datavorous/entrance-exam-dataset huggingface dataset:
Among the following, the set of parameters that represents path functions, is:
i) q+w
ii) q
iii) w
iv) H-TS

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

The correct answer is:
(ii) and (iii)

Solution:
State function:

It is a property of which value does not depend on the path only depends upon the initial and final point

Like: Internal energy U, Gibb's free energyG

Path function: Does not depend upon the initial and final point like heat energy q and work done w.

A q+w=ΔU←definite quantity
B q→ Path function
C w→ Path function
D H-TS=G→ State function.

∴ Answer is 1

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2019 (09 Apr Shift 1)
Among the following, the species having the smallest bond is

Options:
A
$\mathrm{NO}^{-}$

B
$\mathrm{NO}^{+}$

C
$\mathrm{O}_2$

D
$\mathrm{NO}$

The correct answer is:
$\mathrm{NO}^{+}$

Solution:
$\mathrm{NO}^{-}(16)-$ B.O. $-2 \quad \mathrm{O}_2(16)-$ B.O. $-2$ $\mathrm{NO}^{+}(14)-$ B.O. $-3 \quad \mathrm{NO}(15)-$ B.O. $-2.5$ Higher the bond order lower is the bond length. Hence $\mathrm{NO}^{+}$will have smallest bond.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2012 (07 May Online)
Among the following, the species that is both paramagnetic and coloured is:

Options:
A MnO42-
B K4VCN6
C VO43-
D CrO2Cl2

The correct answer is:
MnO42-

Solution:
MnO42-; Mn is in + 6 oxidation state. Electronic configuration is Ar183d14s0 . As it contains one unpaired electron it is paramagnetic and green coloured because of d-d transition in visible region. TiCl62-,VO43- and CrO2Cl2 are diamagnetic.

Topic: Chemistry, Coordination Compounds, JEE Main
Among the following, the species with identical bond order are

Options:
A CO and O22-
B O2- and CO
C O22- and B2
D CO and N2+

The correct answer is:
O22- and B2

Solution:
The bond order can be calculated
as

B.O=12Nb-Na

Where, Nb= Electrons in bonding orbitals Na= Electrons in anti bonding orbitals.
(a) CO and O22-

The electronic configuration of CO (14) is

σ1s2σ*1s2σ2s2σ*2s2σ*2pz2π2px2π2p2y

∴ B.O=12(10-4)=62=3

(b) O2- and CO

The electronic configuration of 02-(17) is

σ1s2σ*1s2σ2s2σ*2s2σ2pz2π2px2

π2py2π*2px2π*2py1

B.O=12(10-7)=32=1.5

B.O. of CO is 3

[as calculated in option (a)]

(c) B.O.. of O22- is 1

[as calculated in option (a)]

The electronic configuration of B2(10) is

σ1s2σ*1s2σ2s2σ*2s2π2px1π2py1

B.O=12[6-4]=22=1

(d) B .O. of C O is 3

[as calculated in option (a)]

Electronic configuration of N2+ (13) is

σ1s2σ*1s2σ2s2σ*2s2π2px2π2py2σ2pz1

B.O=12[9-4]=52=2.5

Thus, option (c) is correct.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Among the following the strongest nucleophile is

Options:
A $\mathrm{C}_2 \mathrm{H}_5 \mathrm{SH}$
B $\mathrm{CH}_3 \mathrm{COO}$
C $\mathrm{CH}_3 \mathrm{NH}_2$
D $\mathrm{NCCH}_2^{-}$

The correct answer is:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{SH}$

Solution:
Nucleophiles are those substances which can donate a pair of electrons. These can be neutral or negative. The nucleophilic power depends on the tendency of species to donate electrons. This is more, when an electron pushing group (+I group) is present. Among the alkyl groups, those having higher number of C-atoms will push more, hence ethyl > methyl.

