datavorous/entrance-exam-dataset huggingface dataset:
Argument of $-1-i \sqrt{3}$ is

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

The correct answer is:
$-\frac{2 \pi}{3}$

Solution:
The argument of a complex number in rectangular form, $z=a+b i$, is the angle, denoted by $\theta$, measured from the positive $x$-axis in the complex plane, to the point $z$. We can find it using the following formula:
$\theta=\tan ^{-1}\left(\frac{b}{a}\right)$
In this case, $a=-1$ and $b=-\sqrt{3}$ :
$\theta=\tan ^{-1}\left(\frac{-\sqrt{3}}{-1}\right)=\frac{5 \pi}{3}$
However, the argument of a complex number is not unique and has multiple values that differ by $2 \pi n$, where $n$ is any integer. Because $-1-i \sqrt{3}$ lies in Quadrant III, its argument must be in the interval $\left(-\pi,-\frac{\pi}{2}\right)$.
Converting $\frac{5 \pi}{3}$ to the interval $\left(-\pi,-\frac{\pi}{2}\right)$ by subtracting the nearest multiple of $2 \pi$, we get:
$\frac{5 \pi}{3}-2 \pi=-\frac{2 \pi}{3}$
Therefore, the argument of $-1-i \sqrt{3}$ is $-\frac{2 \pi}{3}$.

Topic: Mathematics, Complex Number, JEE Main
Aromatic compounds burn with sooty flame because

Options:
A They have a ring structure of carbon atoms
B They have a relatively high percentage of hydrogen
C They have a relatively high percentage of carbon
D They resist reaction with oxygen of air

The correct answer is:
They have a relatively high percentage of carbon

Solution:
They have a relatively high percentage of carbon

Topic: Chemistry, Hydrocarbons, JEE Main
Aromatisation of $n$-heptane by passing over $\left(\mathrm{Al}_2 \mathrm{O}_3+\mathrm{Cr}_2 \mathrm{O}_3\right)$ catalyst at $773 \mathrm{~K}$ gives

Options:
A Benzene
B Toluene
C Mixture of both
D Heptylene

The correct answer is:
Toluene

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Arrange $\mathrm{Ce}^{+3}, \mathrm{La}^{+3}, \mathrm{Pm}^{+3}$ and $\mathrm{Yb}^{+3}$ in increasing order of their ionic radii

Options:
A
$\mathrm{Yb}^{+3} < \mathrm{Pm}^{+3} < \mathrm{Ce}^{+3} < \mathrm{La}^{+3}$
B
$\mathrm{Ce}^{+3} < \mathrm{Yb}^{+3} < \mathrm{Pm}^{+3} < \mathrm{La}^{+3}$
C
$\mathrm{Yb}^{+3} < \mathrm{Pm}^{+3} < \mathrm{La}^{+3} < \mathrm{Ce}^{+3}$
D
$\mathrm{Pm}^{+3} < \mathrm{La}^{+3} < \mathrm{Ce}^{+3} < \mathrm{Yb}^{+3}$

The correct answer is:
$\mathrm{Yb}^{+3} < \mathrm{Pm}^{+3} < \mathrm{Ce}^{+3} < \mathrm{La}^{+3}$

Solution:
According to their positions in the periods, these values are in the order :
This is due to lanthanide contraction

Topic: Chemistry, d and f Block Elements, JEE Main
Arrange $\mathrm{Ce}^{+3}, \mathrm{La}^{+3}, \mathrm{Pm}^{+3}$ and $\mathrm{Yb}^{+3}$ in increasing order of their ionic radii

Options:
A
$\mathrm{Yb}^{+3} < \mathrm{Pm}^{+3} < \mathrm{Ce}^{+3} < \mathrm{La}^{+3}$
B
$\mathrm{Ce}^{+3} < \mathrm{Yb}^{+3} < \mathrm{Pm}^{+3} < \mathrm{La}^{+3}$
C
$\mathrm{Yb}^{+3} < \mathrm{Pm}^{+3} < \mathrm{La}^{+3} < \mathrm{Ce}^{+3}$
D
$\mathrm{Pm}^{+3} < \mathrm{La}^{+3} < \mathrm{Ce}^{+3} < \mathrm{Yb}^{+3}$

The correct answer is:
$\mathrm{Yb}^{+3} < \mathrm{Pm}^{+3} < \mathrm{Ce}^{+3} < \mathrm{La}^{+3}$

Solution:
According to their positions in the periods, these values are in the order :
This is due to lanthanide contraction

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2002
Arrange following compounds in decreasing order of rate of electrophilic substitution.

