Skip to main content
Back to STRUCTURE OF ATOM

Your Progress

0 / 592 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →
Showing 121-150 of 592 questions

Preparing practice and saved progress…

121 Medium
121 of 592
Understanding +4 | -1
When 3d-orbital is complete, the newly entering electron goes into:
View answer and explanation
Correct option: C
4p
Option C (4p) is correct because after filling the 3d orbitals, the next electron enters the 4p orbitals according to the Aufbau principle and Hund's rule.
122 Easy
122 of 592
Remembering +4 | -1
The total number of nodes are given by
View answer and explanation
Correct option: C
(n−1)(n-1)
Option C ((n−1)(n-1)) is correct because the total number of nodes in an atomic orbital is given by (n-1), where n is the principal quantum number.
123 Easy
123 of 592
Remembering +4 | -1
The quantum number which is responsible for the size of electron cloud is
View answer and explanation
Correct option: C
Principal
Option C (Principal) is correct because the principal quantum number (n) determines the energy level and hence the size of the electron cloud.
124 Hard
124 of 592
Application +4 | -1
The uncertainty in momentum of an electron is 1×10−51 \times {10^{ - 5}}  kg m/s. the uncertainty in its position will be (h=6.62×10−34(h = 6.62 \times {10^{ - 34}}  kg m2/s){\mathbf{m}^2}/\mathbf{s})
View answer and explanation
Correct option: B
5.25 ×10−30\times {10^{ - 30}} m
Option B (5.25 ×10−30\times {10^{ - 30}} m) is correct because it is the result of applying the Heisenberg uncertainty principle with the given uncertainty in momentum.
125 Easy
125 of 592
Remembering +4 | -1
Probability of finding an electron at the nodal surface is:
View answer and explanation
Correct option: D
zero
Option D (zero) is correct because the probability of finding an electron at a nodal surface is zero, as it's a region where the electron density is zero.
126 Medium
126 of 592
Understanding +4 | -1
Statement – I: Kinetic energy of photoelectrons increases with increase in the frequency of incident radiation.