Topic: Chemistry, General Organic Chemistry, JEE Main
Among the following, the surface that will form micelles in aqueous solution at the lowest molar concentration at ambient condition is:

Options:
A CH 3 CH 2 15 N + CH 3 3 Br −
B CH 3 CH 2 11 OSO 3 −   Na +
C CH 3 CH 2 6 COO − Na +
D CH 3 CH 2 11 N + CH 3 3 Br −

The correct answer is:
CH 3 CH 2 15 N + CH 3 3 Br −

Solution:
Larger the hydrophobic fragment of surfactant molecule easier will be micellisation, smaller the CMC value i.e. CH 3 CH 2 15 N + CH 3 3 Br − will have lowest CMC due to longest hydrophobic chain.

Topic: Chemistry, Surface Chemistry, JEE Main
Among the following, the total number of compounds containing at least one sp3 hybridized carbon atom is/are -.
Acetylene, dimethyl ether, propan - 1 - ol, ethene, 2- chlorobutane

Options:

The correct answer is:
3.00

Solution:
Hybridization of C -atom in given compounds is follows:
Acetylene (sp), dimethyl ether (sp3),
Propan-1-ol(sp3), ethene (sp2),
2-chlorobutane (sp3).

Topic: Chemistry, General Organic Chemistry, JEE Main
Among the following, the weakest Lewis base is

Options:
A $\mathrm{H}^{-}$
B $\mathrm{OH}^{-}$
C $\mathrm{Cl}^{-}$
D $\mathrm{HCO}_3^{-}$

The correct answer is:
$\mathrm{Cl}^{-}$

Solution:
It is a conjugate base of the strong acid.

Topic: Chemistry, Ionic Equilibrium, JEE Main
Among the following the weakest one is

Options:
A Metallic bond
B Ionic bond
C Van der Waal's force
D Covalent bond

The correct answer is:
Van der Waal's force

Solution:
Vander waal's forces is the weakest force of attraction.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Among the following, total number of meta directing functional groups is (Integer based)
-OCH3,-NO2,-CN,-CH3-NHCOCH3, -COR,-OH,-COOH,-Cl

Options:

The correct answer is:
4

Solution:
Due to the presence of some groups, the electron density decreases on ortho and para positions. So, the electron density on meta position is relatively higher than ortho and para positions. Thus, these groups directs the electrophilic substitution on meta position.

If the atom attached to the benzene ring contains a pi bond, then the atom containing group acts as meta directing group.
Meta directing groups among the given groups are:

-NO2,-C≡N,-COR,-COOH
are meta directing.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Among the following, total numbers of radioactive elements are
In, Ac, At, Ba, Tc, Pm, Ta, Xe

Options:

The correct answer is:
4

Solution:
Ac, At, Tc, Pm are radioactive elements

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Among the following transformations, the hybridisation of the central atom remains unchanged in

Options:
A CO2→HCOOH
B BF3→BF4-
C NH3→NH4+
D PCl3→PCl5

The correct answer is:
NH3→NH4+

Solution:
Thus, the hybridisation of the central atom remains unchanged in NH3→NH4+.

Topic: Chemistry, Chemical Bonding and Molecular Structure, KVPY, KVPY 2019 (SB/SX)
Among the following vitamins the one whose deficiency causes rickets (bone deficiency) is :

Options:
A
Vitamin A

B
Vitamin B

C
Vitamin D

D
Vitamin C

The correct answer is:
Vitamin D

Solution:
Deficiency of vitamin D causes rickets.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2013 (25 Apr Online)
Among the following, which is not the π-bonded organometallic compound

Options:
A $\left(\mathrm{CH}_3\right)_4 \mathrm{Sn}$
B $\mathrm{K}\left[\mathrm{PtCl}_3\left(\eta^2-\mathrm{C}_2 \mathrm{H}_4\right)\right]$
C $\mathrm{Fe}\left(\eta^5-\mathrm{C}_5 \mathrm{H}_5\right)_2$
D $\operatorname{Cr}\left(\eta^6-\mathrm{C}_6 \mathrm{H}_6\right)_2$

