datavorous/entrance-exam-dataset huggingface dataset:
Which one of the following lanthanoids does not form MO2? [M is lanthanoid metal]

Options:
A Yb
B Dy
C Nd
D Pr

The correct answer is:
Yb

Solution:
Yb is the only element that does not form MO2 type oxide.

Ytterbium is a soft, malleable and rather ductile element that exhibits a bright silvery luster. A rare earth, the element is easily attacked and dissolved by mineral acids, slowly reacts with water, and oxidizes in air. The oxide forms a protective layer on the surface. Compounds of ytterbium are rare.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (26 Feb Shift 1)
Which one of the following measures of marks is the most suitable one of central location for computing intelligence of students

Options:
A Mode
B Arithmetic mean
C Geometric mean
D Median

The correct answer is:
Median

Solution:
Median tells about middle of the class interval

Topic: Mathematics, Statistics, JEE Main
Which one of the following measures of marks is the most suitable one of central location for computing intelligence of students?

Options:
A Mode
B Arithmetic mean
C Geometric mean
D Median

The correct answer is:
Median

Solution:
Median tells about middle of the class interval

Topic: Mathematics, Statistics, JEE Main
Which one of the following metal ions is coloured?

Options:
A $\mathrm{Ti}^{+}$
B $\mathrm{Cu}^{+}$
C $\mathrm{V}^{4+}$
D $\mathrm{Zn}^{2+}$

The correct answer is:
$\mathrm{V}^{4+}$

Solution:
Atomic number of $\mathrm{V}$ is $23 .$
Electronic configuration of $\mathrm{V}=[\mathrm{Ar}] 4 s^{2} 3 d^{3}$
$$
\mathrm{V}^{4+}=[A \mathrm{r}] 3 d^{1}
$$
$\because$ It contains one unpaired electron, hence it is coloured. While $\mathrm{Ti}^{+}, \mathrm{Cu}^{+}$and $\mathrm{Zn}^{2+}$ do not have unpaired electron.

Topic: Chemistry, d and f Block Elements, JEE Main
Which one of the following metal oxides is antiferromagnetic in nature

Options:
A $\mathrm{MnO}_2$
B $\mathrm{TiO}_2$
C $\mathrm{VO}_2$
D $\mathrm{CrO}_2$

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

Solution:
$\mathrm{MnO}_2$ is antiferromagnetic.

Topic: Chemistry, Solid State, JEE Main
Which one of the following metalic hydroxides does not dissolve in sodium hydroxide solution

Options:
A $\mathrm{Zn}(\mathrm{OH})_2$
B $\mathrm{Al}(\mathrm{OH})_3$
C $\mathrm{Fe}(\mathrm{OH})_3$
D $\mathrm{Pb}(\mathrm{OH})_2$

The correct answer is:
$\mathrm{Fe}(\mathrm{OH})_3$

Solution:
$\mathrm{Fe}(\mathrm{OH}) 3$ is soluble in sodium hydroxide solution.

Topic: Chemistry, s Block Elements, JEE Main
Which one of the following metals will give blue ash when its salt is heated with $\mathrm{Na}_2 \mathrm{CO}_3$ solid and $\mathrm{Co}\left(\mathrm{NO}_3\right)_2$ on a charcoal piece

Options:
A $\mathrm{Cu}$
B $M g$
C $A l$
D $\mathrm{Zn}$

The correct answer is:
$A l$

Solution:
$\mathrm{CoAl} \mathrm{O}_2$ is formed which is blue. Follow cobalt nitrate-charcoal test.

Topic: Chemistry, Practical Chemistry, JEE Main
Which one of the following methods is commonly used to maintain the genetic traits of a given plant?

Options:
A By propagating through seed germination
B By propagating through vegetative multiplication
C By generating hybrids through intergeneric pollination
D By treating the seeds with gamma radiations

The correct answer is:
By propagating through vegetative multiplication

Solution:
The propagation through vegetative multiplication is used to maintain the genetic traits of a given plant. It gives rise to genetically uniform population or clone. In case of plants propagated through seeds, variations creep in due to chance segregation of genes during meiosis and their chance combination during fertilisation.

