datavorous/entrance-exam-dataset huggingface dataset:
Which of the following compounds is expected to be coloured?

Options:
A $\mathrm{ZnSO}_4$
B $\mathrm{MgCl}_2$
C $\mathrm{CuCl}_2$
D $\mathrm{AgCl}$

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

Solution:
Copper chloride $\left(\mathrm{CuCl}_2\right)$ as a yellow-brown powder (in anhydrous form) or a green crystalline solid (in dihydrate form) due to copper exist as divalent $\mathrm{Cu}^{2+}$ ions with $3 d^9$ electronic configuration. Due to presence of 1 unpaired electron in $3 d$-orbital, it is coloured.
Remaining $\mathrm{ZnSO}_4, \mathrm{MgCl}_2$ and $\mathrm{AgCl}$ are not coloured compounds respectively.
Hence, only $\mathrm{CuCl}_2$ is expected to be coloured.

Topic: Chemistry, d and f Block Elements, AP EAMCET, AP EAMCET 2021 (25 Aug Shift 2)
Which of the following compounds is expected to be coloured?

Options:
A CuF2
B CuCl
C Ag2SO4
D MgF2

The correct answer is:
CuF2

Solution:
The electronic configurations of cations in the given salts are as follows
Cu+(3d10),Cu2+(3d9),Ag+(4d10),Mg2+(2s22p6)
Hence only Cu2+ ion has one unpaired electron in 3d orbital it means d-d electronic transition is possible, so its salt is coloured.

Topic: Chemistry, d and f Block Elements, JEE Main
Which of the following compounds is likely to be a product Y in this case ?

Options:
A
B
C
D

The correct answer is:

Solution:
Here the product Y is which is formed as follows

Topic: Chemistry, Amines, JEE Main
Which of the following compounds is likely to show both Frenkel and Schottky defects in its crystalline form?

Options:
A AgBr
B CsCl
C KBr
D ZnS

The correct answer is:
AgBr

Solution:
AgBr can exhibit both Schottky and Frenkel defect.

Schottky defect in AgBr is exhibited due to the precipitation of both cations and anions. In AgBr, Ag+ ion is small in size and when removed from the lattice point, they can occupy the interstitial sites and, hence, shows both Frenkel and Schottky defects.

Topic: Chemistry, Solid State, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Which of the following compounds is metallic and ferromagnetic?

Options:
A MnO2
B TiO2
C CrO2
D VO2

The correct answer is:
CrO2

Solution:
Chromium Oxide (CrO2) is metallic compound and it shows Ferromagnetic property. It is a black synthetic magnetic solid and was once widely used in magnetic tape emulsion.

Topic: Chemistry, Solid State, JEE Main
Which of the following compounds is most reactive to an aqueous solution of sodium carbonate?

Options:
A

B

C

D

The correct answer is:

Solution:
Since anion is Aromatic, the reactant will have a high affinity for the reaction and hence, will be most reactive.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2017 (09 Apr Online)
Which of the following compounds is not an antacid?

Options:
A Ranitidine
B Aluminium hydroxide
C Cimetidine
D Phenelzine

The correct answer is:
Phenelzine

Solution:
Ranitidine, Cimetidine and metal hydroxides like aluminium hydroxide can be used as an antacid but not phenelzine. Phenelzine is not an antacid. It is an antidepressant. Antacids are a type of medication that can control the acid levels in the stomach.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2015 (04 Apr)
Which of the following compounds is not an antacid?

Options:
A Ranitidine
B Aluminium hydroxide
C Cimetidine
D Phenelzine

The correct answer is:
Phenelzine

Solution:
Ranitidine, Cimetidine and metal hydroxides like aluminium hydroxide can be used as an antacid but not phenelzine. Phenelzine is not an antacid. It is an antidepressant. Antacids are a type of medication that can control the acid levels in the stomach.

Topic: Chemistry, Environmental Chemistry, JEE Main
Which of the following compounds is not aromatic?

Options:
A

B

C

D

The correct answer is:

Solution:
Cyclic conjugation
Lone pair involved in aromaticity
Having =6πe- Aromatic
Cyclic conjugation and by Huckel rule =6πe-
So aromatic
Cyclic conjugation 6πe-
Cyclic conjugation lp of -N is involved in delocalization
Having =6πe- Aromatic
Cyclic conjugation and by Huckel rule =4πe-
So ,Anti-aromatic

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (09 Jan Shift 2)
Which of the following compounds is not coloured?

