JEE Main Previous year questions (2016-22): Structure of Atom - Notes | Study Chemistry for JEE - JEE

Q.1. For diatomic molecules, the correct statement(s) about the molecular orbitals formed by the overlap of two 2p_z orbitals is(are)                (JEE Advanced 2022)
(a) σ orbital has a total of two nodal planes.
(b) σ∗ orbital has one node in the xz-plane containing the molecular axis.
(c) π orbital has one node in the plane which is perpendicular to the molecular axis and goes through the center of the molecule.
(d) π∗ orbital has one node in the xy-plane containing the molecular axis.

Ans. a, d

Q.2. Given below are the quantum numbers for 4 electrons.
A. n = 3, 1 = 2, m₁ = 1, ms = +1/2
B. n = 4, 1 = 1, m₁ = 0, ms = +1/2
C. n = 4, 1 = 2, m₁ = −2, ms = −1/2
D. n = 3, 1 = 1, m₁ = −1, ms = +1/2
The correct order of increasing energy is              (JEE Main 2022)
(a) D < B < A < C
(b) D < A < B < C
(c) B < D < A < C
(d) B < D < C < A

Ans. b
Energy of the sub-shell is given by, (n + l) rule.
A ⇒ 3 d ⇒ n + 1 = 5
B ⇒ 4 p ⇒ n + λ = 5
C ⇒ 4 d ⇒ n + ℓ ⇒ 6
D ⇒ 3 s ⇒ (n + ℓ) = 4
D < A < B < C

Q.3. Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R
Assertion A: Zero orbital overlap is an out of phase overlap.
Reason R: It results due to different orientation/direction of approach of orbitals.
In the light of the above statements, choose the correct answer from the options given below:              (JEE Main 2022)
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true

Ans. a
Zero overlapping is something in which there is no overlapping between two orbitals. The first condition is that the two orbitals should not be symmetrical and the second condition is that both orbitals should be in different planes. 

Q.4. Identify the incorrect statement from the following.             (JEE Main 2022)
(a) A circular path around the nucleus in which an electron moves is proposed as Bohr's orbit.
(b) An orbital is the one electron wave function (ψ) in an atom.
(c) The existence of Bohr's orbits is supported by hydrogen spectrum.
(d) Atomic orbital is characterised by the quantum numbers n and l only.

Ans. d
Atomic orbital is characterised by the quantum numbers n, l and m.
Hence option D is incorrect. 

Q.5. Outermost electronic configurations of four elements A, B, C, D are given below:
(A) 3s²
(B) 3s²3p¹
(C) 3s²3p³
(D) 3s²3p⁴
The correct order of first ionization enthalpy for them is:            (JEE Main 2022)
(a) (A) < (B) < (C) < (D)
(b) (B) < (A) < (D) < (C)
(c) (B) < (D) < (A) < (C)
(d) (B) < (A) < (C) < (D)

Ans. b

Orbitals with fully filled and half-filled electronic configuration are stable, and require more energy for ionization
Elements with greater electronegativity require more energy for ionisation
Hence the correct order is (B) < (A) < (D) < (C)

Q.6. The correct decreasing order of energy for the orbitals having, following set of quantum numbers:
(A) n = 3, l = 0, m = 0
(B) n = 4, l = 0, m = 0
(C) n = 3, l = 1, m = 0
(D) n = 3, l = 2, m = 1
is:            (JEE Main 2022)
(a) (D) > (B) > (C) > (A)
(b) (B) > (D) > (C) > (A)
(c) (C) > (B) > (D) > (A)
(d) (B) > (C) > (D) > (A)

Ans. a
(A) n + ℓ = 3 + 0 = 3
(B) n + ℓ = 4 + 0 = 4
(C) n + ℓ = 3 + 1 = 4
(D) n + ℓ = 3 + 2 = 5
Higher n + ℓ valuc, higher the encrgy & if same n + ℓ value, then higher n value, higher the energy.
Thus: D > B > C > A. 

Q.7. Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Energy of 2s orbital of hydrogen atom is greater than that of 2s orbital of lithium.
Reason R: Energies of the orbitals in the same subshell decrease with increase in the atomic number.
In the light of the above statements, choose the correct answer from the options given below.           (JEE Main 2022)
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is NOT the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Ans. a
As the atomic number increases then the potential energy of electrons present in same shell becomes more and more negative. And therefore total energy also becomes more negative.

∴ Energies of the orbitals in the same subshell decreases with increase in atomic number.

Q.8. Which of the following sets of quantum numbers is not allowed?          (JEE Main 2022)
(a)
(b)
(c)
(d)

Ans. c
If n = 3, then possible values of l = 0, 1, 2
But in option (C), the value of l is given ' 3 ', this is not possible.

Q.9. The number of radial nodes and total number of nodes in 4p orbital respectively are:          (JEE Main 2022)
(a) 2 and 3
(b) 2 and 2
(c) 3 and 4
(d) 4 and 4

Ans. a
For 4p
n = 4, l = 1
Total number of nodes = n − 1 = 4 − 1 = 3
Total radial nodes = n − l − 1 = 4 − 1 − 1 = 2

Q.10. Which of the following is the correct plot for the probability density ψ² (r) as a function of distance 'r' of the electron from the nucleus for 2s orbital?         (JEE Main 2022)
(a)
(b)
(c)
(d)

Ans. b
Formula for number of radial nodes in n^th orbital = n - l - 1
For 2s, number of radial nodes = 2 – 0 – 1 = 1 and value of ψ² is always positive.

