datavorous/entrance-exam-dataset huggingface dataset:
The coefficient of $x^n$, where $n$ is any positive integer, in the expansion of $\left(1+2 x+3 x^2+\ldots \ldots \ldots\right)^{1 / 2}$ is

Options:
A 1
B $\frac{n+1}{2}$
C $2 n+1$
D $\mathrm{n}+1$

The correct answer is:
1

Solution:
$$
s=1+2 x+3 x^2 .
$$
.$\infty$
Hints : $\frac{x S=x+2 x^2+\ldots \ldots \ldots \ldots}{s(1-x)=1+x+x^2+\ldots \ldots \ldots . \infty}$
$$
s=\frac{1}{(1-x)^2}
$$
$$
f(x)=\frac{1}{1-x}, f(x)=(1-x)^{-1}=1+x+x^2+x^3 \ldots \ldots \ldots \ldots . . .0=1
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2009
The coefficient of \(x^n\) in the expansion of \(\frac{e^{7 x}+e^x}{e^{3 x}}\) is

Options:
A \(\frac{4^{\mathrm{n}-1}-(-2)^{\mathrm{n}-1}}{\lfloor\mathrm{n}}\)
B \(\frac{4^{n-1}-2^{n-1}}{\lfloor\text { n }}\)
C \(\frac{4^{\mathrm{n}}-2^{\mathrm{n}}}{\lfloor\mathrm{n}}\)
D \(\frac{4^{\mathrm{n}}+(-2)^{\mathrm{n}}}{\underline{n}}\)

The correct answer is:
\(\frac{4^{\mathrm{n}}+(-2)^{\mathrm{n}}}{\underline{n}}\)

Solution:
Hints: \(\frac{e^{7 x}+e^x}{e^{3 x}}=e^{4 x}+e^{-2 x}\)
Co-efficient of \(x^n\)
\(\frac{(4)^{\mathrm{n}}}{\mathrm{n} !}+\frac{(2)^{\mathrm{n}}}{\mathrm{n} !}(-1)^{\mathrm{n}}=\frac{4^{\mathrm{n}}+(-2)^{\mathrm{n}}}{\mathrm{n} !}\)

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2011
The common chord of the circles $x^{2}+y^{2}-4 x-4 y=0 \quad$ and $\quad 2 x^{2}+2 y^{2}=32$ subtends at the origin an angle equal to

Options:
A $\frac{\pi}{3}$
B $\frac{\pi}{4}$
C $\frac{\pi}{6}$
D $\frac{\pi}{2}$

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

Solution:
Given, equation of circles are $x^{2}+y^{2}-4 x-4 y=0$ and $2 x^{2}+2 y^{2}=32$
or $x^{2}+y^{2}-4 x-4 y=0$
and $x^{2}+y^{2}=16$
$\therefore$ Equation of common chord is
$\left(x^{2}+y^{2}-4 x-4 y\right)-\left(x^{2}+y^{2}-16\right)=0$
$\Rightarrow \quad-4 x-4 y+16=0$
$\Rightarrow \quad x+y=4$
This common chord passes through (2,2)
i.e. centre of first circle. Also, (0,0) is at the circumference of the first circle.
$\therefore$ Common chord will subtent $\frac{\pi}{2}$ angle at (0,0)

Topic: Mathematics, Circle, WBJEE, WBJEE 2017
The complex number $z$ satisfying the equation $|z-i|=|z+1|=1$ is

Options:
A 0
B $1+i$
C $-1+i$
D $1-i$

The correct answers are:
0, $-1+i$

Solution:
We have, $|z-i|=|z+1|=1$
Let, $\quad z=x+iy$
$\therefore \quad|z-i|=1$
$\Rightarrow \quad|x+iy-i \mid=1$
$\Rightarrow|x+(y-1) i|=1$
$\Rightarrow \quad x^{2}+(y-1)^{2}=1$
Also, $\quad|z+1|=1$
$\Rightarrow \quad|x+ iy+1 \mid=1$
$\Rightarrow \quad|(x+1)+i y|=1$
$\Rightarrow \quad(x+1)^{2}+y^{2}=1$
From Eqs. (i) and (ii), we get $x=y=0$ and $x=-1, y=1$
$\therefore \quad z=0,-1+i$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2017
The compound that will have a permanent dipole moment among the following is

Options:
A I
B II
C III
D IV

The correct answer is:
I

Solution:
Symmetrical molecules and symmetrical trans-alkene have a net dipole moment zero. $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ is not a symmetrical molecule, thus it will have a permanent dipole.

