datavorous/entrance-exam-dataset huggingface dataset:
Match the column I with column II and mark the appropriate choice.

Column I	Column II
(p)	Liquid ⇌ vapour	(i)	Saturated solution
(q)	Solid ⇌ liquid	(ii)	Boiling point
(r)	Solid ⇌ vapour	(iii)	Sublimation point
(s)	Solute(s) ⇌ solute (solution)	(iv)	Melting point

Options:
A (p)-(i), (q)-(iii), (r)-(ii), (s)-(iv)
B (p)-(ii), (q)-(iv), (r)-(iii), (s)-(i)
C (p)-(iv), (q)-(ii), (r)-(i), (s)-(iii)
D (p)-(iii), (q)-(iv), (r)-(ii), (s)-(i)

The correct answer is:
(p)-(ii), (q)-(iv), (r)-(iii), (s)-(i)

Solution:
(p) Liquid ⇌ vapour equilibrium exists at the boiling point
(q) Solid ⇌ liquid equilibrium exists at the melting point
(r) Solid ⇌ vapour equilibrium exists at the sublimation point
(s) Solute (s) ⇌ Solute (solution) equilibrium exists in a saturated solution.

Topic: Chemistry, Chemical Equilibrium, JEE Main
Match the column I with column II and mark the appropriate choice.

Column I	Column II

(p)	NaH	(i)	Interstitial hydride

(q)	CH4	(ii)	Molecular hydride

(r)	VH0.56	(iii)	Ionic hydride

(s)	B2H6	(iv)	Electron-deficient hydride

Options:
A (p)-(iii), (q)-(iv), (r) –(ii), (s)-(i)
B (p)-(ii), (q)-(iv), (r) –(iii), (s)-(i)
C (p)-(i), (q)-(ii), (r) –(iv), (s)-(iii)
D (p)-(iii), (q)-(ii), (r) –(i), (s)-(iv)

The correct answer is:
(p)-(iii), (q)-(ii), (r) –(i), (s)-(iv)

Solution:
Column I	Column II

(p)	NaH	(iii)	Ionic hydride

(q)	CH4	(ii)	Molecular hydride

(r)	VH0.56	(i)	Interstitial hydride

(s)	B2H6	(iv)	Electron-deficient hydride

Topic: Chemistry, s Block Elements, JEE Main
Match the column I with column II and mark the appropriate choice.

Column I	Column II
(p)	Natural polymer	(i)	Rayon
(q)	Addition polymer	(ii)	Bakelite
(r)	Copolymer	(iii)	Silk
(s)	Semi-synthetic polymer	(iv)	Neoprene

Options:
A (p)-(i), (q)-(ii), (r)-(iv), (s)-(iii)
B (p)-(iii), (q)-(iv), (r)-(ii), (s)-(i)
C (p)-(ii), (q)-(iii), (r)-(i), (s)-(iv)
D (p)-(iv), (q)-(i), (r)-(iii), (s)-(ii)

The correct answer is:
(p)-(iii), (q)-(iv), (r)-(ii), (s)-(i)

Solution:
Column I	Column II
(p)	Natural polymer	(iii)	Silk
(q)	Addition polymer	(iv)	Neoprene
(r)	Copolymer	(ii)	Bakelite
(s)	Semi-synthetic polymer	(i)	Rayon

Topic: Chemistry, Polymers, JEE Main
Match the column I with column II and mark the appropriate choice.

Column I	Column II

(p)	P	(i)	6 unpaired electrons

(q)	Cr	(ii)	2 unpaired electrons

(r)	Fe	(iii)	3 unpaired electrons

(s)	Si	(iv)	4 unpaired electrons

Options:
A (p)-(ii), (q)-(i), (r)-(iii), (s)-(iv)
B (p)-(i), (q)-(iii), (r)-(ii), (s)-(iv)
C (p)-(iii), (q)-(i), (r)-(iv), (s)-(ii)
D (p)-(iv), (q)-(ii), (r)-(i), (s)-(iii)

The correct answer is:
(p)-(iii), (q)-(i), (r)-(iv), (s)-(ii)

Solution:

Topic: Chemistry, Structure of Atom, JEE Main
Match the column with respect to organelles and their respective function.

