datavorous/entrance-exam-dataset huggingface dataset:
Considering the above reaction, X and Y respectively are:

Options:
A
B
C
D

The correct answer is:

Solution:
Initially aniline forms benzene diazonium chloride on reaction with nitrous acid which our major product known as X. Which later on reaction with N,N-dimethyl aniline Shows coupling reaction which form major product Y

Reaction is shown below:

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Considering the basic strength of amines in an aqueous solution, which one has the smallest pKb value?

Options:
A CH3NH
B CH3NH2
C CH33N
D C6H5NH2

The correct answer is:
CH3NH

Solution:
Since in aromatic amines the lone pair undergoes delocalisation, aliphatic amines are more basic than aromatic amines.
Ph-NH2<R-NH2
In an aliphatic amine, due to the +I effect of -CH3 the order will be 3°>2°>1°.
In an aqueous solution, H-bonding takes place. So, the order will be 1°>2°>3°.
However, the resultant basic order is due to a mixture of both the effects(+I and hydrogen bond). So, the order is 2°>3°>1°.
So, the final basic order is

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2014 (06 Apr)
Considering the longitudinal section of a root, starting from the tip, identify the correct order for the arrangement of the four different zones?

Options:
A Root cap zone, region of cell division, region of cell enlargement, region of cell maturation.
B Root cap zone, region of cell division, region of cell maturation, region of cell enlargement.
C Region of cell division, region of cell enlargement, region of cell maturation, root cap zone.
D Region of cell division, region of cell maturation, region of cell enlargement, root cap zone.

The correct answer is:
Root cap zone, region of cell division, region of cell enlargement, region of cell maturation.

Solution:
The root system of a plant develops from the hypocotyl of the embryo. The root shows the following four regions which usually merge into one another, without distinct boundaries between them. The four zones, successively from the apex to the base are:
Root cap zone: The apex of each root is covered by a cushion of thin-walled cells known as root cap.
Region of cell division: It lies just behind the root cap. It is the main growing region of the root where active cell divisions take place.
Region of elongation: The region of elongation is responsible for growth in length of the root.
Region of maturation: Epidermal cells of this region give out small, thin, cylindrical unicellular outgrowth, known as root hairs. These are the main absorbing organs of the root.

Topic: Biology, Morphology of Flowering Plants, NEET
Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation cos-1x-2sin-1x=cos-12x is equal to

Options:
A 0
B 1
C 12
D -12

The correct answer is:
0

Solution:
Given,

cos-1x-2sin-1x=cos-12x

⇒cos-1x-2π2-cos-1x=cos-12x

⇒cos-1x-π+2cos-1x=cos-12x

⇒3cos-1x=π+cos-12x ...1

Taking cos both side we get,

⇒cos3cos-1x=cosπ+cos-12x

⇒coscos-14x3-3x=-coscos-12x

⇒4x3-3x=-2x

⇒4x3=x

⇒x=0,±12

All will satisfy the original equation,

So, sum of all the solution will be 0+12-12=0

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2022 (28 Jul Shift 1)
"Constant environments promote niche specialization and lead to greater species diversity." This statement is true for

Options:
A Tundra region
B Temperate region
C Tropical region
D All of these

The correct answer is:
Tropical region

Solution:
The process of formation of new species takes time. The environment should be relatively constant to have more species formation. Temperate regions were subjected to frequent glaciations, are more seasonal and unpredictable. Tropical regions have remained relatively undisturbed for millions of years and thus, had a long evolutionary time for species diversification. Tropical environments are less seasonal, relatively more constant and predictable. Such constant environments promote niche specialisation and lead to a greater species diversity. There is more solar energy available in the tropics, which contributes to higher productivity; this in turn might contribute indirectly to greater diversity.