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2014
The compound, which evolves carbon dioxide on treatment with aqueous solution of sodium bicarbonate at $25^{\circ} \mathrm{C},$ is

Options:
A $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$
B $\mathrm{CH}_{3} \mathrm{COCl}$
C $\mathrm{CH}_{3} \mathrm{CONH}_{2}$
D $\mathrm{CH}_{3} \mathrm{OOOC}_{2} \mathrm{H}_{5}$

The correct answer is:
$\mathrm{CH}_{3} \mathrm{COCl}$

Solution:
$\mathrm{CH}_{3}-\mathrm{C}O-\mathrm{Cl}$ hydrolyses to form $\mathrm{CH}_{3} \mathrm{COOH}$
even at $25^{\circ} \mathrm{C},$ which subsequently reacts with $NaHCO_{3}$ (sodium bicarbonate) present in the same medium to form carbon dioxide $\left(\mathrm{CO}_{2}\right)$.
The reaction involved is given below. $\mathrm{CH}_{3} \mathrm{COCl}+\mathrm{NaHCO}_{3} \longrightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{HCl}+\mathrm{CO}_{2}$
Thus, the option (b) is correct answer.

Topic: Chemistry, Carboxylic Acid Derivatives, WBJEE, WBJEE 2019
The compounds A and B are respectively

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2021
The compound(s), capable of producing achiral compound on heating at $100^{\circ} \mathrm{C}$ is/are

Options:
A
B
C
D

The correct answers are:

Solution:
Thus, compound (d), capable of producing achiral compound on heating at $100^{\circ} \mathrm{C}$.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2019
The compounds $\mathrm{X}$ and $\mathrm{Y}$ are respectively

Options:
A
B
C
D

The correct answers are:

Solution:

Topic: Chemistry, Amines, WBJEE, WBJEE 2021
The compressibility factor $(Z)$ of one mole of a van der Waals' gas of negligible ' $a$ ' value is

Options:
A 1
B $\frac{b p}{R T}$
C $1+\frac{b p}{R T}$
D $1-\frac{b p}{R T}$

The correct answer is:
$1+\frac{b p}{R T}$

Solution:
van der Waals' equation is $\left(p+\frac{a}{V^{2}}\right)(V-b)=R T$
(for 1 mole)
If $a$ is negligible, $p+\frac{a}{V^{2}} \approx p$
$$
p(V-b)=R T \Rightarrow p V-p b=R T
$$
or $\frac{p V}{R T}-\frac{p b}{R T}=1 \Rightarrow \frac{p V}{R T}=1+\frac{p b}{R T}$

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2014
The condition for a reaction to occur spontaneously is

Options:
A $\Delta H$ must be negative
B $\Delta S$ must be negative
C $(\Delta H-T \Delta S)$ must be negative
D $(\Delta H+T \Delta S)$ must be negative

The correct answer is:
$(\Delta H-T \Delta S)$ must be negative

Solution:
For a spontaneous reaction, $\Delta G$ = negative and $\Delta S=$ positive
$$
\Delta G=\Delta H-T \Delta S=-v e
$$
Hence, $\Delta H-T \Delta S=-$ ve only when $\Delta S=+$ ve and
$$
\Delta H=-v e
$$

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2016
The condition of spontaneity of a process is

Options:
A lowering of enthropy at constant temperature and pressure
B lowering of Gibbs free energy of system at constant temperature and pressure
C increase of entropy of system at constant temperature and pressure
D increase of Gibbs free energy of the universe at constant temperature and pressure

The correct answer is:
lowering of Gibbs free energy of system at constant temperature and pressure

Solution:
From Gibbs Helmholtz equation,
$$
\Delta G=\Delta H-T \Delta S
$$
A process will be spontaneous, if
$$
\Delta G=-\text { ve }
$$

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2013
The conducting loop in the form of a circle is placed in a uniform magnetic field with its plane perpendicular to the direction of the field. An emf will be induced in the loop, if

Options:
A it is translated parallel to itself
B it is rotated about one of its diameters
C it is rotated about its own axis which is parallel to the field
D the loop is deforned from the original shape

The correct answers are:
it is rotated about one of its diameters, the loop is deforned from the original shape

Solution:
Whenever, there is a change in the magnetic flux crossing through the loop, an emf induced in it and hence there is an amount of induced current through the loop. If loop is rotated about one of its diameter or the loop is deformed the flux through it varies and causing the emf induced in it.

