datavorous/entrance-exam-dataset huggingface dataset:
In this figure the resistance of the coil of galvanometer G is 2 Ω. The emf of the cell is 4 V. The ratio of potential difference across C1 and C2 is

Options:
A 1
B 45
C 54
D 34

The correct answer is:
45

Solution:
Current flowing through both capacitors will be zero at steady state.

The potential difference across capacitor C1 will be the sum of the potential difference across the resistor 6 Ω and the galvanometer resistance. Mathematically,

VC1=i6Ω+RG ...1

Similarly, the potential difference across capacitor C2 will be the sum of the potential difference across the resistor 8 Ω and the galvanometer resistance. Mathematically,

VC2=iRG+8Ω ...2

Divide equation (1) by equation (2) to obtain the required ratio.

VC1VC2=i6Ω+RGiRG+8 Ω= 6Ω+2 Ω2 Ω+8 Ω= 45

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (08 Apr Shift 1)
In this molecule
X=no . of 1o- carbon atoms
Y=no . of 2o- carbon atoms
Z=no . of 3o- carbon atoms
M=no . of 4o - carbon atoms
What is the value of X + Y + Z + M 2 ?

Options:

The correct answer is:
8

Solution:
X = 7
Y = 5
Z = 3
M = 1
= X + Y + Z + M 2
= 16 2
=8

Topic: Chemistry, Hydrocarbons, JEE Main
In this sequence z is mainly –

Options:
A Isobutylene
B Isobutane
C Isobutyl acetate
D Ethyl tert. butyl ether

The correct answer is:
Isobutylene

Solution:
CH2CH2OHâNaCH3CH2ONa âE2(CH3)3C-Br CH3 |CH3-C=CH2

Topic: Chemistry, Hydrocarbons, NEET
In Thomson's experiment to determine $\frac{e}{m}$ of an electron, it is found that an electron beam having a kinetic energy of $45.5 \mathrm{eV}$ remains undeflected, when subjected to crossed electric and magnetic fields. If $E=1 \times 10^3 \mathrm{Vm}^{-1}$, the value of $B$ is (mass of the electron is $9.1 \times 10^{-31} \mathrm{~kg}$ )

Options:
A $2.5 \times 10^{-3} \mathrm{~Wb} \mathrm{~m}^{-2}$
B $5.0 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$
C $2.5 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$
D $1.0 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$

The correct answer is:
$2.5 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$

Solution:
The kinetic energy
$K=\frac{1}{2} m v^2$
$v^2=\frac{2 K}{m}$
$v^2=\frac{2 \times 45.5 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$
$v^2=16 \times 10^{12}$
$v=4 \times 10^6$
Again velocity, $v=\frac{E}{B}$
$4 \times 10^6=\frac{1 \times 10^3}{B}$
$B=\frac{1 \times 10^3}{4 \times 10^6}$
$B=2.5 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$

Topic: Physics, Magnetic Effects of Current, AP EAMCET, AP EAMCET 2010
In Thomson's experiment to determine $\frac{e}{m}$ of an electron, it is found that an electron beam having a kinetic energy of $45.5 \mathrm{eV}$ remains undeflected, when subjected to crossed electric and magnetic fields. If $E=1 \times 10^3 \mathrm{Vm}^{-1}$, the value of $B$ is (mass of the electron is $9.1 \times 10^{-31} \mathrm{~kg}$ )

Options:
A $2.5 \times 10^{-3} \mathrm{~Wb} \mathrm{~m}^{-2}$
B $5.0 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$
C $2.5 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$
D $1.0 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$

The correct answer is:
$2.5 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$

Solution:
The kinetic energy
$K=\frac{1}{2} m v^2$
$v^2=\frac{2 K}{m}$
$v^2=\frac{2 \times 45.5 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$
$v^2=16 \times 10^{12}$
$v=4 \times 10^6$
Again velocity, $v=\frac{E}{B}$
$4 \times 10^6=\frac{1 \times 10^3}{B}$
$B=\frac{1 \times 10^3}{4 \times 10^6}$
$B=2.5 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$

Topic: Physics, Magnetic Effects of Current, TS EAMCET, TS EAMCET 2010
In throwing of two dice, what is the number of exhaustive events ?

