datavorous/entrance-exam-dataset huggingface dataset:
The correct order of solubility in water for \(\mathrm{He}, \mathrm{Ne}, \mathrm{Ar}, \mathrm{Kr}, \mathrm{Xe}\) is :

Options:
A \(\mathrm{He} > \mathrm{Ne} > \mathrm{Ar} > \mathrm{Kr} > \mathrm{Xe}\)
B \(\mathrm{Xe} > \mathrm{Kr} > \mathrm{Ar} > \mathrm{Ne} > \mathrm{He}\)
C \(\mathrm{Ne} > \mathrm{Ar} > \mathrm{Kr} > \mathrm{He} > \mathrm{Xe}\)
D \(\mathrm{Ar} > \mathrm{Ne} > \mathrm{He} > \mathrm{Kr} > \mathrm{Xe}\)

The correct answer is:
\(\mathrm{Xe} > \mathrm{Kr} > \mathrm{Ar} > \mathrm{Ne} > \mathrm{He}\)

Solution:
The correct option is \(\mathbf{B} \mathrm{Xe} > \mathrm{Kr} > \mathrm{Ar} > \mathrm{Ne} > \mathrm{He}\)
Noble gases are sparingly soluble in water. As the gas molecules become larger, the strength of the solvent-solute interactions due to London dispersion forces increases, approaching the strength of the solvent-solvent interactions. So, most soluble gas will be \(\mathrm{Xe}\) and least soluble will be He. So correct order is \(\mathrm{Xe} > \mathrm{Kr} > \mathrm{Ar} > \mathrm{Ne} > \mathrm{He}\). Thus option (b) is the correct answer.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The correct order of spin only magnetic moments for the following complex ions is

Options:
A Fe(CN)63-<CoF63-<MnBr42-<Mn(CN)63-
B CoF63-<MnBr42-<Fe(CN)63-<Mn(CN)63-
C Fe(CN)63-<Mn(CN)63-<CoF63-<MnBr42-
D MnBr42-<CoF63-<Fe(CN)63-<Mn(CN)63-

The correct answer is:
Fe(CN)63-<Mn(CN)63-<CoF63-<MnBr42-

Solution:
MnBr42- is a tetrahedral complex. The electronic configuration Mn2+ise2t23 and it has 5 unpaired electrons.

∴μ=5(5+2)=35 B. M.

CoF63- is an octahedral complex and it is high spin complex. The electronic configuration of Co3+ is t2g4eg2, it has 4 unpaired electrons

∴μ=4(4+2)=24 B.M.

Mn(CN)63- is an octahedral complex and it is low spin complex. The electronic configuration Mn3+ is t2g4eg0 it has two unpaired electrons

∴μ=2(2+2)=8 B.M

[Fe(CN)6]3- is an octahedral complex and it is low spin complex. The electronic configuration of Fe3+ is t2g5eg0 and it has one unpaired electron.

∴μ=1(1+2)=23 B.M

Hence the correct answer is option 3

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (08 Apr Shift 1)
The correct order of spin-only magnetic moments among the following is: (Atomic number: $\mathrm{Mn}=$ $25, \mathrm{Co}=27, \mathrm{Ni}=28, \mathrm{Zn}=30$ )

Options:
A
$\left[\mathrm{ZnCl}_4\right]^{2-}>\left[\mathrm{NiCl}_4\right]^{2-}>\left[\mathrm{CoCl}_4\right]^{2-}>\left[\mathrm{MnCl}_4\right]^{2-}$

B
$\left[\mathrm{CoCl}_4\right]^{2-}>\left[\mathrm{MnCl}_4\right]^{2-}>\left[\mathrm{NiCl}_4\right]^{2-}>\left[\mathrm{ZnCl}_4\right]^{2-}$

C
$\left[\mathrm{NiCl}_4\right]^{2-}>\left[\mathrm{CoCl}_4\right]^{2-}>\left[\mathrm{MnCl}_4\right]^{2-}>\left[\mathrm{ZnCl}_4\right]^{2-}$

D
$\left[\mathrm{MnCl}_4\right]^{2-}>\left[\mathrm{CoCl}_4\right]^{2-}>\left[\mathrm{NiCl}_4\right]^{2-}>\left[\mathrm{ZnCl}_4\right]^{2-}$