Options:
A i > ii > iii > iv
B iii > iv > ii > i
C i > iv > ii > iii
D i > ii > iv > iii

The correct answer is:
i > iv > ii > iii

Solution:
(I)	(II)	(III)	(IV)

More α - H , results in more hyper conjugation, resulting in greater electron density on the benzene ring & hence greater reactivity towards electrophilic substitution.

Topic: Chemistry, General Organic Chemistry, JEE Main
Arrange in increasing order of solubility of AgBr in solutions given here
(i) 0.1 M NH3 (ii) 0.1 M AgNO3 (iii) 0.2 M NaBr (iv) pure water

Options:
A (iii) < (ii) < (iv) < (i)
B (iii) < (ii) < (i) < (iv)
C (iii) < (ii) < (i) < (iv)
D (ii) < (iii) < (iv) < (i)

The correct answer is:
(iii) < (ii) < (iv) < (i)

Solution:
AgBr → in NH3 (Solubility increases, due to stable complex)
→ in 0.1 M AgNO3 (Solubility decreases)
→ in 0.2 M NaBr (Solubility decreases more)
→ in pure water (Normal dissociation)
Hence the order is
(iii) < (ii) < (iv) < (i)

Topic: Chemistry, Ionic Equilibrium, JEE Main
Arrange in the correct order of stability (decreasing order) for the following molecules:
(I)
(II)
(III)
(IV)

Options:
A
(I) $>$ (II) $>$ (III) $>$ (IV)

B
(IV) $>$ (III) $>$ (II) $\approx$ (I)

C
(I) $>$ (II) $\approx$ (III) $>$ (IV)

D
(III) $>$ (I) $\approx$ (II) $>$ (IV)

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

Solution:
No Solution Available

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2013 (22 Apr Online)
Arrange reactivity of given alcohols in decreasing order for dehydration reaction with concentrated H2SO4

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

The correct answer is:
2, 4, 1, 3

Solution:
The reactivity for dehydration reaction depends upon the stability of carbocation i.e. more the stability of carbocation more is reactivity for dehydration

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Arrange reactivity of given compounds in decreasing order for electrophilic addition reaction

Select the correct answer from codes given below

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

The correct answer is:
3, 2, 1, 4

Solution:
The reactivity for electrophilic addition reaction depends upon the stability of carbocation i.e. more the stability of carbocation more is reactivity

Topic: Chemistry, General Organic Chemistry, JEE Main
Arrange reactivity of given compounds in decreasing order for hydrolysis reaction?
(1)
(2)
(3)
(4)
Select the correct answer from the codes given below:

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

The correct answer is:
2, 3, 4, 1

Solution:
Here the decreasing order of hydrolysis is as follows

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Arrange the carbanions, $\left(\mathrm{CH}_3\right)_3 \overline{\mathrm{C}}, \overline{\mathrm{C}} \mathrm{Cl}_3,\left(\mathrm{CH}_3\right)_2 \overline{\mathrm{C}} \mathrm{H}, \mathrm{C}_6 \mathrm{H}_5 \overline{\mathrm{C}} \mathrm{H}_2$, in order of their decreasing stability :

Options:
A
$\mathrm{C}_8 \mathrm{H}_5 \overline{\mathrm{C}} \mathrm{H}_2>\overline{\mathrm{C}} \mathrm{Cl}_3>\left(\mathrm{CH}_3\right)_3 \overline{\mathrm{C}}>\left(\mathrm{CH}_3\right)_2 \overline{\mathrm{C}} \mathrm{H}$

