datavorous/entrance-exam-dataset huggingface dataset:
The de-Broglie wavelength of a ball of mass 120 g moving at a speed of 20 m s-1 is (Planck's constant h=6.6Ã10-34 J s )

Options:
A 3.5 Ã 10-34 m

B 2.8Ã10-34 m

C 1.2Ã10-34 m

D 2.1Ã10-34 m

The correct answer is:
2.8Ã10-34 m

Solution:
de-Broglie wavelength, λ=hmv
=6.6×10-34120×10-3×20
=2.75×10-34 m

Topic: Physics, Dual Nature of Matter, JEE Main
The de-Broglie wavelength of a particle having kinetic energy E is λ. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?

Options:
A E
B 7E9
C 16E9
D E9

The correct answer is:
7E9

Solution:
Recall the formula of de-Broglie wavelength, λ=hp, here p is the linear momentum of the particle, now use the relation between linear momentum and kinetic energy, p=2Em⇒E∝1λ2⇒E=kλ2

and E'=16k9λ2⇒ΔE=7E9.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The de-Broglie wavelength of a particle of mass $6.63 \mathrm{~g}$ moving with a velocity of $100 \mathrm{~ms}^{-1}$ is:

Options:
A $10^{-31} \mathrm{~m}$
B $10^{-25} \mathrm{~m}$
C $10^{-33} \mathrm{~m}$
D $10^{-35} \mathrm{~m}$

The correct answer is:
$10^{-33} \mathrm{~m}$

Solution:
$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$
where, $\lambda=$ de-Broglie wavelength
$\mathrm{m}=$ mass of particle
$\mathrm{v}=$ velocity of particle
$\lambda=\frac{6.625 \times 10^{-34} \times 1000}{6.63 \times 100}=10^{-33} \mathrm{~m}$

Topic: Physics, Dual Nature of Matter, JEE Main
The de-Broglie wavelength of a particle of mass $6.63 \mathrm{~g}$ moving with a velocity of $100 \mathrm{~ms}^{-1}$ is:

Options:
A
$10^{-33} \mathrm{~m}$

B
$10^{-35} \mathrm{~m}$

C
$10^{-31} \mathrm{~m}$

D
$10^{-25} \mathrm{~m}$

The correct answer is:
$10^{-33} \mathrm{~m}$

Solution:
De Broglie wavelength $(\lambda)=\frac{h}{m v}$
$$
\frac{6.63 \times 10^{-34}}{6.63 \times 10^{-3} \times 100}=10^{-33} \mathrm{~m}
$$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2014 (12 Apr Online)
The de-Broglie wavelength of a particle with mass $1 \mathrm{gm}$ and velocity $100 \mathrm{~m} / \mathrm{sec}$ is

Options:
A $6.63 \times 10^{-33} \mathrm{~m}$
B $6.63 \times 10^{-34} \mathrm{~m}$
C $6.63 \times 10^{-35} \mathrm{~m}$
D $6.65 \times 10^{-35} \mathrm{~m}$

The correct answer is:
$6.63 \times 10^{-33} \mathrm{~m}$

Solution:
$\lambda=\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{10^{-3} \times 100}=6.63 \times 10^{-33} \mathrm{~m}$

Topic: Chemistry, Structure of Atom, JEE Main
The de-Broglie wavelength of a proton (charge $=16 . \times 10^{-19} \mathrm{C}$, mass $\left.=16 . \times 10^{-27} \mathrm{~kg}\right)$ accelerated through a potential difference of $1 \mathrm{kV}$ is

Options:
A $600 Å$
B $09 . \times 10^{-12} \mathrm{~m}$
C $7 Å$
D $0.9 \mathrm{~nm}$

