datavorous/entrance-exam-dataset huggingface dataset:
Which of the following mixtures will have the lowest pH at 298 K?

Options:
A $10 \mathrm{mL}$ 0.05 $\mathrm{N}$ $\mathrm{CH}_{3} \mathrm{COOH}+5 \mathrm{mL}$ 0.1 $\mathrm{N}$ $\mathrm{NH}_{4} \mathrm{OH}$
B $5 \mathrm{mL}$ 0.2 $\mathrm{N}$ $\mathrm{NH}_{4} \mathrm{Cl}+5 \mathrm{mL}$. $0.2 \mathrm{N }$ $\mathrm{NH}_{4} \mathrm{OH}$
C $5 \mathrm{mL}$. $0.1 \mathrm{N}$ $\mathrm{CH}_{5} \mathrm{COOH}+10 \mathrm{mL}$ 0.05 $\mathrm{N}$ $\mathrm{CH}_{3} \mathrm{COONa}$
D $5 \mathrm{mL}$ 0.1 $\mathrm{N}$ $\mathrm{CH}_{3} \mathrm{COOH}+5 \mathrm{mL}$ 0.1 $\mathrm{N}$ $\mathrm{NaOH}$

The correct answer is:
$5 \mathrm{mL}$. $0.1 \mathrm{N}$ $\mathrm{CH}_{5} \mathrm{COOH}+10 \mathrm{mL}$ 0.05 $\mathrm{N}$ $\mathrm{CH}_{3} \mathrm{COONa}$

Solution:
Key Point For buffer solutions,
$$
\mathrm{pH}=\mathrm{pK}_{a}+\log \frac{[\text { Salt }]}{[\text { Acid }] /[\text { Base }]}
$$
Moles of $\mathrm{CH}_{3} \mathrm{COOH}=5 \mathrm{mL} \times 0.01 \mathrm{m} \mathrm{mol}=0.5 \mathrm{m} \mathrm{mol}$
Moles of $\mathrm{CH}_{3} \mathrm{COONa}=10 \mathrm{mL} \times 0.05 \mathrm{m} \mathrm{mol}$
$$
=0.5 \mathrm{m} \mathrm{mol}
$$
Total volume of solution $=5+10=15 \mathrm{mL}$
$$
\therefore \quad \mathrm{pH}=\mathrm{p} K_{a}+\log \frac{0.5 / 15}{0.5 / 15} \Rightarrow \mathrm{pH}=\mathrm{p} K_{a}
$$
The value of $\mathrm{p} K_{a}$ is the lowest for the mixture of $\begin{array}{llll}\mathrm{CH}_{3} \mathrm{COOH} & \text { and } \mathrm{CH}_{3} \mathrm{COONa} & \text { (acidic buffer). }\end{array}$
Therefore, it has lowest pH. Hence, pH value for option (c) is the lowest.

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2019
Which of the following orbitals will have zero probability of finding the electron in the yz plane ?

Options:
A $P_x$
B $\mathrm{P}_{\mathrm{y}}$
C $\mathrm{P}_{\mathrm{z}}$
D $\mathrm{d}_{\mathrm{yz}}$

The correct answer is:
$P_x$

Solution:
Hints : $P_x$ orbital lies along $x$-axis only.

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2010
Which of the following orders regarding ionization energy is correct?

Options:
A $\mathrm{N}>\mathrm{O}>\mathrm{F}$
B $\mathrm{N} < \mathrm{O} < \mathrm{F}$
C $\mathrm{N}>\mathrm{O} < \mathrm{F}$
D $\mathrm{N} < \mathrm{O}>\mathrm{F}$

The correct answer is:
$\mathrm{N}>\mathrm{O} < \mathrm{F}$

Solution:
Hints : As IE $1 \mathrm{~N}>\mathrm{O}$ (because of half filled orbitals of $\mathrm{N}$ ) and $\mathrm{O} < \mathrm{F}$ (because of smaller size of $\mathrm{F}$ )

Topic: Chemistry, Classification of Elements and Periodicity in Properties, WBJEE, WBJEE 2010
Which of the following plots represent an exothermic reaction?