Options:
A 6
B 12
C 36
D 18

The correct answer is:
36

Solution:
A dice has six faces. So, in throwing of two dice, the number of exhaustive events is $6 \times 6=36$.

Topic: Mathematics, Probability, JEE Main
In Tollen's test for aldehyde, the overall number of electron(s) transferred to the Tollen's reagent formula AgNH32+ per aldehyde group to form silver mirror is ___.(Round off to the Nearest integer)

Options:

The correct answer is:
2

Solution:
AgNO3+NaOH→AgOH+NaNO3

2AgOH→Ag2O+H2O

Ag2O+4NH3+H2O→2AgNH32++2O⊝H

Tollens' reagent is an alkaline solution of ammoniacal silver nitrate and is used to test for aldehydes. Tollens' test, also known as silver-mirror test, is a qualitative laboratory test used to distinguish between an aldehyde and a ketone. It exploits the fact that aldehydes are readily oxidized (see oxidation), whereas ketones are not.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2021 (18 Mar Shift 2)
In tossing three coins at a time, what is the probability of getting at most one head?

Options:
A $\frac{3}{8}$
B $\frac{7}{8}$
C $\frac{1}{2}$
D $\frac{1}{8}$

The correct answer is:
$\frac{1}{2}$

Solution:
Possible samples are as follows $\{$ HHH, HTH, HHT, THH, TTH, THT, HTT, TTT\}
Let $A$ be the event of getting one head. Let $\mathrm{B}$ be the event of getting no head. Favourable outcome for
$A=\{T T H, T H T, H T T\}$
Favourable outcome for
$$
B=\{T T T\}
$$
Total no. of outcomes $=8$
$$
\therefore \quad P(A)=\frac{3}{8}, P(B)=\frac{1}{8}
$$
$\therefore$ Required probability $=$ Probability of getting one head
+ Probability of getting no head
$$
=P(A)+P(B)=\frac{3}{8}+\frac{1}{8}=\frac{4}{8}=\frac{1}{2}
$$

Topic: Mathematics, Probability, JEE Main
In triangle $A B C$, the value of $\sin 2 A+\sin 2 B+\sin 2 C$ is equal to

Options:
A $4 \sin A \cdot \sin B \cdot \sin C$
B $4 \cos A \cdot \cos B \cdot \cos C$
C $2 \cos A \cdot \cos B \cdot \cos C$
D $2 \sin A \cdot \sin B \cdot \sin C$

The correct answer is:
$4 \sin A \cdot \sin B \cdot \sin C$

Solution:
We know that $A+B+C=180^{\circ}$ (in $\triangle A B C$ )
Now, $\sin 2 A+\sin 2 B+\sin 2 C$
$\begin{aligned}
& =2 \sin (A+B) \cos (A-B)+2 \sin C \cos C \\
& =2 \sin (\pi-C) \cos (A-B)+2 \sin C \cos (\pi-A+B) \\
& =2 \sin C \cos (A-B)-2 \sin C \cos (A+B) \\
& =2 \sin C\{\cos (A-B)-\cos (A+B)\} \\
& =2 \sin C\{2 \sin A \sin B\}=4 \sin A \sin B \sin C .
\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
In two different experiments, an object of mass 5 kg moving with a speed of 25 ms-1 hits two different walls and comes to rest within
(i) 3 second, (ii) 5 seconds, respectively.
Choose the correct option out of the following :

Options:
A Impulse and average force acting on the object will be same for both the cases.
B Impulse will be same for both the cases but the average force will be different.
C Average force will be same for both the cases but the impulse will be different.
D Average force and impulse will be different for both the cases.