Q.11. Which of the following statements are correct?
(A) The electronic configuration of Cr is [Ar] 3d⁵ 4s¹.
(B) The magnetic quantum number may have a negative value.
(C) In the ground state of an atom, the orbitals are filled in order of their increasing energies.
(D) The total number of nodes are given by n − 2.
Choose the most appropriate answer from the options given below:         (JEE Main 2022)
(a) (A), (C) and (D) only
(b) (A) and (B) only
(c) (A) and (C) only
(d) (A), (B) and (C) only

Ans. d
(A) Cr (24) = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
= [Ar] 3d⁵ 4s¹
(B) Magnetic quantum number (m) values ranging from − l to + l including zero.
∴ It can have negative value.
(C) According to Aufbau rule, electrons are filled first in these orbitals which have low energy.
∴ Statement C is correct.
(D) We know,
Number of Radial nodes = n − l − 1
and number of Angular nodes = l
∴ Total nodes = n − l − 1 + 1 = n − 1

Q.12. Consider the following statements :
(A) The principal quantum number 'n' is a positive integer with values of 'n' = 1, 2, 3, ...
(B) The azimuthal quantum number 'l' for a given 'n' (principal quantum number) can have values as 'l' = 0, 1, 2, ...... n
(C) Magnetic orbital quantum number 'm_l' for a particular 'l' (azimuthal quantum number) has (2l + 1) values.
(D) ± 1/2 are the two possible orientations of electron spin.
(E) For l = 5, there will be a total of 9 orbital
Which of the above statements are correct?         (JEE Main 2022)
(a) (A), (B) and (C)
(b) (A), (C), (D) and (E)
(c) (A), (C) and (D)
(d) (A), (B), (C) and (D)

Ans. c
(A) Principle quantum number n is a positive integer and it's possible values are n = 1, 2, 3 ........
∴ A is correct.
(B) Azimuthal quantum number 'l' for a given 'n' can have values as l = 0, 1, 2 ....... (n − 1)
∴ Statement B is wrong.
(C) Magnetic orbital quantum number m_l for particular l has values from − l to + 1 including zero means 2l + 1 values.
∴ Statement C is correct.
(D) ±1/2 are the two possible orientation of electron spin.
∴ Statement D is correct.
(E) For l = 5, there will be a total of 11 orbitals.
l = 0 ⇒ 5 subshell
l = 1 ⇒ p subshell
l = 2 ⇒ d subshell
l = 3 ⇒ f subshell
l = 4 ⇒ g subshell
l = 5 ⇒ h subshell
We know,
Number of orbital in any subshell = 2l + 1.
∴ For h subshell, number of orbitals = 2 × 5 + 1 = 11

Q.13. The number of radial and angular nodes in 4d orbital are, respectively         (JEE Main 2022)
(a) 1 and 2
(b) 3 and 2
(c) 1 and 0
(d) 2 and 1

Ans. a
We know,
Radial nodes = n − l − 1
and Angular nodes = l
For 4d orbital,
n = 4
l = 2
∴ Radial nodes = 4 − 2 − 1 = 1
Angular nodes = 2

Q.14. If the radius of the 3^rd Bohr's orbit of hydrogen atom is r₃ and the radius of 4^th Bohr's orbit is r₄. Then:          (JEE Main 2022)
(a)
(b)
(c)
(d)

Ans. b
We know,

For hydrogen atom, 

Q.15. The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is:
[Given: The threshold frequency of platinum is 1.3 × 10¹⁵ s⁻¹ and h = 6.6 × 10⁻³⁴ J s.]          (JEE Main 2022)
(a) 3.21 × 10⁻¹⁴ J
(b) 6.24 × 10⁻¹⁶ J
(c) 8.58 × 10⁻¹⁹ J
(d) 9.76 × 10⁻²⁰ J

Ans. c
The minimum energy possessed by photons will be equal to the work function of the metal.
∴ E_min = hν₀ J
= 6.6 × 10⁻³⁴ × 1.3 × 10¹⁵
= 8.58 × 10⁻¹⁹ J

Q.16. The pair, in which ions are isoelectronic with AI³⁺ is:          (JEE Main 2022)
(a) Br⁻ and Be²⁺
(b) Cl⁻ and Li⁺
(c) S²⁻ and K⁺
(d) O²⁻ and Mg²⁺

Ans. d
O^2–, Mg²⁺ and Al³⁺ are isoelectronic. All have 10 electrons. 

Q.17. The energy of one mole of photons of radiation of wavelength 300 nm is
(Given: h = 6.63 × 10⁻³⁴ J s, N_A = 6.02 × 10²³ mol⁻¹, c = 3 × 10⁸ m s⁻¹)          (JEE Main 2022)
(a) 235 kJ mol⁻¹
(b) 325 kJ mol⁻¹
(c) 399 kJ mol⁻¹
(d) 435 kJ mol⁻¹

Ans. c
Energy of one photon


= 4.1333 eV
∴ Energy of one mole of photon
= 4.1333 × 6.02 × 10²³eV
= 4.1333 × 6.02 × 1023 × 1.6 × 10⁻¹⁹J

= 399 kJ/mol

Q.18. Consider the following pairs of electrons
(A) (a) n = 3, l = 1, m₁ = 1, m_s =      
        (b) n = 3, 1 = 2, m₁ = 1, m_s =
(B) (a) n = 3, l = 2, m₁ = −2, m_s =
       (b) n = 3, l = 2, m₁ = −1, m_s =
(C) (a) n = 4, l = 2, m₁ = 2, m_s = 
       (b) n = 3, l = 2, m₁ = 2, m_s =
The pairs of electrons present in degenerate orbitals is/are:         (JEE Main 2022)
(a) Only (A)
(b) Only (B)
(c) Only (C)
(d) (B) and (C)

Ans. b
For degenerate orbitals, only the value of m must be different. The value of (n + l) must be the same.
Hence, the pair of electrons with quantum numbers given in (B) are degenerate.