Options:
A
B
C
D

The correct answer is:

Solution:
For exothermic reaction.
$\because$
$K_{p} \propto \frac{1}{T}$
Thus, as $\frac{1}{T}$ increases (i.e. $T$ decreases), $K_{p}$ increases.
$\because$ $\Delta$ H for exothermic reactions is negative.

Topic: Chemistry, Chemical Equilibrium, WBJEE, WBJEE 2016
Which of the following reactions will not result in the formation of carbon-carbon bonds?

Options:
A Cannizzaro reaction
B Wurtz reaction
C Reimer-Tiemann reaction
D Friedel-Crafts acylation

The correct answer is:
Cannizzaro reaction

Solution:
$\mathrm{C}-\mathrm{C}$ bond is not formed in Cannizaro reaction while other reactions result in the formation of $\mathrm{C}-\mathrm{C}$ bond.
Cannizaro reaction

Wurtz reaction
$2 R-X+2 \mathrm{Na} \longrightarrow R-R+2 \mathrm{Na} \times$

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2016
Which of the following real valued functions is/are not even functions?

Options:
A $f(x)=x^{3} \sin x$
B $f(x)=x^{2} \cos x$
C $f(x)=e^{x} x^{3} \sin x$
D $f(x)=x-[x]$, where $[x]$ denotes the greatest integer less than or equal to $x$

The correct answers are:
$f(x)=e^{x} x^{3} \sin x$, $f(x)=x-[x]$, where $[x]$ denotes the greatest integer less than or equal to $x$

Solution:
We know that, if $f(-x)=f(x)$, then function is even and $f(-x)=-f(x)$, then function is odd.
(a) $f(x)=x^{3} \sin x$
$$
\begin{aligned}
f(-x) &=(-x)^{3} \sin (-x) \\
&=-x^{3}(-\sin x) \\
&=x^{3} \sin x=f(x)
\end{aligned}
$$
So, $f(x)$ is even.
(b) $f(x)=x^{2} \cos x$
$$
f(-x)=(-x)^{2} \cos (-x)=x^{2} \cos x=f(x)
$$
So, $f(x)$ is even.
(c) $f(x)=e^{x} x^{3} \sin x$
$$
\begin{array}{r}
f(-x)=e^{-x}(-x)^{3} \sin (-x) \\
=e^{-x} x^{3} \sin x \neq f(x)
\end{array}
$$
$\therefore \quad f(x)$ is not even.
(d) $f(x)=x-[x]$ $f(-x)=(-x)-[-x] \neq f(x)$
$\therefore f(x)$ is not even.

Topic: Mathematics, Functions, WBJEE, WBJEE 2013
Which of the following relation represent Biot-Savart's law?

Options:
A $\mathrm{d} \overline{\mathrm{B}}=\frac{\mu_0}{4 \pi} \frac{\overline{\mathrm{d}} \times \overline{\mathrm{r}}}{\mathrm{r}}$
B $\mathrm{d} \overline{\mathrm{B}}=\frac{\mu_0}{4 \pi} \frac{\overline{\mathrm{d}} \mathrm{l} \times \hat{\mathrm{r}}}{\mathrm{r}^3}$
C $\mathrm{d} \overline{\mathrm{B}}=\frac{\mu_0}{4 \pi} \frac{\overline{\mathrm{d}} \times \overline{\mathrm{r}}}{\mathrm{r}^3}$
D $\mathrm{d} \overline{\mathrm{B}}=\frac{\mu_0}{4 \pi} \frac{\overline{\mathrm{d}} \times \overline{\mathrm{r}}}{\mathrm{r}^4}$

The correct answer is:
$\mathrm{d} \overline{\mathrm{B}}=\frac{\mu_0}{4 \pi} \frac{\overline{\mathrm{d}} \times \overline{\mathrm{r}}}{\mathrm{r}^3}$

Solution:
$d \vec{B}=\frac{\mu_0}{4 \pi} \frac{I(\overrightarrow{d \ell} \times \vec{r})}{r^3}$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2009
Which of the following represents the composition of Carnallite mineral?