Topic: Biology, Organisms and Populations, NEET
Contents of three bottles were found to react
(i)Neither with Fehling's solution nor with Tollen's reagent
(ii)Only with Tollen's reagent but not with Fehling's solution
(iii)With both Tollen's reagent and Fehling's solution.
If they contained either ethanal (acetaldehyde) or propanone (acetone) or benzal (benzaldehyde), which bottle contained which

Options:
A In (i) benzal, in (ii) ethanal and in (iii) propanone
B In (i) benzal, in (ii) propanone and in (iii) ethanal
C In (i) propanone, in (ii) benzal and in (iii) ethanal
D In (i) propanone, in (ii) ethanal and in (iii) benzal

The correct answer is:
In (i) propanone, in (ii) benzal and in (iii) ethanal

Solution:
(i) Neither with Fehling's solution nor with Tollen's reagent it means it should be Ketone
(ii) Only with Tollen's reagent but not with Fehling's solution, it means it should be aromatic aldehyde
(iii) With both Tollen's reagent and Fehling's solution. it means it should be aliphatic aldehyde.
So, we have In (I) propanone, in (II) benzal, and in (III) ethanal

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Contractile vacuole in protozoan Amoeba is meant for

Options:
A respiration
B excretion
C locomotion
D osmoregulation

The correct answer is:
osmoregulation

Solution:
Contractile vacuole is the clear rounded, pulsating body present in the posterior part of endoplasm of Amoeba. It is found only in fresh water forms and is mainly concerned with osmoregulation, ie, removal of excess of water.
Plasmalemma, a trilaminar and selectively permeable membrane covering the body, functions both in excretion (ammonia) and respiration in Amoeba.Pseudopodia (false feet) are finger-like and blunt-extensions of cell body in protozoans. In Amoeba, the type of pseudopodium found in lobopodium, composed of both ectoplasm and endoplasm. It is temporary locomotory structure which is also meant for feeding in Amoeba. Locomotion of Amoeba is known as'amoeboid movement' (Sol-gel theory).

Topic: Biology, Cell - The Unit of Life, JIPMER, JIPMER 2009
Contrapositive of the statement :

'If a function f is differentiable at a, then it is also continuous at a', is

Options:
A If a function f is continuous at a, then it is not differentiable at a.
B If a function f is not continuous at a, then it is not differentiable at a.
C If a function f is not continuous at a. then it is differentiable at a.
D If a function f is continuous at a, then it is differentiable at a.

The correct answer is:
If a function f is not continuous at a, then it is not differentiable at a.

Solution:
By using definition, we have;

Contrapositive of p⇒q is ~q⇒~p.

Which means contrapositive of given statement is;

"If a function f is not continuous at a, then it is not differentiable at a."

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2020 (04 Sep Shift 2)
Contrapositive of the statement "If two numbers are not equal, then their squares are not equal". is :

Options:
A If the squares of two numbers are not equal, then the numbers are equal.
B If the squares of two numbers are equal, then the numbers are not equal.
C If the squares of two numbers are equal, then the numbers are equal.
D If the squares of two numbers are not equal, then the numbers are not equal.

The correct answer is:
If the squares of two numbers are equal, then the numbers are equal.

Solution:
Contrapositive of "If $A$ then $B^{\prime \prime}$ is "If $\sim \mathrm{B}$ then $\sim A^{\prime \prime}$. Hence contrapositive of "If two numbers are not equal, then their squares are not equal" is "If squares of two numbers are equal, then the two numbers are equal".

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2019 (11 Jan Shift 2)
Contribution of CO2 and CH4 in the greenhouse effect is respectively.

Options:
A CO2 60%, CH4 20%
B CO2 20%, CH4 60%
C CO2 80%, CH420%
D CO2 70%, CH4 30%

The correct answer is:
CO2 60%, CH4 20%

Solution:
CO2 and CH4 are two important green house gases which are major cause of global warming. These gases maintains earth temperature by absorbing heat radiation. Increase in concentration these gases in past have resulted in rise of earth temperature by 0.6 degree Celsius.

Topic: Biology, Environmental Issues, NEET
Conversion of benzene diazonium chloride to chlorobenzene is an example of which of the following reactions?