B
$\left(\mathrm{CH}_3\right)_2 \overline{\mathrm{C}} \mathrm{H}>\overline{\mathrm{C}} \mathrm{Cl}_3>\mathrm{C}_6 \mathrm{H}_5 \overline{\mathrm{C}} \mathrm{H}_2>\left(\mathrm{CH}_3\right)_3 \overline{\mathrm{C}}$

C
$\overline{\mathrm{C}} \mathrm{Cl}_3>\mathrm{C}_8 \mathrm{H}_5 \overline{\mathrm{C}} \mathrm{H}_2>\left(\mathrm{CH}_3\right)_2 \overline{\mathrm{C}} \mathrm{H}>\left(\mathrm{CH}_3\right)_3 \overline{\mathrm{C}}$

D
$\left(\mathrm{CH}_3\right)_3 \overline{\mathrm{C}}>\left(\mathrm{CH}_3\right)_2 \overline{\mathrm{C}} \mathrm{H}>\mathrm{C}_6 \mathrm{H}_5 \overline{\mathrm{C}} \mathrm{H}_2>\overline{\mathrm{C}} \mathrm{Cl}_3$

The correct answer is:
$\overline{\mathrm{C}} \mathrm{Cl}_3>\mathrm{C}_8 \mathrm{H}_5 \overline{\mathrm{C}} \mathrm{H}_2>\left(\mathrm{CH}_3\right)_2 \overline{\mathrm{C}} \mathrm{H}>\left(\mathrm{CH}_3\right)_3 \overline{\mathrm{C}}$

Solution:
$2^{\circ}$ carbanion is more stable than $3^{\circ}$ and $\mathrm{Cl}$ is $-1$ effect group.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2009
Arrange the following alcohols in order of increasing reactivity towards sodium metal.
(i) CH3OH
(ii) (CH3)2CH-OH
(iii) CH3CH2OH
(iv) (CH3)3C-OH

Options:
A (iv) < (iii) < (ii) < (i)
B (ii) < (i) < (iii) < (iv)
C (iv) < (ii) < (iii) < (i)
D (i) < (iii) < (ii) < (iv)

The correct answer is:
(iv) < (ii) < (iii) < (i)

Solution:
More the number of methyl groups (electron releasing) lesser will be the acidity. Hence the reactivity order towards sodium metal will be as follows
tertiary (i) < secondary (ii) < primary (iii).
(CH3)3C-OH<(CH3)2CH-OH<CH3CH2OH<CH3OH

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Arrange the following amines in the decreasing order of basicity.

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

The correct answer is:
III>I>II

Solution:
Kb Values are a measure of basicity. Higher the Kb value, greater will be basicity.
(I) Basic character α1% s- character
N-sp2 has more % s-character than N-sp3 so it is less basic as compare to N-sp3 Hence III>I.
(II) In the (ii) structure lone pair present on the N atom is involved in aromaticity so, it is difficult to donate. Hence, least basic
Thus, the correct order will be: III>I>II

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Arrange the following anilines in decreasing order of basicity
1. C6H5NH2
2. o-CH3C6H4NH2
3. m-CH3C6H4NH2
4. p-CH3C6H4NH2

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

The correct answer is:
4 > 3 > 1 > 2

Solution:
-CH3 group is inductively electron-donating and base-strengthening from all position. Moreover, if -CH3 is attached to benzene ring (or if R has benzylic H), it is electron-donating by hyper-conjugation from the ortho and para positions. Because of ortho effect, o-substituted aniline become less basic than aniline. Thus,
p-CH3C6H4NH2 (hyper-conjugation and induction) > m-CH3C6H4NH2 (induction) > C6H5NH2>o-CH3C6H4NH2 (ortho effect).