Q.19. The minimum uncertainty in the speed of an electron in an one dimensional region of length 2a_o (Where a_o = Bohr radius 52.9pm) is _________ kms⁻¹.
(Given: Mass of electron = 9.1 × 10⁻³¹ kg, Planck's constant h = 6.63 × 10⁻³⁴ Js)         (JEE Main 2022)

Ans. 548
Heisenberg's uncertainty principle


= 6.63 × 10⁻³⁴ 
4 × 3.14 × 2 × 52.9 × 10⁻¹² × 9.1 × 10⁻³¹
= 548273 ms⁻¹ 
= 548.273 kms⁻¹
= 548 kms⁻¹

Q.20.  If the wavelength for an electron emitted from H-atom is 3.3 × 10⁻¹⁰ m, then energy absorbed by the electron in its ground state compared to minimum energy required for its escape from the atom, is _________ times. (Nearest integer)
[Given: h = 6.626 × 10⁻³⁴ J s]
Mass of electron = 9.1 × 10⁻³¹          (JEE Main 2022)

Ans. 2






= 2.18 × 10⁻¹⁸ J
= 21.8 × 10⁻¹⁹ J
Total energy absorbed = lonization energy + Kinetic energy
= (21.76 + 21.8) × 10⁻¹⁹
= 43.56 × 10⁻¹⁹ J
≈ 2 times of 21.76 × 10⁻¹⁹ J

Q.21. Consider an imaginary ionThe nucleus contains 'a'% more neutrons than the number of electrons in the ion. The value of 'a' is _______________. [nearest integer]          (JEE Main 2022)

Ans. 4
Number of electrons inis 25.
Number of neutrons = 48 − 22 = 26.
% increase in the number of neutrons over electrons

∴ a = 4

Q.22. The wavelength of an electron and a neutron will become equal when the velocity of the electron is x times the velocity of neutron. The value of x is ____________. (Nearest Integer)
(Mass of electron is 9.1 × 10⁻³¹ kg and mass of neutron is 1.6 × 10⁻²⁷ kg )          (JEE Main 2022)

Ans. 1758





= 0.17582 × 10⁴
≃ 1758

Q.23. When the excited electron of a H atom from n = 5 drops to the ground state, the maximum number of emission lines observed are _____________.          (JEE Main 2022) 

Ans. 10
Maximum number of emission lines

n₂ = 5
n₁ = 1

Hence maximum number of emission lines observed are 10.

Q.24. If the work function of a metal is 6.63 × 10⁻¹⁹J, the maximum wavelength of the photon required to remove a photoelectron from the metal is ____________ nm. (Nearest integer)
[Given: h = 6.63 × 10⁻³⁴ J s, and c = 3 × 10⁸ m s⁻¹]          (JEE Main 2022)

Ans. 300
Given,
Work function = 6.63 × 10⁻¹⁹ J

= 4.14 eV
We know,
E = 1240 / λ(nm)
⇒ 4.14 = 1240 / λ
λ = 300 nm

Q.25. Consider the following set of quantum numbers.
The number of correct sets of quantum numbers is __________.          (JEE Main 2022)

Ans. 2
For A,

Given n = 3 and l = 3

but we know maximum value of l = n − 1.

∴ l can't be equal to n.

So, Set A of quantum numbers is not possible.

For B,

Given n = 3, l = 2, m = − 2

Here, l = 2 which follow the rule l = n − 1.

And we know possible value of m is − l to + l.

here possible value of m = −2 to +2

∴ This Set B is valid set of quantum numbers.

For C,

Given n = 2, l = 1, m = +1

Here l = 1 which follows the rule l = n − 1.

For l = 1 possible value of m = −1 to +1

Here m = +1. So value of m is valid.

∴ Set C is valid set of quantum numbers.

For D,

Given n = 2, l = 2, m = +2

l = 2 does not follow the rule l = n − 1 rule.

∴ Set D is not valid set of quantum numbers.

Q.26. If the uncertainty in velocity and position of a minute particle in space are, 2.4 × 10⁻²⁶ (m s⁻¹) and 10⁻⁷ (m) respectively. The mass of the particle in g is ____________. (Nearest integer)
(Given: h = 6.626 × 10⁻³⁴ Js)         (JEE Main 2022)

Ans. 22
We know from hisenberg uncertainty principle
Δx . Δp = h/4π

⇒ Δx . mΔv = h/4π

Given,

Δx = 10⁻⁷ m

Δx = 2.4 × 10⁻²⁶ m/s

h = 6.626 × 10⁻³⁴ Js

⇒ m = 0.022 kg

⇒ m = 22 gm

Q.27. The longest wavelength of light that can be used for the ionisation of lithium atom (Li) in its ground state is x × 10⁻⁸ m. The value of x is ___________. (Nearest Integer).
(Given: Energy of the electron in the first shell of the hydrogen atom is −2.2 × 10⁻¹⁸ J ; h = 6.63 × 10⁻³⁴ Js and c = 3 × 10⁸ ms⁻¹)         (JEE Main 2022) 

Ans. 4
Bohr model is not valid for lithium atom (Li) as Bohr model is valid for only single electronic species, so it would be valid for Li⁺² but not Li atom.
So this question is BONUS.