Column I	Column II

a.	Golgi apparatus	(i)	Helps in spindle formation

b.	Ribosomes bound to ER	(ii)	Synthesis and storage of fats

c.	Microtubules	(iii)	Secretory proteins

d.	Spherosomes	(iv)	Helps in pseudopodia formation

(v)	Acrosome of sperms

.

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

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

Solution:
Active trafficking from the Golgi apparatus is involved in acrosome formation in sperm. The path of a protein destined for secretion has its origins in the rough endoplasmic reticulum, a membrane-bound compartment in the cell. The microtubule-organising centre (MTOC) is a structure found in eukaryotic cells from this microtubules emerge. MTOCs has two main functions: the organization of eukaryotic flagella and cilia and the organization of the mitotic and meiotic spindle apparatus, which separate the chromosomes during cell division. Spherosomes are small membrane-bound organelles which take part in storage and synthesis of fats.

Topic: Biology, Cell - The Unit of Life, NEET
Match the column-I with column-II and select the correct option from the codes given below:

METHOD	MODE OF ACTION

a	WBC	i	Cardiac disc

b	Adipose tissue	ii	Protects and support neurons

c	Communication junctions	iii	Diapedesis

d	Neuroglia	iv	Store fats

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

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

Solution:
White blood cells circulate into the blood stream but they can cross the blood capillary to reach infected tissues. This process is called as diapedesis.
Adipose tissue is composed of a loose collection of specialized cells, called adipocytes, which are embedded in collagen fibers. Its main function is storage of lipids.
Communicating junctions are channels that allow passage of small molecules and ions between two cells. Such junctions support synchronized contraction of cardiac tissue.
Neuroglia (glia) are cells that support and protect neurons.

Topic: Biology, Structural Organisation in Animals, NEET
Match the column-I with column-II and select the suitable option from the following:

Column I	Column II

i	Appendicular skeleton	a	Ilium, ischium and pubis

ii	F-actin	b	30 bones

iii	Coxal bone	c	Inflammation of joints

iv	Ribs	d	Polymer of âGâ actins

v	Arithritis	e	26 bones

f	Bicephalic

g	Rapid spasms

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

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

Solution:
Appendicular skeleton is composed of 30 bones while axial skeleton consists of 26 bones.
F actins helically wound to each other and is a polymer of monomeric G actins.
The coxal bone (hip bone, pelvic bone) is a large, flattened, irregularly shaped bone and is composed of ilium, the ischium and the pubis.
In vertebrate anatomy, ribs are the long curved bones which form the rib cage, part of the axial skeleton. Ribs are headed on both ends and hence they are called BICEPHALIC.
Arthritis is the swelling and tenderness of one or more joints. The main symptoms of arthritis are joint pain and stiffness.
Spasms of skeletal muscles are most common and are often due to overuse and muscle fatigue.

Topic: Biology, Locomotion and Movement, NEET
Match the columns:

Column - A	Column - B

A.	Hypothyroidism	1.	Sleep awake cycle

B.	Vasopressin	2.	Source pituitary gland

C.	Melatonin	3.	Reduction in water loss from the body

D.	MSH	4.	Irregular menstrual cycle

Options:
A A-3, B-4, C-1, D-2
B A-3, B-4, C-2, D-1
C A-4, B-3, C-1, D-2
D A-4, B-3, C-2, D- 1