Topic: Chemistry, Amines, JEE Main
Arrange the following anilines in decreasing order of basicity
1. \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \)
2. o- \( \mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NH}_{2} \)
3. \( \mathrm{m}-\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NH}_{2} \)
4. \( \mathrm{p}-\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NH}_{2} \)

Options:
A \( 4>1>2>3 \)
B \( 4>3>1>2 \)
C \( 1>2>3>4 \)
D \( 4>3>2>1 \)

The correct answer is:
\( 4>3>1>2 \)

Solution:
-CH3 group is inductively electron-donating and base-strengthening from all position. Moreover, if -CH3 is attached to benzene ring (or if R has benzylic H), it is electron-donating by hyper-conjugation from the ortho and para positions. Because of ortho effect, o-substituted aniline become less basic than aniline. Thus,
p-CH3C6H4NH2 (hyper-conjugation and induction) > m-CH3C6H4NH2 (induction) > C6H5NH2>o-CH3C6H4NH2 (ortho effect).

Topic: Chemistry, Amines, JEE Main
Arrange the following bonds according to their average bond energies in descending order:
C-Cl,C-Br,C-F,C-I

Options:
A C-F>C-Cl>C-Br>C-I
B C-Br>C-I>C-Cl>C-F
C C-I>C-Br>C-Cl>C-F
D C-Cl>C-Br>C-I>C-F

The correct answer is:
C-F>C-Cl>C-Br>C-I

Solution:
Bond energy ∝1Bond  length
Bond length order in carbon halogen bonds arein the order of C – F < C – Cl < C – Br < C – IHence, Bond energy orderC – F > C – Cl > C – Br > C – I

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Arrange the following bonds according to their average bond energies in descending order.
CâCl, CâBr, CâF, CâI

Options:
A C–F>C–Cl>C–Br>C–I
B C–I>C–Br>C–Cl>C–F
C C–Br>C–I>C–Cl>C–F
D C–Cl>C–Br>C–I>C–F

The correct answer is:
C–F>C–Cl>C–Br>C–I

Solution:
Bond energy∝ 1 Bond length
As on moving down the group size increases so bond length increases hence bond energy decreases.
Bond energy CâF>CâCl>CâBr>CâI

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Arrange the following carbocations in decreasing order of stability.

Options:
A A>C>B
B B>A>C
C C>B>A
D C>A>B

The correct answer is:
B>A>C

Solution:
The correct order of stability of the given carbocation is as follows

Carbocation, B is the most stable due to resonance stabilization

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2022 (24 Jun Shift 2)
Arrange the following carbocations in order of stability

benzyl allyl methyl vinyl I II III IV

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

The correct answer is:
I > II > III > IV

Solution:
menthyl ( CH 3 ⊕ ), vinyl (CH 2  =  CH + ). Among the given options benzyl is highly stabilised by resonance and so is allylic but lesser than benzyl. Vinylic has a sp 2 carbocation which is unstable in nature.

Topic: Chemistry, General Organic Chemistry, JEE Main
Arrange the following compounds in decreasing order of their boiling points
(i) CH3Cl
(ii) CH3CH2Cl
(iii) CH3CH2CH2Cl
(iv) CH3CH2CH2CH2Cl

Options:
A (iii) > (iv) > (i) > (ii)
B (iv) > (iii) > (ii) > (i)
C (i) > (iii) > (ii) > (iv)
D (i) > (ii) > (iii) > (iv)

The correct answer is:
(iv) > (iii) > (ii) > (i)

Solution:
Boiling point of the given compounds decreases in the following order as when molecular weight decreases magnitude of Van der Waal's forces also decreases so boiling point also decreases.
CH3CH2CH2CH2Cl>CH3CH2CH2Cl>CH3CH2Cl>CH3Cl

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Arrange the following compounds in increasing order of C-OH bond length: methanol, phenol, p -ethoxyphenol

Options:
A methanol <p -ethoxyphenol < phenol
B phenol < methanol <p -ethoxyphenol
C phenol <p -ethoxyphenol < methanol
D methanol < phenol <p -ethoxyphenol

The correct answer is:
phenol <p -ethoxyphenol < methanol

Solution:
There is not any resonance in CH3–OH. Resonance is poor in p-Ethoxyphenol than phenol . So C-OH bond lengh order is: CH3OH >p-Ethoxyphenol > phenol

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (08 Jan Shift 1)
Arrange the following compounds in increasing order of rate of aromatic electrophilic substitution reaction.