Topic: Physics, Electromagnetic Induction, WBJEE, WBJEE 2015
The conductivity measurement of a coordination compound of cobalt (III) shows that it dissociates into 3 ions in solution. The compound is

Options:
A hexaamminecobalt(III) chloride
B pentaamminesulphatocobalt(III) chloride
C pentaamminechloridocobalt(III) sulphate
D pentaamminechloridocobalt(III) chloride

The correct answer is:
pentaamminechloridocobalt(III) chloride

Solution:
Co-ordinate compound of Co (III) dissociates into 3 ions in the solution means it has two ionisable-ions out side the coordination sphere with oxidation state of cobalt $(\mathrm{Co})=+3$
In option Name : Hexaammine cobalt (III) chloride Formula : $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$
$\because$ Oxidation state of $\mathrm{Co}=(+) 3$
In option (b) Name : Pentaammine sulphatocobalt (III) chloride Formula : $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{SO}_{4}\right)\right] \mathrm{Cl}$
$\because$ Oxidation state of $\mathrm{Co}=+3$
In option (c) Name : Pentaamminechloridocobalt
(III) sulphate Formula : $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{SO}_{4}$
$\because$ Oxidation state of $\mathrm{Co}=+3$
In option (d) Name : Pentaamminechloridocobalt
(III) chloride Formula: $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$
$\because$ Oxidation state of $\mathrm{Co}=+3$
Dissociation of this compound is shown below:
$\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \rightarrow \mathrm{CO} \underbrace{\left.\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]^{2+}+2 \mathrm{Cl}^{-}}_{3 \text { ions }}$
This compound can give total 3 -ions, as follows. one $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]^{2+}$ and two $\mathrm{Cl}^{-}$ ions.
Hence, option (d) is the correct answer.

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2019
The conformations of $n$ -butane, commonly known as eclipsed, gauche and anti-conformations can be interconverted by

Options:
A rotation around $C-H$ bond of a methyl group
B rotation around $\mathrm{C}-\mathrm{H}$ bond of a methylene group
C rotation around C1-C2 linkage
D rotation around C2-C3 linkage

The correct answer is:
rotation around C2-C3 linkage

Solution:
The conformation of $n$ -butane, commonly known as eclipsed, gauche and anti-conformations can be interconverted by rotation around C2-C3 linkage.

Hence, the option (d) is correct answer.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2019
The conversion of $\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{COOH}$ to
can be accomplished by

Options:
A $\mathrm{SOCl}_{2}, \mathrm{LiAlH}_{4},$ ethylene glycol.
B $\mathrm{SOCl}_{2}, \mathrm{KMnO}_{4}, \mathrm{NH}_{2} \mathrm{NH}_{2}$
C $\mathrm{SnCl}_{2}, \mathrm{HCl}, \mathrm{Na}_{2} \mathrm{SO}_{3}$
D $\mathrm{HCl}, \mathrm{SnCl}_{2}$ ethylene glycol

The correct answers are:
$\mathrm{SOCl}_{2}, \mathrm{LiAlH}_{4},$ ethylene glycol., $\mathrm{HCl}, \mathrm{SnCl}_{2}$ ethylene glycol

Solution:
Hence, (a) and (d) are correct answers.

Topic: Chemistry, Carboxylic Acid Derivatives, WBJEE, WBJEE 2017
The conversions(s) that can be carried out by bromine in carbon tetrachloride solvent is/are

Options:
A \(\mathrm{PhCH}=\mathrm{CHCH}_3 \rightarrow \mathrm{PhCHBrCHBrCH}{ }_3\)
B

C \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH} \rightarrow \mathrm{CH}_3 \mathrm{CHBrCOOH}\)
D

The correct answers are:
\(\mathrm{PhCH}=\mathrm{CHCH}_3 \rightarrow \mathrm{PhCHBrCHBrCH}{ }_3\),