The correct answer is:
A-4, B-3, C-1, D-2

Solution:
Hypothyroidism: It commonly results from the deficiency of iodine in our diet and is characterized by the enlargement of the thyroid gland (or goitre).
Hypothyroidism during pregnancy: Cretinism. [In cretinism, after birth the growing baby suffers from stunted growth (cretinism), mental retardation, low intelligence quotient, abnormal skin, deaf-mutism, etc.]
Hypothyroidism in adult women: Irregular menstrual cycle.
Melatonin: 1. Melatonin is very important for the regulation of a 24-hour (diurnal) rhythm of our body.
Example: It helps in maintaining the normal rhythms of sleep-wake cycle, body temperature, etc.
2. It also affects - metabolism, pigmentation, menstrual cycle and our immune system.
The pars distalis region of pituitary, commonly called anterior pituitary, produces growth hormone (GH), prolactin (PRL), thyroid-stimulating hormone (TSH), adrenocorticotrophic hormone (ACTH), luteinizing hormone (LH) and follicle-stimulating hormone (FSH). Pars intermedia secretes only one hormone called melanocyte-stimulating hormone (MSH).
Vasopressin acts mainly at the kidney and stimulates resorption of water and electrolytes by the distal tubules and thereby reduces loss of water through urine (diuresis). Hence, it is also called as anti-diuretic hormone (ADH).

Topic: Biology, Chemical Coordination and Integration, NEET
Match the columns I and II, and choose the correct combination from the options given.

Column I	Column II

(a) Haplontic life-cycle	(i) Gymnosperms and Angiosperms

(b) Diplontic life-cycle	(ii) Spirogyra

(c) Haplo-diplontic life	(iii) Bryophytes and Pteridophytes

Options:
A aâi, bâiii, câii
B Â aâiii,Â bâi,Â câii
C aâii, bâi, câiii
D aâiiÂ bâiii,Â câi

The correct answer is:
aâii, bâi, câiii

Solution:
The main form of the haplontic life cycle is haploid and diploid zygote is formed in this cycle for a short period of time and this type of life cycle is found in Spirogyra.
The main form of a diplontic life cycle is diploid which produces gametes and this type of life cycle is found in Gymnosperm and Angiosperm.
In haplo-diplontic life cycle, multicellular diploid sporophyte is present which produces haploid spores via meiosis and it is found in Bryophyta and Pteridophyta.

Topic: Biology, Plant Kingdom, NEET
Match the columns I and II, and choose the correct combination from the options given.

Column I	Column II

(a)	i. Chara

(b)	ii. Dictyota

(c)	iii. Porphyra

Find the CORRECT match.

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

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

Solution:
Dictyota, a type of brown algae belonging to the class Phaephyceae, mainly found in marine habitat and display great variation in size and form. They may be simple branched, filamentous forms or profusely branched form called a kelps. Chlorophyll a,c, carotenoids are present along with varying range of xanthophyll which gives the various shades of brown colour. Storage form of carbohydrates in them are laminarin or mannitol.
Porphyra belongs to the Rhodophyceae, commonly called as red algae. They marine found in warm climates. Food is stored as floridean starch similar to amylopectin and glycogen. It is used as sea-vegetable as nori in Japan and gim in Korea and are rich source of VitB12 .
Chara is a genus of the green algae group, Chlorophyceae. They are multicellular organisms which resemble land plants due to their leaf-like and stem-like structures. Nitrogen fixing bacteria are known to grown on its surface helping in proving nitrogen as nutrients to Chara.

Topic: Biology, Plant Kingdom, NEET
Match the columns I, II and III and mark the appropriate choice.

Column I	Column II	Column III
(a)	Bromine	(i)	Noble metal	(p)	Amalgam
(b)	Gold	(ii)	Crystalline non-metal	(q)	4s24p5
(c)	Mercury	(iii)	Liquid non-metal	(r)	Transition metal
(d)	Iodine	(iv)	Liquid metal	(s)	Violet

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

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

Solution:
Column I	Column II	Column III
(a)	Bromine	(iii)	Liquid non-metal	(q)	4s24p5
(b)	Gold	(i)	Noble metal	(r)	Transition metal
(c)	Mercury	(iv)	Liquid metal	(p)	Amalgam
(d)	Iodine	(ii)	Crystalline non-metal	(s)	Violet

Topic: Chemistry, d and f Block Elements, JEE Main
Match the columns with their respective matches.

