datavorous/entrance-exam-dataset huggingface dataset:
The correct relationships between unit cell edge length ‘a’ and radius of sphere ‘r’ for face-centred and body-centred cubic structures respectively are:

Options:
A 22r=a and 3r=4a
B r=22a and 4r=3a
C r=22a and 3r=4a
D 22r=a and 4r=3a

The correct answer is:
22r=a and 4r=3a

Solution:
A face-centered cubic (fcc) unit cell contains a component in the center of each face in addition to those at the corners of the cube. The atoms touch each other along the face diagonal.

For FCC: a2=4r

a=22r

A body-centered cubic (bcc) unit cell contains a component in the center of the cube in addition to those at the corners of the cube. The atoms touch each other along the body diagonal.
For BCC : a3=4r

Topic: Chemistry, Solid State, JEE Main, JEE Main 2023 (10 Apr Shift 2)
The correct representation in six membered pyranose form for the following sugar [X] is

Options:
A
B
C
D

The correct answer is:

Solution:
The six-membered cyclic structure of glucose is called pyranose structure (α– or β–), in analogy with pyran. Pyran is a cyclic organic compound with one oxygen atom and five carbon atoms in the ring. The cyclic structure of glucose is more correctly represented by Haworth structure

C2 and C3OH are cis. C3 and C4 are anti to each other.

So,the correct option is B.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2023 (01 Feb Shift 1)
The correct representation of Nernst's equation is

Options:
A $E_{M^{n+} / M}=E^o M^{n+} / M+\frac{0.0591}{n} \log \left(M^{n+}\right)$
B $E_{M^{n+} / M}=E^o M^{n+} / M-\frac{0.0591}{n} \log \left(M^{n+}\right)$
C $E_{M^{n+} / M}=E^o M^{n+} / M+\frac{n}{0.0591} \log \left(M^{n+}\right)$
D None of the above

The correct answer is:
$E_{M^{n+} / M}=E^o M^{n+} / M+\frac{0.0591}{n} \log \left(M^{n+}\right)$

Solution:
$E=E^o-\frac{R T}{n F} \ln \frac{1}{\left[M^{n+}\right]} \quad E=E^o+\frac{R T}{n F} \ln \left[M^{n+}\right]$
Substituting the value of $R, T(298 K)$ and $F$ we get
$E=E^o+\frac{0.0591}{n} \log \left(M^{n+}\right)$

Topic: Chemistry, Electrochemistry, JEE Main
The correct sequence in biogas production is

Options:
A Acidogenesis → Solublisation → Methanogenesis
B Solublisation → Acidogenesis → Methanogenesis
C Methanogenesis → Solublisation → Acidogenesis
D Acidogenesis → Methanogenesis → Solublisation

The correct answer is:
Solublisation → Acidogenesis → Methanogenesis

Solution:
Biogas is a type of biofuel having a mixture of gases which is naturally produced by decomposition of organic matter by anaerobic process. It mainly contains methane and carbon dioxide.
The biogas production involves the following three proceses:
1. Anaerobic digestion: The species of Clostridium and Pseudomonas anaerobically digest the waste and the complex insoluble polymers are convered into monomers. This step is also called solubilization.
2. Acidogenesis: The monomers from previous step is changed to organic acids by acidogenic bacteria
3. Methanogenesis: The last step of biogas production. Methane is produced from the organic acids along with carbon dioxide

Topic: Biology, Microbes in Human Welfare, NEET
The correct sequence of acidic strength of the following aliphatic acids in their decreasing order is: \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}, \mathrm{CH}_3 \mathrm{COOH}, \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH}, \mathrm{HCOOH}\)

Options:
A \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH} > \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH} > \mathrm{CH}_3 \mathrm{COOH} > \mathrm{HCOOH}\)
B \(\mathrm{CH}_3 \mathrm{COOH} > \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH} > \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH} > \mathrm{HCOOH}\)
C \(\mathrm{HCOOH} > \mathrm{CH}_3 \mathrm{COOH} > \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH} > \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH}\)
D \(\mathrm{HCOOH} > \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH} > \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH} > \mathrm{CH}_3 \mathrm{COOH}\)