The correct answer is:
$09 . \times 10^{-12} \mathrm{~m}$

Solution:
According to de-Broglie hypothesis
$$
\begin{aligned}
\lambda &=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}=\frac{h}{\sqrt{2 m q V}} \\
\therefore \quad \lambda &=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times\left(1.6 \times 10^{-27}\right)\left(1.6 \times 10^{-19}\right) \times 1000}} \\
&=\frac{6.6 \times 10^{-34}}{7.16 \times 10^{-22}} \\
&=0.9 \times 10^{-12} \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, JEE Main
The de-Broglie wavelength of an electron moving with a velocity $c / 2$ ( $c=$ velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is

Options:
A $1: 4$
B $1: 2$
C $1: 1$
D $2: 1$

The correct answer is:
$1: 2$

Solution:
de-Broglie wavelength
$$
\lambda=\frac{h}{m v}
$$
Here, $\lambda_{e}=\frac{h}{m_{e} \frac{c}{2}}$ and $\lambda_{p}=\frac{h}{m_{p} c}$
Given, $\quad \lambda_{e}=\lambda_{B}$
So, $\quad \frac{h}{m_{e} \frac{c}{2}}=\frac{h}{m_{p} c}$
$\frac{m_{e}}{m_{p}}=2$
Ratio of KE
$$
\begin{array}{l}
\frac{K_{e}}{K_{p}}=\frac{\frac{1}{2} m_{e} v_{e}^{2}}{\frac{1}{2} m_{p} v_{p}^{2}} \\
\frac{K_{e}}{K_{p}}=\frac{1}{2}
\end{array}
$$

Topic: Physics, Dual Nature of Matter, JEE Main
The de-Broglie wavelength of an electron, Î±- particle and proton all having the same kinetic energy is respectively given as Î»e, Î»Î± and Î»p. Then which of the following is not true?

Options:
A λe<λp
B λp>λα
C λe>λα
D λα<λp<λe

The correct answer is:
λe<λp

Solution:
de-Broglie wavelength
λ=hmv=h2mE ⇒λ∝1m
∴ λe∝1me, λα ∝1mα and λp∝1mp
As we know that me<mp<mα
So, λe >λp>λα
Or λe>λα or λp>λα or λe>λp

Topic: Physics, Dual Nature of Matter, JEE Main
The de-Broglie wavelength of an α-particles at a voltage V is
(Given that α -particle has 2 units positive charge and 4 units mass)

Options:
A λ= 12.3 V A o
B λ= 0.286 V A o
C Î»= 0.101 V A o
D λ= 0.856 V A o

The correct answer is:
λ= 0.101 V A o

Solution:
The de-Broglie wavelength associated with the charged particle as
Î»= h mv
1 2 mv 2 â=âK.E
1 2 mv 2 âÃâ2mâ=â2m K.E =â m 2 v 2
2mÂ· K.E â=âmv
Î» = h 2 m K.E
= h 2 m Î± Â· 2 eV
= 6.626 Ã 1 0 - 3 4 2 Ã 1.66 Ã 1 0 - 2 7 Ã 4 Ã 1.6 Ã 1 0 - 1 9 Ã 2 Ã V
= 6.626 Ã 1 0 - 3 4 V Ã 1 0 - 2 3 Ã 6.519
= 1.018 Ã 1 0 - 1 1 V
= 0.1018 V   A o

Topic: Chemistry, Structure of Atom, JEE Main
The de-Broglie wavelength of the electron in the ground state of the hydrogen atom is ..... (radius of the first orbit of hydrogen atom $=0.53 Å$ ).