The correct answer is:
Impulse will be same for both the cases but the average force will be different.

Solution:
Impulse = Change in momentum

I=ΔP

Therefore, in both the cases the impulse is same.

Average force is defined as,

Favg =ΔPΔt

Since the time in the two cases is different, the average force will be different.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2022 (28 Jul Shift 1)
In two separate collisions, the coefficient of restitutions $e_1$ and $e_2$ are in the ratio $3: 1$. In the first collision the relative velocity of approach is twice the relative velocity of separation, then the ratio between relative velocity of approach and the relative velocity of separation in the second collision is

Options:
A $1: 6$
B $2: 3$
C $3: 2$
D $6: 1$

The correct answer is:
$6: 1$

Solution:
Coefficient of restitution is given by
$e=\frac{\text { relative velocity of separation }}{\text { relative velocity of approach }}$
we have
$e_1=\frac{1}{2}$
and $e_2=\left(\frac{\text { relative velocity of separation }}{\text { relative velocity of approach }}\right)_2$
Given, $\quad \frac{e_1}{e_2}=\frac{3}{1} \Rightarrow \frac{e_2}{e_1}=\frac{1}{3}$
$\Rightarrow \quad\left(\frac{\text { relative velocity of separation }}{\text { relative velocity of approach }}\right)_2$
$=\frac{1}{3} \times \frac{1}{2}=\frac{1}{6}$
$\Rightarrow \quad\left(\frac{\text { relative velocity of approach }}{\text { relative velocity of separation }}\right)_2=\frac{6}{1}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
In two setups of Young's double-slit experiment, if lights used have the ratio of wavelengths as 1:2, then fringes of equal width are observed. If the ratio of slit separation is 2:1 then the ratio of the distances between slit and screen D1:D2 will be

Options:
A 1:1

B 1:4

C 2:1

D 4:1

The correct answer is:
4:1

Solution:
As β=λDd
For equal fringe widths
λ1D1d1=λ2D2d2
Or D1D2=d1d2×λ2λ1=21×21=4
∴D1D2=41

Topic: Physics, Wave Optics, JEE Main
In two similar wires of tension 16 N and T, 3 beats are heard, then T=

Options:
A 49 N

B 25 N
C 64 N
D 80 N

The correct answer is:
49 N

Solution:
Here ,T1=16 N, T2=?
As per the choice given, T2>T1
â´n2>n1, n2-n1=3 â¦(i)
As nâT
â´ n2n1=T16=T4
If n1 corresponds to 4:n2 corresponds to 3+4=7, which is T. Therefore, T=49 N

Topic: Physics, Waves and Sound, JEE Main
In U.C.M., when time interval δ→0 , the angle between change in velocity δv and linear velocity (v) will be

Options:
A 0o
B 90o
C 180o
D 45o

The correct answer is:
90o

Solution:
The direction of change in velocity δv is given by
ϕ=180o-θ2=90o-θ2 …. (i)
This can be shown graphically as

For small interval i.e. δt→0 , then angle between v1 and v2 is very small i.e., θ≈0o .
So, from Eq. (i) we get
ϕ=90o

Topic: Physics, Motion In Two Dimensions, JEE Main
In Van Arkel method, if \( \mathrm{I}_{2} \) is introduced at \( 1800 \mathrm{~K} \) over impure zirconium metal, the product will be

Options:
A iodide of the metal.
B pure metal.
C impurities react with iodine.
D none of these.

The correct answer is:
none of these.

Solution:
Zrimpure+2I2⟶<1800 KZrl4;

Zrl4⟶>1800 KZrpure+2I2

and hence over 1800°C practically no reaction can take place between Zr and I2. Therefore, option 4 is correct.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
In van der Waal's equation of state of the gas law, constant 'a' and 'b' are used. The constant 'b' is a measure of

Options:
A Volume occupied by the molecules
B Intermolecular attraction
C Intermolecular repulsions
D Intermolecular collisions per unit volume

The correct answer is:
Volume occupied by the molecules

Solution:
Vander waal's constant for volume correction b is the measure of the effective volume occupied by the gas molecule.