Topic: Biology, Reproduction in Organisms, NEET
Which one of the following methods is commonly used to maintain the genetic traits of a given plant?

Options:
A By propagating through seed germination
B By propagating through vegetative multiplication
C By generating hybrids through intergeneric pollination
D By treating the seeds with gamma radiations

The correct answer is:
By propagating through vegetative multiplication

Solution:
The propagation through vegetative multiplication is used to maintain the genetic traits of a given plant. It gives rise to genetically uniform population or clone. In case of plants propagated through seeds, variations creep in due to chance segregation of genes during meiosis and their chance combination during fertilization.

Topic: Biology, Reproduction in Organisms, AIIMS, AIIMS 2008
Which one of the following methods is commonly used to maintain the genetic traits of a given plant?

Options:
A By propagating through seed germination
B By propagating through vegetative multiplication
C By generating hybrids through intergeneric pollination
D By treating the seeds with gamma radiations

The correct answer is:
By propagating through vegetative multiplication

Solution:
The propagation through vegetative multiplication is used to maintain the genetic traits of a given plant. It gives rise to genetically uniform population or clone. In case of plants propagated through seeds, variations creep in due to chance segregation and recombination of genes during meiosis and their chance combination during fertilisation.

Topic: Biology, Reproduction in Organisms, AIIMS, AIIMS 2017
Which one of the following methods is commonly used to maintain the genetic traits of a given plant?

Options:
A By propagating through seed germination
B By propagating through vegetative multiplication
C By generating hybrids through intergeneric pollination
D By treating the seeds with gamma radiations

The correct answer is:
By propagating through vegetative multiplication

Solution:
The propagation through vegetative multiplication is used to maintain the genetic traits of a given plant. It gives rise to genetically uniform population or clone. In case of plants propagated through seeds, variations creep in due to chance segregation of genes during meiosis and their chance combination during fertilization.

Topic: Biology, Reproduction in Organisms, NEET
Which one of the following methods is neither meant for the synthesis nor for separation of amines?

Options:
A
Hinsberg method

B
Hofmann method

C
Wurtz reaction

D
Curtius reaction

The correct answer is:
Wurtz reaction

Solution:
No Solution Available

Topic: Chemistry, Amines, JEE Main, JEE Main 2005
Which one of the following methods is neither meant for the synthesis nor for separation of amines

Options:
A Hinsberg method
B Hofmann method
C Wurtz reaction
D Curtius reaction

The correct answer is:
Wurtz reaction

Solution:
Wurtz reaction is used to prepare alkanes from alkyl halides

Topic: Chemistry, Amines, JEE Main
Which one of the following methods is neither meant for the synthesis nor for separation of amines?

Options:
A Curtius reaction
B Wurtz reaction
C Hofmann method
D Hinsberg method

The correct answer is:
Wurtz reaction

Solution:
Hinsberg method:

Wurtz reaction:

Wurtz reaction is for the preparation of hydrocarbons from alkyl halide

RX+2Na+XR→R−R+2NaX

Curtius Reaction:

Hofmann Bromamide:

Topic: Chemistry, Amines, JEE Main
Which one of the following methods is suitable to separate a mixture of $n$-pentane and toluene?

Options:
A Steam distillation
B Simple distillation
C Sublimation
D Vacuum distillation

The correct answer is:
Simple distillation

Solution:
$\because$ Toluene is
$n$-pentane ion $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$ one is aromatic and other is straight chain compound. Therefore, it can be separated by simple distillation.

Topic: Chemistry, General Organic Chemistry, JEE Main
Which one of the following methods is used to prepare Me $_{3}$ COEt with a good yield?

Options:
A Mixing EtONa with MegCCl
B Mixing Me $_{3}$ CONa with EtCl
C Heating a mixture of $(1 ; 1)$ EtOH and $\mathrm{Me}_{3} \mathrm{COH}$ in the presence of conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$
D Treatment of Me_COH with EtMgI

The correct answer is:
Mixing Me $_{3}$ CONa with EtCl

Solution:
\(\mathrm{Me}_{3} \mathrm{CONa} \underset{\left(1^{\circ} \text { alkyl halide) }\right.}{\stackrel{\mathrm{EtCl}}{\longrightarrow}} \mathrm{Me}_{3} \mathrm{COEt}\)
This reaction is known as Williamson's synthesis.
So, the correct option is B

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Which one of the following microbes is the source of acetic acid?

