datavorous/entrance-exam-dataset huggingface dataset:
To neutralize completely $20 \mathrm{~mL}$ of $0.1 \mathrm{M}$ aqueous solution of phosphorous acid $\left(\mathrm{H}_3 \mathrm{PO}_3\right)$, the volume of $0.1 \mathrm{M}$ aqueous $\mathrm{KOH}$ solution required is

Options:
A
$10 \mathrm{~mL}$

B
$60 \mathrm{~mL}$

C
$40 \mathrm{~mL}$

D
$20 \mathrm{~mL}$

The correct answer is:
$40 \mathrm{~mL}$

Solution:
No. of gm equivalents of phosphorous acid $=\mathrm{No}$. of gm equivalents of $\mathrm{KOH}$
$$
\begin{aligned}
& 20 \times 0.1 \times 2(n=\text { factor })=0.1 \times V \\
& =0.1 \times V
\end{aligned}
$$
$\mathrm{V}=\frac{4}{0.1}=40 \mathrm{ml}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2004
To neutralize completely 20 mL of 0.1 M aqueous solution of phosphorus acid, the volume of 0.1 M aqueous NaOH solution required is

Options:
A 20 mL
B 30 mL
C 40 mL
D 60 mL

The correct answer is:
40 mL

Solution:
H3PO3 is a dibasic acid (containing two ionisable protons attached to O directly)
H3PO3â2H++HPO32-
0.1â M H 3 PO 3 â=â0.2â N H 3 PO 3
0.1âM NaOHâ=â0.1âN NaOH
N1V1=N2V2
NaOHH3PO3
0.1ÃV1=0.2Ã20
V 1 =40âmL
Alternative short method
H3PO3+2NaOHâNa2HPO3+2H2O
MAVAxA=MBVBxB â¦(i)
Here xA and xB are balancing coefficient of acid and base respectively.
Now putting all the values in equation (i) we get
20Ã0.11=0.1ÃVB2
V B =40 ml

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
To observe diffraction the size of an obstacle

Options:
A Should be of the same order as wavelength
B Should be much larger than the wavelength
C Have no relation to wavelength
D Should be exactly $\frac{\lambda}{2}$

The correct answer is:
Should be of the same order as wavelength

Solution:
Should be of the same order as wavelength

Topic: Physics, Wave Optics, JEE Main
To obtain a buffer which should be suitable for maintaining a $p H$ of about $4-5$, we need to have in solution, a mixture of

Options:
A A strong base + its salt with a weak acid
B A weak base + its salt with a strong acid
C A strong acid + its salt with a weak base
D A weak acid + its salt with a strong base

The correct answer is:
A weak acid + its salt with a strong base

Solution:
A weak acid and its salt with a strong base maintain $\mathrm{pH} 4$ - 5

Topic: Chemistry, Ionic Equilibrium, JEE Main
To prepare a pure sample of n-hexane using sodium metal as one reactant, the other reactant will be

Options:
A n-propyl bromide
B Ethyl bromide and n-butyl bromide
C Ethyl chloride and n-butyl chloride
D Methyl bromide and n -pentyl chloride

The correct answer is:
n-propyl bromide

Solution:
According to wurtz reaction.
$2 \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+2 \mathrm{Na} \xrightarrow{\text { ether }}$ $\mathrm{CH}_3\left(\mathrm{CH}_2\right)_4 \mathrm{CH}_3+2 \mathrm{NaBr}$

Topic: Chemistry, Hydrocarbons, JEE Main
To produce 200 zygotes, how many meiotic divisions are required in flowering spermatophytes?

Options:
A 125
B 250
C 300
D 400

The correct answer is:
250

Solution:
200 zygote are produced when 200 male gamete fuses with 200 eggs. Micropore mother cell undergoes meiotic division to produce microspore tetrads that dissociate to form 4 pollen grains. Similarly, Megaspore's mother cell undergoes meiotic division to form 4 megaspores but only one remains functional while the other three degenerate. Hence, to produce 200 male gametes 50 meiotic division is required whereas to produce 200 egg 200 meiotic division is required. So the total number of meiotic divisions required to produce 200 zygotes is 250

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
To produce an instantaneous displacement current of $2 \mathrm{~mA}$ in the space between the parallel plates of a capacitor of capacitance $4 \mu \mathrm{F}$, the rate of change of applied variable potential difference $\left(\frac{d V}{d t}\right)$ must be