Topic: Chemistry, States of Matter, JEE Main
In Vander Waal's equation of state for a non-ideal gas, the term that accounts for intermolecular forces is

Options:
A $(V-b)$
B $(R T)^{-1}$
C $\left(P+\frac{a}{V^2}\right)$
D $R T$

The correct answer is:
$\left(P+\frac{a}{V^2}\right)$

Solution:
$\left(P+\frac{a}{V^2}\right)$

Topic: Chemistry, States of Matter, JEE Main
In Vander Waals equation of state of the gas law, the constant ' $b$ ' is a measure of

Options:
A
intermolecular repulsions

B
intermolecular collisions per unit volume

C
Volume occupied by the molecules

D
intermolecular attraction

The correct answer is:
Volume occupied by the molecules

Solution:
In Vander Waals equation ' $b$ ' is the excluded volume i.e. the volume occupied by the molecules

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2004
In vegetative propagation by tubers, which of the following remains constant through generation?

Options:
A Morphology
B Vigour only
C Vigour and morphology
D Morphology vigour and disease resistance

The correct answer is:
Morphology vigour and disease resistance

Solution:
Vegetative propagules are the plant parts or organs that are involved in the process of reproduction even though their primary function is not it. The reproductive process is called vegetative propagation as the non-reproductive parts/organs (leaf/stem/roots) are involved. The ginger plant propagates by rhizome and grows in a horizontal direction.
The main advantage of vegetative propagation is that the genetic material is passed to the next generation, thus, the new propagules are clones of each other bear the same morphology, vigor and disease resistance as the mother plant.

Topic: Biology, Reproduction in Organisms, NEET
In virus-infected plants the meristematic tissues in both apical and axillary buds are free of virus because

Options:
A
the dividing cells are virus resistant

B
meristems have antiviral compounds

C
the cell division of meristems are faster than the rate of viral multiplication

D
viruses cannot multiply within meristem cell(s)

The correct answer is:
the cell division of meristems are faster than the rate of viral multiplication

Solution:
In virus infected plant the meristematic tissues in both apical and axillary buds are free of virus due to faster multiplication of meristematic cell is than replication of viruses.
Gene silencing is the main reason behind it. Neither dividing cells are virus resistant nor the meristems have antiviral compounds.

Topic: Biology, Strategies for Enhancement in Food Production
In weak electrolytic solution, degree of ionization

Options:
A Will be proportional to dilution
B Will be proportional to concentration of electrolyte
C Will be proportional to the square root of dilution
D Will be reciprocal to the dilution

The correct answer is:
Will be proportional to the square root of dilution

Solution:
Ostwald's dilution formula is $\alpha^2=K(1-\alpha) / C$ but for weak electrolyte $\alpha$ is very small. So that $(1-\alpha)$ is neglected for weak electrolytes. So for weak electrolyte the dilution formula is $\alpha=\sqrt{\frac{K}{c}}$.

Topic: Chemistry, Ionic Equilibrium, JEE Main
In what condition does a reaction proceed to near completion?

Options:
A When the concentrations of reactants and products are equal
B When only small amounts of products are formed
C When most of the reactants remain unchanged at equilibrium
D When there is no change in the concentrations of reactants or products

The correct answer is:
When the concentrations of reactants and products are equal

Solution:
Reactions that proceed nearly to completion have concentrations of reactants and products that are comparable when the system is in equilibrium.