Options:
A $1.67 \mathrm{~A}$
B $3.33 Å$
C $1.06 Å$
D $0.53 Å$

The correct answer is:
$3.33 Å$

Solution:
According to Bohr's quantisation of angular momentum
$$
m v r=\frac{n h}{2 \pi}
$$
or
$\frac{\mathrm{h}}{\mathrm{mv}}=\frac{2 \pi \mathrm{r}}{\mathrm{n}} \quad \text{...(i)}$
de- Broglie wavelength
$$
\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad \text{...(ii)}
$$
From Eqs. (i) and (ii), we get
Wavelength $\lambda=\frac{2 \pi \mathrm{r}}{\mathrm{n}}$
$$
\begin{aligned}
&=\frac{2 \times \pi \times 0.53 Å}{1} \\
&=3.33 Å
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, JEE Main
The de-Broglie's wavelength of an electron in the \(4^{\text {th }}\) orbit is \(\qquad\) \(\pi \mathrm{a}_0 \cdot\left(\mathrm{a}_0=\right.\) Bohr's radius \()\)

Options:

The correct answer is:
8

Solution:
$\begin{aligned} & 2 \pi \mathrm{r}_{\mathrm{n}}=\mathrm{n} \lambda_{\mathrm{d}} \\ & 2 \pi \mathrm{a}_0 \frac{\mathrm{n}^2}{\mathrm{Z}}=\mathrm{n} \lambda_{\mathrm{d}} \\ & 2 \pi \mathrm{a}_0 \frac{4^2}{1}=4 \lambda_{\mathrm{d}} \\ & \lambda_{\mathrm{d}}=8 \pi \mathrm{a}_0\end{aligned}$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2024 (04 Apr Shift 1)
The de-Broglie's wavelength of electron present in first Bohr orbit of ' $\mathrm{H}$ ' atom is:

Options:
A
$4 \times 0.529 Å$

B
$2 \pi \times 0.529 Å$

C
$\frac{0.529}{2 \pi} Å$

D
$0.529 Å$

The correct answer is:
$2 \pi \times 0.529 Å$

Solution:
First Bohr orbit of $\mathrm{H}$ atom has radius
$$
\mathrm{r}=0.529 Å
$$
Also, the angular momentum is quantised.
$$
\begin{aligned}
&\mathrm{mvr}=\frac{\mathrm{h}}{2 \pi} \\
&2 \pi \mathrm{r}=\frac{\mathrm{h}}{\mathrm{mv}}=\lambda \\
&\lambda=2 \pi \times 0.529 Å
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The decahydrate form of sodium carbonate i.e. washing soda on standing in air effloresces and crumbles to powder. The number of water molecule(s) present in the compound formed is:

Options:

The correct answer is:
1

Solution:
Hydrated crystals of sodium carbonate exposed to air it losses it is 9 molecules of water of crystallization leaving only powder sodium carbonate with one molecule of water
The process of losing water of crystallization is called efflorescence
Na2CO3.10H2Osdecahydrate→effloresces-9H2ONa2CO3.H2Osmonohydrate

Topic: Chemistry, s Block Elements, JEE Main
The decay constant for a radioactive nuclide is 1.5×10-5 s-1. Atomic weight of the substance is 60 g mole-1, NA=6×1023. The activity of 1.0 µg of the substance is ________ ×1010 Bq.

Options:

The correct answer is:
15

Solution:
It is given that

λ=1.5×10-5 s-1

Number of moles is 1×10-660=10-76

Number of atoms=number of moles ×NA

N0=10-76×6×1023=1016

Hence, at t=0, A=A0=N0λ, the activity can be given by

A=N0λe-λt⇒A=1016×1.5×10-5×1=15×1010 Bq

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The decay constant of a radio isotope is $\lambda$. If $A_1$ and $\mathrm{A}_2$ are its activities at times $\mathrm{t}_1$ and $\mathrm{t}_2$ respectively, the number of nuclei which have decayed during the time $\left(\mathrm{t}_1-\mathrm{t}_2\right)$

Options:
A $\mathrm{A}_1 \mathrm{t}_1-\mathrm{A}_2 \mathrm{t}_2$
B $A_1-A_2$
C $\frac{\left(A_1-A_2\right)}{\lambda}$
D $\lambda\left(A_1-A_2\right)$