Options:
A Clostridium acetobutylicum
B Acetobacter aceti
C Gluconacetobacter liquefaciens
D All of these

The correct answer is:
All of these

Solution:
The main organisms that are responsible for the production of acetic acid are Acetobacter, Clostridium acetobutylicum, Gluconacetobacter, and Gluconobacter. These have a high capacity to oxidize ethanol to acetic acid and shows high resistance to acetic acid released into the fermentative medium.

Topic: Biology, Microbes in Human Welfare, NEET
Which one of the following mineral elements plays an important role in biological nitrogen fixation?

Options:
A copper
B manganese
C zinc
D molybdenum

The correct answer is:
molybdenum

Solution:
Molybdenum is a micronutrient specifically for plants that form root nodules with nitrogenfixing bacteria, though plants that do not form nodules also use trace amounts of it in a protein involved with nitrogen metabolism and uptake.
Related Theory
The main functions of leghaemoglobin are (1) to facilitate oxygen supply to the nitrogen fixing bacteria and (2) to protect the enzyme, nitrogenase from being inactivated by oxygen.

Topic: Biology, Mineral Nutrition, NEET, NEET 2003
Which one of the following mixture does not act as a buffer solution-

Options:
A Boric acid and borax
B Sodium Phosphate & disodium hydrogen phosphate
C Sodium propionate and propionic acid
D Sodium acetate and sodium propionate

The correct answer is:
Sodium acetate and sodium propionate

Solution:
Sodium acetate and sodium propionate are not conjugate acid base pairs

Topic: Chemistry, Ionic Equilibrium, JEE Main
Which one of the following mixtures can be separated into pure components by fractional distillation

Options:
A Benzene - toluene
B Water - ethyl alcohol
C Water-nitric acid
D Water - hydrochloric acid

The correct answer is:
Benzene - toluene

Solution:
Aromatic compound generally separated by fractional distillation. e.g. Benzene + Toluene.

Topic: Chemistry, Solutions, JEE Main
Which one of the following molecules contain no $\pi$-bond?

Options:
A $\mathrm{CO}_2$
B $\mathrm{H}_2 \mathrm{O}$
C $\mathrm{SO}_2$
D $\mathrm{NO}_2$

The correct answer is:
$\mathrm{H}_2 \mathrm{O}$

Solution:
All the molecules have $\mathrm{O}$ atom with lone pairs, but in $\mathrm{H}_2 \mathrm{O}$ the $\mathrm{H}$ atom has no vacant orbital for $\pi$ bonding. That's why it does not have any $\pi$ bond.
In all other given molecules, the central atom because of the presence of vacant orbitals is capable to farm $\pi$ bonds.

Topic: Chemistry, Structure of Atom, JEE Main
Which one of the following molecules is expected to exhibit paramagnetic behaviour?

Options:
A O2
B O2-2
C C2
D N2

The correct answer is:
O2

Solution:
M.O. Electronic configuration.
O 2 =σ1 s 2 < σ * 1 s 2 <σ2 s 2 < σ * 2 s 2 <σ2 p z 2 <π2 p y 2 =π2 p x 2 < π * 2 p y 1 = π * 2 p x 1 Paramagnetic
O2-2=σ1 s 2 < σ * 1 s 2 <σ2 s 2 < σ * 2 s 2 <σ2 p z 2 <π2 p y 2 =π2 p 2x < π * 2 p y 2 = π * 2 p x 2 Diamagnetic
C2 =σ1 s 2 < σ * 1 s 2 <σ2 s 2 < σ * 2 s 2 <π2 p x 2 =π2 p y 2 Diamagnetic
N 2 =σ1 s 2 < σ * 1 s 2 <σ2 s 2 < σ * 2 s 2 <π2 p x 2 =π2 p y 2 <σ2 p z 2 Diamagnetic

O2 has unpaired e- so it is paramagnetic.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2013 (07 Apr)
Which one of the following molecules is paramagnetic?