Topic: Chemistry, Chemical Equilibrium, NEET
In what ratio the line $y-x+2=0$ divides the line joining the points $(3,-1)$ and $(8,9)$

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

The correct answer is:
$2: 3$

Solution:
Given, equation of line $y-x+2=0$ and co-ordinates of points $\left(x_1, y_1\right)=(3,-1)$ and $\left(x_2, y_2\right)=(8,9)$. We know that if the ratio in which a line $a x+b y+c=0$ is divided by points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is $\lambda: 1$, then intersecting point $\left(\frac{\lambda x_2+x_1}{\lambda+1}, \frac{\lambda y_2+y_1}{\lambda+1}\right)$ lies on $a x+b y+c=0$. Thus any point on the line joining $(3,-1)$ and $(8,9)$ dividing it in the ratio $\lambda: 1$ is $\left(\frac{8 \lambda+3}{\lambda+1}, \frac{9 \lambda-1}{\lambda+1}\right)$ and if it lies on $y-x+2=0$, then $\frac{9 \lambda-1}{\lambda+1}-\frac{8 \lambda+3}{\lambda+1}+2=0$ or $9 \lambda-1-(8 \lambda+3)+2(\lambda+1)=0$
$3 \lambda-2=0, \lambda=\frac{2}{3} \text { i.e. ratio is } 2: 3 \text {. }$

Topic: Mathematics, Straight Lines, JEE Main
In what ways lithium shows similarities to magnesium in its chemical behaviour ?

Options:

None

Solution:
Lithium resembles magnesium mainly due to same charge/radius ratio or polarizing power. The main points of similarity are:
(a) Both \(\mathrm{LiOH}\) and \(\mathrm{Mg}(\mathrm{OH})_2\) are weak bases.
(b) Both form ionic nitrides when heated in atmosphere of nitrogen, \(\mathrm{Li}_3 \mathrm{~N}\) and \(\mathrm{Mg}_3 \mathrm{~N}_2\).
(c) The hydroxides and carbonates of both of them decompose on heating.
(d) Both \(\mathrm{Li}\) and \(\mathrm{Mg}\) do not form solid bicarbonates.
(e) \(\mathrm{Li}\) and \(\mathrm{Mg}\) do not form peroxides and super oxides.
(f) Both \(\mathrm{Li}\) and \(\mathrm{Mg}\) nitrates decompose on heating producing \(\mathrm{NO}_2\).
\(4 \mathrm{LiNO}_3 \longrightarrow 2 \mathrm{Li}_2 \mathrm{O}+4 \mathrm{NO}_2+\mathrm{O}_2\)
\(2 \mathrm{Mg}\left(\mathrm{NO}_3\right)_2 \longrightarrow 2 \mathrm{MgO}+4 \mathrm{NO}_2+\mathrm{O}_2\)
(g) The hydroxides, carbonates and fluorides of lithium and magnesium are sparingly soluble in water.
(h) \(\mathrm{LiCl}\) and \(\mathrm{MgCl}_2\) are highly soluble in ethanol.

Topic: Chemistry, s Block Elements
In Wheatstone bridge, four resistors of resistances 15 Ω, 12 Ω, 4Ω and 10 Ω respectively, are connected in cyclic order. The resistance (in Ω) that is to be connected in parallel with the resistance of 10 Ω to balance the network is

Options:

The correct answer is:
10

Solution:
10R10+R×12=15×4

on solving R=10Ω

Topic: Physics, Current Electricity, JEE Main
In which alkyl halide, $S N^2$ mechanism is favoured maximum

Options:
A $\mathrm{CH}_3 \mathrm{Cl}$
B $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Cl}$
C $\left(\mathrm{CH}_3\right)_2 \mathrm{CHCl}$
D $\left(\mathrm{CH}_3\right)_3 \mathrm{C}-\mathrm{Cl}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{Cl}$

Solution:
$S N^2$ Substitution nucleophilic bimolecular order of different alkyl halides. $1^{\circ}\gt2^{\circ}\gt3^{\circ}$
$S N^1$ Substitution nucleophilic unimolecular order of different alkyl halides, $3^{\circ}\gt2^{\circ}\gt1^{\circ}$.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
In which compound does vanadium have an oxidation number of +4?