Solution:
Hint : (A) Addition Reaction
(D) Borodine Hunsdiecker reaction

Topic: Chemistry, Haloalkanes and Haloarenes, WBJEE, WBJEE 2023
The co-ordinate of a point on the auxiliary circle of the ellipse $x^{2}+2 y^{2}=4$ corresponding to the point on the ellipse whose eccentric angle is $60^{\circ}$ will be

Options:
A $(\sqrt{3}, 1)$
B $(1, \sqrt{3})$
C $(1,1)$
D $(1,2)$

The correct answer is:
$(1, \sqrt{3})$

Solution:
$\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$
Equation of auxiliary circle is
$x^{2}+y^{2}=4$
$\therefore$ Point on the auxiliary circle is
$\begin{array}{l}
p\left(2 \cos 60^{\circ}, 2 \sin 60^{\circ}\right) \\
p(1, \sqrt{3})
\end{array}$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2021
The coordinates of a moving point \(p\) are \(\left(2 t^2+4,4 t+6\right)\). Then its locus will be a

Options:
A circle
B straight line
C parabola
D ellipse

The correct answer is:
parabola

Solution:
Hints: \(x=2 t^2+4, y=4 t+6, \quad y=4 t+6 \rightarrow t=\left(\frac{y-6}{4}\right)\)
\(\begin{aligned}
& x=2\left(\frac{y-6}{4}\right)^2+4 \Rightarrow \frac{(y-6)^2}{8}=x-4 \\
& (y-6)^2=4(2)(x-4)
\end{aligned}\)

Topic: Mathematics, Parabola, WBJEE, WBJEE 2011
The coordinates of a point on the line $x+y+1=0,$ which is at a distance $\frac{1}{5}$ unit from the line $3 x+4 y+2=0,$ are

Options:
A (2,-3)
B (-3,2)
C (0,-1)
D (-1,0)

The correct answers are:
(-3,2), (-1,0)

Solution:
Let $(h, k)$ be the point on line $x+y+1=0$
So.
$$
h+k+1=0
$$
$$
h=-1-k
$$
Given, distance between $(h, k)$ and line $3 x+4 y+2=0$ is $\frac{1}{5}$
$\therefore$
$\frac{1}{5}=\left|\frac{(3 h)+(4 k)+2}{\sqrt{3^{2}+4^{2}}}\right|$
$\Rightarrow \quad \frac{1}{5}=\left|\frac{3(-1-k)+4 k+2}{\sqrt{25}}\right|$
$\Rightarrow \quad \frac{1}{5}=\pm\left(\frac{-3-3 k+4 k+2}{5}\right)$
$\Rightarrow$
$\frac{1}{5}=\pm\left(\frac{k-1}{5}\right)$
$\Rightarrow \quad \pm(k-1)=1$
$\Rightarrow \quad k=0$ and 2
$\therefore \quad h=-1, -3$
Hence, the required points are (-1,0) and (-3,2)

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2016
The co-ordinates of the focus of the parabola described parametrically by $x=5 t^2+2, y=10 t+4$ are

Options:
A $(7,4)$
B $(3,4)$
C $(3,-4)$
D $(-7,4)$

The correct answer is:
$(7,4)$

Solution:
Hints: $\mathrm{x}=5 \mathrm{t}^2+2 ; \mathrm{y}=10 \mathrm{t}+4,\left(\frac{y-4}{10}\right)^2=\left(\frac{x-2}{5}\right)$ or, $(\mathrm{y}-4)^2=20(\mathrm{x}-2)$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2009
The co-ordinates of the foot of the perpendicular from $(0,0)$ upon the line $x+y=2$ are

Options:
A $(2,-1)$
B $(-2,1)$
C $(1,1)$
D $(1,2)$

The correct answer is:
$(1,1)$

Solution:
Hints : Let $\mathrm{P}$ be the foot of the perpendicular. P lies on a line perpendicular to $x+y=2$.
$\therefore$ Equation of the line on which P lies is of the form : $x-y+k=0$
But this line passes through $(0,0)$.
$$
\therefore k=0
$$
Hence, co-ordinates of P may be obtained by solving $x+y=2$ and $y=x$
$$
\therefore x=1, y=1
$$
Hence, $\mathrm{P} \equiv(1,1)$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2009
The co-ordinates of the point on the curve $y=x^2-3 x+2$ where the tangent is perpendicular to the straight line $y=x$ are

