datavorous/entrance-exam-dataset huggingface dataset:
Which one of the following has the lowest ionization energy?

Options:
A \(1 s^2 2 s^2 2 p^6\)
B \(1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^1\)
C \(1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^5\)
D \(1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^3\)

The correct answer is:
\(1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^1\)

Solution:
Hints: It's an alkalimetal; hence least I.P

Topic: Chemistry, Classification of Elements and Periodicity in Properties, WBJEE, WBJEE 2011
Which one of the following is a condensation polymer?

Options:
A PVC
B Teflon
C Dacron
D Polystyrene

The correct answer is:
Dacron

Solution:
Dacron is a condensation polymer. It is also known as terylene. It is a polymer obtained by condensation reaction between ethylene glycol and terephthalic acid at $420-460 \mathrm{K}$ in the presence of zinc acetate-antimony trioxide catalyst.

Topic: Chemistry, Polymers, WBJEE, WBJEE 2018
Which one of the following is an example of co-polymer?

Options:
A Buna-S
B Teflon
C PVC
D Polypropylene

The correct answer is:
Buna-S

Solution:
Hints : Buna -S is a co-polymer of butadiene and styrene

Topic: Chemistry, Polymers, WBJEE, WBJEE 2010
Which one of the following is called a carbylamine?

Options:
A $\mathrm{RCN}$
B $\mathrm{RCONH}_2$
C $\mathrm{R}-\mathrm{CH}=\mathrm{NH}$
D R NC

The correct answer is:
R NC

Solution:
No Solution Available

Topic: Chemistry, Amines, WBJEE, WBJEE 2009
Which one of the following is not true at room temperature and pressure

Options:
A \(\mathrm{P}_4 \mathrm{O}_{10}\) is a white solid
B \(\mathrm{SO}_2\) is a coloureless gas
C \(\mathrm{SO}_3\) is a colourless gas
D \(\mathrm{NO}_2\) is a brown gas

The correct answer is:
\(\mathrm{SO}_3\) is a colourless gas

Solution:
Hints: \(\mathrm{SO}_3\) is colorless, crystalline transparent solid at room temperature.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2011
Which one of the following is paramagnetic?

Options:
A \(\mathrm{N}_2\)
B \(\mathrm{NO}\)
C \(\mathrm{CO}\)
D \(\mathrm{O}_3\)

The correct answer is:
\(\mathrm{NO}\)

Solution:
Hints:
, Valence electron \(=11\)

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2011
Which one of the following is s-butyl phynylvinyl methane?

Options:
A

B

C

D

The correct answer is:

Solution:
Hints:

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2011
Which one of the following is the correct set of four quantum numbers $(n, l, m, s)$ ?

Options:
A $\left(3,0,-1,+\frac{1}{2}\right)$
B $\left(4,3,-2,-\frac{1}{2}\right)$
C $\left(3,1,-2,-\frac{1}{2}\right)$
D $\left(4,2,-3,+\frac{1}{2}\right)$

The correct answer is:
$\left(4,3,-2,-\frac{1}{2}\right)$

Solution:
$\mathrm{n}=4, \ell=0,1,2,3$
For $\ell=3, \mathrm{~m}=-3,-2,-1,0,+1,+2,+3$
For $m=-2, s=-\frac{1}{2}$

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2022
Which one of the following is wrong about molecularity of a reaction?

Options:
A It may be whole number or fractional
B It is calculated from reaction mechanism
C It is the number of molecules of the reactants taking part in a single step chemical reaction
D It is always equal to the order of elementary reaction

The correct answer is:
It may be whole number or fractional

Solution:
Molecularity of a reaction can never be fractional. It is always a whole number.

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2012
Which one of the following methods is used to prepare Me $_{3}$ COEt with a good yield?

Options:
A Mixing EtONa with MegCCl
B Mixing Me $_{3}$ CONa with EtCl
C Heating a mixture of $(1 ; 1)$ EtOH and $\mathrm{Me}_{3} \mathrm{COH}$ in the presence of conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$
D Treatment of Me_COH with EtMgI

The correct answer is:
Mixing Me $_{3}$ CONa with EtCl

Solution:
\(\mathrm{Me}_{3} \mathrm{CONa} \underset{\left(1^{\circ} \text { alkyl halide) }\right.}{\stackrel{\mathrm{EtCl}}{\longrightarrow}} \mathrm{Me}_{3} \mathrm{COEt}\)
This reaction is known as Williamson's synthesis.
So, the correct option is B

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2012
Which one of the following produced when acet one is saturated with $\mathrm{HCl}$ gas?