The correct answer is:
$\frac{\left(A_1-A_2\right)}{\lambda}$

Solution:
$$
\begin{aligned}
\mathrm{A}_1 & =\lambda \mathrm{N}_1 \\
\mathrm{~A}_2 & =\lambda \mathrm{N}_2 \\
\mathrm{~N}_1-\mathrm{N}_2 & =\left[\frac{\mathrm{A}_1-\mathrm{A}_2}{\lambda}\right]
\end{aligned}
$$

Topic: Physics, Nuclear Physics, NEET, NEET 2010 (Mains)
The decay constants of a radioactive substance for $\alpha$ and $\beta$ emission are $\lambda_\alpha$ and $\lambda_\beta$ respectively. If the substance emits $\alpha$ and $\beta$ simultaneously, then the average half life of the material will be

Options:
A
$\frac{2 T_\alpha T_\beta}{T_\alpha+T_\beta}$

B
$T_\alpha+T_\beta$

C
$\frac{T_\alpha T_\beta}{T_\alpha+T_\beta}$

D
$\frac{1}{2}\left(T_\alpha+T_\beta \quad\right)$

The correct answer is:
$\frac{T_\alpha T_\beta}{T_\alpha+T_\beta}$

Solution:
$T_{a v}=\frac{T_\alpha T_\beta}{T_\alpha+T_\beta}$
If $\alpha$ and $B$ are emitted simultaneously.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2012 (19 May Online)
The decay of a proton to neutron is :

Options:
A not possible as proton mass is less than the neutron mass
B possible only inside the nucleus
C not possible but neutron to proton conversion is possible
D always possible as it is associated only with β+ decay

The correct answer is:
possible only inside the nucleus

Solution:
It is possible only inside the nucleus and not otherwise as it can not decay to a state of greater mass. Inside a nucleus however, the situation is different, as the binding energy of a nucleus provides the necessary energy to compensate for the neutron-proton mass difference.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2021 (18 Mar Shift 2)
The decay profiles of three radioactive species $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are given below :
These profiles imply that the decay constants $\mathrm{k}_{\mathrm{A}}, \mathrm{k}_{\mathrm{B}}$ and $\mathrm{k}_{\mathrm{C}}$ follow the order

Options:
A $k_{A}>k_{B}>k_{C}$
B $k_{A}>k_{C}>k_{B}$
C $k_{B}>k_{A}>k_{C}$
D $k_{C}>k_{B}>k_{A}$

The correct answer is:
$k_{C}>k_{B}>k_{A}$

Solution:
$\mathrm{C}_{\mathrm{t}}=\mathrm{C}_{0} \mathrm{e}^{-\mathrm{kt}}$

Topic: Chemistry, Chemical Kinetics, JEE Main
The deceleration of a car traveling on a straight highway is a function of its instantaneous velocity $v$ given by $\omega=a \sqrt{v}$, where $a$ is a constant. If the initial velocity of the car is $60 \mathrm{~km} / \mathrm{h}$, the distance of the car will travel and the time it takes before it stops are

Options:
A $\frac{2}{3} m_1 \frac{1}{2} s$
B $\frac{3}{2 a} m, \frac{1}{2 a} s$
C $\frac{3 a}{2} m, \frac{a}{2} s$
D $\frac{2}{3 a} m, \frac{2}{a} s$