Options:
A NH4VO2
B K4VCN6
C VSO4
D VOSO4

The correct answer is:
VOSO4

Solution:
VOSO 4  =  VO 2+  +  SO 4 2− ; x + −2  = + 2 = + 4

Topic: Chemistry, d and f Block Elements, JEE Main
In which direction does the equilibrium shift when a gaseous reaction mixture is compressed to one-half of its original volume?

Options:
A The equilibrium shifts in the reverse direction.
B The equilibrium remains unchanged.
C The equilibrium shifts in the forward direction.
D The equilibrium becomes unstable.

The correct answer is:
The equilibrium shifts in the forward direction.

Solution:
When a gaseous reaction mixture is compressed to one-half of its original volume, the equilibrium shifts in the forward direction because pressure has doubled.

Topic: Chemistry, Chemical Equilibrium, NEET
In which direction does the reactivity of non-metals within a group (e.g., halogens) change?

Options:
A Increases down the group
B Decreases down the group
C Remains constant within the group
D Increases up the group

The correct answer is:
Decreases down the group

Solution:
Within a group of non-metals (e.g., halogens), reactivity decreases down the group.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
In which disease does a symptom of acute chest pain appear due to enough oxygen not reaching the heart muscle?

Options:
A Angina pectoris
B Atherosclerosis
C Arteriosclerosis
D Hypertension

The correct answer is:
Angina pectoris

Solution:
Disorders of the Circulatory system:
Angina (or Angina pectoris): Due to insufficient blood supply to heart muscles, a symptom of acute chest pain appears indicating decreased availability of oxygen to the heart muscles. It can occur in both sexes of any age but its occurrence increases with age.
Coronary Artery Disease (CAD): It is often referred to as atherosclerosis, affects the vessels that supply blood to the heart muscle. In this condition, the lumen of arteries gradually becomes narrower due to the deposits of calcium, fat, cholesterol and fibrous tissues.
Heart failure: When the heart is not pumping blood effectively enough to meet the needs of the body. It leads to the congestion of the lungs (one of the main symptoms) because of which it is also termed as congestive heart failure.
Cardiac arrest: When the heart stops beating.
Heart attack: When the heart muscles are suddenly damaged by an inadequate blood supply.

Topic: Biology, Body Fluids and Circulation, NEET
In which group of plant kingdom gametophyte is depend on sporophyte?

Options:
A Algae
B Bryophyta
C Moss
D Gymnosperm

The correct answer is:
Gymnosperm

Solution:
In gymnosperm main plant body is diploid sporophyte which produces small gametophyte which is not free living and independent structure and is dependent on sporophyte for development and nutrition. Male gametophyte is pollen grain which is liberated after its germination and female gametophyte is endosperm which remain on sporophyte before and after fertilization.

Topic: Biology, Plant Kingdom, NEET
In which interval is the given function $f(x)=2 x^3-15 x^2+36 x+1$ is monotonically decreasing

Options:
A $[2,3]$
B $(2,3)$
C $(-\infty, 2)$
D $(3, \infty)$

The correct answer is:
$(2,3)$

Solution:
$\begin{aligned}& y=f(x)=2 x^3-15 x^2+36 x+1 \\
& \frac{d y}{d x}=f^{\prime}(x)=6 x^2-30 x+36=6\left(x^2-5 x+6\right) \\
& f^{\prime}(x)=6(x-2)(x-3)\end{aligned}$
To be monotonic decreasing, $f^{\prime}(x) \lt 0$
$\Rightarrow(x-2)(x-3) \lt 0 \Rightarrow x \in(2,3)$

Topic: Mathematics, Application of Derivatives, JEE Main
In which molecule, cleavage by HlO4 is not observed?

