datavorous/entrance-exam-dataset huggingface dataset:
A very long charged solid cylinder of radius 'a' contains a uniform charge density $\rho$. Dielectric constant of the material of the cylinder is $\mathrm{k}$. What will be the magnitude of electric field at a radial distance ' $x^{\prime}(x < a)$ from the axis of the cylinder?

Options:
A $\rho \frac{x}{\varepsilon_{0}}$
B $\rho \frac{x}{2 k \varepsilon_{0}}$
C $\rho \frac{\mathrm{x}^{2}}{2 \mathrm{a} \varepsilon_{0}}$
D $\rho \frac{\mathrm{x}^{2}}{2 \mathrm{k}}$

The correct answer is:
$\rho \frac{x}{2 k \varepsilon_{0}}$

Solution:
Hint:
Using Gauss's Law
$\mathrm{E}(2 \pi \times \ell)=\frac{\rho\left(\pi \mathrm{X}^{2} \ell\right)}{\mathrm{k} \varepsilon_{0}} \quad \therefore \mathrm{E}=\frac{\rho \mathrm{x}}{2 \mathrm{k} \varepsilon_{0}}$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2020
A very small circular loop of radius $a$ is initially (at $t=0$ ) coplanar and concentric with a much larger fixed circular loop of radius b. A constant current $I$ flows in the larger loop. The smaller loop is rotated with a constant angular speed $\omega$ about the common diameter. The emf induced in the smaller loop as a function of time $t$ is

Options:
A $\frac{\pi a^{2} \mu_{0} l}{2 b} \omega \cos (\omega t)$
B $\frac{\operatorname{\pi a}^{2} \mu_{0}l}{2 b} \omega \sin \left(\omega^{2} t^{2}\right)$
C $\frac{\pi a^{2} \mu_{0} l}{2 b}$ $\omega sin$ $(\omega t)$
D $\frac{\pi a^{2} \mu_{0}l}{2 b} \omega \sin ^{2}(\omega t)$

The correct answer is:
$\frac{\pi a^{2} \mu_{0} l}{2 b}$ $\omega sin$ $(\omega t)$

Solution:
We know that $\tau=$ NBAosin $\omega t$
where $N=$ number of loops $=1$
$$
\begin{aligned}
B &=\frac{\mu_{0} I}{2 b} \text { newton/amp-m } \\
A &=\pi a^{2} \text { metre }^{2} \\
\begin{aligned}
\therefore &=\frac{\mu_{0} I}{2 b}\left(\pi a^{2}\right) \omega \sin \omega t \\
&=\frac{\pi a^{2} \mu_{0} I}{2 b} \cdot \omega \sin \omega t
\end{aligned}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, WBJEE, WBJEE 2014
A weak acid of dissociation constant $10^{-5}$ is being titrated with aqueous $\mathrm{NaOH}$ solution. The $\mathrm{pH}$ at the point of one-third neutralisation of the acid will be

Options:
A $5+\log 2-\log 3$
B $5-\log 2$
C $5-\log 3$
D $5-\log 6$

The correct answer is:
$5-\log 2$

Solution:
Hints: $\mathrm{K}_{\mathrm{a}}=10^{-5} \Rightarrow \mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}=-\log 10^{-5}=5$
$$
\begin{array}{llcc}
& \mathrm{HA}+\mathrm{NaOH} \longrightarrow \mathrm{NaA}+\mathrm{H}_2 \mathrm{O} \\
\text { Initial } & 1 \mathrm{~mole} 0 & 0 & 0 \\
\text { Final } & (1-1 / 3) \mathrm{mole} & 1 / 3 \mathrm{~mole} & 1 / 3 \mathrm{~mole} \\
& =2 / 3 \mathrm{~mole} & 1 / 3 \mathrm{~mole}
\end{array}
$$
(Assumed weak acid to be monoprotic, since only one dissociation constant value is provided)
Final solution acts as an acidic buffer.
$$
\Rightarrow \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { salt }]}{[\text { Acid }]} \Rightarrow \mathrm{pH}=5+\log \frac{\frac{1}{3}}{\frac{2}{3}}=5+\log \frac{1}{2} \Rightarrow \mathrm{pH}=5-\log 2
$$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2010
A Wheatstone bridge has the resistances $10 \Omega$. $10 \Omega, 10 \Omega$ and $30 \Omega$ in its four arms. What resistance joined in parallel to the $30 \Omega$ resistance will bring it to the balanced condition?

