datavorous/entrance-exam-dataset huggingface dataset:
Which of the following vitamins is water soluble as well as an antioxidant?

Options:
A Vitamin- $\mathrm{B}_1$
B Vitamin- $\mathrm{A}$
C Vitamin- $\mathrm{D}$
D Vitamin- $\mathrm{C}$

The correct answer is:
Vitamin- $\mathrm{C}$

Solution:
$\begin{array}{|c|c|c|c|}
\hline & \text{Vitamin} &\text{Function} &\text{Source} \\
\hline \text{I. Water soluble} \\
\hline (a) & \text{Vitamin- }\mathrm{B}_1 & \text{Anti beri-beri} & \begin{array}{l}
\text{Yeast, whole grains}
\end{array} \\
\hline (b) & \text{Vitamin- }\mathrm{B}_2 & \begin{array}{l}
\text{Maintenance of oral mucosa}
\end{array} & \text{Milk, egg} \\
\hline (c) & \text{Vitamin-C} & \begin{array}{l}
\text{Antioxidant help in} \\
\text{collagen and bone formation}
\end{array} & \text{Citrus fruits} \\
\hline \text{II. Fat soluble} \\
\hline (a) & \text{Vitamin-A} & \begin{array}{l}
\text{Promotes normal vision}
\end{array} & \begin{array}{l}
\text{Carrot, cabbage}
\end{array} \\
\hline (b) & \text{Vitamin-D} & \begin{array}{l}
\text{Maintenance of Ca and P} \\
\text{level in body}
\end{array} & \begin{array}{l}
\text{Cod and shark} \\
\text{liver oil}
\end{array} \\
\hline (c) & \text{Vitamin-E} & \text{Antioxidant} & \begin{array}{l}
\text{Green vegetables, oil}
\end{array} \\
\hline (d) & \text{Vitamin-K} & \begin{array}{l}
\text{Anti-haemorrhagic in function}
\end{array} & \begin{array}{l}
\text{Green vegetables}
\end{array} \\
\hline
\end{array}$

Topic: Biology, Digestion and Absorption, JIPMER, JIPMER 2010
Which of the following volume (V), temperature (T) plots represents the behaviour of one mole of an ideal gas at one atmospheric pressure?

Options:
A
B
C
D

The correct answer is:

Solution:
Find the volume by either
V=RTPPV=RT or P1V1=P2V2 and match it with the values given in graph to find correct.
Volume of 1 mole of an ideal gas at 273 K and 1 atm is 22.4 L and that at 373 K and 1 atm pressure is calculated as
V=RTP=0.082×3731=30.58L≃30.6L

Topic: Chemistry, States of Matter, JEE Main
Which of the following waste products in grasshopper is not excreted and is used in other metabolic activities?

Options:
A Carbon dioxide
B Water
C Uric acid
D Faeces

The correct answer is:
Water

Solution:
Excretory organs of grasshopper are the Malpighian tubules and the main excretory product is uric acid. Malpighian tubules of insects such as grasshoppers not only help to excrete the waste but also function to conserve water, which is then used in other metabolic activities. In the Malpighian tubules, bicarbonates of potassium and sodium, water and uric acid are formed during the process of excretion. A large amount of water and bicarbonates of potassium and sodium are reabsorbed by the cells of Malpighian tubules and then transferred to the blood (haemolymph). Uric acid is carried to the alimentary canal of the insect and is finally passed out through anus.

Topic: Biology, Excretory Products and their Elimination, AIIMS, AIIMS 2015
Which of the following wavelengths lying in the range 500 nm to 700 nm is in hydrogen spectrum?

Options:
A 654 nm
B 730 nm
C 815 nm
D 540 nm

The correct answer is:
654 nm

Solution:
The energy of a photon of wavelength 500 nm is
hc Î» = 1 2 4 2 eV nm 5 0 0 nm = 2 Â· 4 4 eV.
The energy of a photon of wavelength 700 nm is
hc Î» = 1 2 4 2 eV nm 7 0 0 nm = 1 Â· 7 7 eV.
The energy difference between the states involved in the transition should, therefore, be between 1.77 eV and 2.44 eV.
Figures above diagram shows some of the energies of hydrogen states. It is clear that only those trasitions which end at n = 2 may emit photons of energy between 1.77 eV and 2.44 eV. Out of these only n = 3 â n = 2 falls in the proper range. The energy of the photon emitted in the transition n = 3 to n = 2 is ÎE = (3.4 - 1.5) eV = 1.9 eV. The wavelength is
Î» = hc Î E
= 1 2 4 2 eV nm 1 Â· 9 eV
= 654 nm.