Options:
A
Claisen

B
Friedel-craft

C
Sandmeyer

D
Wurtz

The correct answer is:
Sandmeyer

Solution:
The above reaction is known as Sandmayer's reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2014 (12 Apr Online)
Convex lens made up of glass Î¼g = 1.5 and radius of curvature R is dipped into the water. Its focal length will be (Refractive index of water = 43)

Options:
A 4R
B 2R
C R
D R2

The correct answer is:
4R

Solution:
Here, μg=1.5=32, μw ᷿=43
R1=R,⇒R2=-R
From, lens maker's formula
1f=μ-11R1-1R2where μ=μLμM
∴1faₐ=μgμaₐ-11R-1-R
⇒1fa=1R
Also
1fw ᷿=μgμwₐ-11R-1-R
Or 1fw=14R
Or fw ᷿=4R

Topic: Physics, Ray Optics, JEE Main
Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will

Options:
A
increase

B
decrease

C
remain same

D
decrease for some, while increase for others

The correct answer is:
increase

Solution:
The molecular kinetic energy increases, and so temperature increases

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2002
Coordinate covalent compounds are formed by

Options:
A Transfer of electrons
B Sharing of electrons
C Donation of electrons
D None of these process

The correct answer is:
Donation of electrons

Solution:
Correct option is 3. Donation of electrons

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Coordination compound have great importance in biological systems. In this context which of the following statements is incorrect?

Options:
A
Chlorophylls are green pigments in plants and contains calcium

B
Carboxypeptidase $-\mathrm{A}$ is an enzyme and contains zinc

C
Cyanocobalamin is $\mathrm{B}_{12}$ and contains cobalt

D
Haemoglobin is the red pigment of blood and contains iron

The correct answer is:
Chlorophylls are green pigments in plants and contains calcium

Solution:
Chlorophyll contains magnesium instead of calcium

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2004
Coordination isomerism is caused by the interchange of ligands between the

Options:
A Cis and Trans structure
B Complex cation and complex anion
C Inner sphere and outer sphere
D Low oxidation and higher oxidation states

The correct answer is:
Cis and Trans structure

Solution:
Co-ordination isomerism is caused by the interchange of ligands between cis and trans structure.

Topic: Chemistry, Coordination Compounds, JEE Main
COP-21 is also known as

Options:
A The Earth summit
B The Paris agreement
C The Montreal protocol
D The Kyoto protocol

The correct answer is:
The Paris agreement

Solution:
COP stands for Conference of Parties, an annual meeting of all nations that make up the United Nations Framework on Climate Change involving 195 nations in total. This is the 21st meeting (thus COP21), with the first occurring in Berlin in 1995. COP - 21 was held between 30 November to 11 December 2015 in Paris, France. The main purpose is this meeting is to hold the increase in global average temperature to well below 2Â°C.

Topic: Biology, Environmental Issues, NEET
Copper becomes green when exposed to moist air for a long period. This is due to:

Options:
A
the formation of a layer of cupric oxide on the surface of copper.

B
the formation of a layer of basic carbonate of copper on the surface of copper.

C
the formation of a layer of cupric hydroxide on the surface of copper.

D
the formation of basic copper sulphate layer on the surface of the metal.

The correct answer is:
the formation of basic copper sulphate layer on the surface of the metal.

Solution:
Copper when exposed to moist air having $\mathrm{CO}_2$. It gets superficially coated with a green layer of basic carbonate $\mathrm{CuCO}_3$. $\mathrm{Cu}$ $(\mathrm{OH})_2$.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2014 (12 Apr Online)
Copper becomes green when exposed to moist air for a long period. This is due to :

Options:
A The formation of a layer of cupric hydroxide on the surface of copper.
B The formation of a layer of basic carbonate of copper on the surface of copper.
C The formation of basic copper sulphate layer on the surface of the metal.
D The formation of a layer of cupric oxide on the surface of copper.

The correct answer is:
The formation of a layer of basic carbonate of copper on the surface of copper.

Solution:
Due to formation of Basic copper carbonate CuCO3.Cu(OH)2 in atmosphere.

Topic: Chemistry, d and f Block Elements, JEE Main
Copper crystallises in fcc with a unit length of $361 \mathrm{~pm}$. What is the radius of copper atom ?