Options:
A \(\mathrm{K}_2 \mathrm{O} \cdot \mathrm{Al}_2 \mathrm{O}_3 \cdot 6 \mathrm{SiO}_2\)
B \(\mathrm{KNO}_3\)
C \(\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{MgSO}_4 \cdot \mathrm{MgCl}_2 \cdot 6 \mathrm{H}_2 \mathrm{O}\)
D \(\mathrm{KCl} . \mathrm{MgCl}_2 \cdot 6 \mathrm{H}_2 \mathrm{O}\)

The correct answer is:
\(\mathrm{KCl} . \mathrm{MgCl}_2 \cdot 6 \mathrm{H}_2 \mathrm{O}\)

Solution:
Hints: Fact

Topic: Chemistry, General Principles and Processes of Isolation of Metals, WBJEE, WBJEE 2011
Which of the following sets of quantum numbers represents the $19^{\text {th }}$ electron of $\operatorname{Cr}(Z=24) ?$

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

The correct answer is:
$\left(4,0,0,+\frac{1}{2}\right)$

Solution:
Electronic configuration for $\operatorname{Cr}(24)$
$=\left[\mathrm{Ar}]_{18} 4 s^{1} 3 d^{3}\right.$
$\therefore 19^{\text {b }}$ electron enters in $4 s$ -orbital and its set of quantum numbers is $n=4, l=0, m=0, m_{x}=+\frac{1}{2}$

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2017
Which of the following solutions will turn violet when a drop of lime juice is added to it?

Options:
A A solution of Nal
B A solution mixture of $\mathrm{KI}$ and $\mathrm{NaIO}_{3}$
C A solution mixture of Nal and KI
D A solution mixture of $\mathrm{KIO}_{3}$ and $NaIO_{3}$

The correct answer is:
A solution mixture of $\mathrm{KI}$ and $\mathrm{NaIO}_{3}$

Solution:
When drop of line juice is added to mixture of $\mathrm{I}+\mathrm{IO}_{3}^{-}$. The following reaction take place and violet colour appears due to formation of $\mathrm{I}_{2}$
$\mathrm{I}+\mathrm{IO}_{3}^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{I}_{2}+\mathrm{H}_{2} \mathrm{O}$

Topic: Chemistry, Practical Chemistry, WBJEE, WBJEE 2017
Which of the following statements is /are correct for $0 < \theta < \frac{\pi}{2} ?$

Options:
A $(\cos \theta)^{1 / 2} \leq \cos \frac{\theta}{2}$
B $(\cos \theta)^{3 / 4} \geq \cos \frac{3 \theta}{4}$
C $\cos \frac{5 \theta}{6} \geq(\cos \theta)^{5 / 6}$
D $\cos \frac{70}{8} \leq(\cos \theta)^{7 / 8}$

The correct answers are:
$(\cos \theta)^{1 / 2} \leq \cos \frac{\theta}{2}$, $\cos \frac{5 \theta}{6} \geq(\cos \theta)^{5 / 6}$

Solution:
$0 < 0 < \frac{\pi}{2}, cos \theta$ is monotonically decreasing function
Option
(a) When $\theta>\frac{\theta}{2},$ then $\cos \theta \leq \cos \frac{\theta}{2}$
$\therefore \quad \sqrt{\cos \theta} \leq \cos \frac{\theta}{2} \quad$ correct
(b) When $0>\frac{30}{4},$ then $\cos \theta \leq \cos \frac{30}{4}$
$\therefore \quad(\cos \theta)^{3 / 4} \leq \cos \frac{30}{4}$
incorrect|
(c) When $\theta>\frac{5 \theta}{6},$ then $\cos \theta \leq \cos \frac{50}{6}$
$\therefore \quad(\cos \theta)^{5 / 6} \leq \cos \frac{50}{6}$
[correct|
(d) When $\frac{7 \theta}{8} < \theta,$ then $\cos \theta \leq \cos \frac{7 \theta}{8}$
But $(\cos \theta)^{7 / 8} \leq \cos \frac{7 \theta}{8}$
|incorrect|