Options:
A Acetone alcohol
B Phorone
C Mesityl oxide
D Benzene

The correct answer is:
Mesityl oxide

Solution:
Hints:
[Note : Phorone is formed as minor product].

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2010
Which one of the following properties is exhibited by phenol?

Options:
A It is soluble in aq. $\mathrm{NaOH}$ and evolves $\mathrm{CO}_{2}$ with aq. $\mathrm{NaHCO}_{3}$
B It is soluble in aq. NaOH and does not evolve $\mathrm{CO}_{2}$ withe aq. $\mathrm{NaHCO}_{3}$
C It is not soluble in aq. NaOH but evolves $\mathrm{CO}_{2}$ with aq. $\mathrm{NaHCO}_{3}$
D It is insoluble in aq. NaOH and does not evolve $\mathrm{CO}_{2}$ with aq. $\mathrm{NaHCO}_{3}$

The correct answer is:
It is soluble in aq. NaOH and does not evolve $\mathrm{CO}_{2}$ withe aq. $\mathrm{NaHCO}_{3}$

Solution:
Phenol is a weak acid. It reacts with $\mathrm{NaOH}$ to produce salt.
$$
\begin{array}{r}
\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}+\mathrm{NaOH} \longrightarrow \\
\qquad \begin{aligned}
\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{ONa} \\
\text { sodium phenoxide }
\end{aligned}+\mathrm{H}_{2} \mathrm{O}
\end{array}
$$
But it is not sufficiently acidic to evolve $\mathrm{CO}_{2}$ from NaHCO $_{3}$ solution.

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2012
Which one of the following solutions will have highest conductivity?

Options:
A $0.1 \mathrm{M} \quad \mathrm{CH}_{3} \mathrm{COOH}$
B $0.1 \mathrm{M} \quad \mathrm{NaCl}$
C $0.1 \mathrm{M} \quad \mathrm{KNO}_{3}$
D $0.1 \mathrm{M} \quad \mathrm{HCl}$

The correct answer is:
$0.1 \mathrm{M} \quad \mathrm{HCl}$

Solution:
$\mathrm{HCl}$ and $\mathrm{KNO}_{3}$ both are strong electrolyte but $\mathrm{HCl}$ have aqueous component $\mathrm{H}^{+}$, which has highest conductivity. $\therefore \mathrm{HCl}$ will have highest conductivity.

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2021
Which one of the following will show optical isomerism?

Options:
A

B

C

D

The correct answer is:

Solution:
Hints : The central carbon is attached to four different substituents, hence it is chiral carbon, therefore optically active.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2011
Which statement is incorrect

Options:
A Phenol is a weak acid
B Phenol is an aromatic compound
C Phenol liberates $\mathrm{CO}_2$ from $\mathrm{Na}_2 \mathrm{CO}_3$ soln
D Phenol is soluble in $\mathrm{NaOH}$

The correct answer is:
Phenol liberates $\mathrm{CO}_2$ from $\mathrm{Na}_2 \mathrm{CO}_3$ soln

Solution:
Phenol does not liberate $\mathrm{CO}_2$ from $\mathrm{Na}_2 \mathrm{CO}_3$ solution

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2009
Which statement is not correct for ortho and para hydrogen?

Options:
A They have different boiling points
B Ortho-form is more stable than para-form
C They differ in their nuclear spin
D The ratio of ortho to para hydrogen changes with change in temperature

The correct answer is:
Ortho-form is more stable than para-form

Solution:
(i) Ortho and para hydrogens have difterent nuclear spin. In $\mathrm{H}_{2}$ molecule, two protons in two H atoms with pare spin are called ortho-hydrogen

(ii) Ortho-hydrogen is more stable than para form at and above room temperature only, therefore para form always tends to revert in ortho-form. Thus, not correct.
(iii) The per cent composition of ortho and para changes with the lemperature. Thus, their ratio also change.
(iv) They have slightly different boling point.
Hence, (b) is the answer.

Topic: Chemistry, Hydrogen, WBJEE, WBJEE 2016
Which statements are correct for the
peroxide ion?

