datavorous/entrance-exam-dataset huggingface dataset:
Based on molecular orbital theory, the bond order depends on the number of electrons in the bonding and antibonding molecular orbitals. Identify the correct statements about bond order ?

Options:
A Can have a negative quantity
B Has always an integral value
C Is a non zero quantity
D Can assume any positive or integral or fractional value including zero

The correct answer is:
Can assume any positive or integral or fractional value including zero

Solution:
bond order of a molecule is given by Half the difference between the number of electrons in bonding and antibonding electrons, it can have any positive or integral or fractional value including zero

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Based on the equation:
$$
\Delta \mathrm{E}=-2.0 \times 10^{-18} \mathrm{~J}\left(\frac{1}{\mathrm{n}_2^2}-\frac{1}{\mathrm{n}_1^2}\right)
$$
the wavelength of the light that must be absorbed to excite hydrogen electron from level $n=1$ to level $\mathrm{n}=2$ will be: $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{Js}, \mathrm{C}=3 \times 10^8 \mathrm{~ms}^{-1}\right)$

Options:
A
$1.325 \times 10^{-7} \mathrm{~m}$

B
$1.325 \times 10^{-10} \mathrm{~m}$

C
$2.650 \times 10^{-7} \mathrm{~m}$

D
$5.300 \times 10^{-10} \mathrm{~m}$

The correct answer is:
$1.325 \times 10^{-7} \mathrm{~m}$

Solution:
$$
\begin{aligned}
&\Delta E=-2.0 \times 10^{-18} \times\left(\frac{1}{2^2}-\frac{1}{1^2}\right) \\
&=-2.0 \times 10^{-18} \times \frac{-3}{4} \\
&=1.5 \times 10^{-18} \\
&\Delta E=\frac{h c}{\lambda} \\
&\lambda=\frac{h c}{\Delta E}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.5 \times 10^{-18}} \\
&=1.325 \times 10^{-7} \mathrm{~m}
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2014 (11 Apr Online)
Based on the first law of thermodynamics, which one of the following is correct?

Options:
A For an isochoric process, ΔE = −q
B For an adiabatic process, ΔE = −w
C For an isothermal process, q=W
D For a cyclic process, q=−W

The correct answer is:
For a cyclic process, q=−W

Solution:
(a) For isochoric process, ÎVâ=â0
Wâ=âpÎVâ=â0
â´âÎEâ=âq
(b) For adiabatic process, q=0
ÎEâ=âW
(c) For isothermal process, ÎTâ=â0
and ÎEâ=â0
qâ=ââW
(d) For cyclic process, state functions like
ÎEâ=â0
q = −W

Topic: Chemistry, Thermodynamics (C), JEE Main
Based on the given figure, the number of correct statement/s is/are

A. Surface tension is the outcome of equal attractive and repulsion forces acting on the liquid molecule in bulk.

B. Surface tension is due to uneven forces acting on the molecules present on the surface.

C. The molecule in the bulk can never come to the liquid surface.

D. The molecules on the surface are responsible for vapour pressure if the system is a closed system.

Options:

The correct answer is:
2

Solution:
Surface tension is the outcome of unequal attractive and repulsion forces acting on the liquid molecule in bulk. Surface tension is due to uneven forces acting on the molecules present on the surface. The molecule in the bulk can come to the liquid surface- due to brownian movement and convection. The molecules on the surface are responsible for vapour pressure if the system is a closed system.

Thus, B and D options are correct.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Based upon VSEPR theory, match the shape (geometry) of the molecules in List-I with the molecules in List - II and select the most appropriate option.

List-I
(Shape)	List-II
(Molecules)

A	T-shaped	I.	XeF4

B	Trigonal planar	II.	SF4

C	Square planar	III.	ClF3

D	See-saw	IV.	BF3

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

The correct answer is:
A-III,B-IV,C-I,D-II

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2022 (27 Jun Shift 1)
BaTi [Si3O9] is a class of

Options:
A orthosilicate
B cyclic silicate
C chain silicate
D sheet silicate

The correct answer is:
cyclic silicate

Solution:
The primary unit of BaTi( Si3O9) or (Si3O96-) is SiO32- (as n = 3). Thus, it is formed by sharing of two oxygen per Si atom and results in the formation of cyclic silicate.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
B-cells and T-cells, responsible for acquired immune response in the body, are formed in the

Options:
A thymus and bone marrow respectively.
B bone marrow.
C thymus only.
D lymph nodes and bone marrow respectively.