Options:
A $2 \Omega$
B $5 \Omega$
C $10 \Omega$
D $15 \Omega$

The correct answer is:
$15 \Omega$

Solution:
In balanced Wheatstone bridge,
$$
\begin{array}{l}
\frac{1}{10}=\frac{1}{x}+\frac{1}{30} \\
x=15 \Omega
\end{array}
$$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2012
A wire carrying a steady current \(I\) is kept in the \(x\)-y plane along the curve \(y=A \sin \left(\frac{2 \pi}{\lambda} x\right)\). A magnetic field \(B\) exists in the \(z\)-direction. The magnitude of the magnetic force in the portion of the wire between \(x=0\) and \(x=\lambda\) is

Options:
A 0
B \(2 \mathrm{I} \lambda \mathrm{B}\)
C \(\mathrm{I} \lambda \mathrm{B}\)
D \(\mathrm{I} \lambda \mathrm{B} / 2\)

The correct answer is:
\(\mathrm{I} \lambda \mathrm{B}\)

Solution:
Hint :
Magnetic force \(=\mathrm{I} \lambda \mathrm{B}\)

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2023
A wire of initial length $L$ and raidus $r$ is stretched by a length $l$. Another wire of same material but with initial length $2 L$ and radius $2 r$ is stretched by a length $2 l$. The ratio of stored elastic energy per unit volume in the first and second wire is

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

The correct answer is:
$1: 1$

Solution:
Elastic energy per unit volume of wire. $\mu=\frac{1}{2} \times$ Young's modulus $\times(\text { Strain })^{2}$
$\therefore \quad \frac{u_{1}}{u_{2}}=\left\{\frac{(\text { Strain })_{1}}{(\text { Strain })_{2}}\right\}^{2}=\frac{l^{2}}{L^{2}} \times \frac{4 L^{2}}{4 l^{2}}=1: 1$

Topic: Physics, Mechanical Properties of Solids, WBJEE, WBJEE 2013
A wire of resistance $5 \Omega$ is drawn out so that its new length is 3 times its original length. What is the reistance of the new wire?

Options:
A $45 \Omega$
B $15 \Omega$
C $5 / 3 \Omega$
D $5 \Omega$

The correct answer is:
$45 \Omega$

Solution:
$\left(\frac{r_1}{r_2}\right)^2=\left(\frac{\ell_2}{\ell_1}\right)=\frac{3 \ell}{\ell}=3$
$\left(\frac{R_2}{R_1}\right)=\frac{\ell_2}{\ell_1} \times \frac{A_1}{A_2}=3 \times\left(\frac{r_1}{r_2}\right)^2=3 \times 3 \Rightarrow R_2=45$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2009
A wire of resistance $\mathrm{R}$ is elongated $n$-fold to make a new uniform wire. The resistance of new wire

Options:
A $n \mathrm{R}$
B $n^2 \mathrm{R}$
C $2 n \mathrm{R}$
D $2 n^2 \mathrm{R}$

The correct answer is:
$n^2 \mathrm{R}$

Solution:
Hints : $\mathrm{R}^{\prime}=n^2 \mathrm{R}$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2010
A Zener diode having break down voltage $\mathrm{Vz}=6 \mathrm{~V}$ is used in a voltage regulator circuit as shown in the figure. The minimum current required to pass through the Zener to act as a voltage regulator is $10 \mathrm{~mA}$ and maximum allowed current through Zener is $40 \mathrm{~mA}$. The maximum value of $R_{\mathrm{s}}$ for Zener to act as a voltage regulator is

Options:
A $100 \Omega$
B $400 \Omega$
C $0.4 \Omega$
D $950 \Omega$

The correct answer is:
$400 \Omega$

Solution:
Minimum current $=10 \mathrm{~mA}\left(\mathrm{I}_{\mathrm{L}}=0\right)$
$$
\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{V}}{\mathrm{l}}=\frac{4 \mathrm{~V}}{10 \mathrm{~mA}}=400 \Omega
$$