Statement – II: The number of photoelectrons ejected increases with increase in intensity of incident radiation.
View answer and explanation
Correct option: D
Both statement I and statement II are correct
Option D (Both statement I and statement II are correct) is correct because the kinetic energy of photoelectrons increases with the frequency of incident radiation, and the number of photoelectrons increases with the intensity of incident radiation.
127 Easy
127 of 592
Application +4 | -1
The number of orbitals having (n+l)<5(n+l) < 5 is
View answer and explanation
Correct option: D
10
Option D (10) is correct because it accurately counts the number of orbitals satisfying the condition (n+l) < 5 based on the quantum numbers n and l.
128 Medium
128 of 592
Understanding +4 | -1
Which of the following set a of quantum numbers is correct for an electron in 4f orbital?
View answer and explanation
Correct option: C
n=4,l=3,m=+1,s=+1/2n\mathrm{ } = \mathrm{ }4,\mathrm{ }l\mathrm{ } = \mathrm{ }3,\mathrm{ }m\mathrm{ } = \mathrm{ } + 1,\mathrm{ }s\mathrm{ } = \mathrm{ } + 1/2
Option C is correct because for a 4f orbital, n=4, l=3, m can be any value from -3 to +3, and s can be +1/2 or -1/2.
129 Medium
129 of 592
Application +4 | -1
Atoms consist of electrons, protons and neutrons. If the mass attributed to neutron was halved and that attributed to the electrons was doubled, the atomic mass of   6C12{\mathrm{\;}_6}{\mathrm{C}^{12}} would be approximately:
View answer and explanation
Correct option: D
Reduced by 25%
Option D (Reduced by 25%) is correct because halving the neutron mass reduces the total atomic mass from 12 to 9, which is a 25% reduction.
130 Easy
130 of 592
Understanding +4 | -1
Compared with an atom of atomic weight 12 and atom ic number 6 , the atom of atomic weight 13 and atom ic number 6
View answer and explanation
Correct option: A
Contains more neutrons
Option A (Contains more neutrons) is correct because isotopes of an element have the same number of protons but different numbers of neutrons.
131 Hard
131 of 592
Application +4 | -1
When a certain metal was irradiated with light of frequency 4.0×1016 s−14.0 \times 10^{16} \mathrm{~s}^{-1}, the photoelectrons emitted has three times the kinetic energy as the kinetic energy of photoelectrons emitted when the metal was irradiated with light of frequency 2.0×1016 s−12.0 \times 10^{16} \mathrm{~s}^{-1} Calculate the critical frequency (v0)\left(v_0\right) of the metal.
View answer and explanation
Correct option: B
1×1016 s−11 \times 10^{16} \mathrm{~s}^{-1}
Option B (1×1016 s−11 \times 10^{16} \mathrm{~s}^{-1}) is correct because it is the critical frequency calculated using Einstein's photoelectric equation and the given conditions.
132 Medium
132 of 592
Remembering +4 | -1
Which of the following electromagnetic radiation have greater frequency?
View answer and explanation
Correct option: A
γ\gamma-rays
Option A (γ\gamma-rays) is correct because gamma rays have the highest frequency in the electromagnetic spectrum.
133 Medium
133 of 592
Application +4 | -1
An excited state of H\mathrm{H} atom emits a photon of wavelength λ\lambda and returns in the ground state, the principal quantum number of excited state is given by :
View answer and explanation
Correct option: B
λR(λR−1)\sqrt{\frac{\lambda R}{(\lambda R-1)}}
Option B is correct because the principal quantum number is derived from the Rydberg formula for hydrogen atom emission.
134 Medium
134 of 592
Understanding +4 | -1
Choose the incorrect statement(s):
View answer and explanation
Correct option: A
Increasing order of wavelength is Microwave >> Radiowaves >> IR waves >> visble waves >> UV waves
Option A is the incorrect statement because the increasing order of wavelength should be Radiowaves > Microwaves > IR waves > visible waves > UV waves.
135 Medium
135 of 592
Understanding +4 | -1
Number of spectral lines falling in Balmer series when electrons are de-excited from nth n^{\text {th }} shell will be given as
View answer and explanation
Correct option: B
(n−2)(n-2) in visible region
Option B is correct. When electrons de-excite from the nthn^{th} shell to the second shell (nf=2n_f = 2), the spectral lines fall in the visible region (Balmer series), and the number of lines is (n-2).
136 Medium
136 of 592
Application +4 | -1
The total number of protons in 8.4 g8.4 \mathrm{~g} of MgCO3\mathrm{MgCO}_3 is (NA=6.02×1023)\left(N_{\mathrm{A}}=6.02 \times 10^{23}\right)
View answer and explanation
Correct option: B
2.52×10242.52 \times 10^{24}
Option B (2.52×10242.52 \times 10^{24}) is correct because it is the total number of protons in 8.4 g of MgCO₃, calculated by determining the number of moles of MgCO₃ and the number of protons per molecule.
137 Medium
137 of 592
Application +4 | -1
What will be the ratio of the wavelength of the first lineto that of the second line of Paschen series of H{H} atom?
View answer and explanation
Correct option: A
256:175256: 175
Option A (256:175) is correct because it is the correct ratio of the wavelengths calculated using the Rydberg formula for the Paschen series.
138 Medium
138 of 592
Remembering +4 | -1
With respect to atomic spectrum, each line in the Lyman series is due to electrons returning
View answer and explanation
Correct option: C
From a particular higher energy level to n=1n=1
Option C (From a particular higher energy level to n=1n=1) is correct because the Lyman series corresponds to electronic transitions from higher energy levels to the n=1 energy level.
139 Easy
139 of 592
Understanding +4 | -1
In which of the orbit of He+,\mathrm{H}{\mathrm{e}^ + }, the angular momentum of the electron in h/2π?h/2\pi ?
View answer and explanation
Correct option: A
First orbit
Option A (First orbit) is correct because the angular momentum is given by nh/2πnh/2\pi, and for the first orbit, n=1.
140 Medium
140 of 592
Analyzing +4 | -1
The decreasing order (highest first) from the values of e/m{e} / {m} (charge/mass) for electron (e),(e), proton (p)(p), neutron (n)(n) and alpha particle (α)(\alpha) is
View answer and explanation
Correct option: A
e,p,α,ne, p, \alpha, n
Option A is correct because the charge-to-mass ratio (e/m) decreases in the order: electron > proton > alpha particle > neutron.
141 Hard
141 of 592
Application +4 | -1
Uncertainty in the position of an electron (mass=9.1×10−31kg)\left( {\mathrm{mass} = 9.1 \times {{10}^{ - 31}}\mathrm{kg}} \right) moving with a velocity 300 ms−1\mathrm{m}{\mathrm{s}^{ - 1}} accurate upon 0.001% will be
(h=6.63×10−34  Js)\left( {h = 6.63 \times {{10}^{ - 34}}\mathrm{\;Js}} \right)
View answer and explanation
Correct option: C
1.92×10−2m1.92 \times {10^{ - 2}}\mathrm{m}
Option C (1.92 x 10⁻² m) is correct because it is calculated using the Heisenberg uncertainty principle, relating the uncertainty in position and velocity of the electron.
142 Medium
142 of 592
Application +4 | -1
Ratio of the radius of first orbit of Li2+\mathrm{Li}^{2+} to the third orbit of He+\mathrm{He}^{+}ions will be
View answer and explanation
Correct option: D
2:272: 27
Option D (2:27) is correct because it is calculated by using the formula for the radius of Bohr orbits and taking the ratio.
143 Hard
143 of 592
Analyzing +4 | -1
Consider the following statements : 1.Electron density in xy plane in 3dx2−y23{d_{{x^2} - {y^2}}} orbital is zero 2.Electron density in xy plane in 3dz23{d_{{z^2}}} orbital is zero 3.2s orbital has only one spherical node 4.For 2pz2{p_z} orbital yz is the nodal plane The correct statements are
View answer and explanation
Correct option: C
Only 3
The correct answer is Only 3 because only the statement about the 2s orbital having one spherical node is correct.
144 Hard
144 of 592
Application +4 | -1
Positronium is a species consisting of an electron bound to a positron. What would be the radius of its first excited state?
View answer and explanation
Correct option: D
4.24 A∘4.24 ~\mathop {\text{A}}\limits^ \circ
The correct answer is 4.24 A∘4.24 ~\mathop {\text{A}}\limits^ \circ because the radius of the first excited state of positronium is calculated using the formula rn=2a0(n)2r_n = 2 a_0(n)^2 with n=2.
145 Medium
145 of 592
Application +4 | -1
Calculate the energy of Li+2{Li}^{+2} atom for 2nd2^{\text {nd}} excited state.
View answer and explanation
Correct option: A
−13.6 eV-13.6 \mathrm{~eV}
Option A (-13.6 eV) is correct because it is calculated using the formula for energy levels in hydrogen-like atoms, considering the atomic number and the excited state.
146 Easy
146 of 592
Analyzing +4 | -1
Statement (S): 24Cr{ }_{24} \mathrm{Cr} has more paramagnetic nature than 25Mn{ }_{25} \mathrm{Mn}.