The correct answer is:
$\left(\mathrm{CH}_3\right)_4 \mathrm{Sn}$

Solution:
$\left(\mathrm{CH}_3\right)_4 \mathrm{Sn}$ is a $\sigma-$ bonded organometallic compound

Topic: Chemistry, Coordination Compounds, JEE Main
Among the following, which is not the $\pi$-bonded organometallic compound

Options:
A $\left(\mathrm{CH}_3\right)_4 \mathrm{Sn}$
B $\mathrm{K}\left[\mathrm{PtCl}_3\left(\eta^2-\mathrm{C}_2 \mathrm{H}_4\right)\right]$
C $\mathrm{Fe}\left(\eta^5-\mathrm{C}_5 \mathrm{H}_5\right)_2$
D $\mathrm{Cr}\left(\eta^6-\mathrm{C}_6 \mathrm{H}_6\right)_2$

The correct answer is:
$\left(\mathrm{CH}_3\right)_4 \mathrm{Sn}$

Solution:
$\left(\mathrm{CH}_3\right)_4 \mathrm{Sn}$ is an organometallic compound where tin is bonded to four methyl groups via single σ-bonds. It does not involve $\pi$-bonding. The other compounds listed involve π-interactions with ligands like ethylene ($\eta^2$), cyclopentadienyl ($\eta^5$), and benzene (\eta^6).

Topic: Chemistry, Coordination Compounds, JEE Main
Among the following which is the best description of water in the solid phase?

Options:
A
Covalent solid

B
Molecular solid

C
Ionic solid

D
Network solid

The correct answer is:
Molecular solid

Solution:
Molecular solid is the best description of water in the solid phase. For example ice is hydrogen bonded molecular solid.

Topic: Chemistry, Solid State, JEE Main, JEE Main 2012 (07 May Online)
Among the following which is the strongest oxidizing agent ? -

Options:
A $\mathrm{Cl}_2$
B $\mathrm{F}_2$
C $\mathrm{Br}_2$
D $\mathrm{I}_2$

The correct answer is:
$\mathrm{F}_2$

Solution:
$\quad \mathrm{F}_2 \rightarrow$ reduction potential very high so strongest oxidizing agent.

Topic: Chemistry, Redox Reactions, NEET, NEET 2009 (Mains)
Among the following, which is the strongest oxidizing agent?

Options:
A Mn3+
B Ti3+
C Fe3+
D Cr3+

The correct answer is:
Mn3+

Solution:
The substance which will have the highest value of reduction potential will be the strongest oxidising agent. The values of reduction potential are as follow

EMn3+/Mn2+0=+1.57VETi3+/Ti2+0=−0.37VEFe3+/Fe2+0=+0.77VECr3+/Cr2+0=−0.41V

As we can see Mn+3 has the highest value of standard reduction potential, so it will be the strongest oxidising agent.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Among the following which metal is not refined by liquation?

Options:
A \( \mathrm{Si} \)
B \( \mathrm{Pb} \)
C \( \mathrm{Sn} \)
D \( \mathrm{Bi} \)

The correct answer is:
\( \mathrm{Bi} \)

Solution:
Because liquation is done for low melting metals.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Among the following which one has the highest cation to anion size ratio?