Options:
A d, b, c, a
B d, b, a, c
C b, c, a, d
D c, a, b, d

The correct answer is:
c, a, b, d

Solution:
The more nucleophilic group will make ring more prone to aromatic substitution reaction. Benzene becomes more reactive towards Electrophilic Aromatic Substitution when any substituent raises its electron density.

In structure (a) -CH2 group shows +H effect. while in (b) -OH group shows +R effect. In (c) +R effect is shown by functional group and in (d) both -OH and -O- show +R effect.

So Correct increasing order is :

c < a < b < d

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Arrange the following compounds in order of decreasing acidity :

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

The correct answer is:
III > I > II > IV

Solution:
Acidity is directly proportional to conjugated base stability.
Stability order of conjugate bases is III > I > II > IV
Conjugate base in I is stabilised by - I > + M
Conjugate base in II is destabilised by + I and hyperconjugation.
Conjugate base in III is stabilised by strong withdrawing effect of -NO2 by - I & - M effect.
Conjugate base in IV is destabilised by + M effect of-OCH3
acidic character  ∝ −M,−I                                  ∝ 1+M,1+HC,1+I
Electron releasing group (i.e +I effect, +M) decreases the acidic strength
Electron withdrawing group (i.e -I effect,-M) increases acidic strength.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2013 (07 Apr)
Arrange the following compounds in order of their reactivity towards LiAlH4

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

The correct answer is:
II<III<I

Solution:
Reactivity ∝ electron deficiency at c-atom of group.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Arrange the following compounds in the increasing order of their acidic strength :
i. m-nitrophenol
ii. m-cresol
iii. phenol
iv. m-chlorophenol

Options:
A iii \( < \mathrm{ii} < \mathrm{i} < \mathrm{iv} \)
B ii \( < \) iv \( < \mathrm{iii} < i \)
C \( \mathrm{ii} < \mathrm{iii} < \mathrm{iv} < \mathrm{i} \)
D ii \( < \mathrm{iii} < \mathrm{i} < \mathrm{iv} \)

The correct answer is:
\( \mathrm{ii} < \mathrm{iii} < \mathrm{iv} < \mathrm{i} \)

Solution:
Nitro group has both \( -R \) effect and \( -1 \) effect; but \( -R \) effect predominates. Due to stronger electron withdrawing nature of
\( \mathrm{NO}_{2} \) group, phenoxide ion is stabilized. Hence, m-nitrophenol is more acidic than phenol. Methyl group destabilizes the
phenoxide ion by \( +I \) effect and hyperconjugation. Hence m-cresol is weaker acid than phenol. Chlorine have both +R
and \( -1 \) effect but \( -1 \) effect predominates
Hence m-chlorophenol is more acidic than phenol. \( -R \) effect of nitro group is stronger than \( -1 \) effect of chlorine hence, m-
nitrophenol is more acidic than m-chlorophenol.
Therefore, the correct order of acidic strength is \( m \)-nitrophenol \( >m \)-chlorophenol \( >p h e n o l>m \)-cresol.
(II)
(III)
(IV)
(I)

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Arrange the following coordination compounds in the increasing order of magnetic moments. (Atomic numbers: Mn=25;Fe=26)

A. FeF63-

B. FeCN63-

C. MnCl63- (high spin)

D. MnCN63-

Choose the correct answer from the options given below

Options:
A A<B<D<C
B B<D<C<A
C A<C<D<B
D B<D<A<C

The correct answer is:
B<D<C<A

Solution:
Magnetic moment is directly proportional to the number of unpaired in the central metal atom of a complex.

It is calculated by the formula, μ=n(n+2)BM

(A) FeF63-

Fe+3→3d54s0

n=5

(B) FeCN63-

Fe+3→3d54s0

n=1

(C) MnCl63-

Mn+3→3d44s0

n=4

(D) MnCN63-

Mn+3→3d44s0

n=2

Hence, correct order of magnetic moment is A>C>D>B

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Arrange the following cyano complexes in decreasing order of their magnetic moment.