Options:
A It has five completely filled anti-bonding molecular orbitals
B It is diamagnetic
C It has bond order one
D It is isoelectronic with neon

The correct answers are:
It is diamagnetic, It has bond order one

Solution:
The correct statements for peroxide ion $\left(\mathrm{O}_{2}^{2-}\right)$ are that it is diamagnetic and it has bond
order one. Electronic configuration of $\mathrm{O}_{2}^{2-}$ (peroxide ion) is
$\sigma \mathrm{l} s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{z}^{2},\left[\pi 2 p_{x}^{2}=\pi 2 p_{y}^{2}\right\}$
$$
\left[\pi^{\prime} 2 p_{x}^{2}=\pi^{*} 2 p_{y}^{2}\right], \sigma^{*} 2 p_{t}^{0}
$$
$\because$ Bond order $=$ No. of electrons present in bonding No. of electrons present in non - bonding
$\therefore \quad B O=\frac{10-8}{2}=\frac{2}{2}=1$
Number of unpaired electrons $=0$
$\mathrm{So}, \mathrm{O}_{2}^{2-}$ is diamagnetic in nature.

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2018
Which statement(s) is/are applicable above critical temperature?

Options:
A A gas cannot be liquified
B Surface tension of a liquid is very high
C A liq. phase cannot be distinguished from a gas phase
D Density changes continuously with \(\mathrm{P}\) or \(\mathrm{V}\)

The correct answers are:
A gas cannot be liquified

Solution:
Hint : Gas cannot be liquified above critical temperature (fact.)

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2023
Which structure has delocalised $\pi$-electrons?

Options:
A $\mathrm{O}_{3}$
B $\mathrm{CO}$
C $\mathrm{HCN}$
D $\mathrm{O}_{3}$ and $\mathrm{HCN}$

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

Solution:
$\mathrm{O}_{3}$ (ozone) has delocalised $\pi$ electrons

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2021
Which type of Gate the following truth table represents?

Options:
A NOT
B \(\mathrm{AND}\)
C OR
D NAND

The correct answer is:
NAND

Solution:
Hints:

Topic: Physics, Semiconductors, WBJEE, WBJEE 2011
Whichever of the following is/are correct?

Options:
A To evaluate $I_{1}=\int_{-2}^{2} \frac{d x}{4+x^{2}}$, it is possible to put $x=\frac{1}{t}$
B To evaluate $I_{2}=\int_{0}^{1} \sqrt{\left(x^{2}+1\right)} d x$, it is possible to put $x=\sec t$
C To evaluate $I_{2}=\int_{0}^{1} \sqrt{\left(x^{2}+1\right)} \mathrm{dx}$, it is not possible to put $x=\operatorname{cosec} \theta$
D To evaluate $\mathrm{I}_{1}$, it is not possible to put $\mathrm{x}=\frac{1}{\mathrm{t}}$

The correct answers are:
To evaluate $I_{2}=\int_{0}^{1} \sqrt{\left(x^{2}+1\right)} \mathrm{dx}$, it is not possible to put $x=\operatorname{cosec} \theta$, To evaluate $\mathrm{I}_{1}$, it is not possible to put $\mathrm{x}=\frac{1}{\mathrm{t}}$

Solution:
$\mathrm{I}_{1}=\int_{-2}^{2} \frac{\mathrm{dx}}{4+\mathrm{x}^{2}}=2 \int_{0}^{2} \frac{\mathrm{dx}}{4+\mathrm{x}^{2}}>0$
By
$\mathrm{x}=1 / \mathrm{t}, \quad \mathrm{I}_{1}=\int_{-1 / 2}^{1 / 2} \frac{-\mathrm{dt}}{4 \mathrm{t}^{2}+1} < 0$
So not possible
$\mathrm{I}_{2}=\int_{0}^{1} \sqrt{\mathrm{x}^{2}+1} \mathrm{dx}, \mathrm{x}=\operatorname{cosec} \theta$
Since $x \in(0,1)$ and $\operatorname{cosec} \theta \in(-\infty,-1] \cup[1, \infty)$
So not possible

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2021
White phosphorous $P_{4}$ has the following
characteristics

Options:
A $6 \mathrm{P}-\mathrm{P}$ single bonds
B $4 \mathrm{P}-\mathrm{P}$ single bonds
C 4 lone pair of electrons
D $\mathrm{P}-\mathrm{P}-\mathrm{P}$ angle of $60^{\circ}$