Options:
A $200 \mathrm{~V} / \mathrm{s}$
B $400 \mathrm{~V} / \mathrm{s}$
C $800 \mathrm{~V} / \mathrm{s}$
D $500 \mathrm{~V} / \mathrm{s}$

The correct answer is:
$500 \mathrm{~V} / \mathrm{s}$

Solution:
$\begin{aligned} & \text { Displacement current } i_d=c \frac{d V}{d t} \\ & \Rightarrow 2 \times 10^{-3}=4 \times 10^{-6} \times \frac{d V}{d t} \\ & \Rightarrow \frac{2 \times 10^{-3}}{4 \times 10^{-6}}=\frac{d V}{d t} \\ & \Rightarrow \frac{d V}{d t}=500 \mathrm{~V} / \mathrm{s}\end{aligned}$

Topic: Physics, Electromagnetic Waves, JEE Main
To project a body of mass $m$ from earth's surface to infinity, the required kinetic energy is (assume, the radius of earth is $R_E, g=$ acceleration due to gravity on the surface of earth):

Options:
A $2 m g R_E$
B $4 m g R_E$
C $m g R_E$
D $1 / 2 m g R_E$

The correct answer is:
$m g R_E$

Solution:
$\begin{aligned} & \frac{1}{2} \mathrm{mv}_{\mathrm{e}}^2=\frac{\mathrm{GMm}}{\mathrm{R}_{\mathrm{E}}} \\ & \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}_{\mathrm{E}}^2} \\ & \mathrm{~K}=\mathrm{mgR}_{\mathrm{E}}\end{aligned}$

Topic: Physics, Gravitation, JEE Main, JEE Main 2024 (06 Apr Shift 1)
To radiate EM signal of wavelength λ with high efficiency, the antennas should have a minimum size equal to:

Options:
A 2λ
B λ2
C λ4
D λ

The correct answer is:
λ4

Solution:
For efficient radiation and reception, the height of transmitting and receiving antennas should be comparable to a quarter of wavelength of the frequency used. So for efficient transmission of signals of wavelengths λ the minimum length of antenna should be λ4.

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (13 Apr Shift 2)
To raise the temperature of a certain mass of gas by 50 °C at a constant pressure,160 calories of heat is required. When the same mass of gas is cooled by 100°C at constant volume,240 calories of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal)?

Options:
A 5
B 6
C 3
D 7

The correct answer is:
6

Solution:
At constant pressure :

∆Q=nCP∆T ⇒ 160=nCP.50 .....(1)

At constant volume

∆Q=nCv∆T ⇒ 240=nCv.100......(2)

Equation (1) divided by (2)

160240=CPCV·50100

CPCV=43=1+2f

f=6

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2020 (03 Sep Shift 2)
To remove last traces of water from alcohol, the metal used is

Options:
A Sodium
B Potassium
C Calcium
D Aluminium

The correct answer is:
Calcium

Solution:
The metal which is used to remove last traces of water from alcohol is Calcium. Hence, Option "C" is the correct answer.

Topic: Chemistry, s Block Elements, JEE Main
To send 10% of the main current through a moving coil galvanometer of resistance 9 Î©, shunt S required is

Options:
A 81 Ω
B 1 Ω
C 10 Ω
D 9 Î©

The correct answer is:
1 Ω

Solution:
i g Ã G = i - i g Ã S
S=0.10.9ÃG {ig=0.1i}
S = 1 9 Ã 9 = 1 Î©

Topic: Physics, Current Electricity, NEET
To synthesise acetone from a Grignard reagent \( \left(\mathrm{CH}_{3} \mathrm{MgI}\right) \), which of the following is used?

Options:
A Benzonitrile
B Pent-3-ennitrile
C Formonitrile
D Acetonitrile

The correct answer is:
Acetonitrile

Solution:
In Acetonitrile CH3CN, there is an electronegativity difference in between the carbon and nitrogen; hence, the carbon behaves as an electrophile site on which the methyl anion from the Grignard reagent attacks, and forms the ketone as shown below:(here R=CH3, R'=CH3)

Topic: Chemistry, Aldehydes and Ketones, JEE Main
To verify Ohm’s law, a student connects the voltmeter across the battery as shown in the figure. The measured voltage is plotted as a function of the current, and the following graph is obtained:
If Vo is almost zero, identify the correct statement:

Options:
A The emf of the battery is 1.5 V and its internal resistance is 1.5 Ω
B The value of the resistance R is 1.5 Ω
C The potential difference across the battery is 1.5 V when it sends a current of 1000 mA
D The emf of the battery is 1.5 V and the value of R is 1.5 Ω