Options:
A [Cr(CN)6]3->[Mn(CN)6]3->[Fe(CN)6]3->[Co(CN)6]3-
B [Mn(CN)6]3->[Cr(CN)6]3->[Fe(CN)6]3->[Co(CN)6]3-
C [Fe(CN)6]3->[Cr(CN)6]3->[Mn(CN)6]3->[Co(CN)6]3-
D [Co(CN)6]3->[Cr(CN)6]3->[Mn(CN)6]3->[Fe(CN)6]3-

The correct answer is:
[Cr(CN)6]3->[Mn(CN)6]3->[Fe(CN)6]3->[Co(CN)6]3-

Solution:
[Cr(CN)6]3- Cr3+ in strong ligand field =t2g3eg0=3 unpaired electron

[Mn(CN)6]3- Mn3+ in strong ligand field =t2g4eg0=2 unpaired electron

[Fe(CN)6]3- Fe3+ in strong ligand field =t2g5eg0=1 unpaired electron

[Co(CN)6]3- Co3+ in strong ligand field =t2g6eg0=0 unpaired electron

Topic: Chemistry, Coordination Compounds, JEE Main
Arrange the following electromagnetic radiations per quantum in the order of increasing energy:
A: Blue light
B: Yellow light
C: X-ray
D: Radiowave

Options:
A C, A, B, D
B B, A, D, C
C D, B, A, C
D A, B, D, C

The correct answer is:
D, B, A, C

Solution:

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2016 (03 Apr)
Arrange the following elements in the increasing order of number of unpaired electrons in it.
(A) \(\mathrm{Sc}\)
(B) \(\mathrm{Cr}\)
(C) \(\mathrm{V}\)
(D) \(\mathrm{Ti}\)
(E) Mn
Choose the correct answer from the options given below :

Options:
A \((\mathrm{A}) < (\mathrm{D}) < (\mathrm{C}) < (\mathrm{B}) < (\mathrm{E})\)
B \((\mathrm{C}) < (\mathrm{E}) < (\mathrm{B}) < (\mathrm{A}) < (\mathrm{D})\)
C \((\mathrm{A}) < (\mathrm{D}) < (\mathrm{C}) < (\mathrm{E}) < (\mathrm{B})\)
D \(\text { (B) } < \text { (C) } < \text { (D) } < \text { (E) } < \text { (A) }\)

The correct answer is:
\((\mathrm{A}) < (\mathrm{D}) < (\mathrm{C}) < (\mathrm{E}) < (\mathrm{B})\)

Solution:
$\begin{array}{ll}
\text{Unpaired electron} \\
\operatorname{Sc}[\mathrm{Ar}] 4 \mathrm{~s}^2 3 \mathrm{~d}^1 & 1 \\
\mathrm{Cr}[\mathrm{Ar}] 4 \mathrm{~s}^1 3 \mathrm{~d}^5 & 6 \\
\mathrm{V}[\mathrm{Ar}] 4 \mathrm{~s}^2 3 \mathrm{~d}^3 & 3 \\
\mathrm{Ti}:[\mathrm{Ar}] 4 \mathrm{~s}^2 3 \mathrm{~d}^2 & 2 \\
\mathrm{Mn}:[\mathrm{Ar}] 4 \mathrm{~s}^2 3 \mathrm{~d}^5 & 5
\end{array}$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Arrange the following gases in increasing order of van der Waals constant 'a'
(A) Ar
(B) CH4
(C) H2O
(D) C6H6
Choose the correct option from the following.

Options:
A D, C, B and A
B B, C, D and A
C C, D, B and A
D A, B, C and D

The correct answer is:
A, B, C and D

Solution:
The van der Waals' constant ′a′ is directly proportional to the intermolecular force of attraction; hence, the gas with greater value of ′a′ can be most easily liquefied.The values of Vander Waal's constant 'a' for the given gases are:
Gases Vander Waal's Constant 'a'
A. Ar 1.35 L2 atm mol2
B. CH4 2.25L2 atm mol2
C. H2O 5.536L2 atm mol2
D. C6H6 18.57L2 atm mol2
∴ Increasing order of 'a' for the given gases
A<B<C<D