Topic: Physics, Semiconductors, WBJEE, WBJEE 2022
A Zener diode having break-down voltage $5.6 \mathrm{V}$ is connected in reverse bias with a battery of emf $10 \mathrm{V}$ and a resistance of $100 \Omega$ in series. The current flowing through the Zener diode is

Options:
A $88 \mathrm{mA}$
B $0.88 \mathrm{mA}$
C $4.4 \mathrm{mA}$
D $44 \mathrm{mA}$

The correct answer is:
$44 \mathrm{mA}$

Solution:
Breakdown voltage of Zener diode $=5.6 \mathrm{V}$
Voltage of source $=10 \mathrm{V}$
Potential difference $\Delta V=(10-5.6) \mathrm{V}=4.4 \mathrm{V}$
Resistance of circuit $=100 \Omega$
The current passing through the Zener diode
$$
\begin{aligned}
l &=\frac{\Delta V}{R} \\
&=\frac{4.4}{100} \\
&=4.4 \times 10^{-2} \mathrm{A} \\
&=44 \times 10^{-3} \mathrm{A} \\
&=44 \mathrm{mA}
\end{aligned}
$$

Topic: Physics, Semiconductors, WBJEE, WBJEE 2016
$\mathrm{AB}$ is a chord of a parabola $\mathrm{y}^2=4 \mathrm{ax},(\mathrm{a}>0)$ with vertex $\mathrm{A} . \mathrm{BC}$ is drawn perpendicular to $\mathrm{AB}$ meeting the axis at $\mathrm{C}$. The projection of $\mathrm{BC}$ on the axis of the parabola is

Options:
A a unit
B 2a unit
C 8a unit
D 4a unit

The correct answer is:
4a unit

Solution:
$$
\begin{aligned}
&\mathrm{B}\left(a \mathrm{at}^2, 2 a t\right) \therefore \mathrm{D}\left(\mathrm{at}^2, 0\right) \\
&\mathrm{m}_{\mathrm{AB}}=\frac{2 a \mathrm{at}^2}{\mathrm{at}^2}=\frac{2}{\mathrm{t}} \therefore \mathrm{m}_{\mathrm{BC}}=-\frac{\mathrm{t}}{2}
\end{aligned}
$$
$\therefore y-2 a t=-\frac{t}{2}\left(x-a t^2\right)$ [Equation of $B C$ ]
for $y=0,-2 a t=-\frac{t}{2}\left(x-a t^2\right)=-\frac{t}{2} x+\frac{a t^3}{2}$
$$
\begin{aligned}
&\Rightarrow \mathrm{tx}=4 a \mathrm{at}+\mathrm{at}^3 \\
&\Rightarrow \mathrm{x}=4 \mathrm{a}+\mathrm{at}^2 \\
&\therefore \mathrm{c}\left(4 \mathrm{a}+\mathrm{at}^2, 0\right)
\end{aligned}
$$
$\therefore$ Projection of $B C$ on the axis is $D C=A C-A D=4 a+a t^2-a t^2=4 a$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2022
Acceleration due to gravity at a height \(\mathrm{H}\) from the surface of a planet is the same as that at a depth of \(\mathrm{H}\) below of surface. If \(\mathrm{R}\) be the radius of the planet, then \(\mathrm{H}\) vs. \(\mathrm{R}\) graph for different planets will be

Options:
A

B

C

D

The correct answer is:

Solution:
Hint :
\(\begin{aligned} & \frac{G M}{(R+H)^2}=\frac{G M(R-H)}{R^3} \\ & (R+H)^2(R-H)=R^3 \\ & R^3-R H^2+H R^2-H^3=R^3 \\ & H^2-R^2+R H=0 \\ & H=\frac{(\sqrt{5}-1)}{2} R\end{aligned}\)

Topic: Physics, Motion In One Dimension, WBJEE, WBJEE 2023
Acid catalysed hydrolysis of ethyl acetate follows a pseudo-first order kinetics with respect to ester. If the reaction is carried out with large excess of ester, the order with respect to ester will be

Options:
A 1.5
B 0
C 0.5
D 1

The correct answer is:
0

Solution:
$\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}$ $+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$
If above reaction is carried out with large excess of ester, the rate of reaction does not depend upon cencentration of both the reactants. Due to which, the reaction becomes of zero order.