Options:
A N2
B O3
C CO
D NO

The correct answer is:
NO

Solution:
As per Molecular orbital theory,
N2 14 electrons all paired electrons, so diamagnetic.
N2⇒σ1s2<σ*1s2<σ2s2<σ*2s2<π2px2=π2py2<σ2pz2
O3-24 electrons and diamagnetic.â
CO 14 electrons all paired so, diamagnetic.
CO⇒σ1s2<σ*1s2<σ2s2<σ*2s2<π2px2=π2py2<σ2pz2
NO 15 electrons one unpaired electron so, paramagnetic.â
NO⇒σ1s2<σ*1s2<σ2s2<σ*2s2<π2px2=π2py2<σ2pz2<π*2px1=π*2py0

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2014 (19 Apr Online)
Which one of the following molecules is polar?

Options:
A
$\mathrm{XeF}_4$

B
$\mathrm{IF}_5$

C
$\mathrm{SbF}_5$

D
$\mathrm{CF}_4$

The correct answer is:
$\mathrm{IF}_5$

Solution:
The geometry of $\mathrm{IF}_5$ is square pyramidal with an unsymmetric charge distribution therefore this molecule is polar.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Which one of the following molecules is polar?

Options:
A
$\mathrm{XeF}_4$

B
$\mathrm{IF}_5$

C
$\mathrm{SbF}_5$

D
$\mathrm{CF}_4$

The correct answer is:
$\mathrm{IF}_5$

Solution:
The geometry of $\mathrm{IF}_5$ is square pyramidal with an unsymmetric charge distribution therefore this molecule is polar.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2013 (09 Apr Online)
Which one of the following nitrates will leave behind a metal on strong heating?

Options:
A
Copper nitrate

B
Manganese nitrate

C
Silver nitrate

D
Ferric nitrate

The correct answer is:
Silver nitrate

Solution:
$\mathrm{AgNO}_3 \rightarrow \mathrm{Ag}+\mathrm{NO}_2+\frac{1}{2} \mathrm{O}_2$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2003
Which one of the following octahedral complexes will not show geometric isomerism (A and B are monodentate ligands)

Options:
A $\left[M A_5 B\right]$
B $\left[M A_2 B_4\right]$
C $\left[M A_3 B_3\right]$
D $\left[M A_4 B_2\right]$

The correct answer is:
$\left[M A_5 B\right]$

Solution:
Octahedral complexes of the type $\left[M A_4 B_2\right],\left[M A_2 B_4\right],\left[M A_3 B_3\right]$ exhibit geometrical isomerism.

Topic: Chemistry, Coordination Compounds, JEE Main
Which one of the following on treatment with $50 \%$ aqueous sodium hydroxide yields the corresponding alcohol and acid?

Options:
A $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}$
B $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHO}$
C

D $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CHO}$

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

Solution:
According to Cannizzaro's reaction.
$2 \mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}+\mathrm{NaOH} \stackrel{50 \% \mathrm{NaOH}}{\rightleftharpoons} \mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}+\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}$
Related Theory
Aldehydes containing no $\alpha$-hydrogen atom on warming with $50 \% \mathrm{NaOH}$ or $\mathrm{KOH}$ undergo disproportionation i.e. self oxidation reduction known as Cannizzaro's reaction.
Caution
Aldehydes that contains alpha hydrogen atoms are not involved in Cannizzaro reaction.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Which one of the following option gives the correct matching of a disease with its causative organism and mode of infection?

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

The correct answer is:
(b)

Solution:
Bacteria like Streptococcus pneumoniae and Haemophilus influenzae are responsible for the disease pneumonia in humans which infect alveoli (air filled sacs) of the lungs. As a result of infection, the alveoli get filled with fluid leading to severe problem in respiration. A healthy person acquires the infection by inhaling the droplets/aerosols released by an infected person or even by sharing glasses and utensils with an infected person.

Topic: Biology, Human Health and Diseases, NEET, NEET 2011 (Mains)
Which one of the following options gives the correct categorisation of six animals according to the type of nitrogenous wastes (A, B, C), they give out?

