datavorous/entrance-exam-dataset huggingface dataset:
The homologue of ethyne is

Options:
A $\mathrm{C}_2 \mathrm{H}_4$
B $\mathrm{C}_2 \mathrm{H}_6$
C $\mathrm{C}_3 \mathrm{H}_8$
D $\mathrm{C}_3 \mathrm{H}_4$

The correct answer is:
$\mathrm{C}_3 \mathrm{H}_4$

Solution:
$\mathrm{C}_2 \mathrm{H}_2$ and $\mathrm{C}_3 \mathrm{H}_4$ are homologue because they differ by $-\mathrm{CH}_2$ group. Both have triple bond in their molecule.

Topic: Chemistry, Hydrocarbons, JEE Main
The homopolymer formed from 4-hydroxy-butanoic acids is.

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Polymers, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The horizontal component of earth's magnetic field at a place is 3.5×10-5 T. A very long straight conductor carrying current of 2 A in the direction from South east to North West is placed. The force per unit length experienced by the conductor is ________×10-6 N m-1.

Options:

The correct answer is:
35

Solution:
Horizontal component of magnetic field, BH=3.5×10-5 T

Force acting on wire, F=ilBsinθ, where i=2 A

Therefore, Fl=iBsinθ=2×3.5×10-5×12

=35×10-6 N m-1

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (30 Jan Shift 1)
The horizontal component of earth's magnetic field at a place is $\sqrt{ } 3$ times the vertical component. The angle of dip at that place is

Options:
A $\pi / 6$
B $\pi / 3$
C $\pi / 4$
D More than one of the above

The correct answer is:
$\pi / 6$

Solution:
As given the relation between the vertical and horizontal components of the earth's magnetic field is:
$B_H=\sqrt{3} B_V$
Also, $\tan \delta=\mathrm{Bv} / \mathrm{BH}=1 / \sqrt{ } 3$
$\Rightarrow \delta=\tan ^{-1}(1 / \sqrt{ } 3)=30^{\circ}$
So, the angle of dip $\delta=30^{\circ}=\pi / 6$.
So, the correct answer is option (1).

Topic: Physics, Magnetic Properties of Matter, JEE Main
The horizontal component of the Earth's magnetic field is 3.6Ã10-5 T where the dip angle is 60Â°. The magnitude of the Earth's magnetic field is

Options:
A 2.8×10-4 tesla
B 2.1×10-4 tesla
C 7.2×10-5 tesla
D 3.6×10-5 tesla

The correct answer is:
7.2×10-5 tesla

Solution:
Horizontal component of Earth’s field,
H=Bcosθ, since, θ=60°
3.6× 10 −5 =B× 1 2 ⇒B=7.2× 10 −5 tesla

Topic: Physics, Magnetic Properties of Matter, JEE Main
The horizontal range and maximum height attained by a projectile are R and H respectively. If a constant horizontal acceleration a = g/4 is imparted to the projectile due to wind, then its horizontal range and maximum height will be

Options:
A R + H , H 2
B R + H 2 , 2 H
C (R + 2H), H
D (R + H), H

The correct answer is:
(R + H), H

Solution:
T = 2 u y g , H = u y 2 2 g
where T = time of flight of the projectile and ux and uy are respectively the horizontal and vertical components of initial velocity
When a horizontal acceleration is also given to the projectile uy and hence T and H will remain unchanged while the range will become
R = u x T + 1 2 aT 2
= R + 1 2 g 4 4 u y 2 g 2
= R + H

Topic: Physics, Motion In Two Dimensions, NEET
The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

Options:
A $\theta=\tan ^{-1}\left(\frac{1}{4}\right)$
B $\theta=\tan ^{-1}(4)$
C $\theta=\tan ^{-1}(2)$
D $\theta=45^{\circ}$

The correct answer is:
$\theta=\tan ^{-1}(4)$

Solution:
Given, $R=H$
Range $R=\frac{u^2(2 \sin \theta \cos \theta)}{g}$
Height $H=\frac{u^2 \sin ^2 \theta}{2 g}$
Hence, $\frac{u^2(2 \sin \theta \cos \theta)}{g}=\frac{u^2 \sin ^2 a}{2 g}$
$\begin{aligned}
2 \cos \theta & =\frac{\sin \theta}{2} \\
\tan \theta & =4 \\
\theta & =\tan ^{-1}(4)
\end{aligned}$