The correct answer is:
\(\mathrm{HCOOH} > \mathrm{CH}_3 \mathrm{COOH} > \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH} > \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH}\)

Solution:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}, \mathrm{CH}_3 \mathrm{COOH}, \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH}$, $\mathrm{HCOOH}$
The correct order is :
$\begin{aligned} & \mathrm{HCOOH}>\mathrm{CH}_3 \mathrm{COOH}>\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}> \\ & \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH}\end{aligned}$

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (08 Apr Shift 2)
The correct sequence of amino acids present in the tripeptide given below is:

Options:
A Val - Ser - Thr
B Leu - Ser - Thr
C Thr - Ser - Val
D Thr - Ser - Leu

The correct answer is:
Val - Ser - Thr

Solution:
Amino acids are organic compounds that contain amino and carboxyl functional groups, along with a side chain specific to each amino acid. The key elements of an amino acid are carbon, hydrogen, oxygen, and nitrogen, although other elements are found in the side chains of certain amino acids.

Peptides are short chains of amino acids linked by peptide bonds

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The correct sequence of amino acids present in the tripeptide given below is:

Options:
A \( \mathrm{Val}- \) Ser \( -\mathrm{Thr} \)
B Leu - Ser - Thr
C Thr - Ser - Leu
D Thr \( - \) Ser \( - \) Val

The correct answer is:
\( \mathrm{Val}- \) Ser \( -\mathrm{Thr} \)

Solution:
Valine posses isopropyl group.

Serine have -CH2OH group.

Threonine have -CH|CH3-OH group.

Leucine have -CH2-CH|CH3-CH3 group.

Hence, from the given structure, we can identify the correct sequence.

Val-Ser-Thr.

Topic: Chemistry, Biomolecules, JEE Main
The CORRECT sequence of ancestors of mammals starting from the oldest is:

Options:
A Sauropsida → Pelycosaurs → Therapsida
B Synapsids → Pelycosaurs → Therapsida
C Synapsids → Therapsida → Pelycosaurs
D Pelycosaurs → Therapsida → Synapsids

The correct answer is:
Synapsids → Pelycosaurs → Therapsida

Solution:
Mammals evolved from reptiles. There have been many stages of development in these mammals. The main ancestors that evolved from reptiles are the Synapsids and Sauropsids. Sauropsids evolved into turtles and Synapsids into mammals. These then evolved into Pelycosaurs which finally evolved into Therapsids. These then formed the mammals. Thus these three are the main ancestors of mammals.

Topic: Biology, Evolution, NEET
The correct sequence of correct reagents for the following transformation is :-

Options:
A (i) Fe,HCl
(ii) Cl2,HCl,
(iii) NaNO2,HCl,0°C
(iv) H2O/H+
B (i) Fe,HCl
(ii) NaNO2,HCl,0°C
(iii) H2O/H+
(iv) Cl2,FeCl3
C (i) Cl2,FeCl3
(ii) Fe,HCl
(iii) NaNO2,HCl,0°C
(iv) H2O/H+
D (i) Cl2,FeCl3
(ii) NaNO2,HCl,0°C
(iii) Fe,HCl
(iv) H2O/H+

The correct answer is:
(i) Cl2,FeCl3
(ii) Fe,HCl
(iii) NaNO2,HCl,0°C
(iv) H2O/H+

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (27 Jul Shift 2)
The correct sequence of decreasing number of π -bonds in the structure of H2SO3, H2SO4 and H2S2O7 is:

Options:
A H2S2O7>H2SO3>H2SO4
B H2S2O7>H2SO4>H2SO3
C H2SO4>H2S2O7>H2SO3
D H2SO3>H2SO4>H2S2O7

The correct answer is:
H2S2O7>H2SO4>H2SO3

Solution:
Number of π-bonds

H2S2O7=HO−S||||OO−O−S||||OO−OH 4

H2So4=OH−S||||OO−OH 2

H2SO3=HO−S..||O−OH 1

Therefore, the correct sequence of decreasing number of pi-bonds is H2S2O7​(4) >H2SO4 (2) >H2SO3​(1). The numbers in bracket indicate the number of pi bonds.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2017 (09 Apr Online)
The correct sequence of increasing covalent character is represented by:

Options:
A $\mathrm{LiCl} < \mathrm{NaCl} < \mathrm{BeCl}_2$
B $\mathrm{BeCl}_2 < \mathrm{LiCl} < \mathrm{NaCl}$
C $\mathrm{NaCl} < \mathrm{LiCl} < \mathrm{BeCl}_2$
D $\mathrm{BeCl}_2 < \mathrm{NaCl} < \mathrm{LiCl}$

The correct answer is:
$\mathrm{NaCl} < \mathrm{LiCl} < \mathrm{BeCl}_2$

Solution:
According to Fajan's Rule,
1. Smaller the size of cation, the larger the size of the anion, greater is the covalent character of the ionic bond.
2. Greater the charge of cation, greater is the covalent character of the ionic bond.
3. For cations with same charge and size, the one, with $(n-1) d^n n s^{\circ}$ which is found in transition elements have greater covalent character than the cation with $n s^2 n p^6$ electronic configuration, which is commonly found in alkali or alkaline earth metals.
Thus, order of covalent character is:
$$
\mathrm{NaCl} < \mathrm{LiCl} < \mathrm{BeCl}_2
$$
I Caution
Larger the anion, less is the effective nuclear charge that holds the valence electron of the ion in place. Since the last electron is loosely bound in large anions, it can easily be polarised by a cation, thereby making the compound more covalent.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The correct sequence of plants in hydrosere is :

Options:
A Oak $\rightarrow$ Lantana $\rightarrow$ Volvox $\rightarrow$ Hydrilla $\rightarrow$ Pistia $\rightarrow$ Scirpus
B Oak $\rightarrow$ Lantana $\rightarrow$ Scirpus $\rightarrow$ Pistia $\rightarrow$ Hydrilla $\rightarrow$ Volvox
C Volvox $\rightarrow$ Hydrilla $\rightarrow$ Pistia $\rightarrow$ Scirpus $\rightarrow$ Lantana $\rightarrow$ Oak
D Pistia $\rightarrow$ Volvox $\rightarrow$ Scirpus $\rightarrow$ Hydrilla $\rightarrow$ Oak $\rightarrow$ Lantana

The correct answer is:
Volvox $\rightarrow$ Hydrilla $\rightarrow$ Pistia $\rightarrow$ Scirpus $\rightarrow$ Lantana $\rightarrow$ Oak

Solution:
No Solution Available

Topic: Biology, Ecosystem, NEET, NEET 2009 (Mains)
The correct sequence of reactions to convert p-nitrophenol into quinol involves

Options:
A reduction, diazotization and hydrolysis
B hydrolysis, diazotization and reduction
C hydrolysis, reduction and diazotization
D diazotization, reduction and hydrolysis

The correct answer is:
reduction, diazotization and hydrolysis

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
The correct sequence of reagents for the following conversion will be

Options:
A CH3MgBr, H+/CH3OH, AgNH32+OH-
B CH3MgBr, AgNH32+OH-, H+/CH3OH
C AgNH32+OH-, CH3MgBr, H+/CH3OH
D AgNH32+OH-, H+/CH3OH, CH3MgBr

The correct answer is:
AgNH32+OH-, H+/CH3OH, CH3MgBr

Solution:
The correct order of reagents for the following reaction is [Ag(NH3)2]+OH- , H+/CH3OH, CH3MgBr & the reaction as well as mechanism is as follows with the correct order of reagents.