The correct answer is:
$\left[\mathrm{MnCl}_4\right]^{2-}>\left[\mathrm{CoCl}_4\right]^{2-}>\left[\mathrm{NiCl}_4\right]^{2-}>\left[\mathrm{ZnCl}_4\right]^{2-}$

Solution:
The complex having higher number of unpaired electrons will have higher value of spin only magnetic moment.
In all these complexes, the central metal ion is in $+2$ oxidation state.
$\mathrm{Zn}^{2+}$ has $3 d^{10}$ outer electronic configuration with 0 unpaired electron.
$\mathrm{Ni}^{2+}$ has $3 d^8$ outer electronic configuration with 2 unpaired electrons.
$\mathrm{Co}^{2+}$ has $3 d^7$ outer electronic configuration with 3 unpaired electrons.
$\mathrm{Mn}^{2+}$ has $3 d^5$ outer electronic configuration with 5 unpaired electrons
Hence the correct order of spin-only magnetic moments is
$$
\left[\mathrm{MnCl}_4\right]^{2-}>\left[\mathrm{CoCl}_4\right]^{2-}>\left[\mathrm{NiCl}_4\right]^{2-}>\left[\mathrm{ZnCl}_4\right]^{2-} \text {. }
$$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The correct order of stability for the following alkoxides is:

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

The correct answer is:
C>B>A

Solution:
If the negative charge is delocalised with electron withdrawing group like (NO2) then stability increase.
A) Negative charge is delocalised with NO2 group
B) Negative charge is delocalised with carbon of alkene
C) Negative charge is localised

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The correct order of the ability of the leaving group is

Options:
A − OCOC 2 H 5 >− OC 2 H 5 >− OSO 2 Et >− OSO 2 CF 3
B − OC 2 H 5 >− OCOC 2 H 5 >− OSO 2 CF 3 >− OSO 2 Me
C − OSO 2 CF 3 >− OSO 2 Me >− OCOC 2 H 5 >− OC 2 H 5
D − OCOC 2 H 5 >− OSO 3 CF 3 >− OC 2 H 5 >− OSO 2 Me

The correct answer is:
− OSO 2 CF 3 >− OSO 2 Me >− OCOC 2 H 5 >− OC 2 H 5

Solution:
Weaker bases are better leaving group.

Topic: Chemistry, General Organic Chemistry, JEE Main
The correct order of the atomic radii of $\mathrm{C}, \mathrm{Cs}, \mathrm{Al},$ and $\mathrm{S}$ is:

Options:
A $\mathrm{C} < \mathrm{S} < \mathrm{A} 1 < \mathrm{Cs}$
B $\mathrm{S} < \mathrm{C} < \mathrm{Cs} < \mathrm{Al}$
C $\mathrm{S} < \mathrm{C} < \mathrm{Al} < \mathrm{Cs}$
D $\mathrm{C} < \mathrm{S} < \mathrm{Cs} < \mathrm{Al}$

The correct answer is:
$\mathrm{C} < \mathrm{S} < \mathrm{A} 1 < \mathrm{Cs}$

Solution:
On going down the group, size increases while going from left to right in a period size decreases, so order is

$\mathrm{C} < \mathrm{S} < \mathrm{Al} < \mathrm{Cs}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The correct order of the calculated spin-only magnetic moments of complexes A to D is:
A NiCO4
B NiH2O6Cl2
C Na2NiCN4
D PdCl2PPh32

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

The correct answer is:
A≈C≈D<B

Solution:
PdPPh32Cl2,Pd2+=4d8,M=0
NiCO4,Ni=3d84s2SFL,M=0
NiCN42- Ni2+=3d8SFL M=0
NiH2O62+ Ni2+=3d8WFL t2g2, 2, 2,eg1, 1 So, unpaired electron is 2,M=8BM

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2020 (08 Jan Shift 2)
The correct order of the calculated spin-only magnetic moments of complexes A to D is