Topic: Physics, Motion In Two Dimensions, JEE Main
The horizontal range is four times the maximum height attained by a projectile. The angle of projection is

Options:
A $90^{\circ}$
B $60^{\circ}$
C $45^{\circ}$
D $30^{\circ}$

The correct answer is:
$45^{\circ}$

Solution:
$R=4 H \cot \theta$, if $R=4 H$ then $\cot \theta=1 \Rightarrow \theta=45^{\circ}$

Topic: Physics, Motion In Two Dimensions, JEE Main
The hormone capable of replacing the requirement of long photoperiods for flowering is:

Options:
A Ethylene
B Auxin
C Gibberellin
D Cytokinin

The correct answer is:
Gibberellin

Solution:
Gibberellic acid (GA3) was one of the first gibberellins to be discovered and remains the most intensively studied form. All GAs are acidic. They produce a wide range of physiological responses in the plants. Their ability to cause an increase in length of axis is used to increase the length of grapes stalks. Gibberellins, cause fruits like apple to elongate and improve its shape. They also delay senescence. Thus, the fruits can be left on the tree longer so as to extend the market period. GA3 is used to speed up the malting process in brewing industry. Sugarcane stores carbohydrate as sugar in their stems. Spraying sugarcane crop with gibberellins increases the length of the stem, thus increasing the yield by as much as 20 tonnes per acre. Spraying juvenile conifers with GAs hastens the maturity period, thus leading to early seed production. Gibberellins also promotes bolting (internode elongation just prior to flowering) in beet, cabbages and many plants with rosette habit.

Topic: Biology, Plant - Growth and Development, NEET
The ______ hormone helps in the maintenance of electrolytes.

Options:
A Aldosterone
B Androgenic steroids
C Corticoids
D Adrenaline

The correct answer is:
Aldosterone

Solution:
Aldosterone acts mainly at the renal tubules and stimulates the reabsorption of Na+ and water and excretion of K+ and phosphate ions. Thus, aldosterone helps in the maintenance of electrolytes, body fluid volume, osmotic pressure and blood pressure.

Topic: Biology, Chemical Coordination and Integration, NEET
The hormone involved in the metabolism of food material in cereal grain during germination is:

Options:
A Auxins
B Gibberellins
C Cytokinins
D ABA.

The correct answer is:
Gibberellins

Solution:
Gibberellin is a class of plant growth regulators. This hormone was discovered from the fungus Gibberella fujikuroi (Class Deuteromycetes), whose perfect stage was assigned to Fusarium moniliforme. This fungus causes Bakanae disease or Bakane foolish seedling disease of rice. This causes rice plants to grow unusually taller, spindle-shaped, pale green in color and production of rice much lesser.
Chemically, gibberellin is gibbane-ring containing complex chemicals with vivid biological activities in plants. There are more than 30 gibberellins known so far. They all are acidic in nature and are called Gibberellic Acid (GA).
The main physiological role of GA includes:
1. Stem and internode elongation. Thus, by applying GA, a genetically dwarf plant can be made into a tall plant.
2. Inducing parthenocarpy for pome type fruits.
3. Help in overcoming natural dormancy in seeds, tubers, rhizomes, etc.
4. Controlling flowering in the long-day plants.
5. Production of male flowers in cannabis like plants.
6. Breakdown of sugar by activation of alpha-amylase in germinating barley seeds, which promotes germination. This is another characteristic property of GAs.
Auxin helps in increased protein synthesis, rhizogenesis, parthenocarpy induction, etc.
Cytokinins are a class of hormones that induce cell division, break apical dominance, expanding leaves, etc.

Topic: Biology, Plant - Growth and Development, NEET
The __________ hormone is secreted by the __________ cells of kidneys and stimulates the formation of erythrocyte.

Options:
A Erythropoietin, JG
B Erythropoietin, Interstitial fibroblasts
C Renin, JG
D Renin. Macula densa

The correct answer is:
Erythropoietin, Interstitial fibroblasts

Solution:
Erythropoietin is a glycoprotein secreted mainly by the kidney in response to cellular hypoxia; it stimulates red blood cell production (erythropoiesis) in the bone marrow. The interstitial fibroblasts are produced in the kidney are are present associated with the peritubular capillary and proximal convoluted tubule.