Options:
A $\mathrm{CsI}$
B CsF
C $\mathrm{LiF}$
D $\mathrm{NaF}$

The correct answer is:
CsF

Solution:
The order of size of given cations is
$$
\mathrm{Li}^{+} < \mathrm{Na}^{+} < \mathrm{Cs}^{+}
$$
and the order of size of given anions is
$$
\mathrm{I}^{-}>\mathrm{F}^{-}
$$
Thus, when the cation is largest and anion is smallest, the cation to anion size ratio is maximum.
Hence, cation to anion size ratio is maximum for CsF.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2010 (Mains)
Among the following which property is commonly exhibited by a covalent compound

Options:
A High solubility in water
B High electrical conductance
C Low boiling point
D High melting point

The correct answer is:
Low boiling point

Solution:
The correct option is 3 Low melting point
$\triangle$ Covalent compounds are very resistant, they do not conduct heat and electricity, mostly insulators. They have low boiling points as they are gases at room temperature. It is very hard to molten the covalent compounds because of high bond energy. Therefore the covalent compound turns liquid from solid state easily. They show low melting points and are non polar hence insoluble in water.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Among the following which statement is incorrect?

Options:
A CH3-C||O-OH→X2/OH- will not respond to haloform test
B CH2=CH-CH=O→OH⊕/Δ gives cannizaro reaction
C Cl-C-| Cl | Cl CH=O→ OH⊕/Δ does not give cannizzaro reaction
D

The correct answer is:

Solution:
(i) is correct
(ii) CH2=CH-CH=O→OH- give cannizzaro
(iii) CCl3-CHO→H-ΔCHCl3+HCOO-Na+
(iv)

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Among the gases a-e, the gases that cause greenhouse effect are:
aCO2
bH2O
cCFCs
dO2
eO3

Options:
A a,b,c and d
B a,b,c and e
C a and d

D a,c,d and e

The correct answer is:
a,b,c and e

Solution:
CO2,âO3,âH2Oâ and CFCs are green house gases.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2020 (08 Jan Shift 1)
Among the given amino acids, the basic essential amino acid is:

Options:
A Valine
B Lysine
C Alanine
D Arginine

The correct answer is:
Lysine

Solution:
Lysine is essential basic amino acid.

Topic: Chemistry, Biomolecules, JEE Main
Among the given organic compounds, the total number of aromatic compounds is

Options:

The correct answer is:
3

Solution:
Huckel’s rule states, for a compound to be aromatic, the compound should be planar, cyclic, and should have (4n+2)π electrons that should be in continuous delocalization or an uninterrupted cyclic pi-electron cloud should be there. Here n=0,1,2…

So if we put n=0, we get 2π electrons

If we put n=1, we get 6π electrons

If we put n=2, we get 10 π electrons and so on.

Compound (A) is not aromatic due to presence of sp3 carbon between the rings.

Hence the compounds,B,C and D are Aromatic as they have (4n+2) delocalising π electrons associated to the ring.

Hence the number of aromatic compounds is 3

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Among the given oxides of nitrogen ; N2O,N2O3,N2O4 and N2O5, the number of compound/(s) having N-N bond is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
The structures of the given oxides of nitrogen are as follows

Only three of the above have N-N bondN2O, N2O3 and N2O4 have N-N type bond.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2022 (28 Jun Shift 2)
Among the given species the Resonance stabilised carbocations are:

Options:
A (C) and (D) only
B (A), (B) and (D) only
C (A) and (B) only
D (A), (B) and (C) only

The correct answer is:
(A) and (B) only

Solution:
(A) and (B) only in Resonance

In (C) and (D) positive charge is not in conjugation therefore resonance is not possible in these.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Among the Group 17 elements, which element is most abundant in sea water?

Options:
A Fluorine
B Chlorine
C Bromine
D Iodine

The correct answer is:
Chlorine

Solution:
Among the Group 17 elements, chlorine is the most abundant in sea water, which mainly consists of sodium chloride solution.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Among the inequalities below, which ones are true for all natural numbers n greater than 1000 ?
I. $\mathrm{n} ! \leq \mathrm{n}^{\mathrm{n}}$
II. $(n !)^{2} \leq n^{n}$
III. $10^{\mathrm{n}} \leq \mathrm{n} !$
IV. $n^{\mathrm{n}} \leq(2 \mathrm{n}) !$