The correct answer is:
The emf of the battery is 1.5 V and its internal resistance is 1.5 Ω

Solution:
V=E-Ir
when V=V0=0⇒0=E-Ir
∴ E=Ir
when I=0, V=E=1.5 V
∴ r=1.5 Ω.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (12 Apr Shift 1)
To which group and period does the element belong if the electronic configuration of an element in its -2 oxidation state is $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 ?$

Options:
A period 3, group 16
B period 3, group 17
C period 4, group 16
D period 4, group 17

The correct answer is:
period 3, group 16

Solution:
Electronic configuration $=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6$ $=16$ electrons
Sulphur $(Z=16)$ refer to group 16 and period 3 .
Sulphur show -2 oxidation state.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
To which of the following gaseous mixtures is Dalton's law not applicable

Options:
A $\mathrm{Ne}+\mathrm{He}+\mathrm{SO}_2$
B $\mathrm{NH}_3+\mathrm{HCl}+\mathrm{HBr}$
C $\mathrm{O}_2+\mathrm{N}_2+\mathrm{CO}_2$
D $\mathrm{N}_2+\mathrm{H}_2+\mathrm{O}_2$

The correct answer is:
$\mathrm{NH}_3+\mathrm{HCl}+\mathrm{HBr}$

Solution:
$\quad \mathrm{NH}_3$ and $\mathrm{HCl} \& \mathrm{HBr}$ is a reacting gas mixture to produce $\mathrm{NH}_4 \mathrm{Cl} ~\&~ \mathrm{NH}_4 \mathrm{Br}$ so Dalton's law is not applicable.

Topic: Chemistry, States of Matter, JEE Main
To which of the following quantities, the radius of the circular path of a charged particle moving at right angles to a uniform magnetic field is directly proportional?

Options:
A energy of the particle
B magnetic field
C charge of the particle
D momentum of the particle

The correct answer is:
momentum of the particle

Solution:
If velocity of particle, $\mathrm{v}$ is perpendicular to magnetic field, $\mathbf{B}$ i.e. $\theta=90^{\circ},$ then particle will experience maximum magnetic force, i.e. $F_{\max }=q v B .$ This force acts in a direction perpendicular to the motion of charged particle. Therefore the trajectory of the particle is a circle. In this case path of charged particle is circular and magnetic force provides the necessary centripetal force,
i.e.
$$
q v B=\frac{m v^{2}}{r}
$$
$\Rightarrow$ Radius of path, $r=\frac{m v}{q B}$
$$
r=\frac{p}{q B} \quad[\because p=m v]
$$
where, $p=$ momentum of the particle
$\therefore r$ momentum
Hence, option (d) is correct

Topic: Physics, Magnetic Effects of Current, JEE Main
To which of the following the angular velocity of the electron in the $n$ -th Bohr orbit is proportional?

Options:
A $n^{2}$
B $\frac{1}{n^{2}}$
C $\frac{1}{n^{3 / 2}}$
D $\frac{1}{n^{3}}$

The correct answer is:
$\frac{1}{n^{3}}$

Solution:
According to the Bohr's atomic model.
$\text {Angular momentum, } L=m v=\frac{n h}{2 \pi}....(i)$
since, angular velocity $\omega=\frac{v}{r}$
$\Rightarrow v=r \omega$
From Eq. (i), we get
$\begin{aligned}
m(r \omega) r &=\frac{n h}{2 \pi} \\
m \omega r^{2} &=\frac{n h}{2 \pi} \\
\omega &=\frac{n h}{2 \pi m r^{2}}....(ii)
\end{aligned}$
since, the radius of the electron in $n^{t h}$ orbit of
Bohr's atomic model is given as, $r=\frac{n^{2} h^{2} \varepsilon_{0}}{\pi m z e^{2}} \ldots$ (iii)
Squaring the Eq. (iii) and substituting its value in Eq. (ii), we get
$\omega=\frac{n h(\pi n z e)^{2}}{2 \pi m\left(n^{4} h^{4} \varepsilon_{0}^{2}\right)} \Rightarrow \omega \propto \frac{1}{n^{3}}$

Topic: Physics, Atomic Physics, JEE Main
To which of the following types the straight lines represented by $2 x+3 y-7=0$ and $2 x+3 y-5=0$ belong

Options:
A Parallel to each other
B Perpendicular to each other
C Inclined at $45^{\circ}$ to each other
D Coincident pair of straight lines

The correct answer is:
Parallel to each other

Solution:
Let $L_1 \equiv 2 x+3 y-7=0$ and $L_2 \equiv 2 x+3 y-5=0$
Here slope of $L_1=$ slope of $L_2=-\frac{2}{3}$
Hence the lines are parallel.