The reaction involves SN2 mechanism at second step which on further reaction with excess CH3MgBr to give the final product.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The correct sequence of reagents for the following conversion will be

Options:
A CH3MgBr, H+/CH3OH, AgNH32+OH-
B CH3MgBr, AgNH32+OH-, H+/CH3OH
C AgNH32+OH-, CH3MgBr, H+/CH3OH
D AgNH32+OH-, H+/CH3OH, CH3MgBr

The correct answer is:
AgNH32+OH-, H+/CH3OH, CH3MgBr

Solution:
The correct order of reagents for the following reaction is [Ag(NH3)2]+OH- , H+/CH3OH, CH3MgBr & the reaction as well as mechanism is as follows with the correct order of reagents.

The reaction involves SN2 mechanism at second step which on further reaction with excess CH3MgBr to give the final product.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2017 (02 Apr)
The correct sequence of reagents for the preparation of Q and R is:

Options:
A (i) Cr2O3, 770 K, 20 atm; (ii) CrO2Cl2, H3O+; (iii) NaOH; (iv) H3O+
B (i) CrO2Cl2, H3O+; (ii) Cr2O3, 770 K, 20 atm; (iii) NaOH; (iv) H3O+
C (i) KMnO4, OH-; (ii) Mo2O3, Δ; (iii) NaOH; (iv) H3O+
D (i) Mo2O3, Δ; (ii) CrO2Cl2, H3O+; (iii) NaOH; (iv) H3O+

The correct answer is:
(i) Cr2O3, 770 K, 20 atm; (ii) CrO2Cl2, H3O+; (iii) NaOH; (iv) H3O+

Solution:
It is a sequence of reactions where heptane is oxidised to methyl benzene, further oxidised to benzaldehyde, and finally cannizzaro reaction occurs to give the desired product.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2023 (25 Jan Shift 1)
The correct sequence of reagents used in the preparation of 4 -bromo-2-nitroethylbenzene from benzene is :

Options:
A CH3COCl/AlCl3,Br2/AlBr3,HNO3/H2SO4,Zn/HCl
B CH3COCl/AlCl3,Zn-Hg/HCl,Br2/AlBr3,HNO3/H2SO4
C HNO3/H2SO4,Br2/AlCl3,CH3COCl/AlCl3,Zn-Hg/HCl
D Br2/AlBr3,CH3COCl/AlCl3,HNO3/H2SO4,Zn/HCl

The correct answer is:
CH3COCl/AlCl3,Zn-Hg/HCl,Br2/AlBr3,HNO3/H2SO4

Solution:
Benzene is an aromatic compound which shows the characteristic reaction of electrophilic aromatic substitution reactions. This is because the reaction is initiated by an electrophile.

Benzene reacts with CH3Ã¢â¬â¹Cl in presence of anhydrous AlCl3Ã¢â¬â¹ to give Toulene. This is fridel crafts Alkylation and an electrophillic substitution.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The correct sequence of spermatogenetic stages leading to the formation of sperms in a mature human testis is :

Options:
A Spermatogonia-spermatid-spermatocytesperms
B Spermatocyte-spermatogonia-spermatidsperms
C Spermatogonia-spermatocyte-spermatidsperms
D Spermatid-spermatocyte-spermatogoniasperms

The correct answer is:
Spermatogonia-spermatocyte-spermatidsperms

Solution:
No Solution Available

Topic: Biology, Human Reproduction, NEET, NEET 2009 (Mains)
The correct sequence of steps involved in the mechanism of Cannizaro's reaction is

Options:
A nucleophilic attack, transfer of $\mathrm{H}^{-}$and transfer of $\mathrm{H}^{+}$
B transfer of $\mathrm{H}^{-}$, transfer of $\mathrm{H}^{+}$and nucleophilic attack
C transfer if $\mathrm{H}^{+}$, nucleophilic attack and transfer of $\mathrm{H}^{-}$
D electrophilic attack by $\mathrm{OH}^{-}$, transfer of $\mathrm{H}^{+}$and transfer of $\mathrm{H}^{-}$