Q.28. Consider a helium (He) atom that absorbs a photon of wavelength 330 nm. The change in the velocity (in cm s⁻¹) of He atom after the photon absorption is __________.
(Assume: Momentum is conserved when photon is absorbed. Use: Planck constant = 6.6 × 10⁻³⁴ J s, Avogadro number = 6 × 10²³ mol⁻¹, Molar mass of He = 4 g mol⁻¹)      (JEE Advanced 2021)

Ans. 30
Wavelength of photon absorbed, λ = 330 nm = 330 × 10⁻⁹ m
Planck's constant, h = 6.6 × 10⁻³⁴ J s
Molar mass of He, M = 4 g mol⁻¹ = 4 × 10⁻³ kg mol⁻¹
Avogadro number, N_A = 6 × 10²³ mol⁻¹
Mass of one atom of He, m =

Velocity, = V cm/s.
Using de-Broglie equation,
λ = hmv


= 30 cm/s 

Q.29. A 50 watt bulb emits monochromatic red light of wavelength of 795 nm. The number of photons emitted per second by the bulb is x × 10²⁰. The value of x is __________. [Given : h = 6.63 × 10⁻³⁴ Js and c = 3.0 × 10⁸ ms⁻¹]           (JEE Main 2021)

Ans. 2
Energy of photon is given as E = nhc/λ .... (i)
where, E = energy of photon (50 W),
n = number of photon
h = Planck's constant (6.63 × 10⁻³⁴ Js)
c = speed of light (3 × 10⁸ m/s)
λ = wavelength of light (795 × 10⁻⁹ m)
E = 50W = 50 J = energy of photon


= 1998.49 × 1017 = 1.998 × 1020
⇒ ≈ 2 × 10²⁰
∴ x = 2 

Q.30. The value of magnetic quantum number of the outermost electron of Zn⁺ ion is ______________.            (JEE Main 2021)

Ans. 0
Zn⁺ → 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹
Outermost electron is in 4s subshell
m = 0 

Q.31. Ge(Z = 32) in its ground state electronic configuration has x completely filled orbitals with m_l = 0. The value of x is ___________.             (JEE Main 2021)

Ans. 7

Completely filled orbital with m_l = 0 are
= 1 + 1 + 1 + 1 + 1 + 1 + 1
= 7
So, answer is 7. 

Q.32. The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm, in 0.1 second is x × 10¹³. The value of x is _____________. (Nearest integer)
(h = 6.63 × 10⁻³⁴ Js, c = 3.00 × 10⁸ ms⁻¹)              (JEE Main 2021)

Ans. 50
Energy emitted in 0.1 sec.

= 10⁻⁴ J
If 'n' photons of λ = 1000 nm are emitted, then 10⁻⁴ = n × hc/λ

⇒ n = 5.02 × 10¹⁴ = 50.2 × 10¹³
⇒ 50 (nearest integer)

Q.33. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal toThe value of 10x is ___________. (a₀ is radius of Bohr's orbit) (Nearest integer) [Given: π = 3.14]              (JEE Main 2021)

Ans. 3155
 


⇒ x = 315.507
⇒ 10x = 3155 (nearest integer)

Q.34. A metal surface is exposed to 500 nm radiation. The threshold frequency of the metal for photoelectric current is 4.3 × 10¹⁴ Hz. The velocity of ejected electron is ____________ × 10⁵ ms⁻¹ (Nearest integer)
[Use: h = 6.63 × 10⁻³⁴ Js, m_e = 9.0 × 10⁻³¹ kg]               (JEE Main 2021)

Ans. 5
υ : speed of electron having max. K.E.
⇒ from Einstein equation : E = ϕ + K.E._max


⇒ 11.271 × 10⁻²⁰J = 1/2 × 9 × 10⁻³¹ × υ²
⇒ υ = 5 × 10⁵ m/sec.

Q.35. An accelerated electron has a speed of 5 × 10⁶ ms⁻¹ with an uncertainty of 0.02%. The uncertainty in finding its location while in motion is x × 10⁻⁹ m. The value of x is ____________. (Nearest integer)
[Use mass of electron = 9.1 × 10⁻³¹ kg, h =6.63 × 10⁻³⁴ Js, π = 3.14]               (JEE Main 2021)

Ans. 58

 

⇒ x × 10⁻⁹ × 10³ = 0.058 × 10⁻³

Q.36. A source of monochromatic radiation of wavelength 400 nm provides 1000 J of energy in 10 seconds. When this radiation falls on the surface of sodium, x × 10²⁰ electrons are ejected per second. Assume that wavelength 400 nm is sufficient for ejection of electron from the surface of sodium metal. The value of x is ______________. (Nearest integer)
(h = 6.626 × 10⁻³⁴ Js)                (JEE Main 2021)

Ans. 2
Total energy provided by
Source per second = 1000/10 = 100 J
Energy required to eject electron = hc/λ

Number of electrons ejected



= 2.01 × 10²⁰

 Q.37. Number of electrons that Vanadium (Z = 23) has in p-orbitals is equal to _________.               (JEE Main 2021)

Ans. 12
₂₃V: 1s²2s²2p⁶3s²3p⁶3d³4s²
Number of electrons in p-orbitals is equal to 12.00.