The correct answers are:
$6 \mathrm{P}-\mathrm{P}$ single bonds, 4 lone pair of electrons, $\mathrm{P}-\mathrm{P}-\mathrm{P}$ angle of $60^{\circ}$

Solution:
WHite phosphorus $\left(P_{4}\right)$ has $6 P-P$ single bonds, 4 lone pairs of electrons and $P-P-P$ angle of $60^{\circ} .$ It is a translucent white waxy solid. It has large atomic size and less electronegativity. So, it forms single bond instead of $p \pi$ - $p \pi$ multiple bond. It consists of discrete tetrahedral $P_{4}$ molecule as shown in the figure.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2018
Within the list shown below, the correct pair of structures of alanine in $\mathrm{pH}$ ranges $2-4$ and $9-11$ is
I. $\mathrm{H}_{3} \mathrm{N}^{+}-\mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CO}_{2} \mathrm{H}$
II. $\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CO}_{2}^{-}$
III. $\mathrm{H}_{3} \mathrm{N}^{+}-\mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CO}_{2}^{-}$
$\mathrm{IV} . \mathrm{H}_{2} \mathrm{N}-\mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CO}_{2} \mathrm{H}$

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

The correct answers are:
I and II

Solution:
Isoelectric point for neutral amino acids ranges from $\mathrm{pH} 5.5$ to 6.3 . Acidic amino acids have isoelectric point around 3 while basic amino acids have ranges from pH 7.6 to 10.8 . At isoelectric point, amino acids exist as Zwitter ions. At low pH, it exists as cation and at high $\mathrm{pH}$, it exists as anion. Hence, alanine (pH range $2-4$ ) will exist as cation while alanine (pH range $9-11$ ) will exist as anion.
Acidic medium:
$$
\mathrm{H}_{3} \mathrm{N}^{+} -\mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CO}_{2} \mathrm{H}
$$
Basic medium:
$$
\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CO}_{2}^{-}
$$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2015
Without changing the direction of the axes, the origin is transferred to the point (2,3) . Then the equation $x^{2}+y^{2}-4 x-6 y+9=0$ changes to

Options:
A $x^{2}+y^{2}+4=0$
B $x^{2}+y^{2}=4$
C $x^{2}+y^{2}-8 x-12 y+48=0$
D $x^{2}+y^{2}=9$

The correct answer is:
$x^{2}+y^{2}=4$

Solution:
Put
$x=x^{\prime}+2$
and $\quad y=y^{\prime}+3$
$\therefore \quad x^{2}+y^{2}-4 x-6 y+9=0$
$$
\begin{array}{l}
\Rightarrow \quad\left(x^{\prime}+2\right)^{2}+\left(y^{\prime}+3\right)^{2}-4\left(x^{\prime}+2\right) \\
-6(y+3)+9=0 \\
\Rightarrow \quad x^{2}+4+4 x^{2}+y^{2}+9 \\
\quad+6 y^{\prime}-4 x^{2}-8-6 y^{\prime}-18+9=0 \\
\Rightarrow x^{2}+y^{2}-4=0 \\
\Rightarrow \quad x^{2}+y^{2}=4
\end{array}
$$

Topic: Mathematics, Pair of Lines, WBJEE, WBJEE 2018
Work done for a certain spring when it is streched to 1mm from mean position is 10J amount of work that must be done on the spring to stretch it further by $1 \mathrm{mm}$ is

Options:
A $30 \mathrm{J}$
B $40 \mathrm{J}$
C $10 \mathrm{J}$
D $20 \mathrm{J}$

The correct answer is:
$30 \mathrm{J}$

Solution:
The work done in stretching the spring, $W=\frac{1}{2} K x^{2}$
$K=$ force constant of spring
$x=$ extension in length of the spring
$$
\begin{array}{l}
W_{1}=\frac{1}{2} K x_{1}^{2}. \\
W_{2}=\frac{1}{2} K x_{2}^{2} \\
x_{1}=1 \\
x_{2}=1+1=2 \\
\frac{W_{1}}{W_{2}}=\frac{\frac{1}{2} K(1)^{2}}{\frac{1}{2} K(2)^{2}}=\frac{1}{4} \\
W_{2}=4 W_{1} \\
W_{1}=10 \mathrm{J} \\
W_{2}=40 \mathrm{J}
\end{array}
$$
$\therefore$ More work required $=40 \mathrm{J}-10 \mathrm{J}=30 \mathrm{J}$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2016