Topic: Physics, Atomic Physics, JEE Main
Which of the following will be able to show geometrical isomerism ?

Options:
A \( \mathrm{MA}_{3} \mathrm{~B} \) - Square planar
B \( \mathrm{MA}_{2} \mathrm{~B}_{2}- \) Tetrahedral
C MABCD - Square planar
D MABCD - Tetrahedral

The correct answer is:
MABCD - Square planar

Solution:
Tetrahedral complexes do not show geometrical isomerismbecause due to symmetrical structure, relative positions of
the ligands are same with respect to each other. Square planar complexes of the type \( M A_{3} B \) do not show geometrical
isomerism because the possible spatial arrangements are equivalent. MABCD type square planar complexes shows
three isomers which can be obtained by fixing the position of one ligand and placing any of the remaining three ligands
at the trans-position one by one.

Topic: Chemistry, Coordination Compounds, KCET, KCET 2015
Which of the following will be most easily attacked by an electrophile

Options:
A

B

C

D

The correct answer is:

Solution:
Due to mesomeric effect (+) of $-\mathrm{OH}$ group the electron density on benzene ring increase. So the electrophile easily attacked on these electron rich center.

Topic: Chemistry, General Organic Chemistry, JEE Main
Which of the following will be oxidized by HIO4 ?

1. R-C||O-C||O-R

2. R-C||O-CH| OH-R

3. R-CH| OH-CH2-CH| OH-R

4. R-CH| OH-CH| OH-R

Select the correct answer using the codes below

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

The correct answer is:
1, 2 and 4

Solution:
The net result of oxidation is the cleavage of the bond between C's with the attached O's with addition of OH to 'C' of each C-O .
In 1, 2-diols, the net effect is the oxidation is the cleavage of bond and attachment of an OH to each 'C' atom. The gem-diol so formed loses a molecule of H2O .
1. R-C||O-C||O-RâHIO4R-C||O-OH+HO-C||O-R
2. R-C||O-CH| OH-RâHIO4R-C||O-OH+O=CH-R
4. R-CH| OH-CH| OH-RâHIO4R-CH=O+O=HCR

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Which of the following will be the major product when 3-phenylpropene reacts with HBr?

Options:
A C6H5CH2CHBrCH3
B C6H5CHBrCHCH2
C C6H5CH2CH2CH2Br
D C6H5CHBrCH2CH3

The correct answer is:
C6H5CHBrCH2CH3

Solution:
C6H5-CH2-CH=CH2âHBrC6H5-CH2CH+ -CH3
â1,2-HâshiftC6H5-CHâ -CH2-CH3more stable carbocation
→ Br Θ C 6 H 5 − CH | Br − CH 2 − CH 3

Topic: Chemistry, Hydrocarbons, JEE Main
Which of the following will exhibit aromatic character?

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

The correct answer is:
II, IV

Solution:
Since in azulene (IV) two rings share a double bond, therefore, two electrons of this double bond can be counted towards each. As a result, each ring has 6π -electrons and hence azulene is aromatic. Pyrrole (II), the two double bonds + one lone pair of electrons on the N atom).

Topic: Chemistry, General Organic Chemistry, JEE Main
Which of the following will favour the reverse reaction in a chemical equilibrium

Options:
A Increasing the concentration of the reactants
B Removal of at least one of the products at regular intervals
C Increasing the concentration of one or more of the products
D Increasing the pressure

The correct answer is:
Increasing the concentration of one or more of the products

Solution:
$\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}$
According to Le Chatelier's principle, as the concentration of the products are increased, the reaction proceeds in the backward direction.

Topic: Chemistry, Chemical Equilibrium, JEE Main
Which of the following will give a pair of enantiomers?