The correct answer is:
bone marrow.

Solution:
The bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced. The thymus is a lobed organ located near the heart and beneath the breastbone. The thymus is quite large at the time of birth but keeps reducing in size with age and by the time puberty is attained it reduces to a very small size. Both bone-marrow and thymus provide micro-environments for the development and maturation of T-lymphocytes.

Topic: Biology, Human Health and Diseases, NEET
Beats are the result of

Options:
A Diffraction
B Destructive interference
C Constructive and destructive interference
D Superposition of two waves of nearly equal frequency

The correct answer is:
Superposition of two waves of nearly equal frequency

Solution:
Superposition of two waves of nearly equal frequency

Topic: Physics, Waves and Sound, JEE Main
\(\begin{array}{c|c|c} \hline \text {Gas } & \mathrm{a}\left(\mathrm{atm}^2 \mathrm{~mol}^{-2}\right) & \mathrm{b}\left(\mathrm{Lmol}^{-1}\right) \\ \hline \mathrm{W} & 4.0 & 0.027 \\ X & 8.0 & 0.030 \\ Y & 6.0 & 0.032 \\ Z & 12.0 & 0.027 \\ \hline \end{array}\)
Which one of these gases has the highest critical temperature

Options:
A $W$
B $X$
C $Y$
D $Z$

The correct answer is:
$Z$

Solution:
For $Z$ gas of given gases, critical temperature is highest
$T_c=\frac{8 a}{27 R b} \Rightarrow T_c=\frac{8 \times 12}{27 \times .0821 \times .027}=1603.98 \mathrm{~K}$

Topic: Chemistry, States of Matter, JEE Main
\( \begin{array}{l} \mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O} \stackrel{\Delta, 120^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{A} \\ \mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O} \stackrel{\Delta, 200^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{B} \end{array} \) In the above reaction, \( \mathrm{A} \) and \( \mathrm{B} \), respectively, are:

Options:
A \( \mathrm{CaSO}_{4}, \mathrm{CaSO}_{4} \cdot \frac{1}{2} \mathrm{H}_{2} \mathrm{O} \)
B \( \mathrm{CaSO}_{4} \cdot \frac{1}{2} \mathrm{H}_{2} \mathrm{O}, \mathrm{CaSO}_{4} \)
C \( \mathrm{CaSO}_{4} \cdot \frac{1}{2} \mathrm{H}_{2} \mathrm{O} \), Gypsum.
D Gypsum, \( \mathrm{CaSO}_{4} \).

The correct answer is:
\( \mathrm{CaSO}_{4} \cdot \frac{1}{2} \mathrm{H}_{2} \mathrm{O}, \mathrm{CaSO}_{4} \)

Solution:
CaSO4·2H2O at120oC CaSO4·12H2O+112H2O

The commercial name of the compound Calcium sulphate hemihydrate, i.e., Plaster of Paris.

CaSO4·2H2O →∆, 200°C anhydrous CaSO4+2H2O

Anhydrous calcium sulphate formed this way is commonly called dead burnt plaster.

Topic: Chemistry, s Block Elements, JEE Main
Beilstein test is used for the estimation of which one of the following elements?

Options:
A
$\mathrm{S}$

B
$\mathrm{Cl}$

C
$\mathrm{C}$ and $\mathrm{H}$

D
$\mathrm{N}$

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

Solution:
Beilstein's test is used for halogens.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2012 (19 May Online)
Benedict solution provides

Options:
A $\mathrm{Ag}^{+}$
B $L i^{+}$
C $\mathrm{Cu}^{+2}$
D $\mathrm{Ba}^{+2}$

The correct answer is:
$\mathrm{Cu}^{+2}$

Solution:
Benedict solution provides $\mathrm{Cu}^{+2}$ ions

Topic: Chemistry, Biomolecules, JEE Main
Benzamide on reaction with POCl3 gives

Options:
A aniline
B chlorobenzene
C benzyl amine
D benzonitrile

The correct answer is:
benzonitrile

Solution:
Benzamide on treatment with POCl3 gives benzonitrile (phenyl cyanide) because in this reaction POCl3 acts as a dehydrating agent and on dehydration of benzamide, benzonitrile is formed.
C6H5CONH2→ΔPOCl3C6H5CN+H2O

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Benzene and napthalene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm of Hg and 32.06 mm of Hg respectively. What will be the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of naphthalene?