Topic: Biology, Excretory Products and their Elimination, NEET
The hormone released during post-ovulatory phase helps in

Options:
A Maintenance of pregnancy
B Development of secondary sexual characters
C Ovulation
D Increase in production of FSH and LH

The correct answer is:
Maintenance of pregnancy

Solution:
The Graafian follicle releases an egg during ovulation and the remains of the follicle get closed and form corpus luteum. This yellow pigmented corpus luteum secretes progesterone hormone under LH's influence and stimulates the uterus to thicken even more in preparation for implantation of a fertilized egg.

Topic: Biology, Human Reproduction, NEET
The hormone responsible for maintaining the diurnal rhythm and sexual behaviour in animals is

Options:
A FSH
B aldosterone
C androgens
D melatonin

The correct answer is:
melatonin

Solution:
The pineal gland is located on the dorsal side of forebrain. Pineal secretes a hormone called melatonin. Melatonin plays a very important role in the regulation of a 24-hour (diurnal) rhythm of our body. For example, it helps in maintaining the normal rhythms of the sleep-wake cycle, body temperature. In addition, melatonin also influences metabolism, pigmentation, the menstrual cycle, as well as our defense capability.

Topic: Biology, Chemical Coordination and Integration, NEET
The hormone which suppresses the immune system is produced by

Options:
A zona fasciculata of adrenal cortex.
B zona glomerulosa of adrenal cortex.
C zona reticularis of adrenal cortex.
D Both (A) and (C)

The correct answer is:
zona fasciculata of adrenal cortex.

Solution:
Glucocorticoids stimulate gluconeogenesis, lipolysis and proteolysis; and inhibit cellular uptake and utilisation of amino acids. Cortisol is also involved in maintaining the cardiovascular system as well as the kidney functions. Glucocorticoids, particularly cortisol, produce anti-inflammatory reactions and suppresses the immune response. Cortisol stimulates the RBC production.

Topic: Biology, Chemical Coordination and Integration, NEET
The hottest region of Bunsen flame shown in the figure below is:

Options:
A Region 3
B Region 4
C Region 1
D Region 2

The correct answer is:
Region 2

Solution:
The hottest region of the Bunsen flame shown in the figure is the region 2. The hottest part of the Bunsen flame is present on the one third height from the bottom of the flame.

The all regions are represented in this figure.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2016 (03 Apr)
The houses on one side of a road are numbered using consecutive even numbers. The sum of the numbers of all the houses in that row is 170 . If there are at least 6 houses in that row and a is the number of the sixth house, then

Options:
A $2 \leq a \leq 6$
B $8 \leq a \leq 12$
C $14 \leq a \leq 20$
D $22 \leq \mathrm{a} \leq 30$

The correct answer is:
$14 \leq a \leq 20$

Solution:
Let house no are $\alpha, \alpha,+2, \alpha+4, \alpha+6, \alpha+8, \alpha+10, \ldots .$ $\alpha+10=a \Rightarrow \alpha=a-10\quad ......\text {(1)}$
House no. will be $(+)$
$\begin{array}{l}
\Rightarrow \alpha=a-10>0 \\
\Rightarrow \alpha>10 \\
\Rightarrow \alpha \geq 12 \text { as a is each too }\quad ......\text {(2)}
\end{array}$
Now $S_{n}=\frac{n}{2}[2 \alpha+(n-1) d]$
$\begin{array}{l}
170=\frac{n}{2}[2 \alpha+(n-1)(2)] \\
=n(\alpha+(n-1)) \\
=n(a-10+n-1) \\
=n(a-11+n) \\
\Rightarrow n^{2}+n(a-11)-170=0 \\
\Rightarrow n=\frac{(11-a) \pm \sqrt{(a-11)^{2}+680}}{2} \quad ......\text {(3)}\\
\because n \geq 6 \\
\Rightarrow \frac{(11-a) \pm \sqrt{(a-11)^{2}+680}}{2} \geq 6 \\
\Rightarrow a \leq \frac{800}{24}\quad ......\text {(4)}
\end{array}$
From $(2)$ and $(4) \Rightarrow 12 \leq a \leq 32$
Now checking through $(3)$ for $a=12,14, \ldots . .$;
we have $a=18, n=10$ and $S_{n}=170$

Topic: Mathematics, Permutation Combination, JEE Main
The human body cell contains a huge number of enzymes which are biocatalyst and help in smooth metabolic activities. What will be the body temperature and pH value for the enzymes to work at its optimum level?