Options:
A $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]\left[\mathrm{Co}(\mathrm{CN})_6\right]$
B $\left[\mathrm{Co}(e n)_2 \mathrm{Cl}_2\right] \mathrm{Cl}$
C $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4\right]\left[\mathrm{PtCl}_6\right]$
D $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{NO}_2$

The correct answer is:
$\left[\mathrm{Co}(e n)_2 \mathrm{Cl}_2\right] \mathrm{Cl}$

Solution:
$\left[\mathrm{Co}(e n)_2 \mathrm{Cl}_2\right] \mathrm{Cl}$

Topic: Chemistry, Coordination Compounds, JEE Main
Which of the following will have a mesoisomer also

Options:
A 2, 3-Dichloropentane
B 2, 3-Dichlorobutane
C 2-Chlorobutane
D 2-Hydroxypropanoic acid

The correct answer is:
2, 3-Dichlorobutane

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
Which of the following will have least hindered rotation about carbon-carbon bond?

Options:
A

Acetylene

B

Hexachloroethane

C

Ethane

D

Ethylene

The correct answer is:

Ethane

Solution:
Rotation around the C - C bond in ethylene C 2 H 4 and acetylene C 2 H 2 is not possible because of the presence of double and triple bond. Among ethane and hexachloroethane, ethane has the smallest sized group (H) bonded to carbons, hence there will be least hindered rotation about C - Câ¡ bond.

Topic: Chemistry, General Organic Chemistry, JEE Main
Which of the following will have meso-isomer also?

Options:
A
2- chlorobutane

B
2- hydroxyopanoic acid

C
2,3 - dichloropentane

D
2-3- dichlorobutane

The correct answer is:
2-3- dichlorobutane

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2004
Which of the following will have the highest dipole moment?

Options:
A BeF2
B BF3
C H2O
D All have zero dipole moment

The correct answer is:
H2O

Solution:
BeF2:Î¼=0
BF3:Î¼=0
H2O:Î¼=1.84 D

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Which of the following will most readily give the dehydrohalogenation product?

Options:
A
B
C
D

The correct answer is:

Solution:
Here, the dehydrohalogenation goes by E1cb, and the most stable carbanion formation is favoured. Which will be formed in option A compound.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2018 (15 Apr)
Which of the following will most readily give the dehydrohalogenation product?

Options:
A

B

C

D

The correct answer is:

Solution:
Here dehydrohalogenation goes by $\mathrm{E1cB}$ and most stable carbanion formation is favoured in (a).

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
Which of the following will not act as an antacid?

Options:
A sodium hydrogen carbonate
B magnesium hydroxide
C sodium carbonate
D aluminium carbonate

The correct answer is:
sodium carbonate

Solution:
Sodium carbonate known as washing soda is quite alkaline in nature and has a horrid soapy taste. It is not very palatable and could burn the throat. Bicarbonate is much gentler and not as strongly alkaline, and therefor more palatable. Hence, it will not act as Antacid.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Which of the following will not be formed when calcium formate is distilled with calcium acetate?

Options:
A Propanone
B Propanal
C Ethanal
D Methanal

The correct answer is:
Propanal

Solution:
Thus, propanal is not formed here.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Which of the following will NOT be observed when a multimeter (operating in resistance measuring mode) probes connected across a component, are just reversed?

Options:
A Multimeter shows an equal deflection in both cases i.e., before and after reversing the probes if the chosen component is resistor.
B Multimeter shows NO deflection in both cases i.e. before and after reversing the probes if the chosen component is capacitor
C Multimeter shows a deflection, accompanied by a splash of light out of connected component in one direction and NO deflection on reversing the probes if the chosen component is LED
D Multimeter shows NO deflection in both cases i.e. before and after reversing the probes is the chsosen component is metal wire.

The correct answer is:
Multimeter shows NO deflection in both cases i.e. before and after reversing the probes if the chosen component is capacitor

Solution:
Multimeter shows NO deflection in both cases i.e. before and after reversing the probes if the chosen component is capacitor.
Multimeter shows deflection when it connects with capacitor
If we assume that LED has negligiable resistance then multimeter shows no deflection for the forward bias but when it connects in reverse direction, it break down occurs so splash of light out.
The resistance of metal wire may be taken zero, so no deflection in multimeter
No matter, how we connect the resistance across multimeter It shows same deflection.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Which of the following will not exist in zwitter ionic form at $\mathrm{pH}=7$ ?

Options:
A

B

C

D

The correct answer is:

Solution:
The $\mathrm{N}$ atom of amide is not basic.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
Which of the following will not exist in Zwitterionic form at pH=7?