Options:
A $(0,2)$
B $(1,0)$
C $(-1,6)$
D $(2,-2)$

The correct answer is:
$(1,0)$

Solution:
Hints: $y=x^2-3 x+2$
$\frac{d y}{d x}=2 x-3=-1 \Rightarrow x=1$ at $x=1, y=0$
$\therefore$ Point is $(1,0)$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2010
The coordinates of the two points lying on \(x+y=4\) and at a unit distance from the straight line \(4 x+3 y=10\) are

Options:
A \((-3,1),(7,11)\)
B \((3,1),(-7,11)\)
C \((3,1),(7,11)\)
D \((5,3),(-1,2)\)

The correct answer is:
\((3,1),(-7,11)\)

Solution:
Hints : Let \(p(h, 4-h)\)
\(\begin{aligned}
& \left|\frac{4 h+3(4-h)-10}{5}\right|=1 \\
& |h+2|=5 \\
& h=3,-7 ; p=1,1 \\
& (3,1),(-7,11)
\end{aligned}\)

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2011
The correct basicity order of the following lanthanide ions is

Options:
A $\mathrm{La}^{3+}>\mathrm{Lu}^{3 +}>\mathrm{Ce}^{3+}>\mathrm{Eu}^{3+}$
B $\mathrm{Ce}^{3}>\mathrm{Lu}^{3+}>\mathrm{La}^{3+} > \mathrm{Eu}^{3 \mathrm{e}}$
C $\mathrm{Lu}^{3 \cdot}>\mathrm{Ce}^{3}>\mathrm{Eu}^{3+}>\mathrm{La}^{3+}$
D $\mathrm{La}^{3+}>\mathrm{Ce}^{3+}>\mathrm{Eu}^{3+}>\mathrm{Lu}^{3}$

The correct answer is:
$\mathrm{La}^{3+}>\mathrm{Ce}^{3+}>\mathrm{Eu}^{3+}>\mathrm{Lu}^{3}$

Solution:
As size of cation decreases, ionic character decreases, thus basicity also decreases Hence, correct order is $\mathrm{La}^{3+}>\mathrm{Ce}^{3+}>\mathrm{Eu}^{3+}>\mathrm{Lu}^{3+}$

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2017
The correct bond order of $\mathrm{B}-\mathrm{F}$ bond in $\mathrm{BF}_3$ molecule is

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

The correct answer is:
$1 \frac{1}{3}$

Solution:
B.O $=\frac{2+1+1}{3}=1 \frac{1}{3}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2022
The correct dimensional formula for impulse is given by

Options:
A $\mathrm{ML}^{2} \mathrm{T}^{-2}$
B MLT $^{-1}$
C $\mathrm{ML}^{2} \mathrm{T}^{-1}$
D $\mathrm{MLT}^{-2}$

The correct answer is:
MLT $^{-1}$

Solution:
As we know that,
Impulse, $I=$ force $(F) \times$ small time interval
$\begin{array}{ll}
I=m a \times t & {[\because F=m a]} \\
I=[\mathrm{M}]\left[\mathrm{LT}^{-2}\right] \times[\mathrm{T}] \\
I=\left[\mathrm{MLT}^{-1}\right]
\end{array}$
Hence, the correct dimensional formula for impulse is given by $\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-1} 1 .\right.$

Topic: Physics, Units and Dimensions, WBJEE, WBJEE 2019
The correct option for the above reaction is

Options:
A

B

C

D

The correct answer is:

Solution:
Hint :
In acidic medium, only monohalogenation takes place.
\(\begin{aligned}
& \mathrm{Y}: \mathrm{CHBr}_3 \\
& \mathrm{Z}: \mathrm{CH}_3 \mathrm{COONa}
\end{aligned}\)
Second step is bromoform reaction that takes place here on \(\mathrm{CH}_3 \mathrm{COCH}_2-\mathrm{Br}\)

Topic: Chemistry, Haloalkanes and Haloarenes, WBJEE, WBJEE 2023
The correct order of acid strength of the following substituted phenols in water at $28^{\circ} \mathrm{C}$ is

Options:
A $p$-nitrophenol $ < $ p-fluorophenol $ < $ p-chlorophenol
B $p$-chlorophenol $ < $ p-fluorophenol $ < $ p-nitrophenol
C p-fluorophenol $ < $ p-chlorophenol $ < $ p-nitrophenol
D $p$-flurophenol $ < $ p-nitrophenol $ < p$-chlorophenol