The correct answer is:
nucleophilic attack, transfer of $\mathrm{H}^{-}$and transfer of $\mathrm{H}^{+}$

Solution:
The Cannizaro reaction is as
The mechanism of Cannizaro reaction is as Step Ist : Attack of nucleophile $\mathrm{OH}^{-}$to the carbonyl carbon
Step IInd : The transfer of hydride ion from anion (I) to second molecule of aldehyde and finally rapid transfer of proton takes place.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The correct sequence of thermal stability of the following carbonates is:

Options:
A BaCO3<CaCO3<SrCO3<MgCO3
B MgCO3<CaCO3<SrCO3<BaCO3
C BaCO3<SrCO3<CaCO3<MgCO3
D MgCO3<SrCO3<CaCO3<BaCO3

The correct answer is:
MgCO3<CaCO3<SrCO3<BaCO3

Solution:
Carbonates of alkaline earth metals are insoluble in water, and all the carbonates decompose on heating. Beryllium carbonate is unstable and can be kept only in the atmosphere of CO2. The thermal stability increases with increasing cationic size.
BeCO3<MgCO3<CaCO3<SrCO3<BaCO3
So the correct option is B.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The correct sequence which shows decreasing order of the ionic radii of the elements is

Options:
A
$\mathrm{Al}^{3+}>\mathrm{Mg}^{2+}>\mathrm{Na}^{+}>\mathrm{F}^{-}>\mathrm{O}^{2-}$

B
$\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}>\mathrm{O}^{2-}>\mathrm{F}$

C
$\mathrm{Na}^{+}>\mathrm{F}>\mathrm{Mg}^{2+}>\mathrm{O}^{2-}>\mathrm{Al}^{3+}$

D
$\mathrm{O}^{2-}>\mathrm{F}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}$

The correct answer is:
$\mathrm{O}^{2-}>\mathrm{F}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}$

Solution:
For isoelectronic species higher the $\frac{\mathrm{Z}}{\mathrm{e}}$ ratio, smaller the ionic radius
$\frac{z}{e}$ for
$$
\begin{aligned}
& \mathrm{O}^{2-}=\frac{8}{10}=0.8 \\
& \mathrm{~F}^{-}=\frac{9}{10}=0.9 \\
& \mathrm{Na}^{+}=\frac{11}{10}=1.1 \\
& \mathrm{Mg}^{2+}=\frac{12}{10}=1.2 \\
& \mathrm{Al}^{3+}=\frac{13}{10}=1.3
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2010
The correct set from the following in which both pairs are in correct order of melting point is:

Options:
A LiF>LiCl;NaCl>MgO
B LiCl>LiF;MgO>NaCl
C LiCl>LiF;NaCl>MgO
D LiF>LiCl;MgO>NaCl

The correct answer is:
LiF>LiCl;MgO>NaCl

Solution:
Lattice energy is defined as the amount of energy required to separate one mole of solid ionic compound into its gaseous ions. Evidently greater the lattice energy, higher is the melting point of the alkali metals halide and lower is its solubility in water.

L.E.∝M.P.

L.E. : LiF>LiCl, MgO>NaCl

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The correct set of diamagnetic substances among the following sets are

Options:
A aluminum, sodium calcium and oxygen
B bismuth, copper, lead and silicon
C cobalt, nickel, gadolinium and aluminium
D silver, niobium, magnesium and calcium

The correct answer is:
bismuth, copper, lead and silicon

Solution:
Bismuth, copper, lead and silicon all are having diamagnetic property as atoms of these material have resultant magnetic moment zero.