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2013
Addition of excess potassium iodide solution to a solution of mereuric chloride gives the halide complex

Options:
A tetrahedral $\mathrm{K}_{2}\left[\mathrm{HgI}_{4}\right]$
B trigonal $\mathrm{K}\left[\mathrm{Hg} \mathrm{I}_{3}\right]$
C linear $\mathrm{Hg}_{2} \mathrm{I}_{2}$
D square planar $\mathrm{K}_{2}\left[\mathrm{Hg} \mathrm{Cl}_{2} \mathrm{I}_{2}\right.$

The correct answer is:
tetrahedral $\mathrm{K}_{2}\left[\mathrm{HgI}_{4}\right]$

Solution:
$\mathrm{HgCl}_{2}+4 \mathrm{KI} \longrightarrow \mathrm{K}_{2}\left[\mathrm{Hg} \mathrm{I}_{4}\right]+2 \mathrm{KCl}$
$\mathrm{Hg}=[\mathrm{Xe}] 4 t^{14}, 5 d^{10} \mathrm{Gs}^{2}$
$$
\mathrm{Hg}^{2+}=[\mathrm{Xe}] 4{ }^{14}, 5 d^{10}, 6 \mathrm{s}^{\circ}
$$

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2013
Addition of sodium thiosulphate solution to a solution of silver nitrate gives $X$ as white precipitate, insoluble in water but soluble in excess thiosulphate solution to give $Y$. On boiling in water, $Y$ gives $Z . X, Y$ and $Z$ respectively are

Options:
A $\mathrm{Ag}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \cdot \mathrm{Na}_{3}\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right], \mathrm{Ag}_{2} \mathrm{S}$
B $\mathrm{Ag}_{2} \mathrm{SO}_{4}, \mathrm{Na}\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right], \mathrm{Ag}_{2} \mathrm{S}_{2}$
C $\mathrm{Ag}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \cdot \mathrm{Na}_{5}\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{3}\right], \mathrm{AgS}$
D $\mathrm{Ag}_{2} \mathrm{SO}_{3} \cdot \mathrm{Na}_{3}\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right], \mathrm{Ag}_{2} \mathrm{O}$

The correct answer is:
$\mathrm{Ag}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \cdot \mathrm{Na}_{3}\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right], \mathrm{Ag}_{2} \mathrm{S}$

Solution:
The sequence of reactions are

Hence, $X=A g_{2} S_{2} O_{3}$
$Y=\mathrm{Na}_{3}\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right] Z=\mathrm{Ag}_{2} \mathrm{S}$

Topic: Chemistry, Practical Chemistry, WBJEE, WBJEE 2015
ADP and ATP differ in the number of

Options:
A phosphate units
B ribose units
C adenine base
D nitrogen atom

The correct answer is:
phosphate units

Solution:
$\begin{array}{llll}\text {} & \text { ADP } & \text { is } & \text { adenosine } & \text { diphosphate }\end{array}$
(2-phosphate groups) and ATP is adenosine triphosphate ( 3 phosphate groups) Thus, they differ in number of phosphate units.

Topic: Chemistry, Biomolecules, WBJEE, WBJEE 2017
Aline cuts the $x$ -axis at $A(7,0)$ and the $y$ -axis at $B(0,-5)$. A variable line $P Q$ is drawn perpendicular to $A B$ cutting the $x$ -axis at $P(a, 0)$ and the $y$ -axis at $Q(0, b)$. If $A Q$ and BP intersect at $R$, the locus of $R$ is

Options:
A $x^{2}+y^{2}+7 x+5 y=0$
B $x^{2}+y^{2}+7 x-5 y=0$
C $x^{2}+y^{2}-7 x+5 y=0$
D $x^{2}+y^{2}-7 x-5 y=0$