Options:
A
$157 \mathrm{~pm}$

B
$128 \mathrm{~pm}$

C
$108 \mathrm{~pm}$

D
$181 \mathrm{~pm}$

The correct answer is:
$128 \mathrm{~pm}$

Solution:
For FCC,
$r=\frac{\sqrt{2} a}{4}=\frac{a}{2 \sqrt{2}}=0.3535 a$
given $a=361 \mathrm{~pm}$
$\begin{aligned}
r & =0.3535 \times 361 \\
& =128 \mathrm{~pm}
\end{aligned}$

Topic: Chemistry, Solid State, JEE Main, JEE Main 2013 (25 Apr Online)
Copper crystallizes in a face-centred cubic lattice with a unit cell length of $361 \mathrm{pm}$. What is the radius of copper atom in pm?

Options:
A 108
B 128
C 157
D 181

The correct answer is:
128

Solution:
$\mathrm{Cu} \rightarrow$ fcc lattice
$$
\begin{aligned}
&4 r=\sqrt{2} a\\
&\mathbf{r}=\frac{1}{2 \sqrt{2}} \mathbf{a}\\
&r=\frac{1}{2 \sqrt{2}} \times 361=128
\end{aligned}
$$

Topic: Chemistry, Solid State, NEET, NEET 2009 (Mains)
Copper crystallizes in fcc with a unit cell length of $361 \mathrm{pm}$. What is the radius of copper atom ?

Options:
A
$108 \mathrm{pm}$

B
$127 \mathrm{pm}$

C
157 pm

D
$181 \mathrm{pm}$

The correct answer is:
$127 \mathrm{pm}$

Solution:
For FCC,
$\sqrt{2} a=4 r$ (the atoms touches each other along the face- diagonal)
$$
\begin{aligned}
& r=\frac{\sqrt{2} a}{4}=\frac{\sqrt{2} \times 361}{4} \\
& =127 \mathrm{pm}
\end{aligned}
$$

Topic: Chemistry, Solid State, JEE Main, JEE Main 2009
Copper is a divalent metal. The value of its electrochemical equivalent is $3.29 \times 10^{-4} \mathrm{~g} .$ Its atomic mass is

Options:
A $31.74 \mathrm{~g}$
B $63.5 \mathrm{~g}$
C $126.9 \mathrm{~g}$
D $15.87 \mathrm{~g}$

The correct answer is:
$63.5 \mathrm{~g}$

Solution:
Electrochemical equivalent, $Z=\frac{E}{96500}$ $\begin{array}{lr}\text { or } & 3.29 \times 10^{-4}=\frac{E}{96500} \\ \therefore \quad E=31.74\end{array}$
Atomic mass $=2 \times E$ $[\because$ Cu is divalent.]
$$
=2 \times 31.74=63.49 \mathrm{~g}
$$

Topic: Chemistry, Redox Reactions, JEE Main
Copper is extracted from copper pyrites ore by heating in a blast furnace. The method is based on the principle that

Options:
A Copper has more affinity for oxygen than sulphur at high temperature.
B Iron has more affinity for oxygen than copper at high temperature.
C Sulphur has less affinity for oxygen at high temperature.
D Copper has less affinity for oxygen than sulphur at high temperature.

The correct answer is:
Iron has more affinity for oxygen than copper at high temperature.

Solution:
Iron has greater affinity for oxygen than copper. Thus, the copper oxide reacts with iron sulphide to from iron oxide and reproduce copper sulphide.
FeS + Cu 2 O ⟶ FeO + Cu 2 S

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Copper pyrite ore is concentrated by:

Options:
A electromagnetic method
B gravity method
C froth floatation process
D all the above

The correct answer is:
froth floatation process

Solution:
Copper pyrite is also known as chalcopyrite. Chalcopyrite is a natural sulphide of Cu and Fe(CuFeS2) . Also froth floatation process is mainly used for sulphide ores.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
Copper sulphate solution (250 mL) was electrolysed using a platinum anode and a copper cathode. A constant current of 2 mA was passed for 16 min . It was found that after electrolysis the absorbance of the solution was reduced to 50% of its original value. Concentration of copper sulphate in the solution to begin with is x×10-5. Find the valu of x

Options:

The correct answer is:
8

Solution:
The number of faraday's passed = 2 Ã 1 0 - 3 Ã 1 6 Ã 6 0 9 6 5 0 0
= 1.99 Ã 1 0 - 5
â number of gram equivalent of Cu 2 + deposited as Cu(s)
= 1.99 Ã 1 0 - 5
â number of moles of Cu 2 + deposited
= 1.99 2 Ã 10 - 5 â 10 - 5
Absorbance is directly proportional to Cu 2 + .
No. of moles of Cu2+ initially present = 2 x 10-5 moles
Cu 2 + Initial = 2 Ã 1 0 - 5 Ã 1 0 0 0 2 5 0
= 8 x10-5M

Topic: Chemistry, Electrochemistry, JEE Main
Copper sulphate solution does not react with

Options:
A Zinc
B Iron
C Silver
D All of these

The correct answer is:
Silver

Solution:
CuSO4will react with elements placed above it in electrochemical series and it would not react with elements places below it in electrochemical series.
∴ CuSO4 reacts with Zn and Fe placed above it.
CuSO4 + Fe → FeSO4 + Cu
CuSO4 + Zn → ZnSO4 + Cu
∴ CuSO4 does not react with Ag, placed below it.
CuSO4 + Ag → no reaction.

Topic: Chemistry, Electrochemistry, JEE Main
Copper wire test for halogens is known as

Options:
A
Duma's Test

B
Beilstein's Test

C
Liebig's Test

D
Lassigne's Test

The correct answer is:
Beilstein's Test

Solution:
Beilstein's test : Organic compounds containing halogens when heated over $\mathrm{Cu}$ wire loop give blue or green colour flame due to formation of volatile copper halides.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2012 (07 May Online)
Cornea transplant in humans is almost never rejected. This is because

Options:
A It is a non-living layer
B Its cells are least penetrable by bacteria
C It has no blood supply
D It is composed of enucleated cells

The correct answer is:
It has no blood supply

Solution:
Cornea can be easily transplanted because it has no blood supply \& therefore, does not involve immune response.

Topic: Biology, Human Health and Diseases, NEET, NEET 2008 (Mains)
Corpora quadrigemina is found in ...................... portion of .................... .

Options:
A Ventral, Forebrain
B Ventral, Midbrain
C Dorsal, Forebrain
D Dorsal, Midbrain.

The correct answer is:
Dorsal, Midbrain.

Solution:
The midbrain is located between the thalamus/hypothalamus of the forebrain and pons of the hindbrain. A canal called the cerebral aqueduct passess through the midbrain. The dorsal portion of the midbrain consists mainly of four round swellings (lobes) called corpora quadrigemina.

Topic: Biology, Neural Control and Coordination, NEET
Corpus luteum in pregnancy has a long life. However, if fertilisation does not take place, it remains active only for 10-12 days. Explain.

Options:

None

Solution:
This is because of a neural signal given by the maternal endometrium to its hypothalamus in presence of a zygote to sustain the gonadotropin (LH) secretion, so as to maintain the corpus luteum as long as the embryo remains there. In the absence of a zygote, therefore, the corpus luteum can not be maintained longer.

Topic: Biology, Human Reproduction
Correct acidic strength of given acids is :-

Options:
A I > II > III
B II > III > I
C II > I > III
D III > II > I

The correct answer is:
II > III > I

Solution:
CH 3 â + I - C O | | - OH HO - C O | | â - I - C O | | - OH HO - C O | | - â - I CH 2 - CH 2 - C O | | - OH I II III
Acidic strength ∝ - I
∝ 1 + I
Order of acidic strength : II > III > I

Topic: Chemistry, General Organic Chemistry, JEE Main
Correct Bernoulli's equation is (symbols have their usual meaning) :

Options:
A $P+m g h+\frac{1}{2} m v^2=$ constant
B $P+\rho g h+\frac{1}{2} \rho v^2=$ constant
C $P+\rho g h+\rho v^2=$ constant
D $P+\frac{1}{2} \rho g h+\frac{1}{2} \rho v^2=$ constant

The correct answer is:
$P+\rho g h+\frac{1}{2} \rho v^2=$ constant

Solution:
$\mathrm{P}+\rho \mathrm{gh}+\frac{1}{2} \rho \mathrm{V}^2=$ constant

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Correct formula of potassium ferrocyanide is

Options:
A $\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$
B $\mathrm{K}_2\left[\mathrm{Fe}(\mathrm{CN})_6\right] \mathrm{H}_2 \mathrm{O}$
C $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$
D None of these

The correct answer is:
$\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$

Solution:
Potassium ferrocyanide is a well-known compound with the formula $\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$. It consists of a ferric iron $(\mathrm{Fe}^{2+})$ center surrounded by six cyanide ligands. Thus, the correct formula is $\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$.