Options:
A
B
C
D

The correct answer is:

Solution:
The N atom of the amide is not basic, because of -m effect of the carbonyl group.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2018 (15 Apr)
Which of the following will not give iodoform test

Options:
A Ethanal
B Ethanol
C 2-propanone
D 3-pentanone

The correct answer is:
3-pentanone

Solution:
Benzyl alcohol does not have the $\mathrm{CH}_3 \mathrm{CO}-$ group or $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{O}^{-}$so it will not give the positive iodoform test

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Which of the following will not show geometrical isomerism?

Options:
A Coox33-
B Coen2Cl2Cl
C CrNH34Cl2Cl
D Both Coen2Cl2Cl and CrNH34Cl2Cl

The correct answer is:
Coox33-

Solution:
Coox33- does not show geometrical isomerism.
As all three symmetrical bidentate ligands bonded to central Co3+ cation.

Topic: Chemistry, Coordination Compounds, JEE Main
Which of the following will not show mutarotation?

Options:
A
Maltose

B
Lactose

C
Glucose

D
Sucrose

The correct answer is:
Sucrose

Solution:
Sucrose does not contain a free aldehydic or ketonic group hence it does not show mutarotation.

Topic: Chemistry, Biomolecules, JEE Main
Which of the following will not show mutarotation?

Options:
A
Maltose

B
Lactose

C
Glucose

D
Sucrose

The correct answer is:
Sucrose

Solution:
Sucrose does not contain a free aldehydic or ketonic group hence it does not show mutarotation.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2014 (12 Apr Online)
Which of the following will produce the maximum depression in freezing point of its aqueous solution

Options:
A $\quad 0.1 \mathrm{M}$ glucose
B $0.1 M_{\text {sodium chloride }}$
C $0.1 \mathrm{M}$ barium chloride
D $0.1 \mathrm{M}$ magnesium sulphate

The correct answer is:
$0.1 \mathrm{M}$ barium chloride

Solution:
$\mathrm{BaCl}_2$ gives maximum ion hence it shows maximum depression in freezing point.

Topic: Chemistry, Solutions, JEE Main
Which of the following will react with CHCl3+alc. KOH?

Options:
A Adenine and proline
B Thymine and proline
C Adenine and lysine
D Adenine and thymine

The correct answer is:
Adenine and lysine

Solution:
Compounds having 1° amine give carbylamine reaction with CHCI3 & alc. KOH

Topic: Chemistry, Amines, JEE Main, JEE Main 2020 (04 Sep Shift 1)
Which of the following will show aromatic behaviour

Options:
A

B

C

D

The correct answer is:

Solution:
(1) Molecule is planar.
(2) $6 \pi$ electrons are present.

Topic: Chemistry, General Organic Chemistry, JEE Main
Which of the following will show cannizzaro reaction:

Options:
A
B
C
D All of these

The correct answer is:
All of these

Solution:
Carbonyl which does not have α-H can exhibit canizzaro reaction.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Which of the following will show optical isomerism?

Options:
A
B
C
D

The correct answer is:

Solution:
The asterisked carbon in linked with four different group of atoms and hence, is 'chiral' . Therefore, compound 'A' shows optical isomerism.
Rest all do not show optical isomerism as they do not have any chiral carbon atom.

Topic: Chemistry, General Organic Chemistry, JEE Main
Which of the following works on Pascal's law?

Options:
A Aneroid barometer
B Hydraulic lift
C Sprayer
D Venturimeter

The correct answer is:
Hydraulic lift

Solution:
As we know that,
In a closed system pressure remains same everywhere. This is Pascal's law.
Similarly, in case of hydraulic lift
Pressure at side $1=$ Pressure at side 2
$$
\begin{aligned}
& \text { and pressure }(p)=\frac{\text { Force }(F)}{\text { Area }(A)} \\
& \therefore \quad \frac{F_1}{A_1}=\frac{F_2}{A_2}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 1)
Which of the following would be produced when acetylene reacts with $\mathrm{HCl}$

Options:
A $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Cl}$
B $\mathrm{CH}_3 \mathrm{CHCl}_2$
C $\mathrm{CHCl}=\mathrm{CHCl}$
D $\mathrm{CH}_2=\mathrm{CHCl}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{CHCl}_2$

Solution:
$\mathrm{CH} \equiv \mathrm{CH}+\mathrm{HCl} \rightarrow \mathrm{CH}_2=\mathrm{CHCl} \xrightarrow{+\mathrm{HCl}} \mathrm{CH}_3 \mathrm{CHCl}_2$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Which of the following would not give 2-phenylbutane as the major product in a Friedel-Crafts alkylation reaction with benzene?