Options:
A 0 oC and pH 2
B 12 oC and pH 5
C 37 oC and pH 7
D 25 oC and pH 7

The correct answer is:
37 oC and pH 7

Solution:
Enzymes functionality depends on the temperature and pH as these two are factors controlling its activity. The biocatalysts display the highest activity at a particular temperature and pH which is generally called the optimum temperature and pH for that particular enzyme. Since the body temperature of the human body remains constant at 37 oC and approx. pH value is at 7, these values are considered as an optimum conditions for enzymes of the human body to work efficiently.

Topic: Chemistry, Biomolecules, NEET
The human body does not produce:

Options:
A enzymes
B DNA
C vitamins
D hormones

The correct answer is:
vitamins

Solution:
The organic compounds other than carbohydrates and proteins which maintain normal growth and nutrition in the human body (but not produced in human body) are called vitamins.

Topic: Chemistry, Biomolecules, NEET, NEET 2006
The human body does not produce

Options:
A Enzymes
B DNA
C Vitamins
D Hormones

The correct answer is:
Vitamins

Solution:
Vitamins are the group of biomolecules which are used in the normal metabolic process and for life growth and health of human beings and animals. Vitamins are gained by body from the different sources to avoid any disease. Vitamin are found in vegetables, fish, meat, egg, fruits, sunlight etc.

Hence, the human body does not produce vitamins.

Topic: Chemistry, Biomolecules, JEE Main
The hybridisation of the central atom will change when

Options:
A NH3 combines with H+
B H3BO3 combines with OH-
C NH3 forms NH2-
D H2O combines with H+

The correct answer is:
H3BO3 combines with OH-

Solution:
Here hybridisation changes from sp2 to sp3 when H3BO3 combines with OH-
H3BO3sp2+OH-→[B(OH)4]-sp3
In rest other reactions hybridisation does not change
N..H3sp3+H+→NH4+sp3
2NH3sp3⇌2NH2-sp3+NH4+sp3
H2Osp3+H+⇌H3O+sp3

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The hybridisations of the atomic orbitals of nitrogen in NO2-,NO2+ and NH4+ respectively are.

Options:
A sp3,sp2 and sp
B sp, sp2 and sp3
C sp3, sp and sp2
D sp2, sp and sp3

The correct answer is:
sp2, sp and sp3

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (20 Jul Shift 2)
The hybridization and magnetic behaviour of cobalt ion in [Co(NH3)6]3+ complex, respectively is

Options:
A sp3d2 and diamagnetic
B d2sp3 and paramagnetic
C d2sp3 and diamagnetic
D sp3d2 and paramagnetic

The correct answer is:
d2sp3 and diamagnetic

Solution:
In [Co(NH3)6]3+, Co is in +3 oxidation state with the configuration 3d6. In the presence of NH3 a strong field ligand, the 3d electrons undergo pairing and leaving two d-orbitals empty. Hence, the hybridisation is d2sp3 which forms an inner orbital octahedral complex. The complex is diamagnetic as it has no unpaired electrons.

So, the correct option is C

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (24 Jan Shift 2)
The hybridization and magnetic nature of MnCN64- and FeCN63-, respectively are :

Options:
A sp3d2 and diamagnetic
B d2sp3 and paramagnetic
C sp3d2 and paramagnetic
D d2sp3 and diamagnetic

The correct answer is:
d2sp3 and paramagnetic

Solution:
Magnetic property: Due to presence of one unpaired electrons in 3d shell, [Fe(CN)6]3− ion is paramagnetic in nature.

Ã¢â¬â¹Ã¢â¬â¹Ã¢â¬â¹Ã¢â¬â¹

∴ hybridisation is d2sp3
and due to presence of unpaired e- complex is paramagnetic in nature

∴ hybridisation is d2sp3
and due to presence of unpaired e- complex paramagnetic in nature

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (25 Feb Shift 1)
The hybridization of orbitals of $\mathrm{N}$ atom in $\mathrm{NO}_3^{-}, \mathrm{NO}_2^{+}$and $\mathrm{NH}_4^{+}$are respectively :

Options:
A
$s p^2, \mathrm{sp}, \mathrm{sp}^3$

B
$s p, \mathrm{sp}^3, \mathrm{sp}^2$

C
$\mathrm{sp}^2, \mathrm{sp}^3, \mathrm{sp}$

D
$s p, \mathrm{sp}^2, \mathrm{sp}^3$

The correct answer is:
$s p^2, \mathrm{sp}, \mathrm{sp}^3$

Solution:
No Solution Available

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2011
The hybridization of P exhibited in PF5 is spxdy. The value of y is

Options:

The correct answer is:
1

Solution:
The hybridization is sp3d hybridization and phosphorous atom forms five sp3d hybrid orbitals. Five hybrid orbitals will be used to form bonds with five fluorine atoms. There are 5 sigma bonds in this compound.