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

The correct answer is:
I, III and IV only

Solution:
(A) $\frac{\mathrm{n}^{\mathrm{n}}}{\mathrm{n} !}=\left(\frac{\mathrm{n}}{\mathrm{n}}\right)\left(\frac{\mathrm{n}}{\mathrm{n}-1}\right)\left(\frac{\mathrm{n}}{\mathrm{n}-2}\right) \ldots \ldots .\left(\frac{\mathrm{n}}{1}\right) \geq 1$
$\mathrm{n}^{\mathrm{n}} \geq \mathrm{n} !$ (correct)
(C) $\frac{\mathrm{n} !}{10^{\mathrm{n}}}=\frac{(\mathrm{n})(\mathrm{n}-1)(\mathrm{n}-2) \ldots \ldots}{(10)(10) \ldots \ldots \mathrm{n} \text { times }}$
given that $\mathrm{n}>1000$
clearly $\frac{\mathrm{n} !}{10^{\mathrm{n}}} \geq 1$
$\mathrm{n} ! \geq 10^{\mathrm{n}}$
(D)
$\frac{(1.2 .3 \cdot 4 \ldots \ldots . . n)(n+1)(n+2)(n+3) \ldots \ldots(n+n)}{n^{n}}$
$(n !)\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\left(1+\frac{3}{n}\right) \ldots \ldots .\left(1+\frac{n}{n}\right)$
clearly $\geq 1$.
$2 n ! \geq n^{n}$

Topic: Mathematics, Binomial Theorem, JEE Main
Among the isoelectronic ionsO2-, N3-, Mg2+, Na+, the ions with the least and the highest ionic radius are, respectively

Options:
A Mg2+,N3-
B Mg2+,O2-
C Na+,N3-
D Na+,O2-

The correct answer is:
Mg2+,N3-

Solution:
For isoelectronic species,

N3-	O2-	Na+	Mg2+

Number of electrons	10	10	10	10

Number of protons	7	8	11	12

With increase in number of protons in the nucleus the electrons are more attracted towards nucleus, thereby causing the decrease in ionic radius.

Order of ionic radius: Mg2+<Na+< O2-< N3-.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Among the oxides of nitrogen:
N2O3, N2O4 and N2O5; the molecule(s) having nitrogen - nitrogen bond is/are:

Options:
A N2O3 and N2O4
B N2O4 and N2O5
C N2O3 and N2O5
D Only N2O5

The correct answer is:
N2O3 and N2O4

Solution:
The structures of the oxides of nitrogen N2O3,N2O4 andN2O5 are as follows:
The oxide molecules of N2O3 ,N2O4 have nitrogen - nitrogen bond in their structure. N2O5 has nitrogen-oxygen-nitrogen bond in its structure.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2018 (16 Apr Online)
Among the pair of ions given below, for which pair, first ion is more stable than second?

Options:
A CH3-CH⊕-CH3 and CH3-CH⊕-OCH3
B CH3-CH2-CH⊕-CH3 and CH2=CH-CH2-CH2⊕
C and
D
and

The correct answer is:
CH3-CH2-CH⊕-CH3 and CH2=CH-CH2-CH2⊕

Solution:
(i) II > I
CH 3 − CH ⊕ −O− CH 3 ↔ CH 3 −CH= O ⊕ − CH 3
After resonance octet will complete in that compound
(ii) I > II
I→αH=5
II→αH=2
I is more the stability of the compound
(iii) II > I
II is more stable due to resonance
(iv) II > I
Due to octet complete after resonance in the compound.