Q.38. The wavelength of electrons accelerated from rest through a potential difference of 40 kV is x × 10⁻¹² m. The value of x is ___________. (Nearest integer)
Give : Mass of electron = 9.1 × 10⁻³¹ kg
Charge on an electron = 1.6 × 10⁻¹⁹ C
Planck's constant = 6.63 × 10⁻³⁴ Js               (JEE Main 2021)

Ans. 6
Wavelength of electron is given by 

Here q = charge on electron, V = potential difference


x = 6
OR


∴ Ans. is 6

Q.39. The Azimuthal quantum number for the valence electrons of Ga⁺ ion is ___________.
(Atomic number of Ga = 31)               (JEE Main 2021)

Ans. 0
Ga = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p¹
Ga⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²
Azimuthal Quantum number (l) for valence shell electron is 0.

Q.40. In the ground state of atomic Fe(Z = 26), the spin-only magnetic moment is ____________ × 10⁻¹ BM. (Round off to the Nearest Integer). [Given: √3 = 1.73, √2 = 1.41]                (JEE Main 2021)

Ans. 49
₂₆Fe = [Ar]3d⁶4s²
No. of unpaired electrons = 4



= 4.8786 BM
= 48.78 × 10⁻¹ BM

Q.41. A certain orbital has n = 4 and m_L = −3. The number of radial nodes in this orbital is ____________. (Round off to the Nearest Integer).                (JEE Main 2021)

Ans. 0
Number of radial nodes = n – ℓ – 1
n = 4, m_L =–3 so ℓ = 3
radial nodes = 4 – 3 – 1 = 0

Q.42. The number of orbitals with n = 5, m₁ = +2 is ___________. (Round off to the Nearest Integer).                 (JEE Main 2021)

Ans. 3
Given, n = 5, m_l = + 2
For n = 5, possible value of l = 0, 1, 2, 3, 4
For l = 0, m_l = 0
l = 1, m_l = −1, 0, 1
l = 2, m_l = −2, −1, 0, 1, 2
l = 3, m_l = −3, −2, −1, 0, 1, 2, 3
l = 4, m_l = −4, −3, −2, −1, 0, 1, 2, 3, 4
Possible value of m_l for a given value of l
= 0, ± 1, ± 2, ± 3 ..... ± l
So, number of orbitals having n = 5 and m_l = ± 2 are 3.

Q.43. When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is _______^oA. (Round off to the Nearest Integer).
[Use: √3 = 1.73, h = 6.63 × 10⁻³⁴ Js
m_e = 9.1 × 10⁻³¹ kg; c = 3.0 × 10⁸ ms⁻¹; 1eV = 1.6 × 10⁻¹⁹ J]                 (JEE Main 2021)

Ans. 9
λ = 248 × 10⁻⁹ m
w₀ = 3 × 1.6 × 10⁻¹⁹ J
E = w₀ + K.E.
hc/λ = W0 + K.E.

= 3.2 × 10⁻¹⁹ J


p = 7.63 × 10⁻²⁵

⇒ λ = 8.7 × 10⁻¹⁰ = 8.7 ^oA ≈ 9 ^oA

Q.44. A ball weighing 10 g is moving with a velocity of 90 ms⁻¹. If the uncertainty in its velocity is 5%, then the uncertainty in its position is ___________ × 10⁻³³ m. (Rounded off to the nearest integer)
[Given: h = 6.63 × 10⁻³⁴ Js]                 (JEE Main 2021)  

Ans. 1
m = 10 g = 10⁻² Kg
v = 90 m/sec.

m . Δv . Δx ≥ h/4π


x = 1.17 ≃ 1

Q.45. A proton and a Li³⁺ nucleus are accelerated by the same potential. If λLi and λp denote the de Broglie wavelengths of Li³⁺ and proton respectively, then the value of  is x × 10^-1.
The value of x is ______. (Rounded off to the nearest integer)
[Mass of Li³⁺ = 8.3 mass of proton]                (JEE Main 2021)  

Ans. 2
Given, mass of Li³⁺ = 8.3 times of mass of proton formula,
De-Broglie wavelength,
Here, h = Planck's constant = 6.624 × 10⁻³⁴ J-s
m = Mass of atom
q = Charge (or number of electrons)


Now, Eq. (i) divided by Eq. (ii), we get,

We know that m_Li = 8.3 m_p

Hence, x × 10⁻¹, x = 2  

Q.46. Given below are two statements.
Statement I: According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in positive charges on the nucleus as there is no strong hold on the electron by the nucleus.
Statement II: According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in principal quantum number.
In the light of the above statements, choose the most appropriate answer from the options given below:                 (JEE Main 2021)  
(a) Both Statement I and Statement II are false
(b) Both Statement I and Statement II are true
(c) Statement I is false but Statement II is true
(d) Statement I is true but Statement II is false

Ans. c
Velocity of electron in Bohr's atom is given by 

Z = atomic number of atom, corresponds to +ve charge so as Z increase velocity increases so statement I is wrong and as 'n' decreases velocity increases so statement II is correct. 

Q.47. If the Thompson model of the atom was correct, then the result of Rutherford's gold foil experiment would have been:                 (JEE Main 2021) 
(a) All of the α-particles pass through the gold foil without decrease in speed.
(b) α-particles are deflected over a wide range of angles.
(c) All α-particles get bounced back by 180^∘.
(d) α-particles pass through the gold foil deflected by small angles and with reduced speed. 

Ans. d
As in Thompson model, protons are diffused (charge is not centred) α-particles deviate by small angles and due to repulsion from protons, their speed decreases. 