Options:
A 0.0675
B 0.675
C 0.35
D 0.5

The correct answer is:
0.675

Solution:
P°AXAP°BXB=XA'XB'
50.71807832.06100128=XA'1-XA'
XA'=0.675

Topic: Chemistry, Solutions, JEE Main
Benzene and toluene form nearly ideal solutions. At $20^{\circ} \mathrm{C}$, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The parial vapour pressure of benzene at $20^{\circ} \mathrm{C}$ for a solution containing $78 \mathrm{~g}$ of benzene and $46 \mathrm{~g}$ of toluene in torr is

Options:
A 50
B 25
C 37.5
D 53.5

The correct answer is:
50

Solution:
Amount of benzene, \(n_1=\frac{m_1}{M_1}=\frac{78 \mathrm{~g}}{78 \mathrm{gmol}^{-1}}=1 \mathrm{~mol}\)
Amount of toluene, \(n_2=\frac{m_2}{M_2}=\frac{46 \mathrm{~g}}{92 \mathrm{gmol}^{-1}}=0.5 \mathrm{~mol}\)
Amount fraction of benzene, \(x_1=\frac{n_1}{n_1+n_2}=\frac{1}{1.5}\)
Partial pressure of benzene, \(p_1=x_1 p_1^*=\frac{1}{1.5}(75\) Torr \()=50\) Torr

Topic: Chemistry, Solutions, JEE Main
Benzene and toluene form nearly ideal solutions. At $20^{\circ} \mathrm{C}$, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at $20^{\circ} \mathrm{C}$ for a solution containing $78 \mathrm{~g}$ of benzene and $46 \mathrm{~g}$ of toluene in torr is

Options:
A
50

B
25

C
37.5

D
53.5

The correct answer is:
50

Solution:
$\mathrm{P}_{\mathrm{B}}=\mathrm{P}_{\mathrm{B}}^{\circ} \times \mathrm{B}=75 \times \frac{1}{1.5}=50$ torr

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2005
Benzene can be obtained in the reaction

Options:
A Ethene + 1, 3-butadiene
B Trimerisation of ethyne
C Reduction of $\mathrm{PhCHO}$
D All of these

The correct answer is:
Trimerisation of ethyne

Solution:
Benzene can be obtained by polymerisation of acetylene.

Topic: Chemistry, Hydrocarbons, JEE Main
Benzene cannot be iodinated with I2 directly. However, in presence of oxidants such as HNO3 , iodination is possible. The electrophiles formed in the case is

Options:
A I+
B I-
C I +δ ...... O +δ H 2 +
D I −δ ...... O −δ H 2 +

The correct answer is:
I+

Solution:
In the presence of oxidising agents such as HNO3 direct, iodination of benzene is possible. The electrophiles formed in this case is I+ .
H +  +  HNO 3  +  1 2 I 2 → I + Iodine cation  +  NO 2  +  H 2 O

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Benzene carbaldehyde is reacted with conc. NaOH solution to give the products AB . The product A can be used as a food preservative and the product B is an aromatic hydroxy compound, where −OH group is linked to sp 3 -hybridised C-atom next to benzene ring. The products A and B respectively are

Options:
A Sodium benzoate and phenol
B Sodium benzoate and phenyl methanol
C Sodium benzoate and cresol
D Sodium benzoate and picric acid

The correct answer is:
Sodium benzoate and phenyl methanol

Solution:
When benzene carbaldehyde reacts with NaOH, it proceeds via Cannizzaro reaction. It is a disproportionation reaction in which half of the molecules of aldehyde are oxidised to carboxylate state and other half are reduced to alcohol state.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Benzene diazonium chloride on reaction with aniline in the presence of dilute hydrochloric acid gives:

Options:
A
B
C
D

The correct answer is:

Solution:
Aniline reacts with a mixture of sodium nitrite and hydrochloric acid and forms benzene diazonium chloride.The structure of Benzene diazonium chloride is

Benzene diazonium chloride reacts with aniline in presence of dilute hydrochloric acid and undergoes coupling reaction and produces stable azo products. In this reaction, benzene diazonium chloride acts as an electrophile in coupling with activated aniline. In this type of reaction, substitution takes place at the para position. So, the reaction Benzene diazonium chloride with aniline in presence of dilute hydrochloric acid is as follows:

Topic: Chemistry, Amines, JEE Main, JEE Main 2019 (12 Apr Shift 2)
Benzene is obtained by

Options:
A Substitution of three acetylene molecules
B Addition of three $\mathrm{C}_2 \mathrm{H}_2$ molecules
C Polymerisation of three $\mathrm{C}_2 \mathrm{H}_2$ molecules
D Condensation of three $\mathrm{C}_2 \mathrm{H}_2$ molecules

The correct answer is:
Addition of three $\mathrm{C}_2 \mathrm{H}_2$ molecules

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Benzene is prepared in laboratory from which one of the following compounds

Options:
A $\mathrm{C}_6 \mathrm{~N}_5 \mathrm{~N}_2 \mathrm{Cl}$
B $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}$
C $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}$
D $\mathrm{C}_6 \mathrm{H}_5 \mathrm{SO}_3 \mathrm{H}$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}$

Solution:
$\underset{\text { Sodiumbenzoate }}{\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}}+\mathrm{NaOH} \xrightarrow[\text { Benzene }]{\mathrm{CaO}} \underset{6}{ } \mathrm{C}_6 \mathrm{H}_6+\mathrm{Na}_2 \mathrm{CO}_3$

Topic: Chemistry, Hydrocarbons, JEE Main
Benzene is the polymer of

Options:
A Methane
B Ethane
C Ethylene
D Ethyne

The correct answer is:
Ethyne

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Benzene on nitration gives nitrobenzene in presence of HNO3 and H2SO4 mixture, where:

Options:
A both H2SO4 and HNO3 act as a bases
B HNO3 acts as an acid and H2SO4 acts as a base
C both H2SO4 and HNO3 act as an acids
D HNO3 acts as a base and H2SO4 acts as an acid

The correct answer is:
HNO3 acts as a base and H2SO4 acts as an acid

Solution:
Reagent for nitration of Benzene

H2SO4Acid+HNO3Base⇄HSO4⊖+H⊕2NO3

H2 NO3⇄H2O+N⊕O2

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (20 Jul Shift 2)
Benzene on treatment with a mixture of conc. $\mathrm{HNO}_3$ and conc. $\mathrm{H}_2 \mathrm{SO}_4$ at $100^{\circ} \mathrm{C}$ gives

Options:
A Nitrobenzene
B m-dinitrobenzene
C p-dinitrobenzene
D o-dinitrobenzene

The correct answer is:
m-dinitrobenzene

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Benzene react with nitrating mixture (conc. HNO3 and conc H2SO4. ) to produce nitrobenzene. In the nitrating mixture, HNO3 acts as a

Options:
A Base
B Acid
C Reducing agent
D Catalyst

The correct answer is:
Base

Solution:
NO 2 + generated as follows:
H 2 SO 4  +  HNO 3 → HSO 4 −  +  H 2 NO 3 +
O 2 N− OH 2 ⊕ → NO 2 + +H 2 O (nitronium ion)
Here, HNO 3 accept H ⊕ , hence, act as base.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Benzene reacts with $\mathrm{CH}_3 \mathrm{Cl}$ in the presence of anhydrous $\mathrm{AlCl}_3$ to form -

Options:
A Xylene
B Toluene
C Chlorobenzene
D Benzylchloride

The correct answer is:
Toluene

Solution:

Topic: Chemistry, Hydrocarbons, NEET, NEET 2009 (Mains)
Benzene reacts with $\mathrm{CH}_3 \mathrm{COCl}$ in the presence of $\mathrm{AlCl}_3$ to give

Options:
A $\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}$
B $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCl}$
C $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3$
D $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCH}_3$

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

Solution:
Benzene reacts with $\mathrm{CH}_3 \mathrm{COCl}$ in the presence of $\mathrm{AlCl}_3$ to give $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCH}_3$.
This reaction is called Friedel-Crafts Acylation reaction.