(a) Ni(CO)4 (b) [NiH2O)6Cl2 (c) Na2[Ni(CN)4] (d) PdCl2PPh32

Options:
A c<d<b<a
B a≈c<b≈d
C a≈c≈d<b
D c≈d<b<a

The correct answer is:
a≈c≈d<b

Solution:
a Ni(CO)4
Ni= 3d 10 4s 0 S F L ; μ=0
b [NiH2O)6Cl2
Ni 2+ = 3d 8
Weak field ligand
No. of unpaired electron =2
Î¼=nn+2=22+2=8 B.M
c Na2[NiCN)4
Ni2+â3d8S F L
pairing take place (n = 0)
Î¼=0
d PdCl2PPh32
Pd2+=4d8
square planar
Î¼=0

Topic: Chemistry, Coordination Compounds, JEE Main
The correct order of the decreasing ionic radii among the following isoelectronic species is

Options:
A $\mathrm{Ca}^{2+}>\mathrm{K}^{+}>\mathrm{S}^{2-}>\mathrm{Cl}^{-}$
B $\mathrm{Cl}^{-}>\mathrm{S}^{2-}>\mathrm{Ca}^{2+}>\mathrm{K}^{+}$
C $\mathrm{S}^{2-}>\mathrm{Cl}^{-}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}$
D $\mathrm{K}^{+}>\mathrm{Ca}^{2+}>\mathrm{Cl}^{-}>\mathrm{S}^{2-}$

The correct answer is:
$\mathrm{S}^{2-}>\mathrm{Cl}^{-}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}$

Solution:
Key Idea:- lonic radii $\propto$ charge on anion $\propto \frac{1}{\text { charge on cation }}$ During the formation of a cation, the electrons are lost from the outer shell and the remaining electrons experience a great force of attraction by the nucleus, ie, attracted more towards the nucleus. In other words, nucleus hold the remaining electrons more tightly and this results in decreased radii.
However, in case of anion formation, the addition of electron(s) takes place in the same outer shell, thus the hold of nucleus on the electrons of outer shell decreases and this results in increased ionic radii.
Thus, the correct order of ionic radii is
$$
\mathrm{S}^{2-}>\mathrm{Cl}^{-}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}
$$

Topic: Chemistry, Structure of Atom, NEET, NEET 2010 (Screening)
The correct order of the first ionization enthalpies is:

Options:
A Ti<Mn<Zn<Ni
B Ti<Mn<Ni<Zn
C Mn<Ti<Zn<Ni
D Zn<Ni<Mn<Ti

The correct answer is:
Ti<Mn<Ni<Zn

Solution:
As we observe from left to right in a period, there is a gradual increase in ionization potential due to the increase in Nuclear charge.
From left to right, the elements will be Ti22, Mn25,Ni(28),Zn(30) .
Hence, the correct order will be:
Ti<Mn<Ni<Zn.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The correct order of the following compounds showing increasing tendency towards nucleophilic substitution reaction is:

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

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

Solution:
For nucleophile substitution in aromatic halides

Correct order is:

(i) < (ii) < (iii) < (iv)

More No. of NO2 substituted aromatic halide, increase the rate of nucleophile substitution reaction in aromatic halides.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The correct order of the increasing ionic character is

Options:
A $\mathrm{BeCl}_2 \lt \mathrm{MgCl}_2 \lt \mathrm{CaCl}_2 \lt \mathrm{BaCl}_2$
B $\mathrm{BeCl}_2 \lt \mathrm{MgCl}_2 \lt \mathrm{BaCl}_2 \lt \mathrm{CaCl}_2$
C $\mathrm{BeCl}_2 \lt \mathrm{BaCl}_2 \lt \mathrm{MgCl}_2 \lt \mathrm{CaCl}_2$
D $\mathrm{BaCl}_2 \lt \mathrm{CaCl}_2 \lt \mathrm{MgCl}_2 \lt \mathrm{BeCl}_2$

The correct answer is:
$\mathrm{BeCl}_2 \lt \mathrm{MgCl}_2 \lt \mathrm{CaCl}_2 \lt \mathrm{BaCl}_2$

Solution:
$\mathrm{BeC}_2 < \mathrm{MgCl}_2 < \mathrm{CaC}_2 < \mathrm{BaCl} 2$
As we go down the group I.E. decreases. Hence ionic character increases.