The correct answer is:
p-fluorophenol $ < $ p-chlorophenol $ < $ p-nitrophenol

Solution:
The order of electron withdrawing tendency from benzene ring i.e..
$$
-F < -C l < -N O_{2}
$$
$\therefore$ Correct order of acidic strength of substituted phenols will be

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2013
The correct order of acid strengths of benzoic acid $(X),$ peroxybenzoic acid $(Y)$ and $p$-nitrobenzoic acid $(Z)$ is

Options:
A $Y>Z>X$
B $Z>Y>X$
C $Z>X>Y$
D $Y>X>Z$

The correct answer is:
$Z>X>Y$

Solution:
$\because$ In $p$ -nitrobenzoic acid $(Z)-\mathrm{NO}_{2}$ present as electron withdrawing group, thus is most acidic

$(X) \because$ Benzoic acid is stabilised via resonance, thus is more acidic than per oxybenzoic acid.

(Y) peroxy benzoic acid does not show any such resonance.

Thus is least acidic.
Hence, correct order is $Z>X>Y$

Topic: Chemistry, Carboxylic Acid Derivatives, WBJEE, WBJEE 2017
The correct order of acidity for the following compounds is:

Options:
A $II < I V < I < I$
B $II < I < III < IV$
C $II < III < IV < I$
D $III < II < I < IV$

The correct answer is:
$II < I < III < IV$

Solution:
Hint:
Compound IV is a carboxylic acid whereas others are alcohols. So IV is most acidic, amongst the alcohols. Presence of $-\mathrm{NO}_{2}$ increases acidic strength while the presence of $-\mathrm{CH}_{3}$ lowers. Amongst I and III, more acidic is $\mathrm{I}$ as $-\mathrm{NO}_{2}$ exerts it's $-\mathrm{R}$ influence in para position.

Topic: Chemistry, Carboxylic Acid Derivatives, WBJEE, WBJEE 2020
The correct order of acidity of above compounds is

Options:
A II \( > \) IV \( > \) I \( > \) III
B III \( > \) IV \( > \) II \( > \) I
C IV \( > \) II \( > \) III \( > \) I
D IV \( > \) III \( > \) I \( > \) II

The correct answer is:
IV \( > \) III \( > \) I \( > \) II

Solution:
Hint : \(-\mathrm{NO}_2\) shows strong \(-\mathrm{R}\)
\(-\mathrm{CH}=\mathrm{CH}_2\) shows \(-\mathrm{R}\)
\(-\mathrm{CH}_3\) shows hyperconjugation
\(-\mathrm{NMe}_2\) shows \(+\mathrm{R}\)
Donating groups lowers acidity while withdrawing groups raise acidity

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2023
The correct order of acidity of the following hydra acids is

Options:
A $\mathrm{HF}>\mathrm{HCl}>\mathrm{HBr}>\mathrm{HI}$
B $\mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}$
C $\mathrm{HF} < \mathrm{HCl}>\mathrm{HBr}>\mathrm{HI}$
D $\mathrm{HF}>\mathrm{HCl} < \mathrm{HBr}>\mathrm{HI}$

The correct answer is:
$\mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}$

Solution:
$\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}>\mathrm{HF}$
Down the group acidic strength increases.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2022
The correct order of basicity of the following compounds is

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

The correct answer is:
$2 < 1 < 3 < 4$

Solution:
The correct order is $2 < 1 < 3 < 4$. Since, more the number of $2^{\circ}$ - NH- group, more is the basic nature.
(1) Thus, 3 and 4 are more basic than 1 and 2 .
(ii) Between 3 and 4,4 has one two - $\mathrm{NH}_{2}$ group while 3 has only one $-\mathrm{NH}_{2}$ group. $\therefore 4$ is more basic than 3
(iii) Between 1 and 2,1 has one $-\mathrm{NH}_{2}$ group with no resonance, thus is more basic than 2.