The correct answer is:
$\frac{2}{3 a} m, \frac{2}{a} s$

Solution:
Deceleration $\omega=-a \sqrt{v}$.
But $\quad \omega=\frac{d v}{d t} \quad \Rightarrow \frac{d v}{d t}=-a \sqrt{v}$
$$
\begin{gathered}
\frac{-d v}{\sqrt{V}}=a \cdot d t \Rightarrow \int_{v_1}^0 \frac{d v}{\sqrt{V}}=\int a d t \\
|2 \sqrt{v}|_{v_1}^0=a t \quad \Rightarrow t=\frac{2}{a} \sqrt{v_3 S}
\end{gathered}
$$
Again,
$$
\begin{aligned}
\frac{-d v}{d t} & =a \sqrt{v} \Rightarrow \frac{d v}{d x} \cdot \frac{d x}{d t}=-a \sqrt{v} \\
\frac{d v}{d x} \cdot v & =-a \sqrt{v} \Rightarrow d v \sqrt{v}=-a \cdot d x \\
\int_{v_0}^0 \sqrt{v} d v & =-a \int_0^s d s
\end{aligned}
$$
After solving, we get
$$
s=\frac{2}{3 a} \cdot v_0^{\frac{3}{2}}
$$

Topic: Physics, Motion In One Dimension, JEE Main
The decimal equivalent of the binary number $(11010.101)_2$ is

Options:
A 9.625
B 25.265
C 26.625
D 26.265

The correct answer is:
26.625

Solution:
$$
\text { Hints : }(11010.101)=0 \times 2^0+1 \times 2^1+0 \times 2^2+1 \times 2^3+1 \times 2^4+1 \times 2^{-1}+0 \times 2^{-2}+1 \times 2^{-3}=2+8+16+\frac{1}{2}+\frac{1}{8}=26.625
$$

Topic: Physics, Mathematics in Physics, JEE Main
The decimal number equivalent to a binary number 1011001 is

Options:
A 13
B 17
C 89
D 178

The correct answer is:
89

Solution:
Given, binary number $=1011001$ Its equivalent decimal number
$\begin{array}{l}
=2^{\circ}+0+0+2^{3}+2^{4}+2^{6} \\
=1+8+16+64=89
\end{array}$

Topic: Physics, Mathematics in Physics, JEE Main
The decomposition of $\mathrm{N}_2 \mathrm{O}_4$ ta $\mathrm{NO}_2$ is carried out at in chloroform. When equilibrium has been established, $0.2 \mathrm{~mol}$ of $\mathrm{N}_2 \mathrm{O}_4$ and $2 \times 10^{-3} \mathrm{~mol}$ of $\mathrm{NO}_2$ are present in 2 litre solution. The equilibrium constant for reaction $\mathrm{N}_2 \mathrm{O}_4 \rightleftharpoons 2 \mathrm{NO}_2$ is

Options:
A $1 \times 10^{-2}$
B $2 \times 10^{-3}$
C $1 \times 10^{-5}$
D $2 \times 10^{-5}$

The correct answer is:
$1 \times 10^{-5}$

Solution:
$K=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2 \mathrm{O}_4\right]}=\frac{\left[2 \times \frac{10^{-3}}{2}\right]^{-2}}{\left[\frac{2}{2}\right]}=\frac{10^{-6}}{10^{-1}}=10^{-5}$.

Topic: Chemistry, Chemical Equilibrium, JEE Main
The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10–4 mol L– 1 sec–1?
2NH 3 g → N 2 g + 3H 2 g

Options:
A 1.25Ã 10 â4 â M s â1 ,3.75Ã 10 â4 â M s â1
B 3.00Ã 10 â4 â M s â1 ,7.50Ã 10 â4 â M s â1
C 2.50Ã 10 â4 â M s â1 ,â1.25Ã 10 â4 â M s â1
D 3.75Ã 10 â4 â M s â1 ,2.50Ã 10 â4 â M s â1

The correct answer is:
1.25Ã 10 â4 â M s â1 ,3.75Ã 10 â4 â M s â1

Solution:
2NH 3 g → N 2 g + 3H 2 g
Rate = k[NH3]o = k = 2.5 × 10–4 M s–1.
1 2 −d dt NH 3 = 1 1 d dt N 2 = 1 3 d dt H 2
d⁡ d⁡ t N 2 = 2 · 5 × 1 0 - 4 2 = 1 · 2 5 × 1 0 - 4 M s - 1
d⁡ d⁡ t H⁡ 2 = 3 2 × 2 · 5 × 1 0 - 4 = 3 · 7 5 × 1 0 - 4 M s - 1