Topic: Chemistry, s Block Elements, JEE Main
The correct order of the increasing magnetic moments for the following ions is $\mathrm{NiCl}_4^{2-} ; \mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6^{2+} ; \mathrm{Ni}(\mathrm{CN})_4^{2-} ; \mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6^{2+}$

Options:
A $\mathrm{Ni}(\mathrm{CN})_4^{2-} < \mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6^{2+} < \mathrm{NiCl}_4^{2-} < \mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6^{2+}$
B $\mathrm{NiCl}_4^{2-} < \mathrm{Ni}(\mathrm{CN})_4^{2-} < \mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6^{2+} < \mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6^{2+}$
C $\mathrm{Ni}(\mathrm{CN})_4^{2-} < \mathrm{NiCl}_4^{2-} < \mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6^{2+} < \mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6^{2+}$
D $\mathrm{Ni}(\mathrm{CN})_4^{2-} < \mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6^{2+} < \mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6^{2+} < \mathrm{NiCl}_4^{2-}$

The correct answer is:
$\mathrm{Ni}(\mathrm{CN})_4^{2-} < \mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6^{2+} < \mathrm{NiCl}_4^{2-} < \mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6^{2+}$

Solution:
$\mathrm{NiCl}_4^{2-}$ is paramagnetic but $\mathrm{Ni}(\mathrm{CN})_4^{2-}$ is diamagnetic because $\mathrm{CN}$ is strong ligand, in excited state electrons get paired and no electrons of $\mathrm{Ni}$ remain unpaired. While $\mathrm{Cl}$ is weak ligand and two unpaired electrons are remain in exciting state thus, it gives paramagnetic nature. In $\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}, \mathrm{Cu}^{2+}$ has one unpaired electron therefore, it give paramagnetic property but lesser than $\mathrm{NiCl}_4^{2-}$.
In $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}, \mathrm{Fe}^{2+}$ has 4 unpaired electrons and produce strongly paramagnetic nature.
Thus, order of magnetic character is
$$
\mathrm{Ni}(\mathrm{CN})_4^{2-} < \left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} < \mathrm{NiCl}_4^{2-} < \left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}
$$

Topic: Chemistry, d and f Block Elements, TS EAMCET, TS EAMCET 2019 (04 May Shift 2)
The correct order of the ionic radii of O2-,N3-,F-,Mg2+,Na+ and Al3+ is :

Options:
A N3-<O2-<F-<Na+<Mg2*<AI3+
B AI3+<Na+<Mg2+<O2-<F-<N3-
C Al3+<Mg2+<Na+<F-<O2-<N3-
D N3-<F-<O2-<Mg2+<Na+<AI3+

The correct answer is:
Al3+<Mg2+<Na+<F-<O2-<N3-

Solution:
Al3+<Mg2+<Na+<F-<O2-<N3-

All are isoelectronic species so more is the z of less will be the ionic size.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2020 (05 Sep Shift 2)
The correct order of the macromolecules that gets digested in the body is

Options:
A Carbohydrate-fat-protein
B Carbohydrate-protein-fat
C Fat-protein-carbohydrate
D Fat-carbohydrate-protein

The correct answer is:
Carbohydrate-fat-protein

Solution:
During starvation, i.e., when food requirement of body is not fulfilled by ingested food, reserve carbohydrates is used up first by the body. After carbohydrates, fat is used as energy source, and at last when both carbohydrates and fats are completely consumed, proteins are used as energy sources. Proteins are used at last because proteins are main structural component of body.

Topic: Biology, Digestion and Absorption, NEET
The correct order of the number of unpaired electrons in the given complexes is

(A) [Fe(CN)6]3-

(B) FeF63-

(C) CoF63-

(D) Cr( oxalate )33-

(E) Ni(CO)4

Choose the correct answer from the options given below:

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

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

Solution:
The number of unpaired electrons in coordination complexes is determined by the nature of the central metal ion, the coordination number, and the identity and arrangement of the ligands around the metal ion. Based on these factors, the correct order of the number of unpaired electrons in the given complexes is:

A	Fe(CN)63-:Fe3+3d5 (Pairing of electrons takes place)	1

B	FeF63-:Fe3+3d5	5

C	CoF63-:Co3+3d6	4

D	Cr( oxalate )33-:Cr3+3d3	3

E	Ni(CO)4 :Ni3d10(Pairing of electrons takes place)	0

No. of unpaired electrons order: E < A < D < C < B

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (10 Apr Shift 2)
The correct order of the oxidation states of nitrogen in NO, N2O, NO2 and N2O3 is

Options:
A N2O<NO<N2O3<NO2
B NO2<N2O3<NO<N2O
C NO2<NO<N2O3<N2O
D N2O<N2O3<NO<NO2

The correct answer is:
N2O<NO<N2O3<NO2

Solution:
Oxidation states of following
N2O⇒2x+1-2=0⇒x=1
NO⇒x+1-2=0⇒x=2
N2O3⇒2x+3-2=0⇒x=3
NO2⇒x+2-2=0⇒x=4

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2019 (09 Apr Shift 1)
The correct order of the solubility of alkaline-earth metal sulphates in water is:

Options:
A Mg > Ca > Sr > Ba
B Mg > Sr > Ca > Ba
C Mg < Ca < Sr < Ba
D Mg < Sr < Ca < Ba

The correct answer is:
Mg > Ca > Sr > Ba

Solution:
Hydration power energy ∝ q 1 q 2 ( 1 r + + 1 r − )
∴ Solubility order of IIA group element sulphates in water:
MgSO4>CaSO4>SrSO4>BaSO4
Generally bigger size ion form a strong bond with bigger ion so solubility will decrease or smaller the cation, greater the hydration sphere.
Hence, Solubility of SO4-2 ion with alkali or alkaline earth metal decrease top to bottom.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2016 (09 Apr Online)
The correct order of the spin –only magnetic moment of metal ions in the following low-spin complexes, [V(CN)6]4-,[Fe(CN)6]4-,[Ru(NH3)6]3+ and [Cr(NH3)6]2+, is:

Options:
A V2+>Cr2+>Ru3+>Fe2+
B Cr2+>V2+>Ru3+>Fe2+
C Cr2+>Ru3+>Fe2+>V2+
D V2+>Ru3+>Cr2+>Fe2+

The correct answer is:
V2+>Cr2+>Ru3+>Fe2+

Solution:
(i) V+223⟹Ar3d34d0⟹ n=3 ⇒μ33+2=3.87
(ii) Cr+224⟹Ar3d4 n=2 ⇒μ22+2=2.82
(iii) Ru+344⟹kr4d5 (n=1) ⇒μ11+2=3
(iv) Fe+226 ⟹Ar3d6 (n=0) ⇒μ=0
Order of magnetic moment will be ∝ no. of unpaired e-
V+2> Cr+2>Ru+3 >Fe+2
μ=n(n+2) B.M. Hence more the value of n more will be magnitude moment.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The correct order of the spin-only magnetic moments of the following complexes is:
(I) CrH2O6Br2
(II) Na4Fe(CN)6
(III) Na3FeC2O43Δ0>P
(IV) Et4N2CoCl4

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

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

Solution:
As μs=nn+2
Complex (II) →Fe+2⟹S.F.L⟹t2g2,2,2eg0,0⟹μs=0
Complex (I) →Cr+2⇒W.F.L⇒t2g1,1,1eg1,0⇒μs=24B.M
Complex (IV) →Co+2⇒W.F.L⇒eg2,2,t2g^1,1,1,⇒μs=15B.M
Complex (III) →Fe+3⇒S.F.L⇒t2g1,1,1eg0,0⇒μs=3B.M

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (09 Jan Shift 2)
The correct order of the spin-only magnetic moments of the following complexes is

(I) [CrH2O)6Br2

(II) Na4[FeCN)6

(III) Na3FeC2O43Δ0>P

(IV) (Et4N)2CoCl4

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

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

Solution:
CrH2O6 Br2:Cr2+=Ar 4s03d4
t 2g 3 e g 1
n=4,Î¼=n(n+2)=24 BM
Na4[Fe(CN)6]:Fe2+=[Ar]4s03d6
t 2g 6 e g 1
n=0,Î¼=0
Na3FeC2O43:Fe3+=Ar3d5
t 2g 5 e g 0
n=1,Î¼=3BM
Et4N2CoCl4:Co2+=Ar3d7
eg4t2g3
n=3,Î¼=15 BM

Topic: Chemistry, Coordination Compounds, JEE Main
The correct order of the stability of the following compounds based on hyperconjugation is

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

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

Solution:
Hyperconjugation increases the stability of alkene which does not have any other resonance effect. And hyperconjugation effect depends on the number of alpha hydrogen around the alkene. More the number of alpha hydrogen atoms more stable the alkene will be.