Column I	Column II

P.	Inspiratory Capacity	I.	ERV + RV

Q.	Expiratory Capacity	II.	TV + ERV

R.	Functional Residual Capacity	III.	ERV + TV + IRV

S.	Vital Capacity	IV.	TV + IRV

Options:
A P - III, Q - II, R - IV, S - I
B P - IV, Q - I, R - III, S - II
C P - III, Q - IV, R - I, S - II
D P - IV, Q - II, R - I, S - III

The correct answer is:
P - IV, Q - II, R - I, S - III

Solution:
Inspiratory Capacity (IC = TV + IRV)
It is the total volume of air that a person can inspire after a normal expiration.
Expiratory Capacity (EC = TV + ERV)
It is the total volume of air that a person can expire after a normal inspiration.
Functional Residual Capacity (FRC = ERV + RV)
It refers to the volume of air that remains in the lungs after a normal expiration.
Vital Capacity (VC = ERV + TV + IRV)
It is the maximum volume of air that a person can inspire after forced expiration.

Topic: Biology, Breathing and Exchange of Gases, NEET
Match the compound with the metal for which it is used for the process of extraction.

(i) NaCN (a) Titanium

(ii) Iodine (b) Aluminium

(iii) Cryolite (c) Silver ore

Options:
A

(i)-(c),(ii)-(a),(iii)-(b)

B

(i)-(c),(ii)-(b),(iii)-(a)

C

(i)-(a),(ii)-(c),(iii)-(b)

D

(i)-(b),(ii)-(a),(iii)-(c)

The correct answer is:

(i)-(c),(ii)-(a),(iii)-(b)

Solution:
(i) Role of NaCN in the extraction of silver is to do the leaching of silver ore in the presence of air from which the silver is obtained later by replacement 4Ag(s)+8C N â (aq)+2 H 2 O(aq)+ O 2 (g)â Ag (CN) 2 â (aq)+4O H â (aq) 2 Ag (CN) 2 â (aq)+Zn(s)â2Ag(s)+ Zn (CN) 4 2â (aq)
(ii) Ti+ I 2 â Î TiI 4 â Î Ti+ I 2 (Van-Arkel method)
(iii) In the metallurgy of aluminium, cryolite lowers the melting point of the mixture and improves the electrical conductivity of the cell.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Match the compounds given in column I with oxidation states of carbon given in column II and mark the appropriate choice.

Column I	Column II
(p)	C6H12O6	(i)	+3
(q)	CHCl3	(ii)	-3
(r)	CH3CH3	(iii)	+2
(s)	(COOH)2	(iv)	0

Options:
A (p)-(iv), (q)-(iii), (r)-(ii), (s)-(i)
B (p)-(i), (q)-(ii), (r)-(iii), (s)-(iv)
C (p)-(ii), (q)-(iii), (r)-(iv), (s)-(i)
D (p)-(iii), (q)-(ii), (r)-(i), (s)-(iv)

The correct answer is:
(p)-(iv), (q)-(iii), (r)-(ii), (s)-(i)

Solution:
C6H12O6:6x+12+-12=0
x=0
CHCl3:x+1+-3=0
x=+2
CH3CH3:x+3+x+3=0
x=-3
(COOH)2:x+x+-4×2++2=0
x=+3

Topic: Chemistry, Redox Reactions, JEE Main
Match the compounds given in list I with list II and select the suitable option using the code given belows.

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

The correct answer is:
(d)

Solution:
(A) Benzoin condensation When an ethanolic solution of benzaldehyde is heated with strong alkali like $\mathrm{KCN}$ or $\mathrm{NaCN}$, we get benzoin.
(B) Formation of phenolphthalein When phenol is reated with phthalic anhydride in the presence of concentrated $\mathrm{H}_2 \mathrm{SO}_4$, it gives phenolphthalein, an indicator.
(C) Fries rearrangement Phenyl benzoate when heated with anhydrous $\mathrm{AlCl}_3$ in the presence of inert solvent gives ortho-and para-hydroxybenzophenone. In this rearrangement, there is a benzoyl group migration from the phenolic oxygen to an ortho-and para-position.
(D)
Methylsalicylate
(A chief constituent of oil of wintergreen)

Topic: Chemistry, Aldehydes and Ketones, NEET, NEET 2011 (Mains)
Match the compounds given in List I with their characteristic reactions given in List II. Select the correct option.