Q.48. Given below are two statements:
Statement I: Rutherford's gold foil experiment cannot explain the line spectrum of hydrogen atom.
Statement II: Bohr's model of hydrogen atom contradicts Heisenberg' uncertainty principle.
In the light of the above statements, choose the most appropriate answer from the options given below:                 (JEE Main 2021) 
(a) Statement I is false but statement II is true.
(b) Statement I is true but statement II is false.
(c) Both statement I and statement II are false.
(d) Both statement I and statement II are true.

Ans. d
Rutherford's gold foil experiment only proved that electrons are held towards nucleus by electrostatic forces of attraction and move in circular orbits with very high speeds.
Bohr's model gave exact formula for simultaneous calculation of speed & distance of electron from the nucleus, something which was deemed impossible according to Heisenberg. 

Q.49. The spin only magnetic moments (in BM) for free Ti³⁺, V²⁺ and Sc³⁺ ions respectively are
(At.No. Sc : 21, Ti : 22, V : 23)                 (JEE Main 2021)
(a) 3.87, 1.73, 0
(b) 1.73, 3.87, 0
(c) 1.73, 0, 3.87
(d) 0, 3.87, 1.73 

Ans. b

Ti⁺³ = [Ar]3d¹ n = 1 μ = 1.73BM
V⁺² = [Ar]3d³ n = 3 μ = 3.87BM
Sc⁺³ = [Ar]3d⁰4s⁰ n = 0 μ = 0

Q.50. Given below are two statements:
Statement I: Bohr's theory accounts for the stability and line spectrum of Li+ ion.
Statement II: Bohr's theory was unable to explain the splitting of spectral lines in the presence of a magnetic field.
In the light of the above statements, choose the most appropriate answer from the options given below:                 (JEE Main 2021)
(a) Statement I is false but statement II is true.
(b) Both statement I and statement II are true.
(c) Statement I is true but statement II is false.
(d) Both statement I and statement II are false.

Ans. a
Bohr’s theory is applicable for unielectronic species only Li⁺ has two electrons.
Bohr’s theory could not explain the splitting of spectral lines in the presence of external magnetic field (Zeeman effect)
Statement I: false
Statement II: true 

Q.52. A certain orbital has no angular nodes and two radial nodes. The orbital is:                 (JEE Main 2021)
(a) 2s
(b) 3s
(c) 3p
(d) 2p

Ans. b
Angular nodes = ℓ = 0
Radial nodes ⇒ (n – ℓ – 1) = 2
⇒ n – 0 – 1 = 2
⇒ n = 3
So orbital is 3s. 

Q.53. The ionic radius of Na⁺ ions is 1.02 ^oA. The ionic radii (in ^oA) of Mg²⁺ and Al³⁺, respectively, are                 (JEE Main 2021)
(a) 1.05 and 0.99
(b) 0.72 and 0.54
(c) 0.85 and 0.99
(d) 0.68 and 0.72

Ans. b
Order of ionic size Na⁺ > Mg²⁺ > Al³⁺ 

Q.54. What is the spin-only magnetic moment value (BM) of a divalent metal ion with atomic number 25, in it's aqueous solution?                 (JEE Main 2021)
(a) zero
(b) 5.26
(c) 5.0
(d) 5.92

Ans. d
₂₅Mn = [Ar]3d⁵ 4s²
Mn²⁺ = [Ar]3d⁵ 4s⁰
No. of unpaired electron (n) = 5

Q.55. Which of the following forms of hydrogen emits low energy β^- particles?                 (JEE Main 2021)
(a) Tritium
(b) Proton H⁺
(c) Protium
(d) Deuterium

Ans. a
Tritium isotope of hydrogen is radioactive and emits low energy β⁻ particles. It is because of high n/p ratio of tritium which makes nucleus unstable.

Q.56. The orbital having two radial as well as two angular nodes is:                 (JEE Main 2021)
(a) 3p
(b) 5d
(c) 4d
(d) 4f

Ans. b
Number of radial nodes = (n – l – 1)
Number of angular nodes = l
for 5d; n = 5, l = 2
5d orbital has two radial nodes and two angular nodes 

Q.57. In which of the following pairs, the outer most electronic configuration will be the same?                 (JEE Main 2021)
(a) Cr⁺ and Mn²⁺
(b) Ni²⁺ and Cu⁺
(c) V²⁺ and Cr⁺
(d) Fe²⁺ and Co⁺

Ans. a
Cr⁺ → [Ar]3d⁵
Mn²⁺ → [Ar]3d⁵ 

Q.58. According to Bohr's atomic theory: 
(A) Kinetic energy of electron is 
(B) The product of velocity (v) of electron and principal quantum number (n), ′vn′ ∝ Z². 
(C) Frequency of revolution of electron in an orbit is 
(D) Coulombic force of attraction on the electron is
Choose the most appropriate answer from the options given below:
(a) (A), (C) and (D) only
(b) (A) only
(c) (C) only
(d) (A) and (D) only

Ans. d
According to Bohr's theory,

(∴ Correct) 
II. Speed of electron
(Here, Z = atomic number, n = number of shells) 
∴ v × n ∝ Z (∴ Incorrect) 
III. Frequency of revolution of electron = v/2πr 
Frequency(∴ Incorrect)

Q.59. The number of orbitals associated with quantum number n = 5, m_s = +1/2 is    (2020)
(a) 11
(b) 25
(c) 50
(d) 15

Ans. b
For principle quantum number (n) = 5, total of 5 subshells are associated that is s, p, d, f and g-subshells. Each orbital has two spin quantum number (m_s) = + 1/2 and 1/2. Thus, number of orbitals associated with n = 5 and m_s = + 1/2 are equal to the number of orbitals in 5^th shell.