Topic: Mathematics, Straight Lines, JEE Main
To which type of barriers under innate immunity, do the saliva in the mouth and the tears from the eyes, belong?

Options:
A Physical barriers
B Cytokine barriers
C Cellular barriers
D Physiological barriers

The correct answer is:
Physiological barriers

Solution:
Acid in stomach, saliva in mouth, tears from eyes all prevent microbial growth. They are part of innate immunity and grouped under physiological barriers.

Topic: Biology, Human Health and Diseases, NEET, NEET 2008 (Mains)
Tobacco plants resistant to a nematode have been developed by the introduction of DNA that produced (in the host cells)

Options:
A both sense and antisense RNA
B a particular hormone
C an antifeedant
D a toxic protein

The correct answer is:
both sense and antisense RNA

Solution:
Antisense RNA can be produced by inverting a cDNA copy of an mRNA with respect to the promoter in an expression vector. This yields a full length complementary copy of the mRNA sequence. Antisense RNA molecule are thought to interact with mRNA molecule by base pairing to form double stranded RNA.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2012 (Mains)
T.O.Diener discovered a :

Options:
A Bacteriophage
B Free infection DNA
C Free infectious RNA
D Infectious protein

The correct answer is:
Free infectious RNA

Solution:
No Solution Available

Topic: Biology, Biological Classification, NEET, NEET 2009 (Mains)
Tollen's reagent and Fehling solutions are used to distinguish between

Options:
A
acids and alcohols

B
alkanes and alcohols

C
ketones and aldehydes

D
n-alkaens and branched alkanes

The correct answer is:
ketones and aldehydes

Solution:
All aldehydes and ketones show reaction with Tollen's reagent and Fehling solutions, but ketones do not show this reaction.
Note :- Benzaldehyde do not give reaction with Fehling solution.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2012 (26 May Online)
Toluene can be easily converted into benzaldehyde by which of the following reagents?

Options:
A CrO3 /acetic acid, H3O+
B CrO3 /acetic anhydride, H3O+
C KMnO4/HCl,H3O+
D CO, HCl, Anhyd. AlCl3

The correct answer is:
CrO3 /acetic anhydride, H3O+

Solution:
Toluene is oxidised to benzaldehyde in presence of CrO3 /acetic anhydride, H3O+

Toluene is converted to benzylidene diacetate and hydrolysis reaction of benzylidene diacetate gives benzaldehyde.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2022 (26 Jun Shift 2)
Toluene is nitrated and the resulting product is reduced with tin and hydrochloric acid. The product so obtained is diazotised and then heated with cuprous bromide. The reaction mixture so formed contains

Options:
A
mixture of $o-$ and p-bromotoluenes

B
mixture of $o-$ and p-dibromobenzenes

C
mixture of $\mathrm{o}$ - and p-bromoanilines

D
mixture of 0 - and $\mathrm{m}$-bromotoluenes

The correct answer is:
mixture of $o-$ and p-bromotoluenes

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2008
Toluene on oxidation with dilute $\mathrm{HNO}_3$ and alkaline $\mathrm{KMnO}_4$ gives

Options:
A Benzaldehyde
B Phenol
C Nitrotoluene
D Benzoic acid

The correct answer is:
Benzoic acid

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Toluene reacts with excess of $\mathrm{Cl}_2$ in presence of sunlight to give a product which on hydrolysis followed by reaction with $\mathrm{NaOH}$ gives

Options:
A

B

C

D None of these

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Torques $\tau_1$ and $\tau_2$ are required for a magnetic needle to remain perpendicular to the magnetic fields of $B_1$ and $B_2$ at two different places. The ratio of $B_1: B_2$ is equal to

Options:
A $\tau_2: \tau_1$
B $\tau_1: \tau_2$
C $\frac{\tau_1+\tau_2}{\tau_1-\tau_2}$
D $\frac{\tau_1-\tau_2}{\tau_1+\tau_2}$