PF5⇒sp3d hybridisation

(5 sigma bonds, zero lone pair on central atom) Value of y=1

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2022 (28 Jun Shift 1)
The hybridization of the central atom and the shape of $\left[\mathrm{IO}_{2} \mathrm{~F}_{5}\right]^{2-}$ ion respectively, are -

Options:
A
B
C
D

The correct answer is:

Solution:
$\left[\mathrm{IO}_{2} \mathrm{~F}_{5}^{-}\right]^{2-}$ ion
Hybridisation is $\mathrm{sp}^{3} \mathrm{~d}^{3}$ shape is pentagonal bipyramidal
Double bond cause more repulsion so they would be on Axial position $180^{\circ}$ angle to each other so shape is

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The hydration of propyne results in formation of

Options:
A
Acetone

B
Propanol-1

C
Propene

D
Propanal

The correct answer is:
Acetone

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2012 (26 May Online)
The hydride $(\mathrm{H}^-)$ ions are isoelectronic with

Options:
A $\mathrm{He}^{+}$
B $\mathrm{H}^{+}$
C He
D $\mathrm{Be}$

The correct answer is:
He

Solution:
$\mathrm{H}^{-}=1 s^2$ and $\mathrm{He}^{+}=1 s^2$.

Topic: Chemistry, Structure of Atom, JEE Main
The hydride ion $\mathrm{H}^{-}$is a stronger base than its hydroxide ion $\mathrm{OH}^{-}$ Which of the following reactions will occur if sodium hydride $(\mathrm{NaH})$ is dissolved in water

Options:
A $\mathrm{H}^{-}(a q)+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{-}(a q)$
B $\mathrm{H}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{OH}^{-}(a q)+\mathrm{H}_2(g)$
C $\mathrm{H}^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(l) \rightarrow$ No reaction
D None of these

The correct answer is:
$\mathrm{H}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{OH}^{-}(a q)+\mathrm{H}_2(g)$

Solution:
$\underset{\text { base1 }}{\mathrm{H}^{-}(a q)}+\underset{\text { acid 2 }}{\mathrm{H}_2 \mathrm{O}(\mathrm{I})} \rightarrow \underset{\text { base2 }}{\mathrm{OH}^{-}(a q)}+\underset{\text { acid1 }}{\mathrm{H}_2(g)}$

Topic: Chemistry, Hydrogen, JEE Main
The hydride that is NOT electron deficient is:

Options:
A $\mathrm{SiH}_{4}$
B $\mathrm{B}_{2} \mathrm{H}_{6}$
C $\mathrm{GaH}_{3}$
D $\mathrm{AlH}_{3}$

The correct answer is:
$\mathrm{SiH}_{4}$

Solution:
$\mathrm{SiH}_{4}:$ Electron precise hydride $\mathrm{B}_{2} \mathrm{H}_{6}, \mathrm{GaH}_{3}$ and $\mathrm{Al} \mathrm{H}_{3}$ are electron deficient

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2019 (11 Jan Shift 2)
The hydrocarbon of molecular mass 72 gives a single monochloride and two dichlorides on photochlorination is

Options:
A pentane
B 2-methylbutane
C 2,2-dimethylpropane
D none of the above

The correct answer is:
2,2-dimethylpropane

Solution:
The hydrocarbon of mass 72 is C5H12 . Since there is one monoclhoro derivative, the compound has twelve equivalent H's in four equivalent CH3′s . The compound is (CH3)4C (2,2-dimethylpropane) and the chloro derivative are as follows

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The hydrocarbon that cannot be prepared effectively by Wurtz reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
In this reaction alkyl halides are treated with sodium metal in dry ether to produce higher symmetrical alkanes.