Options:
A $\mathrm{Na}_{2}\left[\mathrm{Cu}\left(\mathrm{Cl}_{4}\right]\right.$
B $\mathrm{Na}\left[\mathrm{Cd}(\mathrm{Cl})_{4}\right]$
C $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$
D $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$

The correct answer is:
$\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$

Solution:
We know that complex compound having no unpaired electron is colourless. Among the given complexes, $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ has no unpaired electron as $\mathrm{CN}^{-}$is a strong field ligand and causes pairing of electrons. So, it is colourless.

Topic: Chemistry, Coordination Compounds, JEE Main
Which of the following compounds is not expected to show Lassaignes' test for nitrogen ?

Options:
A
Propanenitrile

B
Hydroxylamine hydrochloride

C
Nitromethane

D
Ethanamine

The correct answer is:
Hydroxylamine hydrochloride

Solution:
Lassaigne's test is used for the detection of nitrogen and given by all nitrogenous compound except diazo
compounds.
This test is shown only by the compounds containing $\mathrm{C}$ and $\mathrm{N}$ both hence hydroxyl amine hydrochloride $\left(\mathrm{H}_2 \mathrm{NOH} . \mathrm{HCl}\right)$ will not perform this test.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2013 (09 Apr Online)
Which of the following compounds is obtained as the main product when phenol is heated with zinc dust?

Options:
A Benzene
B Toluene
C Ethylbenzene
D Phenylacetone

The correct answer is:
Benzene

Solution:
The primary product of the reaction between phenol and zinc dust is benzene.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Which of the following compounds is produced as the main product in the decarboxylation of benzoic acid with soda lime?

Options:
A Toluene
B Benzaldehyde
C Benzene
D Benzoic acid

The correct answer is:
Benzene

Solution:
The primary product of the decarboxylation of benzoic acid with soda lime is benzene.

Topic: Chemistry, Hydrocarbons, NEET
Which of the following compounds is used in anti-knock compositions to prevent the deposition of oxides of lead on spark plug, combustion chamber and exhaust pipe?

Options:
A Benzene
B Glycol
C 1, 2-Dibromoethane
D Glycerol

The correct answer is:
1, 2-Dibromoethane

Solution:
1, 2-dibromoethane compound is used in antiknock compositions to prevent lead from being deposited in the cylinder head. It is sufficiently volatile and can be removed with the exhaust gas from the exhaust pipe.

Topic: Chemistry, Environmental Chemistry, JEE Main
Which of the following compounds is(are) coloured due to charge transfer spectra and not due to d-d transitions?

Options:
A K2Cr2O7
B KMnO4
C CrO3
D All of these

The correct answer is:
All of these

Solution:
K2Cr2O7,KMnO4 and CrO3 are coloured due to charge transfer as in them there is no unpaired d-orbital electron. So d-d electronic transition is not possible.

Topic: Chemistry, d and f Block Elements, JEE Main
Which of the following compounds may not exist as enantiomers

Options:
A $\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CO}_2 \mathrm{H}$
B $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}_2 \mathrm{OH}$
C $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}_3$
D $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHClCH}$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}_3$

Solution:
In $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}_3$ chiral centre is absent.

Topic: Chemistry, General Organic Chemistry, JEE Main
Which of the following compounds of elements in group 14 would you expect to be most ionic in character

Options:
A $\mathrm{CCl}_4$
B $\mathrm{SiCl}_4$
C $\mathrm{PbCl}_2$
D $\mathrm{PbCl}_4$

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

Solution:
$\mathrm{PbCl}_2$ is most ionic because on going down the group the metallic character increases and also the inert pair effect predominates

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Which of the following compounds produces an optically inactive compound on hydrogenation?

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2020 (03 Sep Shift 1)
Which of the following compounds shows geometrical isomerism?

Options:
A 2-methylpent-2-ene
B 4-methylpent-2-ene
C 4-methylpent-1-ene
D 2-methylpent-1-ene

The correct answer is:
4-methylpent-2-ene

Solution:
Can show G.I. because both doubly bonded carbon have two different groups.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (06 Sep Shift 1)
Which of the following compounds shows optical isomerism?

Options:
A
$\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{+2}$

B
$\left[\mathrm{ZnCl}_4\right]^{-2}$

C
$\left[\mathrm{Cr}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{-3}$

D
$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{-3}$

The correct answer is:
$\left[\mathrm{Cr}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{-3}$

Solution:

Topic: Chemistry, Coordination Compounds, JEE Main
Which of the following compounds shows optical isomerism?

