution:
B2H6, known as diborane, is an example of an electron-deficient hydride. HF, or hydrogen fluoride, is an example of an electron-rich hydride. Electron-rich hydrides are molecules that contain more electrons than what would be expected based on the normal valence electron count for the atoms involved.

CH4 is electrons precise hydride since, there are exact number of electrons to form normal covalent bonds for completing octet.

MgH2 is salline hydride, as this is the hydride that formed when hydrogen reacts with s-Block elements.

A	Electron deficient hydride	III	B2H6

B	Electron rich hydride	II	HF

C	Electron precise hydride	IV	CH4

D	Saline hydride	I	MgH2

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Match List I with List II.

List I With Ore	List II Composition

A.Siderite	I.FeCO3

B.Malachite	II.CuCO3·Cu(OH)2

C.Sphalerite	III.ZnS

D.Calamine	IV.ZnCO3

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-E, D-I

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

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

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

Solution:
Siderite - FeCO3

Malachite is a copper carbonate hydroxide mineral with the formula Cu2CO3OH2

Sphalerite is a sulphide mineral with the chemical formula ZnS

Calamine is ZnCO3

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Match List I with List II

List II	List II

(A)	Tranquilizers	(I)	Anti blood clotting

(B)	Aspirin	(II)	Salvarsan

(C)	Antibiotic	(III)	Antidepressant drugs

(D)	Antiseptic	(IV)	Soframicine

Choose the correct answer from the options given below:

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

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

Solution:
Tranquilizers are a class of chemical compounds used for the treatment of stress and mild or even severe mental diseases. For example, noradrenaline is one of the neurotransmitters that play a role in mood changes. If the level of noradrenaline is low for some reason, then the signal-sending activity becomes low, and the person suffers from depression. In such situations, antidepressant drugs are required.

Aspirin inhibits the synthesis of chemicals known as prostaglandins which stimulate inflammation in the tissue and cause pain. These drugs are effective in relieving skeletal pain such as that due to arthritis. These drugs have many other effects such as reducing fever (antipyretic) and preventing platelet coagulation. Because of its anti-blood clotting action, aspirin finds use in the prevention of heart attacks.

Antibiotics are used as drugs to treat infections because of their low toxicity for humans and animals. Paul Ehrlich, a German bacteriologist, investigated arsenic-based structures in order to produce less toxic substances for the treatment of syphilis. He developed the medicine, arsphenamine, known as salvarsan.

Antiseptics are applied to the living tissues such as wounds, cuts, ulcers and diseased skin surfaces. Examples are furacine, soframicine, etc.

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

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Match List I with List II

List	List II

A	Young's Modulus (Y)	I	[ML–1 T –1]

B	Co-efficient of Viscosity (η)	II	[ML2 T –1]

C	Planck's Constant (h)	III	[ML–1 T –2]

D	Work Function ϕ	IV	[ML2 T–2]

Choose the correct answer from the options given below:

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

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

Solution:
(A) Young's modulus is stated asY=StressStrain=FA∆ll=FlA∆l=MLT-2LL2L=ML-1T-2(B) Coefficient of viscosity is stated as η=F6πrv, here, v is velocity.

η=MLT-2LLT-1=ML-1T-1

(C) Planck's Constant is stated as h=EnergyFrequency=ML2T-2T-1=ML2T-1

(D) Work function has the same dimension as that of energy, so ϕ=ML2T-2

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

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Match List I with List II.

List-I
(Anion)	List-II
(gas evolved on reaction with dil. H2SO4)

(A)	CO32-	(I)

Colourless gas which turns lead acetate paper black.

(B)	S2-	(II)

Colourless gas which turns acidified potassium dichromate solution green.

(C)	SO32-	(III)	Brown fumes which turns acidified KI solution containing starch blue.

(D)	NO2-	(IV)	Colourless gas evolved with brisk effervescence, which turns lime water milky.

Choose the correct answer from the options given below

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

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

Solution:
When small amount of carbonate salt is treated with dilute H2SO4 or dilute HCl, A colourless, odourless gas CO2 is evolved with brisk effervescence

When small amount of the following salts are treated with dilute H2SO4 the following observations are seen.

Carbonate salt CO32−	Brisk effervescence of CO2 gas, CO32−+H2SO4→SO42−+H2O+CO2↑

Sulphide salt S-2	A colourless gas H2S with rotten egg smell

Na2S+H2SO4→Na2SO4+H2S↑

NO2 – on reaction with dilute H2SO4 will give brown NO2(g) will turn KI solution blue.

SO32-on reaction with dilute H2SO4 will give SO2g, which will turn acidified potassium dichromate solution green.

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Match List - I with List - II

LIST-I
(Atomic number)	LIST-II
(Block of periodic
table)

(A)	37	I.	p-block

(B)	78	II.	d-block

(C)	52	III.	f-block

(D)	65	IV.	s-block

Choose the correct answer from the options given below:

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

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

Solution:
The orbital where the last electron enters defines the block in the periodic table where the elements belong to:

Atomic number	Block	Electronic Configuration

37 K	s-block	[Ar] 4s1

78 Pt	d-block	[Xe]4f145d96s1

52 Te	p-block	[Kr] 4d10 5s25p4

65 Tb	f-block	[Xe] 4f96s2

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Match List - I with List - II :

List-I (Class of Chemicals)	List-II (Example)

(a)	Antifertility drug	(i)	Meprobamate

(b)	Antibiotic	(ii)	Alitame

(c)	Tranquilizer	(iii)	Norethindrone

(d)	Artificial Sweetener	(iv)	Salvarsan

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

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

Solution:
(A) Antifertility drug → (iii) Nor ethindrone
(B) Antibiotic → (iv) Salvarsan
(C) Tranquilizer → (i) Meprobamate
(D) Artificial sweetener → (ii) Alitame

A-iii,B-iv,C-i,D-ii

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Match List I with List II

LIST-I
Complex	LIST-II
Colour

A.	MgNH4PO4	I.	brown

B.	K3CoNO26	II.	white

C.	MnOOH2	III.	yellow

D.	Fe4FeCN63	IV.	blue

Choose the correct answer from the options given below:

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

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

Solution:
The formula Mg(NH4)PO4 represents magnesium ammonium phosphate it is white in colour. K3CoNO26 is known as potassium hexanitritocobaltate(III). It is yellow in colour. The yellow colour is due to the presence of the cobalt ion (Co3+) in the complex, which absorbs certain wavelengths of light and reflects yellow light. MnOOH2is brown in colour. Fe4FeCN63 is blue in colour.

Complex	Colour

MgNH4PO4	White

K3CoNO26	Yellow

MnOOH2	Brown

Fe4FeCN63	Blue

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (11 Apr Shift 2)