Topic: Chemistry, General Organic Chemistry, JEE Main
Among the properties (a) reducing (b) oxidising (c) complexing, the set of properties shown by $\mathrm{CN}^{-}$ion towards metal species is

Options:
A
$a, b, c$

B
$a, b, c$

C
$c, a$

D
$\mathrm{b}, \mathrm{c}$

The correct answer is:
$c, a$

Solution:
No Solution Available

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2004
Among the reactions a-d, the reaction(s) that does/do not occur in the blast furnace during the extraction of iron is/are:
a CaO+SiO2→CaSiO3
b 3Fe2O3+CO→2Fe3O4+CO2
c FeO+SiO2→FeSiO3
d FeO→Fe+12O2

Options:
A a
B a and d
C c and d
D d

The correct answer is:
c and d

Solution:
In blast furnace (metallugy of iron) involved
reactions are
(a) CaO+ SiO 2 → CaSiO 3
(b) 3 Fe 2 O 3 +CO→2 Fe 3 O 4 + CO 2

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Among the reactions a-d , the reaction(s) that does/do not occur in the blast furnace during the extraction of iron is/are
a âCaOâ+â SiO 2 â CaSiO 3
b  3  Fe 2 O 3  + CO→2  Fe 3 O 4  +  CO 2
c âFeO+ SiO 2 â FeSiO 3
d âFeOâFe+ 1 2 O 2

Options:
A c and d
B a and d
C a
D d

The correct answer is:
c and d

Solution:
CaO+SiO2→CaSiO3
3Fe2O3+CO→2Fe3O4+CO2
Used as flux.
Reduction is done by CO.
Hence these two reactions are possible butthe reactions given below are not possible
FeO→Fe+12O2
FeO+SiO2→FeSiO3

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Among the relations
S=a,b:a,b∈R-0,2+ab>0 and T=a,b:a,b∈R,a2-b2∈Z,

Options:
A S is transitive but T is not
B both S and T are symmetric
C neither S not T is transitive
D T is symmetric but S is not

The correct answer is:
T is symmetric but S is not

Solution:
Given,

a, b ∈ I and relation T is defined as a2-b2∈I

Now if a2-b2∈I so b2-a2 will also be integer, for example if 52-42=9 is integer then 42-52=-9 is also a integer,

Now checking transitive,

If 42-32=7 is an integer, 32-22=5 is an integer then 42-22=12 is also an integer, hence we can say that T is symmetric and transitive,

Now checking relation S which is defined as 2+ab>0,

So, if we replace ab by -19 then 2+ab>0 is true but we take ba=-9 for symmetric we get 2+ba=2-9=-7≯0 hence, the relation is not symmetric,

Now checking transitive, now if 2+ab>0⇒ab>-2 and bc>-2 then we cannot say that ac>-2,

For example if we take 41>-2, 1-1>-2 then 4-1≯-2, hence it is not transitive,

Hence, we can say that T is symmetric and S is not.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Among the statement (I – IV), the correct ones are:

(I) Be has smaller atomic radius compared to Mg.

(II) Be has higher ionization enthalpy than AI.

(III) Charge/radius ratio of Be is greater than that of Al.

(IV) Both Be and Al form mainly covalent compounds

Options:
A (I), (II) and (IV)
B (I), (III) and (IV)
C (II), (III) and (IV)
D (I), (II) and (III)

The correct answer is:
(I), (II) and (IV)

Solution:
Charge/radius ratio of Be and Al is same because of diagonal relationship. Remaining statements are correct.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Among the statements a-d, the correct ones are:
(a) Lithium has the highest hydration enthalpy among the alkali metals.
(b) Lithium chloride is insoluble in pyridine.
(c) Lithium cannot form ethynide upon its reaction with ethyne.
(d) Both lithium and magnesium react slowly with H2O .

Options:
A (a),(b) and (d) only
B (a),(c) and (d) only
C (b) and (c) only
D (a) and (d) only

The correct answer is:
(a),(c) and (d) only

Solution:
Lithium has highest hydration enthalpy among
alkali metals due to its small size.
LiCl is soluble in pyridine because LiCl have
more covalent character.
Li does not form ethynide with ethyne.
Both Li and Mg reacts slowly with H 2 O

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2020 (09 Jan Shift 2)
Among the statements a-d , the incorrect ones are:
a Octahedral CoIII complexes with strong field ligands have very high magnetic moments
b When ∆0<P , the d-electron configuration of CoIII in an octahedral complex is teg4eg2
c Wavelength of light absorbed by Coen33+ is lower than that of CoF63-
d If the ∆0 for an octahedral complex of CoIII is 18,000cm-1, the ∆t for its tetrahedral complex with the same ligand will be 16,000cm-1.