Topic: Chemistry, Chemical Kinetics, JEE Main
The decomposition of $\mathrm{O}_3(g)$ follows first order kinetics and is given by
$$
\mathrm{O}_3(9) \longrightarrow \mathrm{O}_2(9)+\mathrm{O}(9)
$$
The rate constant for this reaction is $1.0 \times 10^{-3} \mathrm{~s}^{-1}$. The initial pressure of $\mathrm{O}_3(g)$ is $100 \mathrm{~atm}$.
What will be the partial pressure (in atm) of $\mathrm{O}_3, \mathrm{O}_2, \mathrm{O}$ respectively after 38.38 minutes?

Options:
A $95,5,5$
B $10,90,0$
C $10,90,90$
D $10,0,90$

The correct answer is:
$10,90,90$

Solution:
Decomposition of $\mathrm{O}_3$,
$$
\mathrm{O}_3(g) \longrightarrow \mathrm{O}_2(g)+\mathrm{O}(g)
$$
Initial pressure $100 \mathrm{~atm} \quad 0 \quad 0$
Partial pressure $100-x \quad x \quad x$ after $38.38 \mathrm{~m}$ in
$$
\begin{aligned}
p_i(\text { initial pressure }) & =100 \mathrm{~atm} \\
\left.p_f \text { (final pressure }\right) & \text { after } 38.38 \mathrm{~m} . \\
& =100-x
\end{aligned}
$$
From first order reaction,
Given, $\quad k=1 \times 10^{-3} \mathrm{~s}^{-1}, t=38.38 \mathrm{~m}$.
$$
\begin{aligned}
k & =\frac{2.303}{t} \log \frac{p_i}{p_f} \\
1 \times 10^{-3} \mathrm{~s}^{-1} & =\frac{2.303}{38.38 \times 60} \log \frac{100(\mathrm{~atm})}{100-x(\mathrm{~atm})} \\
\log \frac{100}{100-x} & =\frac{1 \times 10^{-3} \mathrm{~s}^{-1} \times 38.38 \times 60}{2.303}
\end{aligned}
$$
$$
x=90 \mathrm{~atm}
$$
Pressure of $\mathrm{O}_3$ after $38.38 \mathrm{~m}$.
$$
p \mathrm{o}_3=p_f=100-x=100-90 \mathrm{~atm}=10 \mathrm{~atm}
$$
Pressure of $\mathrm{O}_2(g)(x)=90 \mathrm{~atm}$
Pressure of $\mathrm{O}(g)(x)=90 \mathrm{~atm}$
Hence, option (3) is correct.

Topic: Chemistry, Chemical Kinetics, JEE Main
The decrease in the number of sperms in semen is called ______X______and is due to the insufficient amount of ________Y_______. Select the correct option for X and Y.

Options:
A X-Azoospermia; Y-Hyaluronidase
B X-Oligospermia; Y-Acid phosphatase
C X-Oligospermia; Y-Testosterone
D X-Oligospermia; Y-Inhibin

The correct answer is:
X-Oligospermia; Y-Testosterone

Solution:
Oligospermia-It refers to fall or decrease the number of sperms in semen, which is mainly due to deficiency of male sex hormone testosterone. This leads to infertility in male.
Azospermia-This refers to absence of live sperms in semen.

Topic: Biology, Human Reproduction, NEET
The decreasing order of basicity of the following amines is:

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

The correct answer is:
III>II>I>IV

Solution:
Basic strength depends upon availability of lone pairs. Greater the resonance of lone pairs lesser the basic strength. In (II) and (III) lone pair is localised. (III) is more basic due to +I effect of carbon chain whereas in II, negative mesomeric effect is operating. In (I) and (IV) lone pair is delocalised, so their basic strength is less that former two amines. If (IV) donates its lone pair then its aromaticity will get lost therefore, its basic strength is less than that of I.