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

The correct answer is:
(c)

Solution:
(A) $1^{\circ}$ amines give a bad smelling compound, called carbylamine, with alcoholic $\mathrm{KOH}$ and $\mathrm{CHCl}_3$. The reaction is called carbylamine reaction.
$$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2 \stackrel{\mathrm{CHCl}_3+\text { alc } \mathrm{KOH}}{\longrightarrow} \\
& \mathrm{CH}_3\left(\mathrm{CH}_2\right)_3 \mathrm{NC} \\
& \text { bad smelling } \\
&
\end{aligned}
$$
(B) Terminal alkynes like $\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CH}$ give white precipitate with ammoniacal $\mathrm{AgNO}_3$.
$$
\begin{array}{r}
\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CH}+\mathrm{AgNO}_3+\mathrm{NH}_4 \mathrm{OH} \longrightarrow \\
\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CAg} \\
\text { white ppt }
\end{array}
$$
(C) Ester $\left(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOCH}_3\right)$ undergoes alkaline hydrolysis.
$$
\begin{array}{r}
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOCH}_3+\mathrm{NaOH} \longrightarrow \\
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COONa}+\mathrm{CH}_3 \mathrm{OH}
\end{array}
$$
(D) $2^{\circ}$ alcohols (the alcohols that have $\mathrm{CH}_3 \mathrm{CHOH}$ - group) give cloudiness with Lucas reagent (anhy $\mathrm{ZnCl}_2+$ conc $\mathrm{HCl}$ ) within 5 minutes.

Topic: Chemistry, Amines, JEE Main
Match the compounds given in List I with their characteristic reactions given in List II. Select the correct option.

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

The correct answer is:
(c)

Solution:
(A) $1^{\circ}$ amines give a bad smelling compound, called carbylamine, with alcoholic $\mathrm{KOH}$ and $\mathrm{CHCl}_3$. The reaction is called carbylamine reaction.
$$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2 \stackrel{\mathrm{CHCl}_3+\text { alc } \mathrm{KOH}}{\longrightarrow} \\
& \mathrm{CH}_3\left(\mathrm{CH}_2\right)_3 \mathrm{NC} \\
& \text { bad smelling } \\
&
\end{aligned}
$$
(B) Terminal alkynes like $\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CH}$ give white precipitate with ammoniacal $\mathrm{AgNO}_3$.
$$
\begin{array}{r}
\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CH}+\mathrm{AgNO}_3+\mathrm{NH}_4 \mathrm{OH} \longrightarrow \\
\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CAg} \\
\text { white ppt }
\end{array}
$$
(C) Ester $\left(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOCH}_3\right)$ undergoes alkaline hydrolysis.
$$
\begin{array}{r}
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOCH}_3+\mathrm{NaOH} \longrightarrow \\
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COONa}+\mathrm{CH}_3 \mathrm{OH}
\end{array}
$$
(D) $2^{\circ}$ alcohols (the alcohols that have $\mathrm{CH}_3 \mathrm{CHOH}$ - group) give cloudiness with Lucas reagent (anhy $\mathrm{ZnCl}_2+$ conc $\mathrm{HCl}$ ) within 5 minutes.