The correct answer is:
$x^{2}+y^{2}-7 x+5 y=0$

Solution:
Hint:
$\mathrm{P}$ is orthocentre of $\triangle \mathrm{ABQ}$
$\mathrm{m}_{\mathrm{BR}} \times \mathrm{m}_{\mathrm{AR}}=-1$
$\Rightarrow\left(\frac{\mathrm{k}+5}{\mathrm{~h}}\right) \times\left(\frac{\mathrm{k}}{\mathrm{h}-7}\right)=-1$
$\Rightarrow x^{2}+y^{2}-7 x+5 y=0$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2020
Among Be, B, Mg and Al, the second ionisation potential is maximum for

Options:
A B
B Be
C Ma
D $A$

The correct answer is:
B

Solution:
Group II A Group III A
$\quad \quad \quad \mathrm{Be} \quad \quad \quad \mathrm{B}$
$\quad \quad \quad \mathrm{Mg} \quad \quad \quad \mathrm{Al}$
The electronic configuration of boron is $2 s^{2}2p$. And electronic configuration of $\mathrm{B}^{+}$ is $2 \mathrm{s}^{2}$. Hence it is difficult to remove second electron ty $2 s^{2}$ shell because half-filled and fully orbitals are more stable than others.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, WBJEE, WBJEE 2015
Among $\mathrm{Me}_{3}\mathrm{N}, \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}$ and MeCN
(Me $=$ methyl group) the electronegativity of $\mathrm{N}$ is in the order

Options:
A $\mathrm{MeCN}>\mathrm{C}_{\mathrm{g}} \mathrm{H}_{\mathrm{s}} \mathrm{N}>\mathrm{Me}_{3} \mathrm{N}$
B $\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{N}>\mathrm{Me}_{3} \mathrm{N}>\mathrm{MeCN}$
C $\mathrm{Me}_{3} \mathrm{N}>\mathrm{MeCN}>\mathrm{C}_{3} \mathrm{H}_{\mathrm{s}} \mathrm{N}$
D Electronegativity same in all

The correct answer is:
$\mathrm{MeCN}>\mathrm{C}_{\mathrm{g}} \mathrm{H}_{\mathrm{s}} \mathrm{N}>\mathrm{Me}_{3} \mathrm{N}$

Solution:
$\because$ In Me $_{3} \mathrm{N} \rightarrow \mathrm{N}$ is $s p^{3}$ -hybrid

and in $\mathrm{MeCN} \rightarrow \mathrm{N}$ is $s p$ -hybrid Also electronegativity of any element $\alpha \%$ $s$ -character in hybrid orbitals. Thus, order of electronegativity for nitrogen is
$$
\mathrm{MeCN}>\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} < \mathrm{Me}_{3} \mathrm{N}
$$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, WBJEE, WBJEE 2017
Among the alkenes which one produces tertiary bytyl alcohol on acid hydration

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

The correct answer is:
\(\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CH}_2\)

Solution:
Hints:

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2011
Among the following carbocations :
(I) \(\mathrm{Ph}_{2}\stackrel{+}{\mathrm{C}} \mathrm{CH}_{2} \mathrm{Me}(\mathrm{II}) \mathrm{Ph} \mathrm{CH}_{2} \mathrm{CH}_{2} \stackrel{+}{\mathrm{C}} \mathrm{HPh}\)
(III) \(\mathrm{Ph}_{2} \mathrm{CH} \stackrel{+}{\mathrm{C}} \mathrm{HMe}\) (IV) \(\mathrm{Ph}_{2} \mathrm{C}(\mathrm{Me}) \stackrel{+}{\mathrm{C}} H_{2}\) the order of stability is :

Options:
A IV $>\mathrm{II}>\mathrm{I}>\mathrm{III}$
B $I>\mathrm{II}>\mathrm{III}>\mathrm{IV}$
C II $>I>\mathrm{IV}>\mathrm{III}$
D $\mathrm{I}>\mathrm{TV}>\mathrm{III}>\mathrm{II}$

The correct answer is:
$I>\mathrm{II}>\mathrm{III}>\mathrm{IV}$

Solution:
The order of stability of carbocation is $3^{\circ}$ benzylic $>2^{\circ}$ benzylic $>2^{*}>1^{*}$ Thus, the correct order of stability will be

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2012
Among the following chlorides the compounds which will be hydrolysed most easily and most slowly in aqueous $\mathrm{NaOH}$ solution are respectively.
1.Methoxymethyl chloride
2.Benzyl chloride
3.Neopentyl chloride
4.Propyl chloride