Options:
A
$\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{+2}$

B
$\left[\mathrm{ZnCl}_4\right]^{-2}$

C
$\left[\mathrm{Cr}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{-3}$

D
$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{-3}$

The correct answer is:
$\left[\mathrm{Cr}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{-3}$

Solution:

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2005
Which of the following compounds undergoes nucleophilic substitution reaction most easily?

Options:
A

B

C

D

The correct answer is:

Solution:
has electron withdrawing group $-\mathrm{NO}_2$ which reduces the double bond character between
carbon of benzene ring and chlorine. Thus, the correct order of nucleophilic substitution reactions

Topic: Chemistry, Haloalkanes and Haloarenes, NEET, NEET 2011 (Mains)
Which of the following compounds undergoes nucleophilic substitution reaction most easily?

Options:
A
B
C
D

The correct answer is:

Solution:
It has electron withdrawing group - $\mathrm{NO}_{2}$ which reduces the double bond character between carbon of benzene ring and chlorine. Hence, the correct order of nucleophilic substitution reactions are,

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Which of the following compounds undergoes nucleophilic substitution reaction most easily?

Options:
A

B

C

D

The correct answer is:

Solution:
has electron withdrawing group $-\mathrm{NO}_2$ which reduces the double bond character between
carbon of benzene ring and chlorine. Thus, the correct order of nucleophilic substitution reactions

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Which of the following compounds will be most easily attacked by an electrophile?

Options:
A
B
C
D

The correct answer is:

Solution:
Strong Electron releasing group favours ESR effectively.

Topic: Chemistry, General Organic Chemistry, JEE Main
Which of the following compounds will be suitable for Kjeldahl's method for nitrogen estimation?

Options:
A
B
C
D

The correct answer is:

Solution:
Kjeldahl method is not applicable to compounds containing nitrogen in nitro and azo groups and nitrogen present in the ring (e.g. pyridine) as nitrogen of these compounds does not change to ammonium sulphate under these conditions.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2018 (08 Apr)
Which of the following compounds will behave as reducing sugar in an aqueous KOH solution?

Options:
A
B
C
D

The correct answer is:

Solution:
Sugar has an ester functional group -O-CO-R, which gets hydrolyzed in an alkaline medium to give hemi-acetal, which then acts as a reducing sugar.

Topic: Chemistry, Biomolecules, JEE Main
Which of the following compounds will behave as reducing sugar in an aqueous KOH solution?

Options:
A
B
C
D

The correct answer is:

Solution:
Sugar has an ester functional group -O-CO-R, which gets hydrolyzed in an alkaline medium to give hemi-acetal, which then acts as a reducing sugar.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2017 (02 Apr)
Which of the following compounds will exhibit cis-trans (geometrical) isomerism?

Options:
A 1-Butanol
B 2-Butene
C 2-Butanol
D 2-Butyne

The correct answer is:
2-Butene

Solution:
$\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$
Can exists as
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2009 (Mains)
Which of the following compounds will exhibit geometrical isomerism?

Options:
A 1, 1 - Diphenyl - 1 - propane
B 1 - Phenyl - 2 - butene
C 3 - Phenyl - 1 - butene
D 2 - Phenyl - 1 - butene

The correct answer is:
1 - Phenyl - 2 - butene

Solution:
To exhibit geometric isomerism, the double bonded carbon should have two different substitutents. Terminal alkene will not show geometric isomerism

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2015 (04 Apr)
Which of the following compounds will form significant amount of meta product during mono-nitration reaction?

Options:
A
B
C
D

The correct answer is:

Solution:
In a strong acidic medium acid-base reaction takes place so NH2 group is converted to N+H3 group. Due to the formation of anilinium ion, significant meta nitroaniline is also formed.
The net reaction can be written as follows,

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2017 (02 Apr)
Which of the following compounds will form the precipitate with aq. AgNO3 solution most readily?

Options:
A
B
C
D

The correct answer is:

Solution:
It is an example of SN1 reaction.

Rate of SN1 reaction α stability of carbocation

In option 1 carbocation formed after removing halogen is destabilized by -I effect.

In option 2 carbocation formed after removing halogen is stabilized by +M effect since lone pair is present at oxygen atom.

In option 3 carbocation formed after removing halogen is stabilized by +M effect of lone pair present on nitrogen atom but lone pair is also in conjugation with benzene ring therefore +M effect is less.