The correct answer is:
$\tau_1: \tau_2$

Solution:
Given that torques $\tau_1$ and $\tau_2$ are required for a magnetic needle to remain perpendicular to the magnetic field $B_1$ and $B_2$.
We know that, torque acting on a magnet in uniform magnetic field.
$$
\therefore \quad \tau=M B \sin \theta
$$
where, $\theta=90^{\circ}$ (Given)
Now, $\frac{\tau_1}{\tau_2}=\frac{M B_1 \sin 90^{\circ}}{M B_2 \sin 90^{\circ}}$ Hence, $B_1 / B_2=\tau_1 / \tau_2$

Topic: Physics, Magnetic Effects of Current, AP EAMCET, AP EAMCET 2021 (24 Aug Shift 1)
Total internal reflection of a ray of light is possible when the ($i_c=$ critical angle, $i=$ angle of incidence)

Options:
A Ray goes from denser medium to rarer medium and $i \lt i_c$
B Ray goes from denser medium to rarer medium and $i\gt i_c$
C Ray goes from rarer medium to denser medium and $i\gt i_c$
D Ray goes from rarer medium to denser medium and $i \lt i_c$

The correct answer is:
Ray goes from denser medium to rarer medium and $i\gt i_c$

Solution:
Total internal reflection is a phenomenon when the incident ray experiences complete reflection on a refracting surface. The three conditions in refraction when critical angle is given are,
1. When $i \lt i$, then the incident refracts and the angle it makes with the normal is called the angle of refraction.
2. When $i=i_e$ then the incident ray refracting and coincides with the refracting. The angle of refraction is $90^{\circ}$ in this case.
3. When $i \gt i_0$, then the incident ray is totally reflected back and this incident is termed as total internal reflection.

Topic: Physics, Ray Optics, JEE Main
Total internal reflection of light is possible when light enters from

Options:
A Air to glass
B Vacuum to air
C Air to water
D Water to air

The correct answer is:
Water to air

Solution:
For total internal reflection light must travel from denser medium to rarer medium.

Topic: Physics, Ray Optics, JEE Main
Total normal of aromatic compounds are ?

Options:
A
B
C
D

The correct answer is:

Solution:
The kjeldahl’s method for nitrogen estimation is not applicable to compounds containing nitrogen in the ring (e.g., pyridine, quinoline, etc.) and compounds containing nitrogen directly linked to oxygen atom (e.g. NO2) or another nitrogen atom i.e. azo(—N = N—) compounds since during digestion, there are not completely converted into (NH4)2SO4.

Topic: Chemistry, General Organic Chemistry, JEE Main
Total number of 6- digit numbers in which only and all the five digits 1,3,5,7 and 9 appears, is

Options:
A 126!
B 6!
C 56
D 526!

The correct answer is:
526!

Solution:
As we have to make a six-digit number by using all the digits 1, 3, 5, 7, 9, hence exactly one of the digits is repeating.

For the digit to repeat we have 5C1=5 choices.
And, six digits in which one is repeating can be arranged in 6!2! ways.
Hence, total such numbers =5×6!2!=526!.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2020 (07 Jan Shift 1)
Total number of acidic oxides among N2O3,NO2, N2O,Cl2O7,SO2,CO,CaO,Na2O and NO is _____ .

Options:

The correct answer is:
4

Solution:
Acidic oxides are oxides of non-metals. Thus out of the given compounds, acidic oxides are : N2O3,NO2,Cl2O7,SO2

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2023 (29 Jan Shift 2)
Total number of aromatic compounds among the following compounds is ______.

Options:

The correct answer is:
1

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Total number of aromatic compounds from below is

Options:
A 2
B 3
C 4
D 5

The correct answer is:
3

Solution:
According to the Huckle's rule,

The compound is aromatic which is planar and has the 4n+2π electrons. The compound is aromatic in nature because it is a planar compound with 4n+2π electrons. Aniline, naphthalene and pyridine are aromatic compounds.

Compounds :-

Hence, the three compounds are aromatic in nature.

Topic: Chemistry, General Organic Chemistry, JEE Main
Total number of atoms represented by the compound $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ is

Options:
A $27$
B $21$
C $5$
D $8$

The correct answer is:
$21$

Solution:
Total number of atoms in $\mathrm{CuSO} 4 \cdot 5 \mathrm{H} 2 \mathrm{O}=1+1+4+5 \times 3=21$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Total number of boron atoms in anionic part of borax which is/are sp 3 hybridised?