Topic: Chemistry, Amines, NEET, NEET 2010 (Mains)
Match the compounds listed in Column I with characteristic listed in Column II

Column I	Column II

(1)	BeO(s)	(p)	Amphoteric in nature

(2)	NaHCO3 (crystalline)	(q)	Imparts characteristic colour to Bunsen Flame

(3)	BeCl2	(r)	Produce H2O2 and O2 on reaction with H2O

(4)	CsO2	(s)	Shows H-bonding

(t)	Have a chain structure

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

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

Solution:
(1) BeO(s) Amphoteric in nature
(2) NaHCO3 (crystalline) Amphoteric in nature, Imparts characteristic colour to Bunsen Flame,
Shows H-bonding and Have a chain structure
(3) BeCl2 Have a chain structure
(4) CsO2 Imparts characteristic colour to Bunsen Flame and Produces H2O2 and O2 on reaction with H2O

Topic: Chemistry, s Block Elements, JEE Main
Match the corresponding entries of column 1 with column 2. [Where m is the magnification produced by the mirror]

Column 1	Column 2

(A)	m=-2	(a)	Convex mirror

(B)	m=-12	(b)	Concave mirror

(C)	m=+2	(c)	Real image

(D)	m=+12	(d)	Virtual image

Options:
A A → b and c; B → b and c; C → b and d; D → a and d
B A → a and c; B → a and d; C → a and b; D → c and d
C A → a and d; B → b and c; C → b and d; D → b and c
D A → c and d; B → b and d; C → b and c; D → a and d

The correct answer is:
A → b and c; B → b and c; C → b and d; D → a and d

Solution:
m=-Vu=ff-u
m=-2 then "V" and "u" same given
-2=ff-u=-2f+2u=f
=3f=2u
+3f2=u
For the concave mirror, f negative, u negative use same for all.

Topic: Physics, Ray Optics, JEE Main
Match the disease in Column I with the appropriate items (pathogen/prevention/treatment) in Column II

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

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

Solution:
No Solution Available

Topic: Biology, Human Health and Diseases, NEET, NEET 2008 (Mains)
Match the following.

(1)	Energy of ground state of He+	(p)	6.04 eV

(2)	Potential energy of 1 orbit of H atom	(q)	-27.2 eV

(3)	Kinetic energy of II excited state of He+	(r)	54.4 eV

(4)	Ionisation potential of He+	(s)	-54.4 eV

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

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

Solution:
Energy of G.S. of He+=EHÃZ2
=â13.6Ã4=â54.4âeV
Ionisation potential of He+=Eâ-E1
=â â54.4 =54.4âeV
T.E. of H of 1 orbit =â13.6âeV
2 T.E. = P.E. and K.E=− T. E
P.E.â=â2Ã13.6=â27.2âeV
T.E. of H of II excited state =-13.69Ã4
of Heâ=â6.04âeV
K.E.=â(T.E.)=â â6.04âeV =+6.04âeV

Topic: Chemistry, Structure of Atom, JEE Main
Match the following.

Options:
A $\begin{array}{llll}1 & 4 & 3 & 5\end{array}$
B $\begin{array}{llll}2 & 4 & 1 & 5\end{array}$
C $\begin{array}{llll}5 & 1 & 4 & 2\end{array}$
D $\begin{array}{llll}5 & 1 & 3 & 2\end{array}$

The correct answer is:
$\begin{array}{llll}2 & 4 & 1 & 5\end{array}$

Solution:
Thus, these two are tautomers.
Eclipsed and staggered ethane are two conformations of ethane.
(+)2-butanol and (-)2-butanol are enantiomers as these are non-superimposable mirror images.
Due to difference in the nature of alkyl groups attached to the same functional group, these are called metamers.

Topic: Chemistry, General Organic Chemistry, JEE Main
Match the following

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

The correct answer is:
option (c) is correct

Solution:
Potential of hydrogen electrode at $\mathrm{pH}=10$ is $-0.591 \mathrm{~V}$.
Reduction potential of $\mathrm{Cu}^{2+} / \mathrm{Cu}$ is $0.36 \mathrm{~V}$.
Oxidation potential of $\mathrm{Zn} / \mathrm{Zn}^{2+}$ is $0.76 \mathrm{~V}$.
$\frac{2.303 R T}{F}=\frac{2.303 \times 8.314 \times 298}{96500}=0.059$