So, stability order will be IV>III>II>I.

Topic: Chemistry, General Organic Chemistry, JEE Main
The correct order of the thermal stability of hydrogen halides $(H-X)$ is

Options:
A $\mathrm{HI} \gt \mathrm{HBr} \gt \mathrm{HCl} \gt \mathrm{HF}$
B $\mathrm{HF} \gt \mathrm{HCl} \gt \mathrm{HBr} \gt \mathrm{HI}$
C $\mathrm{HCl} \lt \mathrm{HF} \lt \mathrm{HBr} \lt \mathrm{HI}$
D $\mathrm{HI} \gt \mathrm{HCl} \lt \mathrm{HF} \lt \mathrm{HBr}$

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

Solution:
$\mathrm{HF} \gt \mathrm{HCl} \gt \mathrm{HBr} \gt \mathrm{HI}$ (Thermal stability)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The correct order of the thermal stability of hydrogen halides (H – X) is

Options:
A
HI > HBr > HCl > HF

B
HF > HCl > HBr > HI

C
HCl < HF > HBr < HI
D
HI > HCl < HF < HBr

The correct answer is:
HF > HCl > HBr > HI

Solution:
No Solution Available

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2005
The correct order of their reactivity towards hydrolysis at room temperature is:

Options:
A (A)>(B)>(C)>(D)
B (D)>(A)>(B)>(C)
C (D)>(B)>(A)>(C)
D (A)>(C)>(B)>(D)

The correct answer is:
(A)>(B)>(C)>(D)

Solution:
+M group attached with carbonyl bond will decrease reactivity of acid derivative towards hydrolysis. For halogen -I dominate over +M therefore, acid chloride is most reactive. As extent of +M effect increases, reactivity decreases. Overall, order of reactivity is:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2021 (20 Jul Shift 2)
The correct order of thermal stability of hydroxides is

Options:
A MgOH2<SrOH2<CaOH2<BaOH2
B MgOH2<CaOH2<SrOH2<BaOH2
C BaOH2<SrOH2<CaOH2<MgOH2
D BaOH2<CaOH2<SrOH2<MgOH2

The correct answer is:
MgOH2<CaOH2<SrOH2<BaOH2

Solution:
MgOH2<CaOH2<SrOH2<BaOH2

Thermal stability increases down the group due to the decrease in the charge density of the metal cation. All group 2 metal ions have the same charge (2+) but the smaller ones at the top of the group have it all packed into a smaller space, so they polarise the O-H bond more (pull electrons towards the oxygen). This means the O-H bond is already partly broken, so it requires less heat to break fully. As you go down the group, the O-H is less polarised by the larger group 2 cations with a lower charge density, so the thermal stability increases.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2015 (10 Apr Online)
The correct order of viscosity of the following liquids will be :

Options:
A
Water $ < $ methyl alcohol $ < $ dimethyl ether $ < $ glycerol
B
methyl alcohol $ < $ glycerol $ < $ water $ < $ dimethyl ether
C
dimethyl ether $ < $ methyl alcohol $ < $ water $ < $ glycerol
D
glycerol $ < $ dimethyl ether $ < $ water $ < $ methyl alcohol

The correct answer is:
dimethyl ether $ < $ methyl alcohol $ < $ water $ < $ glycerol

Solution:
The correct order of viscosity of the given liquids is dimethyl ether $ < $ methyl alcohol $ < $ water $ < $ glycerol.

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2013 (23 Apr Online)
The correct orientation of dipoles in pyrrole and pyridine is

Options:
A
B
C
D

The correct answer is:

Solution:
lone pair in pyrrole is delocalized in conjugation so N acquires positive charge so the dipole will be like
while pyridine does not do so dipole will be like .