Options:
A a and d only
B c and d only
C a and b only
D b and c only

The correct answer is:
a and d only

Solution:
So if there are strong field ligands in complex, pairing of electrons will take place, due to which number of unpaired electrons decrease due to which magnetic moment decrease.
(c) Splitting power of ethylenediamine (en) is greater than fluoride F − ligand therefore more energy absorbed by Co en 3+ as compared to CoF 6 3−
So wave length of light absorbed by Co en 3+ is lower than that of CoF 6 3−
(d) Δt=49Δ0
so if Δ0=18,000 cm−1
Δt=49×18000=8000 cm−1
Statement (a) and (d) are incorrect.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Among the statements

S1: p⇒q∨~p∧q is a tautology

S2: q⇒p⇒~p∧q is a contradiction

Options:
A Neither S1 and S2 is True
B Both S1 and S2 are True
C Only S2 is True
D Only S1 is True

The correct answer is:
Neither S1 and S2 is True

Solution:
Plotting the Truth table of the given expression we get,

p	q	~p	~p∧q	p⇒q	q⇒p	p⇒q∨~p∧q	q⇒p∨~p∧q

T	T	F	F	T	T	T	T

T	F	F	F	F	T	F	T

F	T	T	T	T	F	T	T

F	F	T	F	T	T	T	T

Hence, q⇒p∨~p∧q is a tautology, and p⇒q∨~p∧q is neither tautology and contradiction,

So both statements are false.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (06 Apr Shift 2)
Among the sulphates of alkaline earth metals, the solubilities of BeSO4 and MgSO4 in water, respectively, are :

Options:
A poor and poor
B high and poor
C high and high
D poor and high

The correct answer is:
high and high

Solution:
BeSo4MgSO4CaSO4SrSO4BaSO4↓On moving down Solubility decreases

So, BeSO4 and MgSO4 have high solubility in water.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2020 (06 Sep Shift 1)
Among the triatomic molecules/ions, BeCl2,N3-,N2O,NO2+,O3,SCI2,ICI2-,I3- and XeF2 , the total number of linear molecule(s)/ion(s) where the hybridization of the central atom does not have contribution from the d-orbital(s) is

[Atomic number : S=16, Cl=17, I= 53 and Xe=54 ]

Options:

The correct answer is:
4

Solution:
Cl−Be−Cl N −1 = N +1 = N −1 N≡Ν→O O= N + =O Linear sp hybridisation

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Among them intensive property is

Options:
A $\Delta S$
B $\Delta G$
C $\Delta H$
D $\Delta Q$

The correct answer is:
$\Delta H$

Solution:
Surface tension is an intensive property which do not depend upon the quantity of matter present in the system.

Topic: Chemistry, Thermodynamics (C), JEE Main
Amongst FeCl3.3H2O, K3FeCN6 and CoNH36Cl3, the spin-only magnetic moment value of the inner-orbital complex that absorbs light at shortest wavelength is B.M. [nearest integer]

Options:

The correct answer is:
2

Solution:
Complex	EC	Hybridisation	No. of unpaired e-

(i)	FeCl3·3H2O	Fe3+=3 d5	sp3d2	5

(ii)	K3FeCN6	Fe3+=3 d5	d2sp3	1

(iii)	CoNH363+	Co3+=3 d6	d2sp3	0

CN- is strong ligand than NH3 so splitting in K3[FeCN6] is higher and it absorb light of shortest wave length

μspinonly=nn+2=3=1.73BM≈2BM

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2022 (25 Jun Shift 2)