Topic: Chemistry, Hydrocarbons, JEE Main
Benzene reacts with chlorine to form benzene hexachloride in presence of

Options:
A Nickel
B $\mathrm{AlCl}_3$
C Bright sunlight
D Zinc

The correct answer is:
Bright sunlight

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Benzoic acid can be prepared by reacting phenyl magnesium bromide with

Options:
A N, N-dimethylformamide
B carbon dioxide
C formaldehyde
D ethyl chloroformate

The correct answer is:
carbon dioxide

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Benzoic acid gives benzene on being heated with $X$ and phenol gives benzene on being heated with $Y$. Therefore $X$ and $Y$ are respectively

Options:
A Sodalime and copper
B $\mathrm{Zn}$ dust and $\mathrm{NaOH}$
C $Z n$ dust and sodalime
D Sodalime and zinc dust

The correct answer is:
Sodalime and zinc dust

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Benzoic acid gives benzene on heating with \( \mathrm{X} \) and phenol gives benzene on heating with \( \mathrm{Y} \). Therefore, \( \mathrm{X} \) and \( \mathrm{Y} \), respectively are

Options:
A soda lime and copper.
B \( \mathrm{Zn} \) dust and \( \mathrm{NaOH} \).
C Zn dust and soda lime.
D soda lime and Zn dust.

The correct answer is:
soda lime and Zn dust.

Solution:
When sodium salt of a carboxylic acid is treated with NaOH in the presence of CaO, it undergoes decarboxylation via formation of intermediate carbanion.

When phenol reacts with Zn dust, a reduction reaction will take place. Here, zinc metal is the source of electrons, and it is the electron transfer reduction.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Benzoic acid is a organic compound. Its crude sample can be purified by crystallization from hot water. What characteristic differences in the properties of benzoic acid and the impurity make this process of purification suitable?

Options:
A Benzoic acid can be purified by hot water because impurities present in benzoic acid are either insoluble in water or more soluble in water to such an extent that they remain in solution as the mother liquor upon crystallization.
B Benzoic acid has high temperature.
C Benzoic acid forms a colored solution in hot water, allowing for easy separation.
D Benzoic acid and the impurity have identical properties, but the process relies on temperature control.

The correct answer is:
Benzoic acid can be purified by hot water because impurities present in benzoic acid are either insoluble in water or more soluble in water to such an extent that they remain in solution as the mother liquor upon crystallization.

Solution:
Correct Option is : (A)
Benzoic acid can be purified by hot water because impurities present in benzoic acid are either insoluble in water or more soluble in water to such an extent that they remain in solution as the mother liquor upon crystallization.

Topic: Chemistry, General Organic Chemistry, NEET
Benzoin is

Options:
A $\alpha$-hydroxy aldehyde
B $\alpha$-hydroxy ketone
C compound containing an aldehyde and a ketonic group
D $\alpha, \beta$ - unsaturated acid

The correct answer is:
$\alpha$-hydroxy ketone

Solution:
Benzoin is a $\alpha$-hydroxy ketone.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Benzoyl chloride is prepared from benzoic acid by

Options:
A Cl2, hv
B SO2Cl2
C SOCl2 , pyridine
D Cl2,H2O

The correct answer is:
SOCl2 , pyridine

Solution:
SOCl2 easily donates Cl and gets to SO2.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Benzyl alcohol and sodium benzoate is obtained by the action of sodium hydroxide on benzaldehyde. This reaction is known as

Options:
A Perkin's reaction
B Cannizzaro's reaction
C Sandmeyer's reaction
D Claisen condensation

The correct answer is:
Cannizzaro's reaction

Solution:
Benzaldehyde on treatment with $50 \%$ aqueous or ethanolic alkali solution undergoes Cannizzaro's reaction like $\mathrm{HCHO}$ (no $\alpha$-hydrogen atom) i.e., one molecule is oxidised and one is reduced with the formation of benzoic acid and benzyl alcohol respectively.
$2 \mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO} \xrightarrow{\mathrm{NeOH}} \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}+\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}$

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Benzyl alcohol is obtained from benzaldehyde by