Options:
A 1-butene + HF
B 2-butanol + H2SO4
C Butanoylchloride + AlCl3 then Zn-Hg/HCl
D Butyl chloride + AlCl3

The correct answer is:
Butanoylchloride + AlCl3 then Zn-Hg/HCl

Solution:
The Friedal â crafts alkylation reaction will give propyl phenyl ketone which on further Clemmensonâs reduction will give butyl benzene.
C 6 H 6  +  CH 3 CH 2 CH 2 COCl → AlCl 3 C 6 H 5 COCH 2 CH 2 CH 3 → Zn-Hg/HCl C 6 H 5 CH 2 CH 2 CH 2 CH 3 Butylbenzene
Rest other will give 2-phenylbutane as in them rearrangement of carbocation is possible.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Which of the following would not give 2-phenylbutane as the major product in a Friedel-Crafts alkylation reaction ?

Options:
A
1-butene $+\mathrm{HF}$

B
2-butanol $+\mathrm{H}_2 \mathrm{SO}_4$

C
Butanoyl chloride $+\mathrm{AlCl}_3$ then $\mathrm{Zn}, \mathrm{HCl}$

D
Butyl choloride $+\mathrm{AlCl}_3$

The correct answer is:
Butanoyl chloride $+\mathrm{AlCl}_3$ then $\mathrm{Zn}, \mathrm{HCl}$

Solution:
The Friedal-crafts alkylation reaction will give propyl phenyl ketone which further on Clemmenson's reduction will give butyl benzene

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2013 (22 Apr Online)
Which of the following xenon-oxo compounds may not be obtained by hydrolysis of xenon fluorides?

Options:
A
$\mathrm{XeO}_2 \mathrm{~F}_2$

B
$\mathrm{XeOF}_4$

C
$\mathrm{XeO}_3$

D
$\mathrm{XeO}_4$

The correct answer is:
$\mathrm{XeO}_4$

Solution:
$\mathrm{XeF}_6+\mathrm{H}_2 \mathrm{O} \stackrel{\text { Partial }}{\longrightarrow} \mathrm{XeOF}_4+2 \mathrm{HF}$
$\mathrm{XeF}_6+2 \mathrm{H}_2 \mathrm{O} \stackrel{\text { Partial }}{\longrightarrow} \mathrm{XeO}_2 \mathrm{~F}_2+4 \mathrm{HF}$
$$
\mathrm{XeF}_6+3 \mathrm{H}_2 \mathrm{O} \stackrel{\text { Compute }}{\longrightarrow} \mathrm{XeO}_3+6 \mathrm{HF}
$$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2014 (12 Apr Online)
Which of the four resistances $P, Q, R$ and $S$ generate the greatest amount of heat when a current flows from A to B ?

Options:
A
$\mathrm{Q}$

B
$\mathrm{S}$

C
$\mathrm{P}$

D
$\mathrm{R}$

The correct answer is:
$\mathrm{S}$

Solution:
$$
\begin{aligned}
& \mathrm{R}_1=\mathrm{P}+\mathrm{Q}=2 \Omega+4 \Omega=6 \Omega \\
& \mathrm{R}_2=\mathrm{R}+\mathrm{S}=1 \Omega+2 \Omega=3 \Omega \\
& \mathrm{I}_1 \mathrm{R}_1=\mathrm{I}_2 \mathrm{R}_2 \\
& \mathrm{I}_1=\frac{\mathrm{R}_2}{\mathrm{R}_1} \mathrm{I}_2=\frac{3}{6} \mathrm{I}_2=\frac{\mathrm{I}_2}{2}
\end{aligned}
$$
or $\mathrm{I}_2=2 \mathrm{I}_1$
Heat flow $\mathrm{H}=\mathrm{I}^2 \mathrm{Rt}$
For $\mathrm{Q}, \mathrm{H}_{\mathrm{Q}}=\mathrm{I}_1^2 \mathrm{Qt}=\frac{\mathrm{I}_2^2}{4} \times 4 \mathrm{t}=\mathrm{I}_2^2 \mathrm{t}$
For $\mathrm{S}, \mathrm{H}_{\mathrm{S}}=\mathrm{I}_2^2 \mathrm{St}=\mathrm{I}_2^2 \cdot 2 \mathrm{t}=2 \mathrm{I}_2^2 \mathrm{t}$
$\therefore$ Greatest amount of heat generated by $\mathrm{S}$.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2013 (23 Apr Online)
Which of the given below is best method of preparing alkyl chloride