Options:

The correct answer is:
2

Solution:
Borax is 2Na+[B4O5(OH)4]2-.8H2O

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Total number of Boron atoms in anionic part of borax which participate in back bonding.

Options:
A 3
B 4
C 2
D None of these

The correct answer is:
2

Solution:
Borax 2Na+[B4O5(OH)4]2-.8H2O
sp2 hybridised boron atoms only participate in pπ-pπ back bonding as they have vacant p-orbitals.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Total number of chain isomers possible for C 7 H 16 are

Options:

The correct answer is:
4

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
Total number of compounds with Chiral carbon atoms from following is

CH3-CH2-CHNO2-COOH

CH3-CH(I)-CH2-NO2

CH3-CH2-CHBr-CH2-CH3

CH3-CH2-CH(OH)-CH2OH

Options:

The correct answer is:
5

Solution:
The chiral carbon atoms are carbon atoms that are attached to four different atoms or groups of atoms losses all symmetry, that are placed at the corners of a tetrahedron. The configuration of such a tetrahedral unit is chiral. The structure may exist in either a right-handed configuration or a left-handed configuration. This type of configurational stereoisomers is termed enantiomorphism.

Amongst the given compounds, these compounds have chiral carbon in it.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Total number of covalent bonds in C3O2 is x and y is the number of sp-hybridised atom. Find the sum of x + y?

Options:

The correct answer is:
11

Solution:
O=Csp=Csp=Csp=O
There are eight covalent bonds in C3O2, so x = 8 and y = 3 as 3 C-atoms are sp-hybridised.
So x + y = 8 + 3 = 11.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Total number of deactivating groups in aromatic electrophilic substitution reaction among the following is

Options:

The correct answer is:
2

Solution:
Atoms or groups that make the benzene molecule more reactive by increasing the ring's electron density are called activating groups.

An atom or group that makes the benzene molecule less reactive by removing electron density from the ring acts as a deactivating group. Deactivating groups direct incoming electrophiles to the meta position. Deactivating groups decreases the rate of reaction relative to H. If the atom of the group attached to benzene ring has pi bond then that group act as deactivating group.

Hence, -CN, -COCH3 are deactivating groups.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (01 Feb Shift 1)
Total number of electrons present in \(\left(\pi^*\right)\) molecular orbitals of \(\mathrm{O}_2, \mathrm{O}_2{ }^{+}\) and \(\mathrm{O}_2^{-}\) is _______.

Options:

The correct answer is:
6

Solution:
$\begin{aligned}
& \mathrm{O}_2(16 \mathrm{e}):\left(\sigma_{1 \mathrm{~s}}\right)^2\left(\sigma_{1 \mathrm{~s}}^*\right)^2\left(\sigma_{2 \mathrm{~s}}\right)^2\left(\sigma_{2 \mathrm{~s}}^*\right)^2 \\
& \left(\sigma_{2 \mathrm{p}}\right)^2\left[\left(\pi_{2 \mathrm{p}}\right)^2=\left(\pi_{2 \mathrm{p}}\right)^2\right],\left[\left(\pi_{2 \mathrm{p}}^*\right)^1=\left(\mathrm{m}_{2 \mathrm{p}}^*\right)^1\right]
\end{aligned}$
Number of $\mathrm{e}^{-}$present in $\left(\pi^*\right)$ of $\mathrm{O}_2=2$
Number of $\mathrm{e}^{-}$present in $\left(\pi^*\right)$ of $\mathrm{O}_2^{+}=1$
Number of $\mathrm{e}^{-}$present in $\left(\pi^*\right)$ of $\mathrm{O}_2^{-}=3$
So total $\mathrm{e}^{-}$in $\left(\pi^*\right)=2+1+3=6$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Total number of electrophiles present in the following are
NO 2 + ,  SO 3 ,  CN − ,  F − ,  CCl 4 ,  NH 3 , :CCl 2

Options:

The correct answer is:
3

Solution:
NO 2 + ,  SO 3 ,  :CCl 2 are electrophiles.

Topic: Chemistry, General Organic Chemistry, JEE Main
Total number of elements which are present in a row on the periodic table between those elements, whose wavelength of KÎ± lines are equal to 250 and 179 pm are (Rydberg constant = 1.097 Ã 107 m-1)

Options:

The correct answer is:
3

Solution:
Wavelength of KÎ± line is given by
Î»=43RZ-12
For Î», between 250 and 179 pm three elements are possible
( Z = 24, 25 and 26 ).

Topic: Physics, Atomic Physics, JEE Main