Topic: Mathematics, Trigonometric Equations, WBJEE, WBJEE 2015
Which of the following statement(s) is (are) correct when a mixture of $\mathrm{NaCl}$ and $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ is gently warmed with conc. $\mathrm{H}_{2} \mathrm{SO}_{4} ?$

Options:
A A deep red vapour is evolved
B The vapour when passed through NaOH solution, gives a yellow solution
C Chiorine gas is also evolved
D Chromyl chioride is formed

The correct answers are:
A deep red vapour is evolved, The vapour when passed through NaOH solution, gives a yellow solution, Chromyl chioride is formed

Solution:
(A) When $\mathrm{NaCl}$ and $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ warmed with $\mathrm{H}_{2} \mathrm{SO}_{4}$ (i.e. in acidic medium), they produce deep red vapours of chromyl chloride $\left(\mathrm{CrO}_{2} \mathrm{Cl}_{2}\right)$
(B) When NaOH is passod, the product $CrO_{3}$, formed In step (a) will react with NaOH and gives yellow colour solution
(C) Chlorine gas is not evolved, thus false
(D) In the given reaction, chromyl chloride $\left(\mathrm{CrO}_2\mathrm{Cl}_{2}\right)$ is formed. All reactions are as follows
(i) $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{KHSO}_{4}+2 \mathrm{CrO}_{3}+\mathrm{H}_{2} \mathrm{O}$
(ii) $\mathrm{NaCl}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{NaHSO}_{4}+\mathrm{HCl}$
(iii) $\mathrm{CrO}_{3}+2 \mathrm{HCl} \longrightarrow \mathrm{CrO}_{2} \mathrm{Cl}_{2}$
(iv) $\mathrm{CrO}_{2} \mathrm{Cl}_{2}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{CrO}_{4}+2 \mathrm{HCl}$

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2016
Which of the following statements is incorrect?

Options:
A \(\left[\mathrm{VF}_6\right]^{3-}\) is paramagnetic with 2 unpaired electrons.
B \(\left[\mathrm{CuCl}_4\right]^{2-}\) is paramagnetic with 1 unpaired electron.
C \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}\) is diamagnetic.
D \(\left[\mathrm{CoF}_6\right]^{3-}\) is paramagnetic with 2 unpaired electrons.

The correct answer is:
\(\left[\mathrm{CoF}_6\right]^{3-}\) is paramagnetic with 2 unpaired electrons.

Solution:
Hint : \(\left[\stackrel{\text{III}}{\mathrm{Co}}\mathrm{F}_6\right]^{3-} \mathrm{As} \mathrm{F}^{-}\)is weak field ligand.
Oxidation Number of \(\mathrm{Co}=+3\)
\(\mathrm{Co}^{3+}\left(4 \mathrm{~s}^0 3 \mathrm{~d}^6\right)\)
Number of unpaired \(\mathrm{e}^{-}=4\)

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2023
Which of the following statements is not correct

Options:
A Silicon is extensively used as a semiconductor
B Carborundum is $\mathrm{SiC}$
C Silicon occurs in free state in nature
D Mica contains the element silicon

The correct answer is:
Silicon occurs in free state in nature

Solution:
Hints : Silicon exist in nature in combined state as $\mathrm{SiO}_2$

Topic: Chemistry, p Block Elements (Group 13 & 14), WBJEE, WBJEE 2009
Which of the following statements is not true for the reaction $2 \mathrm{~F}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 4 \mathrm{HF}+\mathrm{O}_{2} ?$