Options:
A ROH+SOCl2→
B ROH+PCl5→
C ROH+PCl3→
D ROH+HCl→anhy.ZnCl2

The correct answer is:
ROH+SOCl2→

Solution:
The chlorination of alcohol by SOCl2 (thionyl chloride) is the best method for the preparation of alkyl halides as in this method all the other product are gaseous and thus halides are obtained on quite pure state R−OH +  SOCl 2 → Δ R−Cl + HCl +  SO 2 ↑

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Which of the given compounds can enhance the efficiency of hydrogen storage tank?

Options:
A Li/P4
B SiH4
C NaNi5
D Di-isobutylaluminium hydride

The correct answer is:
NaNi5

Solution:
Dihydrogen gas is converted into liquid state by cooling to 20K. This would require expensive insulated tanks. Tanks are metal alloy like NaNi5, Ti-TiH2, Mg-MgH2, etc. are in use for storage of dihydrogen.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Which of the given elements is radioactive?

Options:
A Uranium
B Aluminum
C Magnesium
D Cobalt

The correct answer is:
Uranium

Solution:
Uranium is a radioactive element.

Topic: Physics, Nuclear Physics, JEE Main
Which of the given reactions is not an example of disproportionation reaction?

Options:
A 2H2O2→2H2O+O2
B 2NO2+H2O→HNO3+HNO2
C MnO4-+4H++3e-→MnO2+2H2O
D 3MnO4+2-4H+→2MnO4+-MnO2+2H2O

The correct answer is:
MnO4-+4H++3e-→MnO2+2H2O

Solution:
Disproportionation reaction, is a type of redox reaction involving simultaneous reduction and oxidation of atoms of the same element from one oxidation state (OS) to two different oxidation states.

2H2O2-1⟶2H2O-2+O20 : Disproportionation

2NO2+H2O→HNO3+HNO2 : Disproportionation

MnO4-+4H++3e-→MnO2+2H2O : reduction only

3MnO42-+4H+→2MnO4-+MnO2+4+2H2O : Disproportionation

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2022 (26 Jul Shift 1)
Which of the given statement is correct?

Options:
A Boiling point of cis-But-2-ene > trans-But-2-ene
B Boiling point of trans-But-2-ene > cis-But-2-ene
C Boiling point of cis-But-2-ene = trans-But-2-ene
D	Boiling point cannot be predicted

The correct answer is:
Boiling point of cis-But-2-ene > trans-But-2-ene

Solution:
Dipole moment of cis> trans. More polar the molecule more is the Van der Waal force of attraction, hence greater is the boiling point.

Topic: Chemistry, General Organic Chemistry, JEE Main
Which of the given statement is not correct

Options:
A Phenol is more acidic than acetic acid
B Ethanol is less acidic than phenol
C Ethanol has lower boiling point than ethane
D Ethyne is a non-linear molecule

The correct answer is:
Phenol is more acidic than acetic acid

Solution:
Phenols are much more acidic than alcohols but less so than carboxylic acids or even carbonic acid. This is indicated by the values of ionisation constants. The relative acidity follows the order:
Ka10-5RCOOH>10-7H2CO3>10-10C6H5OH>10-14HOH>10-18ROH

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Which of the given statement is not true?

Options:
A Copper pyrites also contain Fe2S3
B Zinc blende mainly contain zinc chloride
C Gold is found in native state
D Silver glance mainly contains silver sulphide

The correct answer is:
Zinc blende mainly contain zinc chloride

Solution:
Zinc blende is ZnS not ZnCl2

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
Which of the given statements is incorrect about glycogen?

Options:
A It is present in animal cells.
B It is a straight chain polymer similar to amylose.
C It is present in some yeast and fungi
D Only α- linkages are present in the molecule.

The correct answer is:
It is a straight chain polymer similar to amylose.

Solution:
Glycogen is an animal starch. It consists of α- amylose and amylopectin. Amylopectin is branched chain polysaccharide. Hence statement 1 is incorrect.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2019 (12 Apr Shift 2)