Topic: Chemistry, Amines, WBJEE, WBJEE 2016
The correct order of boiling points of N-ethylethanamine (I), ethoxyethane (II) and butan-2-ol (III) is

Options:
A III \( < \) II \( < \) I
B II \( < \) III \( < \) I
C II \( < \) I \( < \) III
D III \( < \) I \( < \) II

The correct answer is:
II \( < \) I \( < \) III

Solution:
Hint : Butan-2-ol shows stronger H-bonding than N-ethylenthanamine. Ethoxyethane involves dipole association, weaker than \(\mathrm{H}\)-bonding

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2023
The correct order of decreasing acidity of nitrophenols will be

Options:
A m-Nitrophenol \( > \) p-Nitrophenol \( > \) o-Nitrophenol
B o-Nitrophenol \( > \) m- Nitrophenol \( > \) p-Nitrophenol
C p-Nitrophenol \( > \) m- Nitrophenol \( > \) o-Nitrophenol
D p-Nitrophenol \( > \) o-nitrophenol \( > \) m-Nitrophenol

The correct answer is:
p-Nitrophenol \( > \) o-nitrophenol \( > \) m-Nitrophenol

Solution:
Hints:
Due to \(-\mathrm{I}\) and \(-\mathrm{R}\) influence, \(\mathrm{NO}_2\) in ortho-postion should have raised the acidity to the maximum extent. But it is due to intramolecular \(\mathrm{H}\) - bonding, ortho-nitrophenol is less acidic than para - nitrophenol.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2011
The correct order of decreasing $\mathrm{H}-\mathrm{C}-\mathrm{H}$ angle in the following molecule is

Options:
A I>II>III
B II>I>III
C III>II>I
D I>III>II

The correct answer is:
II>I>III

Solution:
$\mathrm{H}-\mathrm{C}-\mathrm{H}$ bond angle is about $120^{\circ}$, when $\mathrm{C}$ is bonded through a double bond but in case of $\mathrm{CH}_{4}$ the $\mathrm{H}-\mathrm{C}-\mathrm{H}$ bond angle is only $109^{\circ} \mathrm{2B}^{\prime}$
Cyclic compounds have a $\mathrm{H}-\mathrm{C}-\mathrm{H}$ bond angle Feater than $109^{\circ} 28^{\prime}$ due to hindrance. Thus, the correct order of $1 \mathrm{H}-\mathrm{C}-\mathrm{H}$ bond angle is

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2014
The correct order of decreasing length of the bond as indicated by the arrow in the following structures is

Options:
A I>II>III
B II>I>III
C III>II>I
D I>III>II

The correct answer is:
III>II>I

Solution:
The length of carbon-carbon single bond is always greater than that of the carbon-carbon double bond.
In III, positive charge is not in conjugation of double bond, so the bond does not acquire a double bond character. However, chances of acquiring double bond character fof the indicated bond) is much more in I than that in case of II.
Thus, the correct order of decreasing bond length is

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2014
The correct order of equivalent conductances at infinite dilution in water at room temperature for $\mathrm{H}^{+}, \mathrm{K}^{+}, \mathrm{CH}_{3} \mathrm{COO}^{-}$ and $\mathrm{HO}^{-}$ ions is

Options:
A $\mathrm{HO}^{-}>\mathrm{H}^{+}>\mathrm{K}^{*}>\mathrm{CH}_{3} \mathrm{COO}^{-}$
B $H^{+}>\mathrm{HO}^{-}>\mathrm{K}^{+}>\mathrm{CH}_{3} \mathrm{COO}^{-}$
C $H^{+}>K^{+}>H O^{-}>C H_{3} C O O^{-}$
D $H^{+}>K^{+}>C H_{3} C OO^{-}>H O^{-}$

The correct answer is:
$H^{+}>\mathrm{HO}^{-}>\mathrm{K}^{+}>\mathrm{CH}_{3} \mathrm{COO}^{-}$

Solution:
Equivalent conductance $\left(\Lambda_{\mathrm{eq}}\right)$ is defined as the conducting power of all the ions produced by one gram equivalent of an electrolyte in a given solution.in case of weak electrolytes (as $\mathrm{CH}_{3} \mathrm{COOH}, \mathrm{NH}_{4} \mathrm{OH}, \mathrm{AgCl}$ etc. $),$ ionisation is very small compared to strong electrolytes (as KCl, NaOH etc.), hence $\Lambda_{\text {oq }}$ of weak electrolyte is low. Moreover, ionisation of $\mathrm{H}^{*}$ is maximum among given ions. Thus, correct of order of $\Lambda_{\mathrm{eq}}$ is $\mathrm{H}^{+}>\mathrm{OH}^{-}>\mathrm{K}^{+}>\mathrm{CH}_{3} \mathrm{COO}^{-}$

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2013