Overall order of basic strength is III>II>I>IV

Topic: Chemistry, Amines, JEE Main, JEE Main 2020 (09 Jan Shift 2)
The decreasing order of boiling points of the following alcohols is

Options:
A 3-methylbutan-2-ol > 2-methylbutan-2-ol > pentan-1-ol
B pentan-1-ol > 3-methylbutan-2-ol > 2-methylbutan-2-ol
C 2-methylbutan-2-ol > 3-methylbutan-2-ol > pentan-1-ol
D 2-methylbutan-2-ol > pentan-1-ol > 3-methylbutan-2-ol

The correct answer is:
pentan-1-ol > 3-methylbutan-2-ol > 2-methylbutan-2-ol

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The decreasing order of bond angles in BF3, NH3, PF3 and I3- is:

Options:
A I3-> BF3> NH3> PF3
B BF3> I3-> PF3> NH3
C BF3> NH3> PF3>I3-
D I3-> NH3> PF3> BF3

The correct answer is:
I3-> BF3> NH3> PF3

Solution:
Bond angle =120o, sp2
Bond angle =107o, sp3 (1 lone pair)
Bond angle ≈ 92o
When central atom size increases, bond angle decreases
sp3, NH3>PF3
Bond angle =180o, sp3d

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2018 (15 Apr)
The decreasing order of bond angles in $\mathrm{BF}_3, \mathrm{NH}_3, \mathrm{PF}_3$ and $\mathrm{I}_3^{-}$is :

Options:
A
$\mathrm{I}_3^{-}>\mathrm{BF}_3>\mathrm{NH}_3>\mathrm{PF}_3$

B
$\mathrm{BF}_3>\mathrm{I}_3^{-}>\mathrm{PF}_3>\mathrm{NH}_3$

C
$\mathrm{BF}_3>\mathrm{NH}_3>\mathrm{PF}_3>\mathrm{I}_3^{-}$

D
$\mathrm{I}_3^{-}>\mathrm{NH}_3>\mathrm{PF}_3>\mathrm{BF}_3$

The correct answer is:
$\mathrm{I}_3^{-}>\mathrm{BF}_3>\mathrm{NH}_3>\mathrm{PF}_3$

Solution:
$\text { B is } s p^2 \text {, Bond angle }=120^{\circ}$
$\mathrm{N} \text { is } s p^3 \text { with } 1 ~l p, \text { Bond angle }=107^{\circ}$
$\mathrm{P} \text { is } s p^3 \text { with } 1 ~l p$ , when central atom
size $\uparrow$, bond angle $\downarrow, \quad \therefore \mathrm{NH}_3>\mathrm{PF}_3$
$\mathrm{I}$ is $s p^3 d$ (linear), Bond angle $=180^{\circ}$
$\therefore \quad$ Decreasing order of bond angle
$\mathrm{I}_3^{-}>\mathrm{BF}_3>\mathrm{NH}_3>\mathrm{PF}_3$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The decreasing order of ease of alkaline hydrolysis for the following esters is

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

The correct answer is:
III>II>I>IV

Solution:
The rate of hydrolysis will depend on how likely a molecule is to accepting OH- ions. Rate of hydrolysis Î±-M effect or -I of ester. -M>-I. -NO2 is a strong EWG (Electron withdrawing group),-Cl is a weak EWG and OCH3 is an EDG(Electron donating group).

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2019 (10 Jan Shift 1)
The decreasing order of electrical conductivity of the following aqueous solutions is:
A 0.1 M Formic acid,
B 0.1 M Acetic acid,
C 0.1 M Benzoic acid.