In option 4 carbocation formed after removing halogen is stabilized by +M effect of lone pair present on nitrogen atom and +M effect of nitrogen atom is more than that of oxygen atom. Hence, option 4 is most stable.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (04 Sep Shift 2)
Which of the following compounds will give a yellow precipitate with iodine and alkali?

Options:
A Acetophenone
B Methyl acetate
C Acetamide
D 2-hydroxypropane

The correct answer is:
2-hydroxypropane

Solution:
$$
\text { Out of the given compounds, only }
$$
yellow precipitate with iodine and alkali (iodoform test). But
do not perform this test because $\alpha-\mathrm{H}$ is not active due to $+M$ effect of nearby group.

Topic: Chemistry, Aldehydes and Ketones, NEET, NEET 2012 (Mains)
Which of the following compounds will give silver mirror with ammoniacal silver nitrate?
A. Formic acid
B. Formaldehyde
C. Benzaldehyde
D. Acetone
Choose the correct answer from the options given below :

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

The correct answer is:
A, B and C only

Solution:
also gives silver mirror test with ammonical silver nitrate.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Which of the following compounds will have the highest melting point here?

Options:
A III
B II
C I
D They have equal melting point

The correct answer is:
II

Solution:
Compound II is most symmetrical because it has both CH3 groups and Br atoms are apposite to each other therefore, it fits in the crystal lattice better than the other two isomers and hence it has high melting point.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Which of the following compounds will have the longest C=C bond?

Options:
A CH3-CH=CH2
B
C CH2=C=CH2
D

The correct answer is:

Solution:
Have more α-H more conjugating structure. Show more hyperconjugation so bond length double length is longest
(A) CH3-CH=CH2 , α H=3
(B) , α H=0
(C) CH2=C=CH2 , α H=0
(D) , α H=6

Topic: Chemistry, General Organic Chemistry, JEE Main
Which of the following compounds will most readily be dehydrated to give alkene under acidic condition?

Options:
A 4 - Hydroxypentan - 2-one
B 3- Hydroxypentan-2-one
C 1-Pentanol
D 2-Hydroxycyclopentanone

The correct answer is:
4 - Hydroxypentan - 2-one

Solution:
3-Hydroxypentan-2-one and 4-Hydroxypentan-2-one will most readily be dehydrated to give alkene under acidic condition. This will lead to formation of α−β unsaturated ketone (conjugated system). But out of both 4-Hydroxypentan-2-one will most readily dehydrate to give alkene under acidic condition since the carbocation formed will be more stable than in the case of 3-Hydroxypentan-2-one.

Dehydration of 2-Hydroxycyclopentanone will lead to C=C double bond in 5 member ring that will cause strain and product is less likely to be formed.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2018 (16 Apr Online)
Which of the following compounds will not be soluble in sodium bicarbonate?

Options:
A 2, 4, 6 - Trinitrophenol
B Benzene sulphonic acid
C o - Nitrophenol
D Benzoic acid

The correct answer is:
o - Nitrophenol

Solution:
o-Nitrophenol is insoluble in sodium hydrogen carbonate because it is less acidic than carbonic acid.While, others being more acidic than H2CO3 dissolve in sodium bi carbonate.

Hence, the given reaction will not be spontaneous.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Which of the following compounds will not be soluble in sodium bicarbonate?

Options:
A 2, 4, 6 - Trinitrophenol
B Benzene sulphonic acid
C o - Nitrophenol
D Benzoic acid

The correct answer is:
o - Nitrophenol

Solution:
o-Nitrophenol is insoluble in sodium hydrogen carbonate because it is less acidic than carbonic acid.While, others being more acidic than H2CO3 dissolve in sodium bi carbonate.

Hence, the given reaction will not be spontaneous.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2014 (19 Apr Online)
Which of the following compounds will NOT give the positive test for Tollen's reagent?

Options:
A Propanone
B Propanal
C Ethanal
D Both (b) and (c)

The correct answer is:
Propanone

Solution:
Only aldehyde gives Tollen's test. Ketone do not give this test. Therefore, propanone does not give this test,

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Which of the following compounds will not undergo Friedel Crafts reaction with benzene?