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

The correct answer is:
(c)

Solution:
On the basis of nitrogenous metabolic waste the animals are of four types :
(i) Ammonotelic Excreted ammonia, e.g., aquatic animals, bony fishes, frog, toad and crocodile, etc.
(ii) Aminotelic Excreted amino acid, e.g., Mollusca and echinoderms.
(iii) Ureotelic Excreted urea, e.g., frog, toad and mammals.
(iv) Uricotelic Excreted uric acid, e.g., insects, land reptiles and birds.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2012 (Mains)
Which one of the following options is correct for the spontaneity of the reaction?

Options:
A ∆G= positive (+Ve); ∆H= positive (+Ve)
B ∆H= positive (+ve); ∆S= negative (-Ve)
C ∆G= negative (-Ve); ∆S= negative (-Ve)
D ∆G= negative (-Ve); ∆S= positive (+Ve)

The correct answer is:
∆G= negative (-Ve); ∆S= positive (+Ve)

Solution:
∆G=∆H-T∆S
Where, ∆G= Gibbs free energy change
∆H= Enthalpy change
T= Temperature in k and
∆S= Entropy change
A reaction is spontaneous, when ∆G is negative. If ∆S is positive, T∆S become positive, so that ∆H-T∆S can become negative.

Topic: Chemistry, Thermodynamics (C), JEE Main
Which one of the following order represents the correct sequence of the increasing basic nature of the given oxides ?

Options:
A
$\mathrm{MgO} < \mathrm{K}_2 \mathrm{O} < \mathrm{Al}_2 \mathrm{O}_3 < \mathrm{Na}_2 \mathrm{O}$
B
$\mathrm{Na}_2 \mathrm{O} < \mathrm{K}_2 \mathrm{O} < \mathrm{MgO} < \mathrm{Al}_2 \mathrm{O}_3$
C
$\mathrm{K}_2 \mathrm{O} < \mathrm{Na}_2 \mathrm{O} < \mathrm{Al}_2 \mathrm{O}_3 < \mathrm{MgO}$
D
$\mathrm{Al}_2 \mathrm{O}_3 < \mathrm{MgO} < \mathrm{Na}_2 \mathrm{O} < \mathrm{K}_2 \mathrm{O}$

The correct answer is:
$\mathrm{Al}_2 \mathrm{O}_3 < \mathrm{MgO} < \mathrm{Na}_2 \mathrm{O} < \mathrm{K}_2 \mathrm{O}$

Solution:
Across a period metallic strength decreases $\&$ down the group it increases

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2011
Which one of the following orders is not in accordance with the property stated against it?

Options:
A $\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$ : bond dissociation energy
B $\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$ : oxidising power
C $\mathrm{HI}>\mathrm{HBr}>\mathrm{HCI}>\mathrm{HF}$; acidic property in water
D $\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$ : electronegativity

The correct answer is:
$\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$ : bond dissociation energy

Solution:
Bond dissociation energy of halogen family decreases down the group as the size of atom increases. The decreasing order for the bond dissociation enthalpy of halogen molecules is
$\mathrm{Cl}_2(57 \mathrm{kcal} / \mathrm{mol})>\mathrm{Br}_2(45.5 \mathrm{kcal} /$ $\mathrm{mol})>\mathrm{F}_2(38 \mathrm{kcal} / \mathrm{mol})>\mathrm{I}_2(35.6$ $\mathrm{kcal} / \mathrm{mol}$ ).
A Caution
A halogen molecule having larger atoms should have low dissociation energy and vice versa. Fluorine is an exception because of interelectronic repulsion is present in small atom of fluorine

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Which one of the following orders of acid strength is correct

Options:
A $\mathrm{RCOOH}\gt\mathrm{HC} \equiv \mathrm{CH}\gt\mathrm{HOH}\gt\mathrm{ROH}$
B $\mathrm{RCOOH}\gt\mathrm{ROH}\gt\mathrm{HOH}\gt\mathrm{HC} \equiv \mathrm{CH}$
C $\mathrm{RCOOH}\gt\mathrm{HOH}\gt\mathrm{ROH}\gt\mathrm{HC} \equiv \mathrm{CH}$
D $\mathrm{RCOOH}\gt\mathrm{HOH}\gt\mathrm{HC} \equiv \mathrm{CH}\gt\mathrm{ROH}$

The correct answer is:
$\mathrm{RCOOH}\gt\mathrm{HOH}\gt\mathrm{ROH}\gt\mathrm{HC} \equiv \mathrm{CH}$

Solution:
Carboxylic acid is stronger than alcohol and water because after removal of proton, carboxylate ion is stabilized by resonance

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Which one of the following ores is best concentrated by froth floatation method?