Q.60. For the Balmer series in the spectrum of H atom,  the correct statements among (I) to (IV) are    (2020)
(I) As wavelength decreases, the lines in the series converge
(II) The integer n₁ is equal to 2
(III) The lines of longest wavelength correspond to n₂ = 3
(IV) The ionization energy of hydrogen can be calculated from wave number of these lines
(a) (I), (III), (IV)
(b) (I), (II), (III)
(c) (I), (II), (IV)
(d) (II), (III), (IV)

Ans. b
For Balmer series in hydrogen spectrum, as the wavelength decreases, the lines in the series get converge. For Balmer series, n₁ is 2 and for longest wavelength n₂ is 3.

Q.61. The radius of the second Bohr orbit, in terms of the Bohr radius, a₀, in Li²⁺ is    (2020)
(a) 2a₀/9
(b) 4a₀/9
(c) 4a₀/3
(d) 2a₀/9

Ans. c
We know,
Radius of Bohr’s orbit = a₀n²/Z
For Li²⁺ (Z = 3), the radius of second orbit

Q.62. The de Broglie wavelength of an electron in the 4^th Bohr orbit is    (2020)
(a) 2πa₀
(b) 4πa₀
(c) 6πa₀
(d) 8πa₀

Ans. d
From the relation,
2πr = nλ

For hydrogen atom,

⇒ λ = 8πa₀

Q.63. For emission line of atomic hydrogen from n_i = 8 to n_f = n, the plot of wave number against (1/n²) will be (The Rydberg constant, R_H is in wave number unit)    (2019)
(a) Linear with intercept - R_H
(b) Non linear
(c) Linear with slope R_H
(d) Linear with slope - R_H

Ans. d
As we know,

After putting the values, we get


Comparing to y = mx + c, we get

Q.64. Which of the following combination of statements is true regarding the interpretation of the atomic orbitals?
(a) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum.
(b) For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number.
(c) According to wave mechanics, the ground state angular momentum is equal to  h/2π
(d) The plot of ψ vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value.    (2019)
(a) (a), (d)
(b) (a), (b)
(c) (a),(c)
(d) (b), (c)

Ans. a
(a) Angular momentum (L) = nh/2π
Therefore, as n increases, L also increases.
(b) 
(c) For n = 1, L = 4/2π
(d) As l increases, the peak of ψ vs r shifts towards higher 'r' value.

Q.65. Which of the graphs shown below does not represent the relationship between incident light and the electron ejected from metal surface?    (2019)
(a)
(b)
(c)
(d)

Ans. c
K.E. = hv - hv₀
where, v = Frequency of incident radiation
v₀ = Threshold frequency
KE is independent of intensity, it depends on frequency of light. Intensity is directly proportional to the no. of electrons emitted.

Q.66. The ground state energy of hydrogen atom is -13.6 eV. The energy of second excited state of He⁺ ion in eV is:    (2019)
(a) -54.4
(b) -3.4
(c) -6.04
(d) -27.2

Ans. c
According to Bohr’s model energy in n^th state

For second excited state, of He⁺, n = 3

Q.67. Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H-atom is suitable for this purpose?
[R_H = 1 x 10⁵ cm^-1. h = 6.6 x 10^-34 Js, c = 3 x 10⁸ ms^-1]    (2019)
(a) Paschen,∞ → 3
(b) Paschen, 5 → 3
(c) Balmer, ∞ → 2    
(d) Lyman, ∞ → 1

Ans. a

Q.68. The de Broglie wavelength (λ) associated with a photo-electron varies with the frequency (v) of the incident radiation as, [v₀ is threshold frequency]:    (2019)
(a)
(b)
(c)
(d)

Ans. d
According to de-Broglie wavelength equation,

According to photoelectric effect,

Q.69. What is the work function of the metal if the light of wavelength 4000 Å generates photo-electrons of velocity 6 x 10⁵ ms^-1 from it?    (2019)
(Mass of electron= 9 x 10^-31 kg
Velocity of light = 3 x 10⁸ ms^-1 
Planck’s constant= 6.626 x 10^-34 Js
Charge of electron = 1.6 x 10^-19 JeV^-1)
(a) 0.9 eV    
(b) 3.1 eV
(c) 2. leV
(d) 4.0 eV

Ans. c



According to photoelectric effect,

Q.70. If the de Broglie wavelength of the electron in n^th Bohr orbit in a hydrogenic atom is equal to 1.5 πa₀ (a₀ is Bohr radius), then the value of n/z is:    (2019)
(a) 0.40
(b) 1.50
(c) 1.0
(d) 0.75

Ans. d

Q.71. The quantum number of four electrons are given below:
I. n = 4, l = 2, m₁ = -2, m_s = -1/2
II. n = 3, l = 2, m₁= 1, m_s = + 1/2
III. n = 4, l = 1, m₁ = 0,  m_s = + 1/2
IV. n = 3, l= 1, m₁= 1, m_s = - 1/2
The correct order of their increasing energies will be:    (2019)
(a) IV<III<II<I
(b) I<II<III<IV
(c) IV<II<III<I
(d) I<III<II<IV

Ans. c

The energy of an atomic orbital increases with increasing n + l. For identical values of n + l, energy increases with increasing n. Therefore the correct order of energy is:

Q.72. If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength λ, then for 1.5p momentum of the photoelectron, the wavelength of the light should be:
(Assume kinetic energy of ejected photoelectron to be very high in comparison to work function):    (2019)

Ans. 4
In photoelectric effect,

Given that KE of ejected photoelectron is very high in comparison to work function w.