Options:
A $\mathrm{F}_{2}$ is more strongly oxidising than $\mathrm{O}_{2}$
B $\mathrm{F}-\mathrm{F}$ bond is weaker than $\mathrm{O}=\mathrm{O}$ bond
C $\mathrm{H}-\mathrm{F}$ bond is stronger than $\mathrm{H}-\mathrm{O}$ bond
D $\mathrm{F}$ is less electronegative than $\mathrm{O}$

The correct answer is:
$\mathrm{F}$ is less electronegative than $\mathrm{O}$

Solution:
Hint:
$\mathrm{F}$ is more electronegative than $\mathrm{O}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, WBJEE, WBJEE 2020
Which of the following statement(s) is/are incorrect:

Options:
A A sink of $\mathrm{SO}_{2}$ pollutant is $\mathrm{O}_{3}$ in the atmosphere.
B FGD is a process of removing $\mathrm{NO}_{2}$ from atmosphere
C NO in fuel gases can be removed by alkaline scrubbing
D The catalyst used to convert $\mathrm{CCl}_{4}$ to $\mathrm{CF}_{4}$ by $\mathrm{HF}$ is $\mathrm{SbF}_{5}$

The correct answers are:
A sink of $\mathrm{SO}_{2}$ pollutant is $\mathrm{O}_{3}$ in the atmosphere., FGD is a process of removing $\mathrm{NO}_{2}$ from atmosphere, The catalyst used to convert $\mathrm{CCl}_{4}$ to $\mathrm{CF}_{4}$ by $\mathrm{HF}$ is $\mathrm{SbF}_{5}$

Solution:
Hint:
Oxides of nitrogen can be removed by alkaline scrubbing.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2020
Which of the following statements regarding Lanthanides is false?

Options:
A All lanthanides are solid at room temperature
B Their usual oxidation state is +3
C They can be separated from one another by ion-exchange method
D Ionic radii of trivalent lanthanides steadily increases with increase in atomic number

The correct answer is:
Ionic radii of trivalent lanthanides steadily increases with increase in atomic number

Solution:
The option (d) is false as lonic radi of trivalent lanthanides decreases with increase in atomic number due to lanthanoid contraction.

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2016
Which of the following statements regarding ozone is not correct?

Options:
A The Ozone molecule is angular in shape
B The Ozone is a resonance hybrid of two structures
C The Oxygen-Oxygen bond length in ozone is identical with that of molecular oxygen
D Ozone is used as germicide and disinfectant for the purification of air.

The correct answer is:
The Oxygen-Oxygen bond length in ozone is identical with that of molecular oxygen

Solution:
Hints : Due to resonance the bond order in ozone is 1.5 , hence $\mathrm{O}-\mathrm{O}$ bond length in $\mathrm{O}_3>\mathrm{O}-\mathrm{O}$ bond length in $\mathrm{O}_2$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2010
Which of the following thermodynamic relation is correct?

Options:
A $\mathrm{dG}=\mathrm{VdP}-\mathrm{SdT}$
B $\mathrm{dE}=\mathrm{PdV}+\mathrm{TdS}$
C $\mathrm{dH}=-\mathrm{VdP}+\mathrm{TdS}$
D $\mathrm{dG}=\mathrm{VdP}+\mathrm{SdT}$

The correct answer is:
$\mathrm{dG}=\mathrm{VdP}-\mathrm{SdT}$

Solution:
$$
\text { Hints: } \begin{gathered}
\mathrm{dG}=\mathrm{dH}-\mathrm{TdS}-\mathrm{SdT}(\text { as } \mathrm{G}=\mathrm{H}-\mathrm{TS} \text { ) } \\
\text { again, } \mathrm{H}=\mathrm{U}+\mathrm{PV} \\
\therefore \mathrm{dH}=\mathrm{dU}+\mathrm{PdV}+\mathrm{VdP} \\
\& \mathrm{dU}=\mathrm{TdS}-\mathrm{PdV}
\end{gathered}
$$
Thus $d G=(T d S-P d V)+P d V+V d P-T d S-S d T$
$$
=\mathrm{VdP}-\mathrm{SdT}
$$

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2010
Which of the following will be dehydrated most readily in alkaline medium?