Topic: Chemistry, Electrochemistry, JEE Main
Match the following

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

The correct answer is:
(a)

Solution:
I. $\int_{-1}^1 x|x| d x=0 \quad[\because x|x|$ is an odd function $]$
II. Let $I=\int_0^{\pi / 2}\left[1+\left(\log \frac{4+3 \sin x}{4+3 \cos x}\right)\right] d x$
$$
\begin{aligned}
& I=\int_0^{\pi / 2}\left(1+\left[\log \frac{4+3 \sin \left(\frac{\pi}{2}-x\right)}{4+3 \cos \left(\frac{\pi}{2}-x\right)}\right]\right) d x \\
& I=\int_0^{\pi / 2}\left(1+\left(\log \frac{4+3 \cos x}{4+3 \sin x}\right)\right) d x \\
& \Rightarrow \quad 2 I=\int_0^{\pi / 2}\left(2+\log \frac{4+3 \sin x}{4+3 \cos x}\right. \\
& \left.+\log \frac{4+3 \cos x}{4+3 \sin x}\right) d x \\
& \Rightarrow \quad 2 I=\int_0^{\pi / 2}\left(2+\log \frac{(4+3 \sin x)(4+3 \cos x)}{(4+3 \cos x)(4+3 \sin x)}\right) d x \\
& \Rightarrow \quad Z I=\int_0^{\pi / 2}(2+\log 1) d x=\int_0^{\pi / 2} 2 d x=\pi \\
& \Rightarrow \quad I=\frac{\pi}{2} \\
&
\end{aligned}
$$
III. $\int_0^a f(x) d x=\int_0^a f(a-x) d x$
$$
\begin{aligned}
\text { IV. } \int_{-a}^a f(x) d x & =\int_0^a f(x) d x+\int_0^a f(-x) d x \\
& =\int_0^a[f(x)+f(-x)] d x
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main
Match the following.
$\begin{array}{llll}A & B & C & D\end{array}$

Options:
A (v)$\quad$(ii)$\quad$(iv)$\quad$(iii)
B (ii)$\quad$(iii)$\quad$(iv)$\quad$(v)
C (ii)$\quad$ (iii)$\quad$ (i) $\quad$ (v)
D (iii)$\quad$ (ii)$\quad$ (iv)$\quad$ (i)

The correct answer is:
(ii)$\quad$(iii)$\quad$(iv)$\quad$(v)

Solution:
I. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
Here Co is present as $\mathrm{Co}^{3+}$ ion. $\mathrm{NH}_3$ being strong field ligand pair up the unpaired electrons of $\mathrm{Co}$.
II.$\left[\mathrm{Ni}(\mathrm{CO})_4\right]$
Here $\mathrm{Ni}$ is present in its ground state. $\mathrm{CO}$ being a strong field ligand, pair up the unpaired electrons of $\mathrm{Ni}$
$$
{ }_{28} \mathrm{Ni}=[\mathrm{Ar}] 3 d^8 4 s^2
$$
III. $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$
Here $\mathrm{Pt}$ is present as $\mathrm{Pt}^{2+}$ ion.
$$
\begin{aligned}
{ }_{78} \mathrm{Pt} & =[\mathrm{Xe}] 5 d^9 6 s^1 \\
\mathrm{Pt}^{2+} & =[\mathrm{Xe}] 5 d^8 6 \mathrm{~s}^0
\end{aligned}
$$
IV.$\left[\mathrm{CoF}_6{ }^{\beta-}\right.]$
Here $\mathrm{Co}$ is present as $\mathrm{Co}^{3+}$ ion, $\mathrm{F}$ being weak field ligand is unable to pair up its unpaired electrons.
$$
\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 d^6 4 s^0
$$
V.$\left[\mathrm{Fe}(\mathrm{CO})_5\right]$
Here $\mathrm{Fe}$ is present in its ground state.
$$
{ }_{26} \mathrm{Fe}=[\mathrm{Ar}] 3 d^6 4 s^2
$$

Topic: Chemistry, Coordination Compounds, JEE Main