Options:
A Fittig's reaction
B Cannizaro's reaction
C Kolbe's reaction
D Wurtz's reaction

The correct answer is:
Cannizaro's reaction

Solution:
Cannizzaro's reaction is given by those aldehyde who doesn't have a carbon. Half of them gets oxidized and half of them gets reduced.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Benzyl chloride when oxidised by $\mathrm{pb} \left(\mathrm{NO}_3\right)_2$ gives

Options:
A Benzoic acid
B Benzaldehyde
C Benzene
D None

The correct answer is:
Benzaldehyde

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Benzyl isocyanide can be obtained by:

(A)

(B)

(C)

(D)

Choose the correct answer from the options given below:

Options:
A A and D
B Only B
C A and B
D B and C

The correct answer is:
A and B

Solution:
When benzyl bromide is treated with AgCN, it forms benzyl isocyanide. This reaction is followed by the SN2 mechanism.

Benzylamine on heating with chloroform and ethanolic potassium hydroxide forms Benzyl isocyanides. This reaction is known as carbylamine reaction or isocyanide test.

2° amine does not give carbylamine reaction.

So, the correct option is C.

Topic: Chemistry, Amines, JEE Main, JEE Main 2023 (30 Jan Shift 1)
BeO reacts with HF in presence of ammonia to give A which on thermal decomposition produces B and ammonium fluoride. Oxidation state of Be in A is

Options:

The correct answer is:
2

Solution:
BeO on reaction with HF in the presence of ammonia, NH42BeF4 is formed which on heating forms BeF2 along with ammonium fluoride.

BeO+2NH3+4HF→NH42BeF4+H2O

Compound A, is NH42BeF4

NH42BeF4→heatBeF2+2NH4F

Compound B is BeF2

In NH42BeF4 the oxidation state of Be is +2.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Bernoulli's equation is conservation of

Options:
A energy
B momentum
C angular momentum
D mass

The correct answer is:
energy

Solution:
Bernoulli's equation is conservation of energy.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Bernoulli's equation is important in the field of

Options:
A electrical circuits.
B magnetism.
C flow of fluids.
D photoelectric effect.

The correct answer is:
flow of fluids.

Solution:
Bernoulli's equation is an indispensable tool for the analysis of flow of liquids.

Bernoulli's equation states that for two points 1 and 2 lying on a streamlined flow, they can be related in terms of the pressure difference between them, their relative height and velocities of the fluid at those points. The formula is given below:

P1-P2=12ρ(v22-v12)+ρg(h2-h1)

where,

P1-P2 is the pressure difference between points 1 and 2,

v1 is the velocity of the fluid at point 1,

v2 is the velocity of the fluid at point 2 and

h2-h1 is the height difference between them.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Bernoulli's theorem is based on the conservation of

Options:
A mass
B momentum
C energy
D All of the above

The correct answer is:
energy

Solution:
According to Bernoulli's theorem,
$$
p+\frac{1}{2} \rho V^2+\rho g h=\text { constant }
$$
where, $p$ is pressure energy per unit volume, $\frac{1}{2} \rho V^2$ is kinetic energy per unit volume and $\rho g h$ is potential energy per unit volume and for a system this value remains constant. Hence, Bernoulli's theorem is based on conservation of energy.

Topic: Physics, Mechanical Properties of Fluids, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 1)
Beryllium and aluminium exhibit many properties which are similar. But the two elements differ in

Options:
A
exhibiting maximum covalency in compound

B
exhibiting amphoteric nature in their oxides

C
forming covalent halides

D
forming polymeric hydrides

The correct answer is:
exhibiting maximum covalency in compound

Solution:
$\because$ Be \& Al have diagonal relationship \& so possess similar properties but Be cannot form polymeric hydrides

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2004
Beryllium differs from rest of the members of its family (Group-IIA) in many ways. The reason for this is its

Options:
A Small size and higher electronegativity
B Small size and lower electronegativity
C Large size and lower iopisation energy
D Large size and largest ionic radius

The correct answer is:
Small size and higher electronegativity

Solution:
(i) Small atomic size.
(ii) High electronegativity
(iii) Absence of $d$ orbitals

Topic: Chemistry, s Block Elements, JEE Main