Options:
A
B
C
D

The correct answer is:

Solution:
In alkaline medium the acidic proton present in the molecule is abstracted to give enolate ion. In the next step hydroxide ion if present at $\beta$ -position leaves giving $\alpha-\beta$ unsaturated carbonyl compound. In (a), (c) and
(d), dehydration cannot take place due to improper placement of groups.

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2016
Which of the following will decrease the $\mathrm{pH}$ of a $50 \mathrm{ml}$ solution of $0.01 \mathrm{M} \mathrm{HCl}$ ?

Options:
A addition of $5 \mathrm{ml}$ of $1 \mathrm{M} \mathrm{HCl}$
B addition of $50 \mathrm{ml}$ of $0.01 \mathrm{M} \mathrm{HCl}$
C addition of $50 \mathrm{ml}$ of $0.002 \mathrm{M} \mathrm{HCl}$
D addition of Mg

The correct answer is:
addition of $5 \mathrm{ml}$ of $1 \mathrm{M} \mathrm{HCl}$

Solution:
$50 \mathrm{ml} 0.01 \mathrm{M} \equiv 50 \times 0.01=0.5$ millimole
$\quad 5 \mathrm{ml} 1(\mathrm{M}) \equiv 5 \times 1=5$ millimole
$\quad$ Total millimoles $=5.5$ millimole
Total volume $=55 \mathrm{ml}$.
Molarity $=\frac{5.5}{55}=0.1(\mathrm{M})=10^{-1}(\mathrm{M})$
$\mathrm{pH}=1$

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2009
Which of the following will show a negative deviation from Raoult's law?

Options:
A Acetone-benzene
B Acetone-ethanol
C Benzene-methanol
D Acetone-chloroform

The correct answer is:
Acetone-chloroform

Solution:
Acetone and chloroform will show a negative deviation due to their association after mixing.

Topic: Chemistry, Solutions, WBJEE, WBJEE 2012
Which of the following would give a linear plot?
(\(k\) is the rate constant of an elementary reaction and \(\mathrm{T}\) is temp. in absolute scale)

Options:
A k vs T
B \(\mathrm{k~vs~} 1 / \mathrm{T}\)
C In k vs T
D \(\ln \mathrm{k~vs~} 1 / \mathrm{T}\)

The correct answer is:
\(\ln \mathrm{k~vs~} 1 / \mathrm{T}\)

Solution:
Hint : Arrhenius equation gives us
\(\begin{aligned}
& k=A e^{-E_a / R T} \\
& \ln k=\ln A-\frac{E_a}{R}\left(\frac{1}{T}\right) \\
& y=c-m x
\end{aligned}\)

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2023
Which of the following would produce enantiomeric products when reacted with methyl magnesium iodide?

Options:
A Benzaldehyde
B Propiophenone
C Acetone
D Acetaldehyde

The correct answers are:
Benzaldehyde, Propiophenone

Solution:
When $\mathrm{R}$ and $\mathrm{R}^{\prime}$ are different and none of them is methyl, we will get enantiomeric product.

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2022
Which of the set of oxides are arranged in the order of proper basic, amphoteric, acidic?