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Arrange the following given acids in decreasing order of their Ka values

FCH2COOH(P) ,ClCH2COOH(Q), O2NCH2COOH(R), NCCH2COOH(S)

Options:
A R>P>S>Q
B P>Q>R>S
C R>S>Q>P
D R>S>P>Q

The correct answer is:
R>S>P>Q

Solution:
Acidic character Î± stability of conjugate base (anion)
As stability of anion ∝−M and âI effect . -Cl and -F have (âI>+Meffect) in which F is more electron
withdrawing due to greater electronegativity. NO2 and CN have both -I and -M effect with NO2 being more electron withdrawing.
Order of -I- effect -NO2>CN>-F>-Cl Hence, order of acidity ka will be
O2NCH2COOH>NC CH2COOH>FCH2COOH>Cl CH2COOH

Topic: Chemistry, General Organic Chemistry, JEE Main
Arrange the following in decreasing order of solubility in water

Options:
A I \( > \) III \( > \) II
B III \( > \) II \( > \) I
C II \( > \) III \( > \) I
D All are equally soluble

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

Solution:
Higher the electron density on \(\mathrm{O}\), stronger is the \(\mathrm{H}\)-bond with water and thus more is the solubility. Thus solubility of the three ethers follow the order

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Arrange the following in increasing order of acidic strength

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

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

Solution:
Acidic nature of sulphonic acid is more than that of carboxylic acid and carboxylic acid is more acidic then an alcohol.
Acidic nature â electron withdrawing group ∝ 1 electron releasing group .
Here the correct order of acidic nature is III < V < IV < II < I.

Topic: Chemistry, General Organic Chemistry, JEE Main
Arrange the following in increasing order of melting point:
\( \mathrm{Li}_{2} \mathrm{O}, \mathrm{LiF}, \mathrm{Li}_{3} \mathrm{~N} \)

Options:
A \( \mathrm{Li}_{3} \mathrm{~N} < \mathrm{Li}_{2} \mathrm{O} < \mathrm{LiF} \)
B \( \mathrm{Li}_{2} \mathrm{O} < \mathrm{Li}_{3} \mathrm{~N} < \mathrm{LiF} \)
C \( \mathrm{LiF} < \mathrm{Li}_{2} \mathrm{O} < \mathrm{Li}_{3} \mathrm{~N} \)
D \( \mathrm{Li}_{3} \mathrm{~N} < \mathrm{LiF} < \mathrm{Li}_{2} \mathrm{O} \)

The correct answer is:
\( \mathrm{Li}_{3} \mathrm{~N} < \mathrm{Li}_{2} \mathrm{O} < \mathrm{LiF} \)

Solution:
As we know, higher the ionic character in compound higher the melting point because of higher intermolecular attraction.

Fajan's rule tells about the covalent character in ionic bond.

Covalent character depends upon charge on cation and anion and size of cation and anion.

Here in given molecules, cations are same, so size and charge same on cation of each compound.

Size of all anions (N-3, O-2 , F-) are same because all are isoelectronic species but charge on all anion are different.

According to Fajan's rule, higher the charge on anion, higher the polarization power of cation and higher the covalent character; so, lower the ionic character.

Hence, higher the charge on anion, higher the ionic character and higher the melting point.

Order of melting point Li3N<Li2O<LiF.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Arrange the following in increasing order of reactivity towards nitration

A. p-xylene

B. bromobenzene

C. mesitylene

D. nitrobenzene

E. benzene

Choose the correct answer from the options given below

Options:
A C<D<E<A<B
B D<B<E<A<C
C D<C<E<A<B
D C<D<E<B<A

The correct answer is:
D<B<E<A<C

Solution:
Nitration is an electrophilic aromatic substitution reaction. The +M effect increases the rate of nitration whereas the -M effect decreases the rate of nitration.

-NO2(-M), -Br(+M but dominant -I) and -CH3(+H)

-NO2 is strongly deactivating

-Br , deactivating

-CH3, activating group

D<B<E<A<C

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2022 (28 Jul Shift 2)