Topic: Chemistry, Hydrocarbons, JEE Main
The hydrocarbon which can react with sodium in liquid ammonia is

Options:
A
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CCH}_2 \mathrm{CH}_2 \mathrm{CH}_3$

B
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}$

C
$\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_3$

D
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CCH}_2 \mathrm{CH}_3$

The correct answer is:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}$

Solution:
$$
\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{C} \equiv \mathrm{CH} \stackrel{\mathrm{Na} / \mathrm{Liq} 9 \mathrm{NH}}{\Delta} \longrightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \stackrel{\ominus}{\mathrm{C}} \mathrm{Na}^{\oplus}
$$
It is a terminal alkyne, having acidic hydrogen.
Note: Solve it as a case of terminal alkynes, otherwise all alkynes react with $\mathrm{Na}$ in liq. $\mathrm{NH}_3$.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2008
The hydrocarbon with seven carbon atoms containing a neopentyl and a vinyl group is:

Options:
A 2,2-dimethyl-4-pentene
B 4,4-dimethylpent-1-ene
C Isopropyl-2-butene
D 2,2-dimethyl-3-pentene

The correct answer is:
4,4-dimethylpent-1-ene

Solution:
Hydrocarbon of 7 carbons having neopentyl and a vinyl group.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2016 (09 Apr Online)
The hydrogen bond is strongest in

Options:
A \( \mathrm{O}-\mathrm{H} \ldots \ldots \mathrm{F} \)
B \( \mathrm{O}-\mathrm{H} \ldots \ldots \mathrm{H} \)
C \( \mathrm{F}-\mathrm{H} \ldots \ldots \mathrm{F} \)
D \( \mathrm{O}-\mathrm{H} \ldots \ldots . \mathrm{O} \)

The correct answer is:
\( \mathrm{F}-\mathrm{H} \ldots \ldots \mathrm{F} \)

Solution:
The strength of hydrogen bond depends upon the coulombic interaction between the electronegativity of the attached atom and hydrogen. Fluorine is the most electronegative element. Thus, hydrogen bond will be strongest in F-H.......F.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The hydrogen electrode is dipped in a solution of pH=3 at 25°C. The potential of the electrode will be -______×10-2 V. 2.303RTF=0.059 V Round off the answer to the nearest integer.

Options:

The correct answer is:
18

Solution:
Given,
pH =3

⇒- log [ H+] = 3[ H+] =10-3

The half cell reaction is

2H(aq.)++2e-→H2g

We know,

the Nernst equation is
Ered=E0red+0.059 log [H+]=0+0.059 log 10−3=+ 0.059 ×(−3) =− 0.177 V

= -17.7×10-2V.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The hydrogen ion concentration in weak acid of dissociation constant $K_{\mathrm{a}}$ and concentration $c$ is nearly equal to

Options:
A $\sqrt{K_a / c}$
B $c / K_b$
C $K_a c$
D $\sqrt{K_a c}$

The correct answer is:
$\sqrt{K_a c}$

Solution:
$\mathrm{HA} \rightleftarrows \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{A}^{-}$
At equilibrium $\quad C(1-\alpha) \quad C \alpha \quad C \alpha$
$$
\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\mathrm{C} \alpha \quad \text { and } \mathrm{K}_{\mathrm{a}}=\frac{\mathrm{C} \alpha \times \mathrm{C} \alpha}{\mathrm{C}(1-\alpha)}=\mathrm{C} \alpha^2 \Rightarrow \alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{c}}}
$$
So $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\mathrm{C} \sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{c}}}=\sqrt{\mathrm{K}_{\mathrm{a}} \mathrm{c}}$
So $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\mathrm{C} \sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{c}}}=\sqrt{\mathrm{K}_{\mathrm{a}} \mathrm{c}}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The hydrolysis of ester in alkaline medium is a

Options:
A First-order reaction with molecularity \( 1 \)
B second-order reaction with molecularity \( 2 \)
C First-order reaction with molecularity \( 2 \)
D Second-order reaction with molecularity \( 1 \)

The correct answer is:
second-order reaction with molecularity \( 2 \)

Solution:
Alkaline hydrolysis of ester (saponification) is:

CH3COOC2H5 + NaOH →C2H5OH +CH3COONa

From a balanced chemical equation, the molecularity of the reaction is 2.

From rate law:

rate =kCH3COOC2H5NaOH

Reaction order is 2.

So it is a second-order bimolecular reaction.

Topic: Chemistry, Chemical Kinetics, JEE Main