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

The correct answer is:
1 and 3

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, WBJEE, WBJEE 2021
Among the following compounds, the onele that gives (give) effervescence with aqueoug $\mathrm{NaHCO}_{3}$ solution is (are)

Options:
A I and II
B I and III
C Only II
D I and IV

The correct answer is:
I and II

Solution:
Only carboxylic acids and their derivatives like anhydride are capable to give effervescence with aqueous $NaHCO_{3}$ solution, because they are more acidic as compared to $\mathrm{H}_{2} \mathrm{CO}_{3}$ (carbonic acidi-
$\mathrm{IICH}_{3} \mathrm{COOH}+\mathrm{NaHCO}_{3}$
$$
\longrightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}
$$
$1\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}+2 \mathrm{NaHCO}_{3}$
$$
\longrightarrow 2 \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}
$$

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2014
Among the following observations, the correct one that differentiates between $\mathrm{SO}_{3}^{2-}$ and $\mathrm{SO}_{4}^{2-}$ is

Options:
A both form precipitate with $\mathrm{BaCl}_{2}, \mathrm{SO}_{3}^{2-}$ dissolves in $\mathrm{HCl}$ but $\mathrm{SO}_{4}^{2-}$ does not
B $\mathrm{SO}_{3}^{2-}$ forms precipitate with $\mathrm{BaCl}_{2}, \mathrm{SO}_{4}^{2-}$ does not
C $\mathrm{SO}_{4}^{2-}$ forms precipitate with $\mathrm{BaCl}_{2}, \mathrm{SO}_{3}^{2-}$ does not
D both form precipitate with $\mathrm{BaCl}_{2}, \mathrm{SO}_{4}^{2-}$ dissolves in HCl but $S O_{3}^{2-}$ does not

The correct answer is:
both form precipitate with $\mathrm{BaCl}_{2}, \mathrm{SO}_{3}^{2-}$ dissolves in $\mathrm{HCl}$ but $\mathrm{SO}_{4}^{2-}$ does not

Solution:
$\mathrm{SO}_{3}^{2-}$ and $\mathrm{SO}_{4}^{2-}$ when treated with $\mathrm{BaCl}_{2}$, give white ppt of $BaSO_{3}$ and $BaSO_{4}$ respectively.
$$
\begin{array}{c}
\mathrm{BaCl}_{2}+\mathrm{SO}_{3}^{2-} \longrightarrow \mathrm{BaSO}_{3}+2 \mathrm{Cl}^{-} \\
\text {Barium sulphide} \\
\mathrm{BaCl}_{2}+\mathrm{SO}_{4}^{2-} \longrightarrow \begin{array}{c}
\text { } \\
\mathrm{BaSO}_{4}+2 \mathrm{Cl}^{-} \\
\text {Barium sulphate }
\end{array}
\end{array}
$$
Out of these $\mathrm{two}, \mathrm{SO}_{3}^{2-}$ is soluble in $\mathrm{HCl}$ but $\mathrm{SO}_{4}^{2-}$ does not.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2014
Among the following statements about the molecules $X$ and $Y$, the one(s) which correct is (are)

Options:
A $X$ and $Y$ are diastereomers
B $X$ and $Y$ are enantiomers
C $X$ and $Y$ are both aldohexoses
D $X$ is a D-sugar and $Y$ is an L-sugar

The correct answers are:
$X$ and $Y$ are enantiomers, $X$ and $Y$ are both aldohexoses, $X$ is a D-sugar and $Y$ is an L-sugar

Solution:
Both the given molecules contain 6 carbon atoms with an aldehyde group, so these are called aldohexoser, In $X$, the $-O H$ attached to second last carbon in on right hand side while in $Y$ it is on left hand side, so $x_{i j}$ D-sugar and $Y$ is L-sugar, $X$ and $Y$ are the mirror images of each other, i.e., these are enantiomers.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2014
Among the following structures the one which is not a resonating structure of others is

Options:
A I
B II
C III
D IV

The correct answer is:
IV

Solution:
Resonating structures involve movement of electrons, but they do not involve movernent of atom or group of atoms.