Options:
A Malachite
B Magnetite
C Siderite
D Galena

The correct answer is:
Galena

Solution:
Froth floatation method is mainly applicable for sulphide ores.
(A) Malachite ores: (Cu(OH)2.CuCO3
(B) Magnetic ore: Fe3O4
(C) Siderite ore: FeCO3
(D) Galena ore: PbS (Sulphide ore)

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
Which one of the following ores is best concentrated by froth - floatation method?

Options:
A
Magnetite

B
Malachite

C
Galena

D
Cassiterite

The correct answer is:
Galena

Solution:
No Solution Available

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2004
Which one of the following ores is best concentrated by froth floatation method?

Options:
A Galena
B Malachite
C Magnetite
D Siderite

The correct answer is:
Galena

Solution:
Froth floatation method is used for concentration of sulphide ores.
Galena â PbS
Malachite is Cu 2 CO 3 2 OH 2
Magnetite is Fe 3 O 4
Siderite is Fe CO 3

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Which one of the following ores is best concentrated by froth flotation method?

Options:
A Galena
B Malachite
C Magnetite
D Siderite

The correct answer is:
Galena

Solution:
Froth flotation method is used for concentration of sulphide ores.
Galena → PbS
Malachite is Cu2CO32OH2
Magnetite is Fe3O4
Siderite is FeCO3

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2016 (03 Apr)
Which one of the following ores is best concentrated by froth-floatation method?

Options:
A Magnetite
B Cassiterite
C Galena
D Malachite

The correct answer is:
Galena

Solution:
Froth floatation process is generally used to concentrate sulphide ore (Galena is PbS)

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Which one of the following ores is known as Malachite:

Options:
A Cu2S
B Cu2O
C Cu(OH)2.CuCO3
D CuFeS2

The correct answer is:
Cu(OH)2.CuCO3

Solution:
Copper glance : Cu2S; Cuprite : Cu2O

Malachite : Cu(OH)2.CuCO3 ; Chalcopyrite : CuFeS2

Malachite is a copper carbonate hydroxide mineral with the formula Cu(OH)2.CuCO3.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2014 (19 Apr Online)
Which one of the following ores is known as Malachite :

Options:
A Cu2S
B Cu2O
C Cu(OH)2.CuCO3
D CuFeS2

The correct answer is:
Cu(OH)2.CuCO3

Solution:
Basic copper carbonate
Cu OH 2 Â· CuCO 3 is Malachite green
[ CuFeS2 - copper pyrite]
[ Cu2S - copper glance ]

Topic: Chemistry, d and f Block Elements, JEE Main
Which one of the following organic compounds decolourizes an alkaline $\mathrm{KMnO}_4$ solution

Options:
A $C S_2$
B $\mathrm{C}_3 \mathrm{H}_6$
C $\mathrm{C}_3 \mathrm{H}_8$
D $\mathrm{CH}_3 \mathrm{OH}$

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

Solution:
$\mathrm{C}_3 \mathrm{H}_6$ is an alkene therefore decolourizes alkaline $\mathrm{KMnO}_4$ solution.

Topic: Chemistry, Hydrocarbons, JEE Main
Which one of the following organisms is correctly matched with its three characteristics?

Options:
A Pea $-\mathrm{C}_3$-pathway, endospermic seed, vexillary aestivation
B Tomato-Twisted aestivation, axile placentation, berry
C Onion-Bulb, imbricate aestivation, axile placentation
D Maize $-\mathrm{C}_3$-pathway, closed vascular bundles, scutellum

The correct answer is:
Onion-Bulb, imbricate aestivation, axile placentation

Solution:
Onion belongs to the family Liliaceae. It is a modified stem which is present in the underground, called bulb. It shows imbricate aestivation in perianth. The gynoecium is in axile placentation. The above characteristics are matched with the characteristics of onion.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2012 (Mains)