Topic: Chemistry, Coordination Compounds, JEE Main
Correct gas equation is

Options:
A $\frac{V_1 T_2}{P_1}=\frac{V_2 T_1}{P_2}$
B $\frac{P_1 V_1}{P_2 V_2}=\frac{T_1}{T_2}$
C $\frac{P_1 T_2}{V_1}=\frac{P_2 V_2}{T_2}$
D $\frac{V_1 V_2}{T_1 T_2}=P_1 P_2$

The correct answer is:
$\frac{P_1 V_1}{P_2 V_2}=\frac{T_1}{T_2}$

Solution:
$\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T} \therefore \frac{P_1 V_1}{P_2 V_2}=\frac{T_1}{T_2}$

Topic: Chemistry, States of Matter, JEE Main
Correct increasing order of first ionistion potential is

Options:
A $\mathrm{Na} \lt \mathrm{Mg}\gt\mathrm{Al} \lt \mathrm{Si}$
B $\mathrm{Na} \lt \mathrm{Mg} \lt \mathrm{Al} \lt \mathrm{Si}$
C $\mathrm{Na}\gt\mathrm{Mg}\gt\mathrm{Al}\gt\mathrm{Si}$
D $\mathrm{Na} \lt \mathrm{Mg} \lt \mathrm{Al}\gt\mathrm{Si}$

The correct answer is:
$\mathrm{Na} \lt \mathrm{Mg}\gt\mathrm{Al} \lt \mathrm{Si}$

Solution:
$\mathrm{Na} \lt \mathrm{Mg}\gt\mathrm{Al} \lt \mathrm{Si}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Correct IUPAC name for
is:

Options:
A 1-Ethoxy-1-ethyl-1-aminopropane
B 1-Ethoxy-1 amino-1-ethylpropane
C 1-Ethoxy-2-butanol
D 2-Ethoxybutan-2-amine

The correct answer is:
2-Ethoxybutan-2-amine

Solution:
It's correct IUPAC name is 2-Ethoxybutan-2-amine

Topic: Chemistry, General Organic Chemistry, JEE Main
Correct IUPAC name for is:

Options:
A 1-Ethoxy-1-ethyl-1-aminopropane
B 1-Ethoxy-1-amino-1-ethylpropane
C 1-Ethoxy-2-butanol
D 2-Ethoxybutan-2-amine

The correct answer is:
2-Ethoxybutan-2-amine

Solution:
Its correct IUPAC name is 2-Ethoxybutan-2-amine

Topic: Chemistry, General Organic Chemistry, JEE Main
Correct order of bond length is

Options:
A $\mathrm{CO}_3^{2-}\gt\mathrm{CO}_2\gt\mathrm{CO}$
B $\mathrm{CO}_2\gt\mathrm{CO}\gt\mathrm{CO}_3^{2-}$
C $\mathrm{CO}\gt\mathrm{CO}_2\gt\mathrm{CO}_3^{2-}$
D None of these

The correct answer is:
$\mathrm{CO}_3^{2-}\gt\mathrm{CO}_2\gt\mathrm{CO}$

Solution:
From valency bond theory, bond order in $\mathrm{CO}$, i.e. $: \overline{\mathrm{C}} \equiv \stackrel{\rightharpoonup}{\mathrm{O}}$ : is 3, that of $\mathrm{O}=\mathrm{C}=\mathrm{O}$ is 2 while that of $\mathrm{CO}_3^{2-}$ ion is 1.33. Since the bond length increases as the bond order decreases, i.e. $\mathrm{CO}_1 \lt \mathrm{CO}_2 \lt \mathrm{CO}_3^{2-}$.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Correct order of Pauling's electronegativity

Options:
A H > B > C
B B > C > H
C C > B > H
D C > H > B

The correct answer is:
C > H > B

Solution:
Electronegativity values H→2.1 , B→2.0 , C→2.5

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main