Explanation (E): Cr\mathrm{Cr} has more number of unpaired electrons than Mn\mathrm{Mn}.
View answer and explanation
Correct option: C
Both S and E are correct and E is correct explanation of S
Option C (Both S and E are correct and E is correct explanation of S) is correct because Chromium (Cr) has more unpaired electrons than Manganese (Mn), leading to a greater paramagnetic nature.
147 Medium
147 of 592
Understanding +4 | -1
The correct set of quantum numbers for 4d4 d-electrons are
View answer and explanation
Correct option: D
4,2,1,−1/24,2,1,-1 / 2
Option D (4,2,1,−1/24,2,1,-1 / 2) is correct because for a 4d electron, n=4, l=2, m can be -2, -1, 0, 1, 2, and s can be +1/2 or -1/2.
148 Easy
148 of 592
Understanding +4 | -1
The number of spherical nodes in 3p orbitals is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the number of spherical nodes in a 3p orbital is calculated using the formula n - l - 1, where n=3 and l=1, resulting in 3-1-1=1.
149 Medium
149 of 592
Analyzing +4 | -1
Energies of 2S orbital in IA group is
View answer and explanation
Correct option: A
E2s(H)>E2s(Li)>E2s(Na)>E2s(K){E_{2s}}\left( H \right) > {E_{2s}}\left( {Li} \right) > {E_{2s}}\left( {Na} \right) > {E_{2s}}\left( K \right)
Option A is correct because as the atomic number increases down the group, the energy of the 2s orbital decreases due to increased nuclear charge and shielding.
150 Easy
150 of 592
Understanding +4 | -1
Co3+,Mn+,Fe2+\mathrm{Co}^{3+}, \mathrm{Mn}^{+}, \mathrm{Fe}^{2+} and Cr\mathrm{Cr} are isoelectronic (24 electrons) in each. Thus, EC are identical in
View answer and explanation
Correct option: A
Mn+,Cr\mathrm{Mn}^{+}, \mathrm{Cr}
Option A is correct because only Mn+ and Cr have the same electronic configuration after accounting for the charge and the exception in Cr's configuration.
Revision notes and key takeaways

Review STRUCTURE OF ATOM in your textbook or class notes, then use this set to check your understanding.

Showing 121-150 of 592 questions

Your practice

0 / 592 answered · 592 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.