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

The correct answer is:
A> C> B

Solution:
Order of acidic strength formic acid > Benzoic acid >

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2019 (12 Apr Shift 2)
The decreasing order of -I effect of the following is:

I. − N ⊕ H 3   II. − N O 2   III. −CN  IV. −COOH

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

The correct answer is:
I > II > III > IV

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
The decreasing order of nucleophilicity among the nucleophiles

Options:
A
(a), (b), (c), (d)

B
(d), (c), (b), (a)

C
(b), (c), (a), (d)

D
(c), (b), (a), (d)

The correct answer is:
(c), (b), (a), (d)

Solution:
No Solution Available

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2005
The decreasing order of nucleophilicity for the following anions is
CH 3 CO 2 − ,  CH 3 O − ,  C 6 H 5 O − ,  NO 3 −

Options:
A CH3CO2->CH3O->C6H5O->NO3-
B CH3O->NO3->C6H5O->CH3CO2-
C CH3O->C6H5O->CH3CO2->NO3-
D C6H5O->CH3O->NO3->CH3CO2-

The correct answer is:
CH3O->C6H5O->CH3CO2->NO3-

Solution:
Since the electron donating atom is same, the nucleophilicity parallels basicity.
A conjugate base of a strong acid is weak and vice-versa.
The order of acidity is
CH3OH<C6H5OH<CH3COOH<HNO3
Thus, the order of basicity and nucleophilicity is
CH3O->C6H5O->CH3CO2->NO3-

Topic: Chemistry, General Organic Chemistry, JEE Main
The decreasing order of rate of reaction for the following compounds towards S N 2Th (bimolecular nucleophilic substitution with tetrahedral intermediate) reaction is

(i)
(ii)
(iii)
(iv)

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

The correct answer is:
II > III > IV > I

Solution:
Reactivity order of compounds towards S N 2Th depend on partial possitve charge present of carbon which is increases due to −I effect hence the order S N 2Th is

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The decreasing order of reactivity of the following compounds towards nucleophilic substitution (SN2) is:

Options:
A (II) > (III) > (I) > (IV)
B (II) > (III) > (IV) > (I)
C (III) > (II) > (IV) > (I)
D (IV) > (II) > (III) > (I)

The correct answer is:
(II) > (III) > (IV) > (I)

Solution:
Rate of SN2 reation α Electrophilicity of alkyl carbon α 1Steric crowding on R-X

SN2 reaction depend upon -I, -M, effect.-I,-M group increases the rate of SN2 reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2020 (03 Sep Shift 2)
The decreasing order of reactivity of the following organic molecules towards AgNO3 solution is:

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

The correct answer is:
B> A> C> D

Solution:
Given reaction is SN1 reaction. In SN1 reaction

Rate of reaction ∝ Stability of C+

(A)

(B)

(C)

(D)

Stability of C+: ii>i>iii>iv

Reactivity order : B>A>C>D

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (04 Sep Shift 1)
The decreasing order of reactivity towards dehydrohalogenation E1 reaction of the following compounds is:

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

The correct answer is:
D>B>C>A

Solution:
E1 reaction proceeds via carbocation formation, therefore greater the stability of carbocation, faster the E1 reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The decreasing order of reactivity towards electrophilic substitution of the following compounds is

Options:
A $1>3>4>2$
B $4>1>3>2$
C $4>1>2>3$
D $1>4>3>2$

The correct answer is:
$4>1>2>3$

Solution:
Due to $+I$-effect of $-\mathrm{CH}_{3}$ group (4) has higher electron density, than benzene (1). Due to strong $-I$-effect of - Cl group and strong $-I$ and $-M$-effect of $-\mathrm{NO}_{2}$ group, both (2) and (3) have lower electron density, than benzene (l).
As - I-effect of -Cl group is weaker, than that of $-\mathrm{NO}_{2}$ (3) has much lower electron density than (2).
Hence, nitrobenzene will have high reactivity towards electrophilic substitution due to lower electron density on benzene ring.
Thus, the overall reactivity order towards electrophilic substitution is $4>1>2>3$.

Topic: Chemistry, Hydrocarbons, JEE Main