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The correct reaction profile diagram for a positive catalyst reaction.

Options:
A

B
C
D

The correct answer is:

Solution:
Most of the chemical reactions have energy barrier. A positive catalyst in a chemical reaction increases the rate of reaction by lowering the energy barrier. It does not alter the energy of reactant as well as product.

In the above graph, the curve in the presence of catalyst is having less activation energy. The enthalpy change is constant after addition of the positive catalyst

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The correct relation between γ=CPCV and temperature T is :

Options:
A γ∝1T
B γ∝T0
C γ∝1T
D γ∝T

The correct answer is:
γ∝T0

Solution:
The ratio of molar specific heat capacity at constant pressure to the molar specific heat capacity at constant volume is

CPCV=γ

where γ is constant for a given gas.

Therefore,

CPCV∝T0

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2023 (31 Jan Shift 1)
The correct relation between α (ratio of collector current to emitter current) and β (ratio of collector current to base current) of a transistor is :

Options:
A β=α1+α
B α=β1-α
C β=11-α
D α=β1+β

The correct answer is:
α=β1+β

Solution:
α=ICIE,β=ICIB

IE=IB+IC

α=ICIB+IC=1IBIC+1

α=11β+1

α=β1+β

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (18 Mar Shift 2)
The correct relation between the degrees of freedom f and the ratio of specific heat γ is:

Options:
A f=2γ-1
B f=2γ+1
C f=γ+12
D f=1γ+1

The correct answer is:
f=2γ-1

Solution:
γ=1+2f

f=2γ-1

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (20 Jul Shift 2)
The correct relationship between free energy change in a reaction and the corresponding equilibrium constant $K_c$ is

Options:
A $\Delta G=R T \ln K_2$
B $-\Delta G=R T \ln K_c$
C $\Delta G^{\circ}=R T \ln K$
D $-\Delta G^{\circ}=R T \ln K$

The correct answer is:
$-\Delta G^{\circ}=R T \ln K$

Solution:
$\because \Delta \mathrm{G}=\Delta \mathrm{G}^{\circ}+\mathrm{RTIn} \mathrm{Q}$, where $\mathrm{Q}$ is reaction quotient at equilibrium, $\Delta \mathrm{G}=0$ and $\mathrm{Q}=\mathrm{K}_c$ $\therefore-\Delta \mathrm{G}^{\circ}+\mathrm{RTIn} \mathrm{Q}$

Topic: Chemistry, Thermodynamics (C), JEE Main
The correct relationship between free energy change in a reaction and the corresponding equilibrium constant $\mathrm{K}_{\mathrm{c}}$ is

Options:
A
$-\Delta \mathrm{G}=\mathrm{RT} \ln \mathrm{K}_{\mathrm{c}}$

B
$\Delta \mathrm{G}^0=\mathrm{RT} \ln \mathrm{K}_{\mathrm{c}}$

C
$-\Delta \mathrm{G}^0=\mathrm{RT}$ In $\mathrm{K}_{\mathrm{c}}$

D
$\Delta \mathrm{G}=\mathrm{RT} \ln \mathrm{K}_{\mathrm{c}}$

The correct answer is:
$-\Delta \mathrm{G}^0=\mathrm{RT}$ In $\mathrm{K}_{\mathrm{c}}$

Solution:
No Solution Available

Topic: Chemistry, Thermodynamics (C), JEE Main
The correct relationship between free energy change in a reaction and the corresponding equilibrium constant $\mathrm{K}_{\mathrm{c}}$ is

Options:
A
$-\Delta \mathrm{G}=\mathrm{RT} \ln \mathrm{K}_{\mathrm{c}}$

B
$\Delta \mathrm{G}^0=\mathrm{RT} \ln \mathrm{K}_{\mathrm{c}}$

C
$-\Delta \mathrm{G}^0=\mathrm{RT}$ In $\mathrm{K}_{\mathrm{c}}$

D
$\Delta \mathrm{G}=\mathrm{RT} \ln \mathrm{K}_{\mathrm{c}}$

The correct answer is:
$-\Delta \mathrm{G}^0=\mathrm{RT}$ In $\mathrm{K}_{\mathrm{c}}$

Solution:
No Solution Available

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2003