Options:
A
B
C
D

The correct answer is:

Solution:
Formation of carbocation is not possible in case of CH2=CHCl,​because partial double bond character is present in carbon and chlorine bond due to resonance between lone pair of chlorine atom and double bond.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2017 (08 Apr Online)
Which of the following compounds will produce a precipitate with $\mathrm{AgNO}_{3} ?$

Options:
A

B

C

D

The correct answer is:

Solution:
Only those halides give precipitate with $\mathrm{AgNO}_{3}$ which easily ionises to give halide ion and a stable carbocation. Corresponding carbocation from the four bromides are

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2019 (11 Jan Shift 2)
Which of the following compounds will provide a tertiary alcohol on reaction with excess of CH3MgBr followed by hydrolysis ?

Options:
A
B
C
D

The correct answer is:

Solution:
Since the given question is single correct choice the best appropriate option is (A)

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Which of the following compounds will show retention in configuration on nucleophile substitution by OH-ion ?

Options:
A
B
C
D

The correct answer is:

Solution:
Option 1,2 and 3 have chiral carbon in which inversion of configuration takes place during SN2 reaction but 4th option has achiral carbon associated with halogen group so it will show retention after substitution reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2020 (02 Sep Shift 1)
Which of the following compounds will show the maximum 'enol' content?

Options:
A CH3COCH2CONH2
B CH3COCH2COCH3
C CH3COCH2COOC2H5
D CH3COCH3

The correct answer is:
CH3COCH2COCH3

Solution:
Due to intramolecular hydrogen bonding and resonance stabilization, enol content is maximum in CH3COCH2COCH3.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (08 Apr Shift 2)
Which of the following compounds would give the following set of qualitative analysis?

(i) Fehling’s Test: Positive

(ii) Na fusion extract upon treatment with sodium nitroprusside gives a blood red colour but not prussian blue.

Options:
A
B
C
D

The correct answer is:

Solution:
Aromatic aldehydes do not respond Fehling's test. (A), (B), (C) are aromatic aldehydes.

If N and S attached with carbon atom give blood red with sodium nitroprusside gives a blood red colour but not prussian blue.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Which of the following compounds would not be expected to decarboxylate when heated?

Options:
A

B

C

D

The correct answer is:

Solution:
acid having a keto group at $\beta$-position is decarboxylated on heating.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Which of the following conclusions could not be derived from Rutherford's α-particle scattering experiment:

Options:
A Most of the space in the atom is empty.
B The radius of the atom is about 10-10 m, while that of the nucleus is 10-15 m.
C Electrons move in circular paths of fixed energy, called orbits.
D Electrons and the nucleus are held together by electrostatic forces of attraction.

The correct answer is:
Electrons move in circular paths of fixed energy, called orbits.

Solution:
Rutherford's scattering experiment established the idea of a nucleus, while maintaining that the electrons do not reside in the nucleus.

On the basis of his experiment, Rutherford put forward the model of an atom which had the following features:

There is a positively charged centre in an atom called the nucleus. Nearly all the mass of an atom resides in the nucleus.

The size of the nucleus is very small as compared to the size of the atom.

However, Rutherford later suggested that the electrons are revolving in circular orbits.

Topic: Chemistry, Hydrogen, AP EAMCET, AP EAMCET 2020 (23 Sep Shift 1)
Which of the following conditions are correct?

Process	V max value	Km value

a	Competitive inhibition	Increases	Decreases

b	Non-competitive inhibition	Decreases	Remains same

c	Allosteric inhibition	Decreases	Increases

d	Un-competitive inhibition	Increases	Increases

Options:
A a
B b
C c
D d

The correct answer is:
b

Solution:
Any substance that can diminish the velocity of an enzyme is known as an Inhibitor. The types of inhibition are Competitive, Non Competitive and Allosteric inhibition. In Competitive inhibition, the Enzyme and Inhibitor competes to bind to the same active site. Hence if substrate concentration is increased the chances of substrate binding to the enzyme will increase .Thus if (S) increases Km (Michaleis Constant) also increases. In non-competitive inhibition, the substrate and inhibitor binds at different sites hence do not compete with each other. This cannot be reversed by increasing the substrate concentration and hence Km remains constant. This velocity of reaction also decreases.

Topic: Chemistry, Biomolecules, NEET
Which of the following conditions in drinking water causes methemoglobinemia?

Options:
A >50 ppm of nitrate
B >50 ppm of chloride
C >100 ppm of sulphate
D >50 ppm of lead

The correct answer is:
>50 ppm of nitrate

Solution:
Methemoglobinemia disease occur due to NO3-, when it's concentration is higher than 50 ppm in water.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (09 Jan Shift 2)