Topic: Physics, Magnetic Properties of Matter, JEE Main
The correct set of four quantum numbers for outermost electron of potassium $(Z=19)$ is

Options:
A $4,1,0, \frac{1}{2}$
B $3,1,0, \frac{1}{2}$
C $4,0,0, \frac{1}{2}$
D $3,0,0, \frac{1}{2}$

The correct answer is:
$4,0,0, \frac{1}{2}$

Solution:
$\mathrm{K}(19): 1 \mathrm{~s}^{2}, 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6}, 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6}, 4 \mathrm{~s}^{1}$
$4 s^{1}$ is the valence electron in potassium, hence the correct set of four quantum numbers for outermost electron is $4,0,0, \frac{1}{2}$.

Topic: Chemistry, Structure of Atom, JEE Main
The correct set of four quantum numbers for the outermost electron of sodium $(Z=11)$ is

Options:
A $3,1,1, \frac{1}{2}$
B $3,2,1, \frac{1}{2}$
C $3,0,0, \frac{1}{2}$
D $3,1,0, \frac{1}{2}$

The correct answer is:
$3,0,0, \frac{1}{2}$

Solution:
$$
{ }_{11} \mathrm{Na}=1 \mathrm{~s}^{2}, 2 \mathrm{~s}^{2}, 2 \mathrm{p}^{6}, 3 \mathrm{~s}^{1}
$$
For $3 s^{1}$,
$$
\mathrm{n}=3, \quad \mathrm{l}=0, \quad \mathrm{~m}=0, \quad \mathrm{~s}=\frac{1}{2}
$$

Topic: Chemistry, Structure of Atom, JEE Main
The correct set of four quantum numbers for the valence electrons of rubidium atom Z=37 is

Options:
A 5, 0, 0, + 1 2
B 5, 1, 0, + 1 2
C 5, 1, 1, + 1 2
D 5, 0, 1, + 1 2

The correct answer is:
5, 0, 0, + 1 2

Solution:
The electronic configuration can be written as,
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s1
The outermost electron has i.e., Kr365s1
Principal quantum number=5
Azimuthal quantum number=0 (corresponds to s)
Magnetic quantum number=0
Spin quantum number = + 1 2 (positive spin)

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2014 (06 Apr)
The correct set of species responsible for the photochemical smog is:

Options:
A CO2,NO2,SO2 and hydrocarbons
B NO2,O2,O3 and hydrocarbons
C NO2,NO2 and hydrocarbons
D NO,NO2,O3 and hydrocarbons

The correct answer is:
NO,NO2,O3 and hydrocarbons

Solution:
Smog is derived from smoke + fog, this is most common example of air pollution, photochemical smog occurs in warm, dry and sunny climate also known as oxidizing smog.
Unburnt hydrocarbons + Nitric oxide
NO2g→hνNOg+Og
O2g+Og→O3g
CH4+2O3→3CH2OFormaldehyde+3H2O
So the common component of photochemical smog are Ozone, No, acrolein, PAN, formaldehyde.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The correct shape and I-I-I bond angles respectively in I3- ion are:

Options:
A T-shaped; 180° and 90°
B Distorted trigonal planar; 135° and 90°
C Trigonal planar; 120°
D Linear; 180°

The correct answer is:
Linear; 180°

Solution:
The central iodine atom undergoes sp3d hybridization and has 3 lone pairs of electrons and 2 bond pairs of electrons.

Thus, I3- has trigonal bipyramidal geometry and linear shape .

Shape : Linear, I-I-I Bond angle ⇒180°

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The CORRECT Sidgwick structure for a ferrocyanide complex is

Options:
A
B
C
D

The correct answer is:

Solution:
Ferrocyanide ion is FeCN64-.

The central metal is Fe2+ and six ligands are CN-, so the net charge over the complex will be -4.

The ligands are lone pair donors, so the arrow will be pointed towards the central metal.