Options:
A
B
C
D

The correct answer is:

Solution:
All those compounds which have
groups are oxidised by periodic acid HIO4. Thus is not oxidised.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In which of the folloiwng species the interatomic bond angle is $109^{\circ} 28^{\prime} ?$

Options:
A
$\mathrm{NH}_3,\left(\mathrm{BF}_4\right)^{-1}$

B
$\left(\mathrm{NH}_4\right)^{+}, \mathrm{BF}_3$

C
$\mathrm{NH}_3, \mathrm{BF}_4$

D
$\left(\mathrm{NH}_2\right)^{-1}, \mathrm{BF}_3$

The correct answer is:
$\mathrm{NH}_3,\left(\mathrm{BF}_4\right)^{-1}$

Solution:
No Solution Available

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2002
In which of the following arrangements the given sequence is not strictly according to the property indicated against it?

Options:
A $\mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}$ : increasing acidic strength
B $\mathrm{H}_2 \mathrm{O} < \mathrm{H}_2 \mathrm{~S} < \mathrm{H}_2 \mathrm{Se} < \mathrm{H}_2 \mathrm{Te}$ : increasing $\mathrm{p} K_a$ values
C $\mathrm{NH}_3 < \mathrm{PH}_3 < \mathrm{AsH}_3 < \mathrm{SbH}_3$ : increasing acidic character
D $\mathrm{CO}_2 < \mathrm{SiO}_2 < \mathrm{SnO}_2 < \mathrm{PbO}_2$ : increasing oxidising power

The correct answer is:
$\mathrm{H}_2 \mathrm{O} < \mathrm{H}_2 \mathrm{~S} < \mathrm{H}_2 \mathrm{Se} < \mathrm{H}_2 \mathrm{Te}$ : increasing $\mathrm{p} K_a$ values

Solution:
(a,c) As we move from top to bottom in a group, acidic strength of hydrides increases. Therefore, order of acidic strength of hydrides of V A and VII A group elements is
$$
\begin{aligned}
& \mathrm{NH}_3 < \mathrm{PH}_3 < \mathrm{AsH}_3 < \mathrm{SbH}_3 \\
& \mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI} \\
&
\end{aligned}
$$
(b) As we move from top to bottom in a group, acidic nature $\left(K_a\right)$ increases. Therefore, $\mathrm{p} K_a$ decreases. Thus, order of $\mathrm{p} K_a$ values of hydrides of VI A group elements is
$$
\mathrm{H}_2 \mathrm{O}>\mathrm{H}_2 \mathrm{~S}>\mathrm{H}_2 \mathrm{Se}>\mathrm{H}_2 \mathrm{Te}
$$
(d) On moving from top to bottom, oxidising power of oxides increases. Thus, order of oxidising power of oxides of IVA group elements is
$$
\mathrm{CO}_2 < \mathrm{SiO}_2 < \mathrm{SnO}_2 < \mathrm{PbO}_2
$$

Topic: Chemistry, Hydrogen, NEET, NEET 2012 (Mains)
In which of the following arrangements the order is NOT according to the property indicated against it?

Options:
A
$$
\mathrm{Al}^{3+} < \mathrm{Mg}^{2+} < \mathrm{Na}^{+} < \mathrm{F}^{-}
$$
Increasing ionic size
B
$$
\mathrm{B} < \mathrm{C} < \mathrm{N} < \mathrm{O}
$$
Increasing first ionization enthalpy
C
$$
\mathrm{l} < \mathrm{Br} < \mathrm{F} < \mathrm{Cl}
$$
Increasing electron gain enthalpy (with negative sign)
D
$\mathrm{Li} < \mathrm{Na} < \mathrm{K} < \mathrm{Rb}$
Increasing metallic radius

The correct answer is:
$$
\mathrm{B} < \mathrm{C} < \mathrm{N} < \mathrm{O}
$$
Increasing first ionization enthalpy

Solution:
B < C < O < N

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2005