Options:
A $\mathrm{SO}_{2}, \mathrm{P}_{2} \mathrm{O}_{5}, \mathrm{CO}$
B $\mathrm{BaO}, \mathrm{Al}_{2} \mathrm{O}_{3} \cdot \mathrm{SO}_{2}$
C $\mathrm{CaO} . \mathrm{SiO}_{2}, \mathrm{Al}_{2} \mathrm{O}_{3}$
D $\mathrm{CO}_{2}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{CO}$

The correct answer is:
$\mathrm{BaO}, \mathrm{Al}_{2} \mathrm{O}_{3} \cdot \mathrm{SO}_{2}$

Solution:
The correct set of oxides arranged in the proper order of basic, amphoteric, acidic are $\mathrm{BaO}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SO}_{2}$
Oxides of non-metallic elements are acidic such $\mathrm{CO}_{2} \mathrm{NO}_{2}, \mathrm{SO}_{2}$ etc. Oxides of less electropositive elements (such as BeO, $A l_{2} O_{3}, B i_{2} O_{3}$, ZnO etc.) are amphoteric i.e. these behaves as acids toward strong bases and as bases towards strong acids. Oxides of elcctropositive elements $\left(\mathrm{Na}_{2} \mathrm{O}, \mathrm{CaO}, \mathrm{T}_{2} \mathrm{O}, \quad \mathrm{Ba} 0\right.$
etc.) are basic and contain discrete $\mathrm{O}^{2-}$ ions.

Topic: Chemistry, p Block Elements (Group 13 & 14), WBJEE, WBJEE 2018
Which of the statements are incorrect?

Options:
A $\mathrm{pH}$ of a solution of salt of strong acid and weak base is less than 7 .
B $\mathrm{pH}$ of a solution of a weak acid and weak base is basic if $\mathrm{K}_{\mathrm{b}} < \mathrm{K}_{\mathrm{a}}$.
C $\mathrm{pH}$ of an aqueous solution of $10^{-8}(\mathrm{M}) \mathrm{HCl}$ is 8 .
D Conjugate acid of $\mathrm{NH}_2^{-}$is $\mathrm{NH}_3$.

The correct answers are:
$\mathrm{pH}$ of a solution of a weak acid and weak base is basic if $\mathrm{K}_{\mathrm{b}} < \mathrm{K}_{\mathrm{a}}$., $\mathrm{pH}$ of an aqueous solution of $10^{-8}(\mathrm{M}) \mathrm{HCl}$ is 8 .

Solution:
* $\mathrm{pH}$ of a solution of strong acid and weak base $\left(\right.$ say $\left.\mathrm{NH}_4 \mathrm{Cl}\right)$ is less than $7 .$
* For hydrolysis of salt of weak acid and weak base.
When $\mathrm{K}_{\mathrm{a}}>\mathrm{K}_{\mathrm{b}}, \quad \mathrm{pH} < 7$
* $\mathrm{pH}$ of $10^{-8} \mathrm{M} \mathrm{HCl}(\mathrm{aq})$ will be less than 7 .
* Conjugate acid of $\mathrm{NH}_2^{-}$is $\mathrm{NH}_3$. This is a true statement.

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2022
Which of these species will have non-zero magnetic moment?

Options:
A $\mathrm{Na}^{+}$
B Mg
C $F^{-}$
D $A r^{+}$

The correct answer is:
$A r^{+}$

Solution:
Key point The species which has one or more unpaired electrons have non-zero magnetic moment.
Electronic configuration of
$$
\begin{array}{l}
\mathrm{Na}^{+}(Z=11)=1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 p^{6} \\
\mathrm{Mg}(Z=12)=1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 p^{6} 3 \mathrm{s}^{2} \\
\mathrm{F}^{-}(Z=9)=1 \mathrm{s}^{2} 2 s^{2} 2 p^{6} \\
\quad \mathrm{Ar}^{+}(Z=18)=1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 p^{6} 3 \mathrm{s}^{2} 3 p^{5}
\end{array}
$$
As, Ar $^{+}$ has an unpaired electron, it has non-zero magnetic moment. Hence, (d) is the correct option.

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2019
Which one is not a constituent of nucleic acid?

Options:
A Uracil
B Guanidine
C Phosphoric acid
D Ribose sugar

The correct answer is:
Guanidine

Solution:
Hints: Guanine is the constituent of nucleic acid and not guanidine.