New wavelength

Q.73. For any given series of spectral lines of atomic hydrogen, let   be the difference in maximum and minimum frequencies in cm^-1. The ratio of  is:   (2019)
(a) 4 : 1
(b) 9 : 4
(c) 5 : 4
(d) 27 : 5

Ans. b

Q.74. Which one of the following about an electron occupying the 1s orbital in a hydrogen atom is incorrect? (The Bohr radius is represented by a₀).   (2019)
(a) The probability density of finding the electron is maximum at the nucleus.
(b) The electron can be found at a distance 2a₀ from the nucleus.
(c) The magnitude of the potential energy is double that of its kinetic energy on an average.
(d) The total energy of the electron is maximum when it is at a distance a₀ from the nucleus.

Ans. d
The total energy of the electron is minimum at a distance of a₀ from the nucleus for 1s orbital.

Q.75. The graph between |ψ|² and r (radial distance) is shown below. This represents:    (2019)
(a) 3s orbital
(b) 2s orbital
(c) 1s orbital
(d) 2p orbital

Ans. b
The given probability density curve is for 2s orbital due to the presence of only one radial node. 1s and 2p orbital do not have any radial node and 3s orbital has two radial nodes. Hence, option (2) is correct.

Q.76. The isoelectronic set of ions is:    (2019)
(a) N^3-, O^2-, F^- and Na⁺ 
(b) N³ Li⁺, Mg²⁺ and O^2- 
(c) F^-, Li⁺, Na⁺ and Mg²⁺ 
(d) Li⁺, Na⁺, O^2- and F^-

Ans. a
Atomic numbers of N, O, F and Na are 7, 8, 9 and 11 respectively. Therefore, total number of electrons in each of N^3-, O^2-, F^-, and Na⁺ are 10 and hence they are isoelectronic.

Q.77. The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The spectral series are:    (2019)
(a) Lyman and Paschen
(b) Balmer and Brackett
(c) Brackett and P fund 
(d) Paschen and P fund

Ans. a
For determined shortest wavelength, n₂ = ∞

Q.78. The group number, number of valence electrons, and valency of an element with atomic number 15, respectively, are:    (2019)
(a) 16, 5 and 2
(b) 15, 5 and 3
(c) 16, 6 and 3
(d) 15, 6 and 2

Ans. a
Phosphorus has atomic number 15. Its group number is 15, number of valence electrons is 5 and valency is 3.

Q.79. The electrons are more likely to be found:    (2019)

(a) in the region a and c
(b) in the region a and b
(c) only in the region a
(d) only in the region c

Ans. a
Probability of finding an electron will have maximum value at both 'a' and 'c'. There is zero probability of finding an electron at 'b'.

Q.80. Ejection of the photoelectron from metal in the photoelectric effect experiment can be stopped by applying 0.5 V when the radiation of 250 nm is used. The work function of the metal is:    (2018)
(a) 4.5 eV
(b) 4 eV
(c) 5.5 eV
(d) 5 eV

Ans. a

Q.81. Which of the following statements is false?    (2018)
(a) Splitting of spectral lines in electrical field is called Stark effect
(b) Frequency of emitted radiation from a black body goes from a lower wavelength of higher wavelength as the temperature increases
(c) Photon has momentum as well as wavelength
(d) Rydberg constant has unit of energy    

Ans. b, d
When temperature is increased, black body emit high energy radiation, from higher wavelength to lower wavelength.
Rydberg constant has unit length^–1 (i.e. cm^–1)

Q.82. The radius of the second Bohr orbit for hydrogen atom is
(Planck's Const. h = 6.6262 × 10^–34 Js;
mass of electron = 9.1091 × 10^–31 kg;
charge of electron e = 1.60210 × 10^–19 C;
permittivity of vacuum ε₀ = 8.854185 × 10^–12 kg^–1 m^–3 A²)    (2017)
(a) 1.65 Å 
(b) 4.76 Å 
(c) 0.529 Å 
(d) 2.12 Å 

Ans. d

Q.83. If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelength in Paschen series of He⁺ is:    (2017)
(a) 5A/9
(b) 36A/7
(c) 36A/5
(d) 9A/5

Ans. b
Shortest wavelength is corresponding to best ine
∴ n_L = 1 (Lyman series)
n_H = ∞ (infinite)
 

Q.84. The electron in the hydrogen atom undergoes transition from higher orbitals to orbital of radius 211.6 pm. This transition is associated with:    (2017)
(a) Lyman series
(b) Balmer series
(c) Brackett series
(d) Paschen series

Ans. b
R = 211.6 pm = 2.11 Å
R = 0.529 x n²/Z = 2.11 Å
n² = 4 ⇒ n = 2

Q.85. A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h/l (where l is wavelength associated with electron wave) is given by:    (2016)
(a) me V 
(b) 2me V 
(c)
(d)

Ans. d

Q.86. The total number of orbitals associated with the principal quantum number 5 is:    (2016)
(a) 5
(b) 20
(c) 25
(d) 10

Ans. c
Given, n = 5
Possible subshells are
5s, 5p, 5d, 5f, 5g
Total number of orbital = 1 + 3 + 5 + 7 + 9 = 25

Q.87. Aqueous solution of which salt will not contain ions with the electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁶?    (2016)
(a) NaCl
(b) CaI₂ 
(c) NaF
(d) KBr

Ans. c
NaF: Na⁺ = 1s² 2s² 2p⁶ 
F^– = 1s² 2s² 2p⁶