Thus, structure IV is not a resonating structure of others.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2014
Among the following, the extensive variables
are

Options:
A $\mathrm{H}$ (Enthalpy)
B $p$ (Pressure)
C E (internal energy)
D $\vee$ (volume)

The correct answers are:
$\mathrm{H}$ (Enthalpy), E (internal energy), $\vee$ (volume)

Solution:
Among the given options, the e
xtensive variables are $H$ (enthalpy), $E$ (internal energy). $V$ (volume).These variables have values that depends upon the quantity or size of matter present in the system. Other examples are heat
capacity, entropy, free energy, length and mass.

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2018
Among the following, the ion which will be more effective for flocculation of $\mathrm{Fe}(\mathrm{OH})_{3}$ sol. is

Options:
A $\mathrm{PO}_{4}^{3-}$
B $\mathrm{SO}_{4}^{2-}$
C $\mathrm{SO}_{3}^{2-}$
D $\mathrm{NO}_{3}^{-}$

The correct answer is:
$\mathrm{PO}_{4}^{3-}$

Solution:
Hint:
$\mathrm{Fe}(\mathrm{OH})_{3}$ is a positive sol. $\mathrm{PO}_{4}^{3-}$ carrying the highest -ve charge amongst the given ions is most effective

Topic: Chemistry, Surface Chemistry, WBJEE, WBJEE 2020
Among the following, the one which is not a "green house gas", is

Options:
A $\mathrm{N}_{2} \mathrm{O}$
B $\mathrm{CO}_{2}$
C $\mathrm{CH}_{4}$
D $\mathrm{O}_{2}$

The correct answer is:
$\mathrm{O}_{2}$

Solution:
Oxygen is not a green house gas

Topic: Chemistry, Environmental Chemistry, WBJEE, WBJEE 2014
Amongst $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}, \quad\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2} ]\right.$, $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right.]$ and $\left[\mathrm{Ni}\left(\mathrm{CN}_{4}\right)\right]^{2-}$ the paramagnetic species are

Options:
A $\left[\mathrm{NiCl}_{4}\right]^{2-},\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \cdot\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$
B $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right],\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right],\left[\mathrm{NiCl}_{4}\right]^{2-}$
C $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-} \cdot\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+},\left[\mathrm{NiCl}_{4}\right]^{2-}$
D $\left[\mathrm{Nl}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right\}\left[\mathrm{Ni}(\mathrm{CO})_{4}\right\}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$

The correct answer is:
$\left[\mathrm{NiCl}_{4}\right]^{2-},\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \cdot\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$

Solution:
Species having unpaired electrons are paramagnetic
$$
\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{d}^{8}
$$

(i) In $[\left.\mathrm{NiCl}_{4}\right]^{2-}$ and $[\left.\operatorname{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ ligands $\mathrm{Cl}^{-}$ and
$\mathrm{H}_{2} \mathrm{O}$ are weak ligands, therefore no pairing will be possible. Thus, there are two unpained electrons.
(II) In [Ni $\left.\left(P h_{3}\right)_{2} C l_{2}\right\}$. although $P P h_{3}$, has d-acceptance nature but presence of Cl, makes electrons unpaired.

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2013
Amongst the following compounds, the one that will not respond to Cannizzaro reaction upon treatment with alkali is

Options:
A $\mathrm{Cl}_{3} \mathrm{CCHO}$
B $\mathrm{Me}_{3} \mathrm{CCHO}$
C $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}$
D HCHO

The correct answer is:
$\mathrm{Cl}_{3} \mathrm{CCHO}$

Solution:
For a carbonyl group to give Cannizaro reaction, it should not have an $\alpha$ -hydrogen atom. Among the given compound $\mathrm{Ne}_{3} \mathrm{CCHO}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}$ and HCHO respond
to Cannizaro reaction but $\mathrm{Cl}_{3} \mathrm{CCHO}$ do not due to the following reason.

When hydroxyl group attacks the carbonyl group a tetrahedral intermediate forms. This tetrahedral intermediate will revert to a carbonyl compound by
expelling the best leaving group. The CCl $_{3}$ carbonium is resonance stabilised therefore it is better leaving group than OH. These initial products exchange a proton to give ion of carboxylic acid and trichloromethane ($CHCl_{3}$).

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2016