Topic: Chemistry, Biomolecules, WBJEE, WBJEE 2011
Which one of the following characteristics belongs to an electrophile?

Options:
A It is any species having electron deficiency which reacts at an electron rich C-centre
B It is any species having electron enrichment, that reacts at an electron deficient C-centre
C It is cationic in nature
D It is anionic in nature

The correct answer is:
It is any species having electron deficiency which reacts at an electron rich C-centre

Solution:
Electrophiles are electron deficient species (neutral or cationic). They attack on electron rich C-atom.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2012
Which one of the following contains $\mathrm{P}-\mathrm{O}-\mathrm{P}$ bond ?

Options:
A Hypophosphorus acid
B Phosphorus acid
C Pyrophosphoric acid
D Orthophosphoric acid

The correct answer is:
Pyrophosphoric acid

Solution:
Hints:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2010
Which one of the following corresponds to a photon of highest energy?

Options:
A $\lambda=300 \mathrm{mm}$
B $v=3 \times 10^{4} \mathrm{s}^{-1}$
C $\bar{v}=30 \mathrm{cm}^{-1}$
D $\varepsilon=6.626 \times 10^{-27} \mathrm{J}$

The correct answer is:
$\lambda=300 \mathrm{mm}$

Solution:
$\because E=h v=\frac{h c}{\lambda}=h c \cdot \bar{v} \quad\left(\because \frac{1}{\lambda}=\bar{v}\right)$
where $E=$ energy of photon
$c=$ velocity of photon (-light)
$\lambda=$ wavelength of photon
$h=$ plank's constant.
$\therefore$ For $(\mathrm{a})$
$E=6.63 \times 10^{-34} \times 3 \times 10^{4} / 300 \times 10^{-9}$
$E=1.98 \times 10^{-25} \cdot \mathrm{J} / 300 \times 10^{-9}$
$\therefore \quad E=\frac{1.98 \times 10^{-2} \mathrm{J}}{300 \times 10^{-9} \mathrm{m}}$
$(a) \rightarrow E=6.6 \times 10^{-19} \mathrm{J}$
For (b) $E=h v=6.63 \times 10^{-34} \times 3 \times 10^{4}$
$(b) \rightarrow E=1.98 \times 10^{-35} \mathrm{J}$
For (c) $E=h c \times v$
$\because \frac{1}{3}$
$$
E=6.63 \times 10^{-4} \times 3 \times 10^{*} \times 30 \times 10^{2}
$$
(c) $\Rightarrow E=5.96 \times 10^{\circ}$
For $(d) d \rightarrow E=6.62 \times 10^{-27} J$
Hence, highest energy for photon is in (a).

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2017
Which one of the following electronic arrangements is absurd?

Options:
A $n=3, l=1, m=-1$
B $n=3, l=0, m=0$
C $n=2,l=0, m=-1$
D $n=2, l=1, m=0$

The correct answer is:
$n=2,l=0, m=-1$

Solution:
The three quantum numbers $n, l$ and $m$ that describe an orbital has integer values of 0,1,2,3 and so on.

The electronic arrangement of option (c) is absurd because
$$
\begin{array}{l}
n=2 \\
l=0 \\
m=0
\end{array}
$$
To sum up, where $n=2, l=0, m=0$ Hence, for $l=0,$ value of $m$ should be zero.

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2019
Which one of the following formulae does not represent an organic compound?

Options:
A $\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}_4$
B $\mathrm{C}_4 \mathrm{H}_8 \mathrm{O}_4$
C $\mathrm{C}_4 \mathrm{H}_7 \mathrm{ClO}_4$
D $\mathrm{C}_4 \mathrm{H}_9 \mathrm{O}_4$

The correct answer is:
$\mathrm{C}_4 \mathrm{H}_9 \mathrm{O}_4$

Solution:
Unsaturation factor $=0,1,1,0.5 \quad$ Hence $(D)$

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2009