Topic: Chemistry, Coordination Compounds, JEE Main
The correct stability order of following species is:

Options:
A x > Y > w > z
B y > x > w > z
C x > w > z > y
D z > x > y > w

The correct answer is:
x > w > z > y

Solution:
X is a conjugated diene system.
W is an isolated diene system
Z is a cumulated diene system
Y is antiaromatic system
Hence stability order is X >W >Z >Y

Topic: Chemistry, Hydrocarbons, JEE Main
The correct stability order of the following resonance structures of \(\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}\) is

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

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

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (09 Apr Shift 2)
The correct stability order of the following species is

Options:
A c < a < b
B c = b < a
C c < a = b
D a = b = c

The correct answer is:
c < a = b

Solution:
(a) Trans-1,2-dimethyl cyclohexane (e,e)
(b) Trans-1,2-dimethyl cyclohexane (e,e)
(c) Trans-1,2-dimethyl cyclohexane (axial, axial)

Topic: Chemistry, General Organic Chemistry, JEE Main
The correct statement about ICl5 and ICl4- is:

Options:
A ICl5 is trigonal bipyramidal and ICl4- is tetrahedral.
B Both are isostructural.
C ICl5 is square pyramidal and ICl4- is tetrahedral.
D ICl5 is square pyramidal and ICl4- is square planar.

The correct answer is:
ICl5 is square pyramidal and ICl4- is square planar.

Solution:
ICl4-=lp+bp=2+4=sp3d2

Square planar

ICl5 =1 lp+5 bp=6=sp3d2

Square pyramidal.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2019 (08 Apr Shift 2)
The correct statement about orthoboric acid is

Options:
A It is a strong monobasic acid
B It is not a proton donor, but a weak Lewis acid
C It is a tribasic acid
D It is harmful for eyes

The correct answer is:
It is not a proton donor, but a weak Lewis acid

Solution:
Orthoboric acid is a weak mono basic acid with Ka=1.0 x 10-9. It does not act as protonic acid (i.e. proton donor) but behaves as Lewis acid by accepting a pair of electrons from OH- ion.
B OH 3 + 2 H - O - H → B OH 4 - + H 3 O +

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The correct statement about probability density (except at infinite distance from nucleus) is :

Options:
A It can be zero for 1s orbital
B It can be negative for 2p orbital
C It can be zero for 3p orbital
D It can never be zero for 2s orbital

The correct answer is:
It can be zero for 3p orbital

Solution:
From the following Ψ2 function graph (Ψ2= probability density)
Ψ2 can be zero for 3 p orbital other than infinity.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2020 (05 Sep Shift 2)
The correct statement about the magnetic properties of [Fe(CN)6]3- and [FeF6]3- is : (Z = 26).

Options:
A Both are paramagnetic.
B Both are diamagnetic.
C [Fe(CN)6]3- is paramagnetic,
[FeF6]3- is diamagnetic.
D [Fe(CN)6]3- is diamagnetic,
[FeF6]3- is paramagnetic.

The correct answer is:
Both are paramagnetic.

Solution:
In both complexes Fe is in +3 oxidation state.
F- ion is weak field ligand so no pairing of electrons in inner 3d-orbitals & outer orbital complex. sp3d2 hybridisation is formed.
have 5 unpaired electrons, so paramagnetic
μ[FeF6]−3=5(5+2)=35 B.M
ââ
One unpaired Electron, CN- ion strong ligand field. So will pair the d-orbital electrons & form inner orbital hybridization d2sp3 .
μ Fe CN 6 3 - = 1 1 + 2 (1 unpaired e- so para magnetic)
= 3 B.M .

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2014 (09 Apr Online)
The correct statement about the synthesis of erythritol CCH2OH4 used in the preparation of PETN is

Options:
A The synthesis requires three aldol condensations and one Cannizzaro reaction.
B Alpha hydrogen of ethanol and methanol are involved in this reaction.
C The synthesis requires two aldol condensation and two Cannizzaro reactions.
D The synthesis requires four aldol condensations between methanol and ethanol.

The correct answer is:
The synthesis requires three aldol condensations and one Cannizzaro reaction.

Solution:
The first three steps are Aldol condensations followed by one Cannizzaro reaction step in presence of base. In final step a salt of formic acid will also be formed as side product.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2016 (10 Apr Online)